gate electromagnetics by kanodia

ELECTROMAGNETICS

GATE CLOUD

R. K. KanodiaAshish Murolia

JHUNJHUNUWALA

ELECTROMAGNETICS

JAIPUR

GATE CLOUD

GATE CLOUD ELECTROMAGNETICS, 1e

R. K. Kanodia, Ashish Murolia

CC1015

Information contained in this book has been obtained by author, from sources believes to be reliable.

However, neither Jhunjhunuwala nor its author guarantee the accuracy or completeness of any

information herein, and Jhunjhunuwala nor its author shall be responsible for any error, omissions,

or damages arising out of use of this information. This book is published with the understanding that

Jhunjhunuwala and its author are supplying information but are not attempting to render engineering

or other professional services.

Copyright by Jhunjhunuwala

ISBN 978-8192-34838-4

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GATE CLOUD caters a versatile collection of Multiple Choice Questions to the students who are preparing

for GATE (Gratitude Aptitude Test in Engineering) examination. This book contains over 1200 multiple

choice solved problems for the subject of Electromagnetics, which has a significant weightage in the GATE

examination of Electronics and Communication Engineering. The GATE examination is based on multiple

choice problems which are tricky, conceptual and tests the basic understanding of the subject. So, the

problems included in the book are designed to be as exam-like as possible. The solutions are presented using

step by step methodology which enhance your problem solving skills.

The book is categorized into ten chapters covering all the topics of syllabus of the examination. Each chapter

contains :

Ÿ Exercise 1 : Level 1

Ÿ Exercise 2 : Level 2

Ÿ Exercise 3 : Mixed Questions Taken form Previous Examinations of GATE & IES.

Ÿ Detailed Solutions to Exercise 1, 2 and 3.

Although we have put a vigorous effort in preparing this book, some errors may have crept in. We shall

appreciate and greatly acknowledge the comments, criticism and suggestion from the users of this book

which leads to some improvement.

You may write to us at [email protected] and [email protected].

Wish you all the success in conquering GATE.

Authors

PREFACE

GATE ELECTRONICS & COMMUNICATION ENGINEERING

IES ELECTRONICS & TELECOMMUNICATION ENGINEERING

Networks:

Electromagnetic Theory:

Elements of vector calculus: divergence and curl; Gauss’ and Stokes’ theorems, Maxwell’s

equations: differential and integral forms. Wave equation, Poynting vector. Plane waves:

propagation through various media; reflection and refraction; phase and group velocity; skin depth.

Transmission lines: characteristic impedance; impedance transformation; Smith chart; impedance

matching; S parameters, pulse excitation. Waveguides: modes in rectangular waveguides; boundary

conditions; cut-off frequencies; dispersion relations. Basics of propagation in dielectric waveguide

and optical fibers. Basics of Antennas: Dipole antennas; radiation pattern; antenna gain.

Analysis of electrostatic and magnetostatic fields; Laplace's and Poisson's equations; Boundary

value problems and their solutions; Maxwell's equations; application to wave propagation in

bounded and unbounded media; Transmission lines : basic theory, standing waves, matching

applications, microstrip lines; Basics of wave guides and resonators; Elements of antenna theory.

SYLLABUS

IES ELECTRICAL ENGINEERINGElectromagnetic Theory:

Electric and magnetic fields. Gauss's Law and Amperes Law. Fields in dielectrics, conductors and

magnetic materials. Maxwell's equations. Time varying fields. Plane-Wave propagating in

dielectric and conducting media. Transmission lines.

CONTENTS

*******

CHAPTER 1VECTOR ANALYSIS

2 Vector Analysis Chap 1

GATE CLOUD Electromagnetics By RK Kanodia & Ashish Murolia

For View Only Shop Online at www.nodia.co.in

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EXERCISE 1.1

MCQ 1.1.1 The field lines of vector rB a= q is

MCQ 1.1.2 Given the two vectors 5 2 4M a a ax y z= - + and 8 7 2N a a ax y z=- - + . The unit vector in the direction of (M-N ) will be(A) 0.82 0.36 0.14a a ax y z+ - (B) 0.92 0.36 0.41a a ax y z- +

(C) 0.92 0.36 0.14a a ax y z+ + (D) 0.92 0.36 0.14a a ax y z- - -

MCQ 1.1.3 A vector field is specified as 4 2 2 3xy x zG a a a2x y z

2= + + +^ h . The unit vector in the direction of G at the point ( 2, 1, 3)- will be(A) 0.26 0.39 0.88a a ax y z+ + (B) 0.26 0.39 0.88a a ax y z- + +

(C) 0.36 0.29 0.24a a ax y z- + (D) 0.88 0.29 0.36a a ax y z- +

MCQ 1.1.4 Consider three nonzero vectors ,A B and C . Which of the following is not a correct

Chap 1 Vector Analysis 3


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For View Only Shop Online at www.nodia.co.inrelation between them?(A) 0A A# = (B) A B C A B A C# ## + = +^ ^ ^h h h

(C) ( )A B C A B C# # # #= ^ h (D) ( )B A A B# #=-^ h

MCQ 1.1.5 The tips of three vectors A, B and C drawn from a point define a plane. A B C: #^ h equals to(A) +1

(B) 1-

(C) zero

(D) can’t be determined as A, B and C are not given

MCQ 1.1.6 The component of vector A along vector B is

(A) AA B: (B) B

A B:

(C) A B: (D) ABA B:

MCQ 1.1.7 The vector fields are defined as 2 3A a a az= + +r f and 6B a a aza b= + -r f . If the fields A and B are parallel then the value of a and b are respectively.(A) 2- , 2- (B) 2- , 4-

(C) 4- , 2- (D) 2- , 1-

MCQ 1.1.8 Consider the vectors A 4 2k ka a ax y z= + + and B 4 4a a ax y z= + - . For what value of k the two vectors A and B will be orthogonal ?(A) 0 (B) 1+

(C) 2- (D) 1-

Common Data for Question 9 - 10 :In a cubical region ( 2x < , y 2< , z 2< ) a vector field is defined as

E 9 (2 ) 8 (2 ) 2 (2 )cos sin sinzy x zy x y xa a ax y z2 2= + + .

MCQ 1.1.9 The vector field component Ex will be zero in the plane(A) z 0= (B) y 0=

(C) /x 4p= (D) All of the above

MCQ 1.1.10 In the plane y z4 0- = , the vector field components Ey and Ez are related as(A) E E2 y z= (B) E E 0y z+ =

(C) E E2y z= (D) E Ey z=

MCQ 1.1.11 The plane surface on which the vector field E will be zero is(A) x 0= (B) y 0=

(C) z 0= (D) none of the above




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MCQ 1.1.12 Distance between the points ( , , )P x y z2 3 1= = =- and ( 4, 50, 2)Q zr f= =- = is(A) 3.74 (B) 4.47

(C) 6.78 (D) 8.76

MCQ 1.1.13 The uniform vector field A ay= can also be expressed as (A) sin sina af f+r f

(B) cos sina af f+r f

(C) sin cos cos cos sina a arq f q f f+ +q f

(D) sin cos cos cos sina a arq f q f f+ -q f

MCQ 1.1.14 The vector filed 1F a2 x= can be expressed in spherical coordinates at the point (x 3= , y 2= , z 1=- ) as(A) 8 2 5a a ar - +q f (B) 8 2.2 5.5a a ar - -q f

(C) 8 2.2 5.5a a ar- + +q f (D) 8 2.2 5.5a a ar + +q f

MCQ 1.1.15 The angle formed between 5 10 3A a a az=- + +r f and surface z 5= is(A) 10c (B) 15c

(C) 45c (D) 75c

MCQ 1.1.16 The component of vector 20 4A a a a4 z=- - +r f parallel to the line x 6= , z 2=- at the point ( , , )P 3 90 2c is(A) 4 2a a- +r f (B) a ax y-

(C) 2ay- (D) 2az-

MCQ 1.1.17 In the cartesian co-ordinate system the co-ordinates of a point P is ( , , )a b c .Now consider the whole cartesian system is being rotated by 145c about an axis from the origin through the point (1, 1, 1) such that the rotation is clockwise when looking down the axis towards the origin. What will be the co-ordinates of the point P in the transformed cartesian system ?(A) ( /a 2, /b 2, /c 2) (B) ( a- , b- , c- )

(C) (c , b , a ) (D) (c , a , b)

MCQ 1.1.18 Consider R be the position vector of a point ( , , )P x y z in cartesian co-ordinate system then gradR equals to

(A) 1 (B) RR4

(C) RR (D) R

R2

MCQ 1.1.19 Given the vector filed A (2 ) (4 2 )y xy x z x yza a ax y z2 2 2= + + + + + . The divergence

of the vector field is(A) ( )x y2 + (B) x y z x y6 22 2 2+ + + +

(C) ( )y x z2 + (D) 0



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For View Only Shop Online at www.nodia.co.inMCQ 1.1.20 A vector field F is defined as F sin cosz za a a2 z

2r f r f= + +r f . The F4# at point , / ,P 1 2 2p^ h equals to(A) 6a az+r (B) 2a az- +r

(C) 3 6a az+r (D) 6a a5 z- +r

MCQ 1.1.21 Which one of the following vector function has divergence and curl both zero ?(A) 2xz yz zA a a ax y z

2= - - (B) xy yzB a ax z= -

(C) xy xz yzC a a ax y z= - - (D) yz xz xyD a a ax y z= + +

MCQ 1.1.22 The curl of the unit vectors ar , af and az in cylindrical co-ordinate system is listed below. Which of them is correct ? ar af az ar af az(A) 0 az1

r 0 (B) 0 a1r r af(C) ar f ar 0 (D) ar f a1r r 0

MCQ 1.1.23 In a certain region consider f and g are the two scalar fields where as A is a vector field. Which of the following is not a correct relation ?(A) ( )f f fA A A: : :d dd = -^ ^h h

(B) f f fA A A#d d# #d = -^ ^ ^h h h

(C) gf

gg f f g

2d d

d = -c m

(D) ( ) ( )

g gg gA A A

2#d # #d d= -c m

MCQ 1.1.24 Laplacian of the scalar field 2 4sin cosf z z4 2 2 2r f f r= + + at the point ( , / , )P 3 2 6p is(A) 16 (B) 0

(C) 46 (D) 40

MCQ 1.1.25 A conservative field M is given by M ( ( ) ) ( )cos cosz xz y kx x xza a a5x y z= + + + . The value of k will be(A) 1 (B) 0

(C) 1- (D) /1 2

MCQ 1.1.26 A scalar field g k x y xyz1 5 2= + +^ h will be harmonic at all the points for the value of k equals to(A) /1 2 (B) 0

(C) /1 2- (D) can’t be determined

MCQ 1.1.27 Curl of the gradient of any scalar field is(A) zero everywhere (B) zero at origin only

(C) zero at infinity only (D) it does not exist




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MCQ 1.1.28 A vector field is given as ( 4 ) ( ) (4 3 )x z x z x y czA a a a3 7x y z= + + - + + - . The value of c for which A will be solenoidal is(A) 1 (B) 1-

(C) 0 (D) 3

MCQ 1.1.29 For a vector function (4 ) ( 5 ) (4 2 )x k z k x z x k y zA a a ax y z1 2 3= + + - + - + to be irrotational value of k1, k2 and k3 will be respectively.(A) 5- , 0, 4 (B) 0, 4, 5

(C) 4, 0, 5 (D) 1, 4, 3

MCQ 1.1.30 The unit vector normal to the plane 2 6 7x y z3+ + = is

(A) 2 3a a a241

x y z+ +^ h (B) a a a141 4 4 6x y z+ +^ h

(C) a a a141 2 3x y z+ +^ h (D) 2 4 6a a a

581

x y z+ +^ h

MCQ 1.1.31 Consider C is a certain closed path and dl is the differential displacement along the

path then the contour integral dlC# is

(A) zero

(B) 1

(C) 1-

(D) can’t be determined as C is not defined.

MCQ 1.1.32 Consider C is any closed path and U is a scalar field. So, the contour integral

U dlC

:d̂ h# is

(A) 1

(B) 1-

(C) zero

(D) Can’t be determined as C and U is not given

MCQ 1.1.33 A vector field is defined as 2yz z x xyA a a a3 x y z2= + + .The surface integral of the

field over a closed surface S is(A) 1

(B) 5

(C) zero

(D) can’t be determined as surface S is not given

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EXERCISE 1.2

MCQ 1.2.1 If the edge of a cube is 3 units then the angle formed between it’s body diagonals will be(A) 70.53c (B) .53 70c

(C) .66 21c (D) 61c

Common Data for Question 2 - 4 :Consider a triangle ABC , whose vertex A, B and C are located at the points ( , , )4 2 5- , (16, 20, )3- and ( 14,10, )15- respectively.

MCQ 1.2.2 The unit vector perpendicular to the plane of the triangle is(A) 0.61 0.42 0.37a a ax y z+ -

(B) 0.66 0.38 0.65a a ax y z- +

(C) 0.54 0. 0.19a a a11x y z- + -

(D) 0.43 0.21 0.11a a ax y z- +

MCQ 1.2.3 The unit vector in the plane of the triangle which is perpendicular to AC is(A) 0.55 0.832 0.077a a ax y z- - +

(B) 0.23 0.11 0.43a a ax y z- -

(C) 0.51 0.41 0.76a a ax y z- + +

(D) 0.49 0.23 0.44a a ax y z- -

MCQ 1.2.4 The unit vector in the plane of the triangle which bisects the interior angle at A is(A) 0.11 0.81 0.44a a ax y z- +

(B) 0.21 0.41 0.52a a ay x z- +

(C) 0.23 0.12 0.11a a az x y+ +

(D) 0.37 0.92 0.17a a az y x+ +

MCQ 1.2.5 A vector field x yx yF a a2x y

2 2=++

^ h at the point (P 2r = , /4f p= , .z 0 1= ) is

(A) 2 3a ax y+ (B) 2 3a ax y- -

(C) 0.5ar (D) 0.5a- r




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MCQ 1.2.6 The component of vector sin cosrA a a ar rr34

2

q f= - +q f tangential to the spherical surface r 20= at point ( , , )P 20 150 330c c is(A) 0.043 100a a+q f (B) 0.43 10a a= +q f

(C) 4.3 100a a+q f (D) 0.043 10a a- -q f

Common Data for Question 7 - 8 :A vector field has the value 12 9A a a a4r=- - +q f at the point ( , , )P 9 150 45c c .

MCQ 1.2.7 The vector component of A, that is tangent to the cone 150cq = is(A) 12 9a ar- +f (B) 12 8a ar- - q

(C) 8 9a a- +q f (D) 12 9a ar- + f

MCQ 1.2.8 The unit vector that is perpendicular to A and tangent to the cone 135cq = is

(A) a a101

3+q fa k (B) a a5

1 4 3r + f^ h

(C) a a51 3 4x + f^ h (D) (9 2 )a a

851

r + f

MCQ 1.2.9 Consider R is a separation vector from a fixed point (a , b , c) to a varying point ( , , )x y z in the Cartesian coordinate system. The value of grad R

1^ h equals to

(A) RR

3- (B) RR

6-

(C) R1

2- (D) RR

3

MCQ 1.2.10 A certain hill located in Udaipur is of height h that varies as :

( , )h x y 6 8 36 56 100x y xy x y32 2= + - + - +

Where x is the distance (in miles) in north and y is the distance (in miles) in east of Udaipur railway station. The top of the hill will be located at(A) 3 miles north, 2 miles west of Railway station.

(B) 2 miles south, 3 miles east of Railway station.

(C) 2 miles north, 3 miles east of Railway station.

(D) 6 miles south, 2 miles east of Railway station.

MCQ 1.2.11 At any point ( , , )P x y z a vector field is given by F aR R15

2= , where R is the position vector of the point P . The divergence of the vector field F will be

(A) zero, everywhere (B) zero, at all points excluding origin

(C) R6

3 , everywhere (D) R6

3 , at all points excluding origin



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For View Only Shop Online at www.nodia.co.inStatement for Linked Question 12 - 14 :Given a vector field A 3 3x yz ya a a3x y z

2 2= + + .

MCQ 1.2.12 The line integral of A from origin to the point (2,2,2) by the route ( , , ) ( , , ) ( , , ) ( , , )0 0 0 2 0 0 2 2 0 2 2 2" " " will be(A) 32 units (B) 8 units

(C) 24 units (D) 6 units

MCQ 1.2.13 The line integral of the vector field A from the origin to the point (2, 2, 2) along the direct straight line is(A) 16 units (B) 24 units


MCQ 1.2.14 The line integral of the field A around the closed loop that goes out along the route defined in Question 12 and back along the route defined in Question 13 is(A) 64 units (B) 0 units


MCQ 1.2.15 A wedge defined by 0 5# #r , 18045 c# #f , z 2= is shown in figure

Circulation of A sin cosza az2r f f= +r along the edge L of the wedge is(A) 10 units (B) 5/2 units

(C) /25 4- units (D) /25 4 units

MCQ 1.2.16 Volume integral of the function f z30 2= over the tetrahedron with corners at ( , , )0 0 1- ; ( , , )0 1 0- , ( , , )1 0 0- , ( , , )0 0 0 is(A) 1/2 (B) /1 2-

(C) /31 2 (D) /1 60

MCQ 1.2.17 Total outward flux of a vector field cos zA a a241 2 2 2r f= +r f through the closed

surface of a cylinder z0 2# # , 2r = is

(A) 4p (B) 16p

(C) p (D) 32p




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MCQ 1.2.18 A quarter cylinder of radius 2 and height 5 exists in the first octant of a cartesian coordinate system as shown in the figure below. Surface integral of a vector field

(4 ) 2 6sin sin zA a a a3 z2r r f r f= + + +r f over the surface of the cylinder will be

(A) 12p (B) 30p

(C) 40p (D) 80p

MCQ 1.2.19 A vector function is given by 6 2x y y zG a a a5x y z2 2= - + . If L is a closed path

defined by the sides of a triangle as shown in the figure then, dG lL

:# equals to

(A) 24 units (B) 7 units

(C) 2 units (D) /10 6 units

Common Data for Question 20 - 21 :Consider S1 and S2 are respectively the top and slanting surfaces of an ice cream cone of slant height 2 m and angle 60c as shown in figure, where a vector field F is defined as

F x y

x y z x x y ya a a a4 5 42 2

32 2x y x y2 2

2 2 2

=+

+ + + - +: D



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MCQ 1.2.20 The surface integral of the vector field F over the surface S1 will be(A) 2.3 units (B) 24 units

(C) 0.18 units (D) 1.9 units

MCQ 1.2.21 The surface integral of the vector field over the surface S2 will be(A) /4 3p (B) /3 3p

(C) /4 3 3p (D) 0

Common Data for Question 22 - 23 :Negative gradient of a scalar field f is A f4=- ( ) ( 3 )x z z x y za a a4x y z= + - + - -

MCQ 1.2.22 The vector A is(A) irrotational but not solenoidal (B) both irrotational and solenoidal

(C) solenoidal but not irrotational (D) neither solenoidal nor irrotational

MCQ 1.2.23 The scalar field, f equals to

(A) x xz z2 22 2

+ + (B) x xz yz z2 2 6 22 2

- - + +

(C) xz yz z3 22

- + + (D) x xz yz z2 3 22 2

- - + +

MCQ 1.2.24 Line integral of a vector field ( )y xA a a5 x y= + from a point P (2, 1, 3) to the point Q (8, 2, 3) along the curve /y x 2= will be(A) 42 units (B) 14 units


MCQ 1.2.25 A vector field cosF a2 2r r r1 13 3 f= +^ h exists in the region between the two spherical

shells of radius 1 m and 2 m centred at the origin. The total outward flux of F through the outer spherical surface will be




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(A) p (B) p-

(C) 2p- (D) 2p

MCQ 1.2.26 Two vectors A and B make an angle 30c between them as shown in figure. Magnitude of vector A and B are 4 units and 3 units respectively. If a third vector R is defined such that 6R A B4= - then it’s graphical construction will be

MCQ 1.2.27 A certain, vector R is defined as R A B C##= ^ h. Directions of A, B and C are mentioned in the list below. Which of the following gives the correct direction of R for the given direction of the three vectors. Direction of A Direction of B Direction of C Direction of R(A) North South East West(B) South North West East(C) East West North West(D) West East South South



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For View Only Shop Online at www.nodia.co.inMCQ 1.2.28 Which of the following vectors are equal :

A 3a a ax y z=- + + at , ,1 2 3^ h in cartesian co-ordinates B 3a a az= + +r f at , / ,2 2 3p^ h in cylindrical co-ordinates C 3a a2 z= +r at , / ,3 3 4 9p^ h in cylindrical co-ordinates(A) A and B only (B) B and C only

(C) A and C only (D) all the three vectors A, B and C

MCQ 1.2.29 Two vectors are defined as A 3a a a5x y z= + + and B 3 2a a ax y z= + + . Which of the following vector is perpendicular to A B+^ h

(A) 4a a4 x y- + (B) 4 4a ay z+

(C) a ax z+ (D) none of these

MCQ 1.2.30 The gradient of a scaler function is , ,V x y zd ^ h 1.5 0.5x yz x z x yza a ax y z2 2 3 2 3= + +

The scalar function is

(A) x yz3 2 (B) x yz2

3 2

(C) x yz2

2 3

(D) xy z3 2

MCQ 1.2.31 The equation of the plane tangential to the surface xyz 1= at the point , ,2 4 41

b l is(A) x y z16 32 24+ + = (B) x y z2 32 12+ + =

(C) x y z2 32 12+ + = (D) x y z16 32 24+ + =

MCQ 1.2.32 Consider a volume v is defined as the part of a spherical volume of radius unity

lying in the first octant. The volume integral xdv2v# is equal to

(A) 2p (B) /16p

(C) /4p (D) /8p

MCQ 1.2.33 Consider a vector field cosA a a3r f r= +r f . If C is the contour shown in the figure

then the contour integral dA lC

:# is equal to

(A) 4p+ (B) 2 1p +

(C) 2p+ (D) 2 1p+




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MCQ 1.2.34 A vector field is defined as 2 cosA a a2r f r= +r f . Consider the two contours C1 andC2 as shown in the figure.

The ratio of the contour integrals d

d

A l

A l

C

C

2

1

:

:

#

# is

(A) 91- (B) 9

1

(C) 9 (D) 9-

MCQ 1.2.35 Consider the contour C as shown in the figure

If a vector field is defined as reA ar

= q

-

b l then the contour integral dA lC

:# is

(A) e21p - (B) e2

1p- -

(C) e2 1 1p + -^ h (D) e2 1 1p - -

^ h

MCQ 1.2.36 The divergence of the unit vectors ar ,aq and af in spherical co-ordinate system is listed below. Which among them is correct ?



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For View Only Shop Online at www.nodia.co.in ar aq af(A) /r 2 / tanr q 0(B) /r2 /cot rq 0(C) /r 2 / tanr q 1(D) /r2 /cot rq 1

MCQ 1.2.37 Which of the following vector can be expressed as the gradient of a scalar ?

(A) 2 2 2yz xz xya a ax y z+ + (B) e arr

f

-

(C) cos sina a22r

f f+r f^ h (D) Both (A) and (C)

MCQ 1.2.38 Which of the following vector can be expressed as curl of another vector ?

(A) 2x y xya a a21

x y z2 2- - +^ h (B) e ar

r

f

-

(C) cos sinr ra a2r3 3

q q+ q (D) All of the above

MCQ 1.2.39 The vector field pattern of yA a3 x= is

MCQ 1.2.40 A two dimensional vector field in Cartesian coordinate system is defined as ,x yA^ h A Aa a2x x y y= +The curl and divergence of the vector field are both zero. Which of the following differential equation satisfies Ax and Ay(A) A 0x

2d = (B) A 0y2d =

(C) A A 0x y2 2d d+ = (D) (A) and (B) both




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MCQ 1.2.41 The circulation of x xz yF a a a3x y z2 2= - - around the path shown below is

(A) 31- (B) 6

1

(C) 61- (D) 3

1

MCQ 1.2.42 If x y zR a a ax y z= + + is the position vector of point ( , , )P x y z and R R= then R Rn:d is equal to

(A) nrn (B) ( 3)n rn+

(C) ( 2)n rn+ (D) 0

MCQ 1.2.43 If sinA a a2r f r= +r r , and L is the contour of figure given below, then circulation

dA lL

:# is

(A) 7 2p+ (B) 7 2p-

(C) 7p (D) 0

***********



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EXERCISE 1.3

MCQ 1.3.1 The direction of vector A is radially outward from the origin, with krA n= . where r x y z2 2 2 2= + + and k is a constant. The value of n for which A 0:d = is(A) 2- (B) 2

(C) 1 (D) 0

MCQ 1.3.2 If xy xA a ax y2= + , then dA l

C$#o over the path shown in the figure is

(A) 0 (B) 3

2

(C) 1 (D) 2 3

MCQ 1.3.3 If a vector field V is related to another vector field A through V Ad#= , which of the following is true? (Note : C and SC refer to any closed contour and any surface whose boundary is C . )

(A) d dV l A SC SC

$ $= ###

(B) d dA l V SC SC

$ $= ###

(C) d dV Al SC SC

# : :d #d=^ ^h h###

(D) d dV l V SC SC

: :d# =^ h ###

MCQ 1.3.4 Consider points A, B, C and D on a circle of radius 2 units as in the shown figure below. The items in List II are the values of af at different points on the circle. Match List I with List II and select the correct answer using the code given below the lists :

GATE 2012GATE 2012

GATE 2010

GATE 2009

IES EC 2010




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List I List II

a. A 1. axb. B 2. ayc. C 3. ax-

d. D 4. ( )/a a 2x y+

5. ( )/a a 2x y- +

6. ( )/a a 2x y-

Codes : a b c d(A) 3 4 5 2(B) 1 6 5 2(C) 1 6 2 4(D) 3 5 4 2

MCQ 1.3.5 If sinh cosV x kye2 pz= is a solution of Laplace’s equation, what will be the value of k ?

(A) p1

12+ (B) p1 2+

(C) p1

12- (D) p1 2-

MCQ 1.3.6 The electric field intensity E at a point P is given by 10 10 10a a ax y z+ + where ,a ax y and az are unit vectors in ,x y and z directions respectively. If , ,a b g respectively the angles the E vector makes with ,x y and z axes respectively, they are given by which of the following ?(A) 30ca b g= = = (B) 60ca b g= = =

(C) cos3

11a b g= = - (D) cos 311a b g= = = -

MCQ 1.3.7 Which one of the following potentials does NOT satisfy Laplace’s Equation ?(A) 10V xy= (B) cosV r f=

(C) /V r10= (D) 10cosV r f= +

MCQ 1.3.8 Laplacian of a scalar function V is(A) Gradient of V

(B) Divergence of V

(C) Gradient of the gradient of V

(D) Divergence of the gradient of V

IES EC 2009

IES EC 2007

IES EC 2003

IES EC 2003



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For View Only Shop Online at www.nodia.co.inMCQ 1.3.9 A field ( 2 )x yz x z x y zA a a a5 x y z

2 3 3= + + - can be termed as(A) Harmonic (B) Divergence less

(C) Solenoidal (D) Rotational

MCQ 1.3.10 Laplace equation in cylindrical coordinates is given by

(A) 0V V VzV1 12

2 2

2

2

2

22

22

22

22

r r rr

r fd = + + =e eo o

(B) VxV

yV

zV2

2

2

2

2

2

2

22

22

22d = + +

(C) V v2

er

d =-

(D) 0sin

V r r rr V

r1 12

2

2 2 2d22

22

2q f= + =c m

MCQ 1.3.11 What is the value of the integral dlc# along the curve c ?( c is the curve ABCD in

the direction of the arrow ) ?

(A) 2 ( )/R a a 2x y+ (B) 2 ( )/R a a 2x y- +

(C) 2Rax (D) 2Ray-

MCQ 1.3.12 Given a vector field cosr aA 4 rf= in cylindrical coordinates. For the contour as

shown below, dA l:# is

(A) 1 (B) 1 ( /2)p-

(C) 1 ( /2)p+ (D) 1-

IES EC 2002

IES EC 2001

IES EE 2005

IES EE 2003




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MCQ 1.3.13 If a vector field B is solenoidal, which of these is true ?

(A) 0dB lL

: =# (B) 0dB ss

: =#

(C) 0B#d = (D) 0B:d =Y

MCQ 1.3.14 Which of the following equations is correct ?(A) a a ax x x

2# =

(B) ( ) ( ) 0a a a ax y y x# #+ =

(C) ( ) ( )a a a a a ax y z x z y# # # #=

(D) 0a a a ar r: :+ =q q

MCQ 1.3.15 Match List I with List II and select the correct answer :

List I (Term) List II (Type)

a curl 0F =^ h 1. Laplace equation

b div 0F =^ h 2. Irrotational

c div Grad 0f =^ h 3. Solenoidal

d div div 0f =^ h 4. Not definedCodes :

a b c d(A) 2 3 1 4(B) 4 1 3 2(C) 2 1 3 4(D) 4 3 1 2

MCQ 1.3.16 If A a a a2 r z= + +f , the value of dA l:# around the closed circular quadrant shown in the given figure is

(A) p (B) 42p +

(C) 4p+ (D) 22p +

***********

IES EE 2002

IES EE 2002

IES EE 2002

IES EE 2001



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SOLUTIONS 1.1

SOL 1.1.1 Option (D) is correct.Given the vector field has the only component in aq direction and its magnitude is r so as r increase from origin to the infinity field lines will be larger and directed along aq as shown in option (A).

SOL 1.1.2 Option (B) is correct. M N- 5 2 4 ( 8 7 2 )a a a a a ax y z x y z= - + - - - + 5 8 7 2a a a a a a2 4x y z x y z= - + + + - 13 5 2a a ax y z= + +So, the unit vector in the direction of (M N- ) is

a M NM N= -

-13 5 213 5 2a a aa a ax y z

x y z= + ++ +

0.92 0.36 0.14a a ax y z= + +

SOL 1.1.3 Option (C) is correct.Vector G at ( , , )2 1 3- : G 4( 2)(1) 2(2 ( 2) ) 3(3)a a ax y z

2 2= - + + - + 8 12 27a a ax y z=- + +So, unit vector in the direction of G at Q :

aG GG= a a a

a a a8 12 278 12 27x y z

x y z= - + +- + +

0.26 0. 9 0.88a a a4x y z=- + +

SOL 1.1.4 Option (B) is correct.Option (A), (B), (D) are the properties of vector product.Now we check the relation defined in option (C). Since the triple cross product is not associative in general so, the given relation is incorrect. This inequality can be explained by considering vector A B= and C ,perpendicular to A as shown in the figure.According to right hand rule we determine that B C#^ h points out of the page and so A B C# #^ h points down that has magnitude ABC .But in L.H.S. of the relation, since A B=So we have ( )A B# 0=and hence ( )A B C# # 0=Therefore A B C# #^ h 0 A B C! # #= ^ h




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SOL 1.1.5 Option (B) is correct.As the direction of cross vector is normal to the plane. So, direction of B C# will be normal to the plane defined by the three vectors.Now the dot product of two mutually perpendicular vectors is always zero and since the direction of B C# will be perpendicular to the plane of vector A. So A B C: #^ h will be zero.

SOL 1.1.6 Option (C) is correct.Consider the two vectors A and B are as shown below.

As the angle between the two vectors is a.So component of vector A along B is A1 cos Aa= ^ h

cosine of the angle between the two vectors is defined as

cosa ABA B:=

So, A1 AB AA B:= b l

BA B:=

SOL 1.1.7 Option (C) is correct.Cross product of two parallel vector fields is always zero since the angle between them is q 0c= .i.e. A B# 0=



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a a a1 2 3

6

z

a b -

r f

0=

a a a12 3 3 6 2 zb a b a- - + + + -r f^ ^ ^h h h 0=Solving it we have, 4b =- and 2a =-

SOL 1.1.8 Option (A) is correct.Dot product of the two orthogonal vectors is always zero.i.e. A B: 0= (4)(1) (2 )(4) ( )( 4)k k+ + - 0= 4 8 4k k+ - 0= 4k 4=- k 1=-

SOL 1.1.9 Option (A) is correct.From the given field vector we have the component Ex 2coszy x3 2= .So for the given condition E 0x = We have, 9 (2 )coszy x2 0=This condition met when, z 0=or, y 0=or, cos x2 0= 2x& /2p= x& /4p=Therefore the planes on which field component Ex will be zero are z 0= y 0= and x /4p=

SOL 1.1.10 Option (A) is correct.From the given field vector we have the field components Ey 2sinzy x2=and Ez siny x2 22=Now, in the plane 4y z 0- = y z4& =So, Ey sinz z x8 4 2= ^ h sinz x32 22= Ez sinz x2 4 22= ^ h sinz x32 22=Thus Ey Ez=

SOL 1.1.11 Option (C) is correct.For the given condition 0E = , we must have Ex 0E Ey z= = =i.e. coszy x9 22 8 2 2 2 0sin sinzy x y x2= = =This condition met in the plane y 0= .

SOL 1.1.12 Option (B) is correct.Since the two points are defined in different coordinate system so we represent the point Q in Cartesian system as x ( ) .cos cos4 50 2 57r f= = - =




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y ( ) .sin sin4 50 3 064r f= = - =-and z 2=So, the distance between the two points ( , , )P 2 3 1- and ( . , . , )Q 2 57 3 064 2- is given as PQ ( . ) ( . ) ( )2 57 2 3 064 3 2 12 2 2= - + - - + + 6.78 units=

SOL 1.1.13 Option (A) is correct.Since, the options include spherical as well as cylindrical representation of A, so, we will transform the vector in both the forms to check the result.The components of vector field A are A 1x = , A 0y = and A 0z =Now, we transform the vector components in cylindrical system as

AAAz

r

f

R

T

SSSS

V

X

WWWW

cossin

sincos

AAA0 0

001

x

y

z

ff

ff= -

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

Ar ( )(1)cos cosf f= = Af ( )( )sin sin1f f= - =- Az 0=So, the vector field in cylindrical system is ( , , )zA r f cos sina af f= -r f

Hence, both the options (A) and (B) are incorrect.Again, we transform the vector components in spherical system as

AAA

r

q

f

R

T

SSSS

V

X

WWWW

sin coscos cos

sin

sin sincos sin

cos

cossin

AAA0

x

y

z

q fq f

f

q fq f

f

qq=

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

Ar ( )( )sin cos sin cos1q f q f= = Aq ( )( )cos cos cos cos1q f q f= = Af ( )(1)sin sinf f= - =-So, the vector field in spherical system is ( , , )rA q f sin cos cos cos sina a arq f q q f= + -q f

SOL 1.1.14 Option (C) is correct.We transform the given vector field in spherical system.Since the given vector field is F 10ax=The Cartesian components of the field are 0F 1x = , F 0y = , F 0z = .So, the spherical components of vector field can be determined as

FFF

r

q

f

R

T

SSSS

V

X

WWWW

sin coscos cos

sin

sin sincos sin

cos

cossin

FFF0

x

y

z

q fq f

q

q fq f

f

qq=

--

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

So, we get Fr sin cos10 q f= Fq cos cos10 q f=and Ff sin10 q=-



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For View Only Shop Online at www.nodia.co.inNow, for the given point (x 3= , y 2= , z 1=- ) we have r ( ) ( ) ( )3 2 1 142 2 2= + + - =

q tan zx y1

2 2

= +-

( )( ) ( )

105.5tan1

3 212 2

c= -+ =-

f p

f tan xy1= -a k 33.7tan 3

21 c= =-b l

Putting all the values in the matrix transformation, we have Fr (105.5 ) (33.7 ) 8sin cos5 c c= = Fq 10 (105.5 ) (33.7 ) .cos cos 2 2c c= =- Ff 10 (33.7 ) .sin 5 5c=- =-Therefore, the vector field in spherical coordinate is F F F Fa a ar r= + +q q f f

8 2.2 5.5a a ar= - -q f

SOL 1.1.15 Option (C) is correct.Since z -axis is normal to the surface z 5= , so first of all we will find the angle between z -axis and A which can be easily obtained from the figure shown below :

i.e. cosf AAz=

( ) ( ) ( )5 10 33

1343

2 2 2=

- + +=

f .cos1343 74 98 751 c c.= =-

c m

Therefore, the angle between surface z 5= and vector A is (90 ) 15c cf- = .

SOL 1.1.16 Option (B) is correct.The given line x 6= , z 2=- is parallel to y -axis. So, the component of A parallel to the given line is Ay ( )A a ay y:= ( 2 20 4 ) aa a a az y y:= - + +r f6 @

( 2 20 )sin cos ayf f= - +At point P , 90cf = , so, Ay 2ay=-

SOL 1.1.17 Option (A) is correct.The given point is shown below :




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After 120c rotation looking down the axis the new co-ordinate axes (xl, yl, zl) will be as shown below :

So, the rotation carries z axis into y ; y -axis into x and x into z .therefore the new co-ordinates of point P are : xl z c= = yl x a= = zl y b= =i.e. (c , a , b) is the co-ordinates of point P in the transformed system.

SOL 1.1.18 Option (B) is correct.The position vector can be defined as : R x y za a ax y z= + + R x y z2 2 2= + +

So, gradR xR

yR

zRa a ax y z2

222

22= + +

x y zx

x y zy

x y zza a a2

1 221 2

21 2

x y z2 2 2 2 2 2 2 2 2=

+ ++

+ ++

+ +

x y z

x y za a ax y z2 2 2

=+ +

+ + RR=



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For View Only Shop Online at www.nodia.co.inSOL 1.1.19 Option (D) is correct.

A:d ( ) ( ) ( )xy

yxy x z

zx yz2 4 22 2 2

22

22

22= + + + + +

x y0 2 2= + + ( )x y2= +

SOL 1.1.20 Option (A) is correct.We have the vector field components as sinF r f=r , F z2r=f and cosF zz f=

Now, F#d F F F

a a a1

z

z

z

r

r

r= 2

222

22

r

r

r

f

f

f

cos cos sinzz

z zz

a a1 12

22

22

22

22

r f f rr r r r f r f= + - -r f; ;E E

sinz a1z

2

22

22

r r rr f r f+ -; E

sin cosz za a a1 0 0 1 3 z3 2

r f r r r r f= - - - - + -r f6 6 6@ @ @

( ) (3 )sin cosz za a1z

3

r f r r f=- + + -r

At point ( , / , )P 1 2 2p F#d 1(2 1 1 ) (3 1 2 0)a az3

# # #=- + + -r

3 6a az=- +r

SOL 1.1.21 Option (A) is correct. D yz xz xya a ax y z= + +

D:d ( ) ( ) ( )xyz

yxz

zxy

22

22

22= + + 0=

D#d yz xz xy

a a ax

x

y

y

z

z= 22

22

22

( ) ( ) ( )x x y y z za a ax y z= - - - + - 0=

SOL 1.1.22 Option (D) is correct.Curl of the unit vector ar is

a#d r a a a

1

1 0 00

z

zr

r= =2

222

22

r

r

f

f

The curl of unit vector af is

a#d f a a a

a a1

0 1 0

1z

z zz

22

r

r

rr r

rr= = =2

222

22

r

r

f

f

^ h

and curl of unit vector az is

az#d a a a

1

0 0 10

z

zr

r= =2

222

22

r

r

f

f




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SOL 1.1.23 Option (A) is correct.Options (A), (B) and (C) are properties of d operator where as :

gA

#d c m ( ) ( )

gg gA A

2# #d d= +

SOL 1.1.24 Option (A) is correct.In a cylindrical coordinate system Laplacian of a scalar field is defined as

f24 f fzf1 1

2 2

2

2

2

22

22

22

22

r r r r r f= + +e o

( )sin sin sin cosz z z1 2 8 1 2 622

22

22

r r r f rr

r f f f f= + + - -c m

( )sin cosz

z2 6 2

22 r f f+ +

(2 16 ) (2 6 2 )sin sin cos cosz z z1 1 622 2

r f rr

r f f f= + - + +

cos cosz16 6 6 222

2

fr

f= + -

At point , / ,P 3 2 6p^ h

f2d 16 ( 1) 400 96 36#

#= + - - =

SOL 1.1.25 Option (A) is correct.A vector field is called conservative (irrotational) if its curl is zero.i.e. Md# 0=

cos cosz xz y kx x xz

a a a

2

x

x

y

y

z

z

+22

22

22 0=

(0 0) ( ) (2 ) 0cos sin cos sinxz xz xz xz xz xz ka a a1x y z- - - - + + - = k2 1- 0=

or, k 21=

SOL 1.1.26 Option (B) is correct.For a scalar field to be harmonic, g2d 0=

xg

yg

zg

2

2

2

2

2

2

22

22

22+ + 0=

2 k y1 2+^ h 0= which results in k 21=-

SOL 1.1.27 Option (D) is correct.Consider V is a scalar field. So the gradient of the field is

V4 xV

yV

zV

22

22

22= + +

and the curl of the gradient of the field is



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( )V4# 4 a a ax

x

xV

y

y

yV

z

z

zV

= 22

22

22

22

22

22

R

T

SSSS

V

X

WWWW

z yV

y zV

x zV

z xV

x yV

y xVa a ax y z

2 2 2 2 2 2

222

222

222

222

222

222= - + - + -c c cm m m 0=

So the curl of the gradient of any scalar field is zero everywhere.

SOL 1.1.28 Option (D) is correct.For vector A to be solenoidal its divergence must be zero.i.e. A:d 0=

( 4 ) (2 3 ) (4 3 )xx z

yx z

zx y cz

22

22

22+ + - + + - 0=

c1 0+ - 0= c 1=

SOL 1.1.29 Option (B) is correct.For a vector function to be irrotational its curl must be zero.i.e. A#d 0=

A A A

a a ax

x

x

y

y

y

z

z

z

22

22

22 0=

( ) ( ) ( )x k z k x z x k y z

a a a

4 5 4 2

x

x

y

y

z

z

1 2 3+ - - +22

22

22 0=

( ( 5)) (4 ) ( )k k ka a ax y z3 1 2- - - - - + 0= ( 5)k3- - - 0= & k3 5= 4 k1- 0= & k1 4=and 0k2 - 0= & k2 0=So k1, k2 and k3 are 4, 0 and 5 respectively.

SOL 1.1.30 Option (C) is correct.The unit vector normal to a given plane f 0= is

an ff

dd=

The given equation is x y z2 4 6+ + 7= x y z2 4 6 7+ + - 0=So, f x y z2 4 6 7= + + -and gradient of f is fd 2 6a a a3x y z=- + + fd 2 4 6 562 2 2= + + =

So, an 2 4 6a a a561

x y z= + +^ h 2 3a a a141

x y z= + +^ h




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SOL 1.1.31 Option (D) is correct.Consider the differential displacement, dl dx dy dza a ax y z= + +

So dlC# dx dy dza a a

Cx

Cy

Cz= + +c c cm m m# # #

For a contour the initial and final points are same. So, all the individual integrals described above will be zero. Therefore,

dlC# 0=

SOL 1.1.32 Option (B) is correct.According to stoke’s theorem.

dA lL

:# dA S:#d= ^ h#

So U dlC

:d̂ h# U dS# :d d= ^ h6 @#Since U#d d̂ h 0= (curl of the gradient of a scalar field is always zero)So the contour integral is zero.

SOL 1.1.33 Option (B) is correct.According to the divergence theorem surface integral of vector over a closed surface is equal to the volume integral of its divergence inside the region defined by closed surface.

i.e. dA SS

:# dvAv

:d= ^ h#

Now, A:d xyz

yz x

zxy2 22

22

22

22= + +^ ^ ^h h h

0 0 0= + +

So dA S:# 0=

***********



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SOLUTIONS 1.2

SOL 1.2.1 Option (D) is correct.Consider that the cube has its edges on the ,x y and z - axes respectively as shown in the figure. As the angle between any of the two body diagonals of the cube will be same so we determine the angle q between the diagonals OB and AC of the cube.

From the figure we get the co-ordinates of points A, B and C as :A , ,0 0 3" ^ h B , ,3 3 3" ^ h and C , ,3 3 0" ^ h

So, the vector length, OB 3 3 3a a ax y z= + +and AC 3 3 3a a ax y z= + -For determining the angle q between them, we take their dot product as OB AC:^ ^h h cosOB AC q= 9 9 9+ - cos3 3 3 3 q= ^ ^h h

So, the angle formed between the diagonals is

q 70.53cos 411 c= =-

b l

SOL 1.2.2 Option (C) is correct.For the given points A, B , C , the vector length,




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AB 20 18 10a a ax y z= + -and AC 10 8 15a a ax y z=- + +Since the cross product of two vectors is always perpendicular to the plane of vectors. So the unit vector perpendicular to the plane of triangle is given by

an AB ACAB AC

#

#=

now, AB AC# a a a2010

188

1015

x y z

=-

-

R

T

SSSS

V

X

WWWW

10 8 20 15 ( 10) ( 10)a a18 15 x y# # # #= - - - - - -^ h6 6@ @

10 18 a20 8 z# #+ - -^ h6 @

350 200 340a a ax y z= - +

So, an 350 200 340350 200 340a a aa a ax y z

x y z=- +- +

( ) ( ) ( )350 200 340a a a350 200 340

x y z2 2 2

=+ - +- +

0.664 0.379 0.645a a ax y z= - +

SOL 1.2.3 Option (D) is correct.The unit vector in the direction of vector AC is given by

aAC ACAC=

( ) ( ) ( )10 8 15a a a10 8 15x y z2 2 2

=- + +- + +

0.507 0.406 0.761a a ax y z=- + + 0. 1 0.41 0.76a a a6 x y z=- + +Since the cross product of two vectors is always perpendicular to the plane of vectors. So, the unit vector in the plane of the triangle which is perpendicular to AC is given by cross product of the unit vector perpendicular to the plane of the triangle and the unit vector aAC . i.e. aP a an AC#= 0.550 0.832 0.077a a ax y z=- - +

SOL 1.2.4 Option (A) is correct.Unit vector in the direction of AB is given by

aAB ( ) ( ) ( )a a a

20 18 1010 18 10x y z

2 2 2=

+ + -+ -

0.697 0.627 0.348a a ax y z= + -A non unit vector in the direction of bisector of interior angle at A is defined as

( )a a21

AB AC+ . . . . . .a a a a a a21 0 697 0 627 0 348 0 507 0 406 0 761x y z x y z= + - - + +6 @

0.095 0.516 0.207a a ax y z= + +So the unit vector in the direction of bisector of interior angle at A is given by

abis (0.095) (0.516) (0. )0.095 0.516 0.207a a a

207x y y2 2 2

=+ ++ +

0.168 0.915 0.367a a ax y z= + +



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For View Only Shop Online at www.nodia.co.in 0.17 0.92 0.37a a ax y z= + +

SOL 1.2.5 Option (B) is correct.The vector field F can be written in cartesian system as

( , , )x y zF x yx ya ax y

2 2=++

x yx

x yya ax y2 2 2 2=

++

+

( , , )x y zF cos sina ax y2 2rr f

rr f= + ( ,cos sinx yr f r f= = )

( )cos sina a1x yr f f= +

The components of vector field F are

1 cosFx r f= , sinF 1y r f= and 0Fz =

So the components of vector field F in cylindrical system can be expressed as

FFFz

r

f

R

T

SSSS

V

X

WWWW

cossin

sincos

FFF0 0

001

x

y

z

ff

ff= -

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

Fr cos sin1 2 2

r f f= +6 @ 1r=

Ff ( ) 0cos sin sin cosr1 f f f f= - + =6 @

Fz 0=

So the vector field ( , , )zF r f a1r= r

At the point (P 2r = , /4f p= , .z 0 1= )

F 0.5a a21= =r r

SOL 1.2.6 Option (D) is correct.Any vector field can be represented as the sum of its normal and tangential component to any surface as A A At n= +where At is tangential component and An is normal component to the surface r 20= at point ( , , )P 20 150 330c c .So, An 20ra ar r= =and therefore, At AA n= -

sin cosrra a244

q f=- +q f

0.043 100a a= +q f

SOL 1.2.7 Option (A) is correct.Any vector field can be represented as the sum of its normal and tangential components to any surface as A A At n= +Here At and An are the tangential and normal components to the conical surface




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Since the unit vector normal to the conical surface is aq . So, An 8a=- q

and therefore the tangential component to the cone is At A An= - 12 9a ar=- + f

SOL 1.2.8 Option (B) is correct.Consider the unit vector perpendicular to A and tangent to the cone 150cq = is b b ba ar r= + f f (Tangential component to the cone will have b 0=q )Now the magnitude of unit vector is 1 So, b br

2 2+ f 1= ...(i)and the dot product of mutually perpendicular vectors is zero.So, bA: 0= 9b b6 r- + f 0=

bf b43r= ....(ii)

So, from equation (i) and (ii) we have

b 1 916

r2 +b l 1=

br 53= , b 5

4=f

Therefore, b (3 4 )a a51

r= + f

SOL 1.2.9 Option (D) is correct.The separation vector R can be defined as : R ( ) ( ) ( )x a y b z ca a ax y z= - + - + -

and R ( ) ( ) ( )x a y b z c2 2 2= - + - + -

So, R1db l ( ) ( ) ( )

xx a y b z c a/

x2 2 2 1 2

22= - + - + - -

6 @

( ) ( ) ( )yx a y b z c a/

y2 2 2 1 2

22+ - + - + - -

6 @

( ) ( ) ( )zx a y b z c a/

z2 2 2 1 2

22+ - + - + - -

6 @

( ) ( ) ( )R x a R y b R z ca a a21

21

21/ / /

x y z3 2 3 2 3 2=- - - - - -- - -

^ ^ ^h h h

( ) ( ) ( )

Rx a y b z c

Ra a a R

/x y z

3 2 3=- - + - + - =-

SOL 1.2.10 Option (C) is correct.The gradient of a scalar field at its maxima is zero. So at the top of hill h4 0=or, (12 4 36) (16 4 56)x y y xa ax y- + + - - 0=Therefore both the components will be equal to zeroi.e. x y12 4 36- + 0=and, y x16 4 56- - 0=



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For View Only Shop Online at www.nodia.co.inSolving the two equations, we get, x 2=- , y 3=Thus the top of the hill is located at 2 miles south ( 2- miles north) and 3 miles east of the railway station.

SOL 1.2.11 Option (C) is correct.Consider the position vector of point P is R 2x y za a a5x y z

4= + +So, the magnitude of R is R x y z2 2 2= + +and unit vector in the direction of R is

aR x y z

x y za a ax y z2 2 2

=+ +

+ +

Therefore, the vector field F at point P is

F Ra10R2= 10R x y zx y za a ax y z

2 2 2 2=

+ ++ +

e o x y zx y za a a

10 /x y z2 2 2 3 2= + ++ +

^ h= G

The divergence of the field F is given as

F:d x x y z

xy x y z

yz x y z

z10 / / /2 2 2 3 2 2 2 2 3 2 2 2 2 3 222

22

22=

+ ++

+ ++

+ +^ ^ ^h h h> H

10( ) ( )

( )( )x y z x y z

x xx y z

123 2 1

/ / /2 2 2 3 2 2 2 2 5 2 2 2 2 3 2=+ +

-+ +

++ +=

23( )

(2 )( )

123( )

(2 )x y zy y

x y z x y zz z

2 2 2 5/2 2 2 2 3/2 2 2 2 5/2-+ +

++ +

-+ + G

10( )( )

R x y zx y z3 3 /3 2 2 2 5 2

2 2 2

= -+ ++ +

> H

10R R3 3

3 3= -: D 0=

But at origin (x 0= , y 0= , z 0= ) the position vector R 0= and so the expression for field F blows up. Therefore, F:d is infinite at origin and zero else where.

SOL 1.2.12 Option (D) is correct.The circulation of A around the route is given by

dA l:# dA l321

:= + +c m###where the route is broken into segments numbered 1 to 3 as described below1st segment : ( , , ) ( , , )0 0 0 2 0 0"

x changes from 0 to 2, y 0= , z 0=

So, dA l1

:# x dx x3 3 32

0

2 3

0

2

= = b l# 8= (d dxl ax= )

2nd segment : ( , , ) ( , , )2 0 0 2 2 0"

x 2= , y changes from 0 to 2, z 0= .

dA l2

:# yz dy6 00

2= =# (d dyl ay= )

3rd segment : ( , , ) ( , , )2 2 0 2 2 2"




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x 2= , y 2= , z changes from 0 to 2.

dA l3

:# ( ) 24y dz z3 3 42

0

2

0

2

#= = =# (d dzl az= )

So total line integral will be :

dA l:# 8 0 24 32 intu s= + + =

SOL 1.2.13 Option (A) is correct.For the straight line from origin to the point (2, 2, 2) we have the relation between the coordinates as x y z2= =or, dx dy dz= =and the line integral along straight line is given as dl dx dy dza a ax y z= + +Therefore, the line integral of the vector field along the straight line is given as

dA l:# x dx yzdy z dz3 6 32 2= + +# # #

x dx x dx x dx3 6 32 2 2= + + ###

x dx12 2= # ( )x x12 3 43

0

23

0

2= =b l

8 intu s3 24#= =

SOL 1.2.14 Option (C) is correct.For the closed path defined,the line integral in forward path = 32 unitsthe line integral in return path = 32- units.So, total integral in the closed path is :

dA l:# 32 32 0 units= - =

SOL 1.2.15 Option (A) is correct.The circulation of A around the path L can be given as

dA l:# dA l321

:= + +c m###where the route is broken into segments numbered 1 to 3 as shown in figure below :

1st segment : ( 30cf = , z 2= , 0 5# #r ) and d dl ar= r



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So, dA l1

:# sin d0

5r f r= # d20

5 r r= # 425= ( 30cf = )

2nd segment : ( 5r = , z 2= , 30 180c c# #f ) and d dl af= f

So, dA l2

:# d0 0f= =#3rd segment : ( 180cf = , z 2= , 5 0# #r ) and d dl ar= r

dA l3

:# sin d5

0r f r= # 0= ( 180cf = )

Therefore, the circulation of vector field along the edge L is

dA lTotal

:# 0 0 units425

425= + + =

SOL 1.2.16 Option (D) is correct.Volume integral of the function is given by

V f dxdydz= ### z dxdydz30 2= ###The surface of the tetrahedron will have a slope x y z+ + 1=-So, for a given value of y and z , x varies from 0 to ( 1 )y z- - - and x integral will be

dx( )y z

0

1- - -

# 1 y z=- - -

again for a given value of z , y ranges from 0 to ( )z1- - . So y -integral will be :

( )y z dy1( )z

0

1

- - -- -

# ( )z y y1 2

( )z2

0

1

= - - -- -

; E

( 1 )( ) ( )

zz z

21

212

2 2

= - - - - - = - -

z z21

23

2

= + +

Now there is only one remaining variable z that ranges from 1- to 0. So we have the volume integral of the function as

V z z z dz30 21

22

2

1

0= + +

-b l#

z z z30 6 4 103 4 5

1

0

= + +-

: D

30 0 61

41

101= + - +: D

30 201

23

#= =

SOL 1.2.17 Option (D) is correct.The net outward flux through the closed cylindrical surface will be summation of the fluxes through the top(in az direction), bottom(in az- direction) and the curved surfaces(in ar direction) as shown in the figure.




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Since, the vector field has no z -component so, the outward flux through the top and bottom surfaces will be zero. Therefore, the total outward flux through the closed cylindrical surface will be only due to the field component in ar direction(flux through the curved surfaces) which is given as

dA S:# A d dzz 0

2

0

2r f= r

f

p

==^ ^h h##

cos d dz4z

2 2

0

2

0

2 r f r f=f

p

==c ^m h##

At 2r = , dA S:# cos d dz42 3

2

0

2

0

2f f=

p^ h; ;E E# #

2 2 4# #p p= =

SOL 1.2.18 Option (A) is correct.According to divergence theorem surface integral of a vector field over a closed surface is equal to the volume integral of its divergence inside the closed region:

i.e. dA S:# ( )dvA:d= #Divergence of vector A is

A:d A AzA1 1z2

222

22

r r r r f= + +r f^ h

( ( )) ( )sin sinzz1 4 2 1 2 62

22

22

22

r r r r r f r f r f= + + +^ h

8 4 2 ( ) 6sin cos 22f f= + + +

8 4 2 2 6sin cos sin2 2 2f f f= + + - + 16=So the surface integral is

dA S:# ( )dvA:d= # ( ) d d dz16 r r f= ###

16 d d dz/

0

2

0

5

0

2r r f=

p

# ## 16 2 2 5# # #p=

80p=



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For View Only Shop Online at www.nodia.co.inNote :The surface integral can also be evaluated directly without using divergence theorem but it will be much complicated as there are 5 different surfaces over which we will have to integrate the given vector field.

SOL 1.2.19 Option (C) is correct.According to stoke’s theorem, line integral of a vector function along a closed path is equal to the surface integral of its curl over the surface defined by the closed path.

i.e. dG lL

:# dG S#d= ] g#Curl of the vector field is G#d 6x az2=-and the differential surface vector dS ( )dxdy az= -So the line integral of the given vector field is

dG lL

:# dG S#d= ] g# x dxdy6 2=- -##

x dydx x dydx6 6xx 2 2

0

2

1

2

00

1= +

-

####

( )x xdx x x dx6 6 22 2

1

2

0

1= + -##

6 4 6 32

4x x x4 3 4

0

1

1

2

= + -: :D D

316

416

32

416 4

1 0 6= - + - - -b b bl l l; E 7= units

SOL 1.2.20 Option (D) is correct.The relationship between cartesian and spherical co-ordinates is : r x y z2 2 2= + + , sinr x y2 2q = + x sin cosr q f= , y sin sinr q f=We put these values in the given expression of vector field as

F x y

x y z x x y ya a a a4 42 2 2 2x y x y2 2

2 2 2

=+

+ + + - +9 C

( ) ( )sinsin cos sin cos sinr

r r a a22 x yqq f f f f= - + +6 @

( ) ( )cos sin cos sinr a ax yf f f f= - + +6 @

Now we transform the vector field from cartesian system to spherical system :

FFF

r

q

f

R

T

SSSS

V

X

WWWW

sin coscos cos

sin

sin sincos sin

cos

cossin

FFF0

x

y

z

q fq f

q

q fq f

f

qq=

--

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

Fr cos sin sin cos sin sin cos sinr rf f q f q f f f= - + +^ ^ ^ ^h h h h

sinr q= Fq cos sin cos cos cos sin cos sinr rf f q f f f q f= - + +^ ^ ^h h h




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cosr q= Ff cos sin sin cos sin cosr rf f f f f f= - - + +^ ^ ^h h h

r=i.e. F sin cosr r ra a arq q= + +q f

The differential surface vector over the surface S1 is dS sinr d d ar2 q q f=and the surface S1 is defined in the region r 2= , 0 30< < cq , 0 2< <f pSo, surface integral through out the surface S1 will be :

dF SS1

:# sinr d d3 2

0

30

0

2q q f=

cp

##

sind d80

22

0

30f q q=

cp

# # 4 3 23p p= -; E (at r 2= )

.2 276= .2 3=

SOL 1.2.21 Option (B) is correct.The vector function in spherical form as calculated in previous question is : F sin cosr r ra a ar

2q q= + +q f

The differential surface vector over the surface S2 is dS sinr d draq f= q

and the surface S2 is defined in the region r0 2# # , 0 2# #f p , 30cq =So, the surface integral of the field over the surface S2 is :

dF SS2

:# sin cosr d dr2

0

2

0

2q q f=

p

## 34 3p=

SOL 1.2.22 Option (C) is correct.For a vector function to be irrotational its curl must be zero. Now we check it for vector A.

A#d ( ) ( ) ( ) ( )yx y z

zz

zx z

xx y za a3 3 3x y2

222

22

22= - - - - + + - - -; :E D

( ) ( )x

zyx z a3 z2

222+ - - +; E

( 3 3) (1 1) (0 0)a a ax y z= - + + - + - 0=So, vector A is irrotational.again for a vector to be solenoidal its divergence must be zero. So we take the divergence of the vector A as

A:d ( ) ( ) ( )xx z

yz

zx y z3 3

22

22

22= + + - + - -

1 0 1 0= + - =So, vector A is solenoidalThus the vector A is both irrotational and solenoidal.Note: Since the curl of the gradient of a scalar field is zero. So, we can have directly the result A4# ( )f4# 4= - ( )f4 4#=-= 0=

SOL 1.2.23 Option (A) is correct.We have



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A fd=- xf

yf

zfa a ax y z2

222

22=- - -

Comparing it with the given vector we get :

xf

22 ( )x z=- + ( , )f x xz f y z2

2

1& =- - +

yf

22 z3= ( , )f yz f x z3 2& = +

zf

22 ( )x y z3=- - - ( , )f xz yz z f x y3 2

2

3& =- + + +

In conclusion, from all the three results, we get

f x xz yz z2 3 22 2

=- - + +

SOL 1.2.24 Option (D) is correct.Given the vector field, A 3( )y xa ax y= +The differential line vector in the cartesian coordinate system is dl dx dy dzaa ax y z= + +

So, dA l:# ydx xdy3 3= +# #The given curve is, y /x 2=So, we put x y2 2= and dx ydy4= in the line integral

dA l:# y dy y dy12 62

1

22

1

2= +# # y3

18 312= 6 @

6 units5 30#= =

SOL 1.2.25 Option (B) is correct.

Given the vector field F cos cosr r

a a1 2 2r r3 3

2f f= + =

and the differential surface vector over the outer spherical surface is dS ( )sinr d d ar2 q q f= (for r 2= ,0 # #i r ,0 2# #f p)So the surface integral over the outer spherical surface is

dF S:# cos sinr

r d d23

22

00

2 f q q f=pp

c ^m h## 2p=-

SOL 1.2.26 Option (D) is correct.Consider that the vector A is in ax direction as shown in the figure.




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So we can write the vectors in cartesian form as A 4ax= A 4=^ h

and B 30 30cos sinB Ba ax yc c= + -^ h

a a23 3

23

x y= - B 3=^ h

Now the resultant vector, R 6 8A B= - 3.22a a6x y= +So, R .3 22 122 2= +^ ^h h 12.43 units=and angle that R makes with x -axis is

q ..cos 12 43

3 221= -b l

75c=So the graphical representation of vector R is

SOL 1.2.27 Option (A) is correct.We go through all the options to check the direction of the vector R for the corresponding directions of A, B and C .Option (A)

Since the direction of cross product is normal to the plane of vectors and determined by right hand rule. So B C# has the direction in which thumb indicates when the curl of the finger directs from B to C . Thus B C# will be directed out of the paper and so we get direction of A B C# #^ h toward east. So the given direction



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For View Only Shop Online at www.nodia.co.inof R is incorrect.

In Option (B) : Direction of B C#^ h is out of the paper so, R A B C# #= ^ h will be directed toward west.

In Option (C): Direction of B C#^ h is into the paper so, R A B C# #= ^ h will be directed toward north.

In Option (D) : Direction of B C#^ h is into the paper so, R A B C# #= ^ h will be directed toward south. So the given direction is correct.

SOL 1.2.28 Option (A) is correct.As the vectors B and C are defined in cylindrical system. So, we transform the vector in cartesian form as belowGiven the vector field B 3a a az= + +r f

the cylindrical components B 1=r , B 1=f , B 3z =So the cartesian components of vector B is

BBB

x

y

z

R

T

SSSS

V

X

WWWW

cossin

sincos

BBB0 0

001 z

ff

ff=

- r

f

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

Bx cos sinB Bf f= -r f cos sinf f= - B B 1= =r f^ h

By sin cosB Bf f= +r f sin cosf f= + B B 1= =r f^ h

Bz B 3z= =




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and so the vector field in cartesian system is B 3cos sin sin cosa a ax y zf f f f= - + + +^ ^h h

So at the point , ,2 2 3pa k

B 3a a ax y z=- + +now we transform the vector field C 3a a2 z= +r in cartesian system.the cylindrical components, Cr 2= , C 0=f , C 3z =So the cartesian components of vector C is

CCC

x

y

z

R

T

SSSS

V

X

WWWW

cossin

sincos

CCC0 0

001 z

ff

ff=

- r

f

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

So, Cx cos2 f= C 2=r

Cy sin2 f=and Cz C 3z= = C 3cos sina a a2 2x y zf f= + +

So, at the point , ,3 43 9p

b l

C 3 3cos sina a a2 43 2 4x y zp p= + +b bl l

3a a a2x y z=- + +So all the three vectors are same at their respective points.

SOL 1.2.29 Option (D) is correct.For checking whether a vector is perpendicular to a given vector or not we take their dot product as the dot product of the two mutually perpendicular vectors is always zero.Now we have A B+ 4 4 4a a ax y z= + +So we take the dot product of ( )A B+ with the all given options to determine the perpendicular vector. In option (A). 4 4 4a a a a a4 4x y x y z:- + + +^ ^h h 16 16 0=- + =In Option (B) a a a a a4 4 4 4 4y z x y z$+ + +^ ^h h 16 16= + 32 0!=In Option (C) a a a a a4 4 4x z x y z$+ + +^ ^h h 4 4 8 0!= + =

SOL 1.2.30 Option (C) is correct.The given gradient is , ,V x y zd ^ h 1.5 0.5x yz x z x yza a ax y z

2 2 3 2 3= + +So,we have the components as,

xV22 . x yz1 5 2 2=



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V . , . ,x yz f y z x yz f y z1 5 3 0 53

21

3 21= + = +^ ^h h

yV22 . x z0 5 3 2=

V . ,x yz f x z0 5 3 22= + ^ h

zV22 x yz3=

V , . ,x yz f y z x yz f y z2 0 53 2

33 2

3= + = +^ ^h h

Thus by comparing all the results we get, V . x yz0 5 3 2=

SOL 1.2.31 Option (B) is correct.Consider the given plane xyz 1= xyz 1- 0=So, function, f xyz 1= - gradient of function, fd yz xz xya a ax y z= + +Since gradient of the function of a plane is directed normal to the plane so the normal vector to the plane at the point , ,2 4 8

1^ h is

fd 8a a a21

41

x y z= + +

Now consider , ,x y z^ h lies in the given surface xyz 1= . So the tangential vector to the given surface at the point , ,2 4 8

1^ h is

T 2x y za a a4 81

x y z= - + - + -^ ^ bh h l

This vector will be perpendicular to fd .So, fT : d^ ^h h 0= (dot product of perpendicular vector)

x y z21 2 4

1 4 8 81- + - + -^ ^ bh h l 0=

2 32x y z4+ + 24=

SOL 1.2.32 Option (A) is correct.As the integral is to be determined in spherical volume so, we transform the function in spherical system as, x2 sin cosr2 q f=and so, we have the integral

xdv2v# 2 sin cos sinr r drd d

//2

r 0

2

00

12q f q q f=

f

p

q

p

===^ ^h h###

sin cosr dr d d2/ /

3

0

12

0

2

0

2

q q f f=p p

; ; ;E E E# # #

2 sin sinr4 2 4

2 //

4

0

1

0

2

02q q f= -

pp

; : 6E D @

2 241

4 8 4p p p

# # #= = =a k




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SOL 1.2.33 Option (B) is correct.Contour integral of the field vector is evaluated in 3 segments as shown below

In Segment (1) dl d ar= r 0 2< #r at 0f =

So, dA l1

:# cos d0

2r f r=

r=# cos0 2

2

0

2r= ^ h; E 2 unit24= =

In segment (2) dl d ar f= f /0 2< #f p at 2r =

So, dA l1

:# d2/

0

2 r r f=f

p

=^ h# 2 /

02f= p

6 @ p=

In segment 3 dl d ar=- r 0 2# #r , at /2f p=

So, dA l3

:# /cos d20

2r p r=-

r=^ ^h h# 0=

So the contour integral is

dA lC

:# dA l321

:= + +; 6E @### units2 p= +^ h

SOL 1.2.34 Option (A) is correct.For the given contour C1

dl d ar f= f 0 2# #f p at 3r =

So, dA lC1

:# d 9 2 180

2

#r r f p p= = =f

p

=^ h#

and for the contour C2

dl d ar f=- f 0 2# #f p at 1r =

So, dA lC2

:# d 20

2r r f p=- =-

f

p

=^ h#

Therefore, the ratio of the contour integral is

d

d

A l

A l

C

C

2

1

:

:

#

#

218 9

pp=

-=-

^ h

SOL 1.2.35 Option (A) is correct.Let us consider a contour abcd as shown in the figure.



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As vector A has only aq component so its integral will not exist along segments ab and cd and so the contour integral for abcd is

dA labcd

:# dA ldacdbcab

:= + + +c m####

For bc segment, r 1= and 0 2# #q p

dl rd aq=- q

and for da segment r d= , and /0 2# #q p dl rd aq= q

So, dA labcd

:# re rd r

e rd// r r

0

2

0

2

q q=- +q

p

q

p - -

==b ^ b ^l h l h##

/e e2 21 p p=- + d- -^ a ^ ^h k h h e e2

1p= - + d- -^ h

As for the given contour C , d tends to zero

So, dA lC

:# lim dA labcd0

:="d

# lim e e20

1p= - +"d

d- -^ h e2 1 1p= - -

^ h

Note : Most of the students do a mistake here by directly integrating the given vector along given contour C but as the vector A includes exponential which is not zero at origin and so at r 0= , 0rdA a: !q f^ ^h h therefore we have taken the contour integral in the form of limits.

SOL 1.2.36 Option (C) is correct.The divergence of unit vector ar is

ar:d 1r r

rrr r

1 1 2 22

222

2= = =^ ^h h

the divergence of unit vector aq is

a:d q sin sin sincos cot

r r r1

22

q q f qq q= = =^ h

and the divergence of unit vector af is

a:d f sinr1 1 0

22

q f= =^ h

SOL 1.2.37 Option (A) is correct.A vector can be expressed as the gradient of a scalar if it’s curl is zero. Now we go through the options.




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Option (A), Curl of the vector yz xz xy

a a a

2 2 20

x

x

y

y

z

z= =22

22

22

Option (B), Curl of the vector 0a a a

1

0 0e

z

z !r

r

r= 2

222

22

r

r

f

f

rr-

^ h

Option (C), Curl of the vector 0cos sin

a a a1

0

z

z2 22

rf

r

f= =2

222

22

r

r

r

f

f

r

So, it can be expressed as gradient of a scalar.

SOL 1.2.38 Option (A) is correct.Any vector for which divergence is zero can be expressed as the curl of another vector. For checking it we go through all the options.

In Option (A), Divergence x

x yyxy

z21 22 2

22

22

22= - - +_ ^ ^i h h& 0

x x 0= - =

In Option (B), Divergence e1 022

r f r= =r-

c m

In Option (C), Divergence cossin

sinr r

rr r r

1 2 12

23 3

2

22

22q

q qq= +b cl m

cossin

sin cosr r1 2 1 24 4q

qq q=- +^ h 0=

So all the vectors can be expressed as curl of another vector.

SOL 1.2.39 Option (B) is correct.for y 0> i.e. above x -axis field will be directed towards ax+ direction and will increase as we go far from the x -axis, since y -increases.For y 0< i.e. below x -axis, field will be directed towards ax- direction and it’s intensity will increases as we go away from the x -axis.

SOL 1.2.40 Option (A) is correct.Given the divergence of the vector field is zeroi.e. A4: 0=

xA

yAx y

22

22+ 0=

xAx22

yAy22=- ...(1)

and the curl of the vector field is zero,i.e. A4# 0=

A A

a a a

0

x

x

x

y

y

y

z

z22

22

22 0=



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zA

zA

xA

yAa a ay

xxx

y xz2

222

22

22- + + -e o 0=

xA

yAy x

22

22- 0= ....(2)

(Since A is only the variable of x and y . So the differentiation with respect to z will be zero).Differentiating equation (ii) with respect to x we get,

xA

x yAy x

2

2 2

22

222- 0=

xA

y xAy x

2

2

22

22

22- b l 0=

xA

y yAy y

2

2

22

22

22- -

e o 0= (from equation (i))

xA

yAy y

2

2

2

2

22

22+ 0=

Ay2d 0=Again differentiating equation (ii) with respect to y we get

x yA

yAy x

2

2

2

222

22- 0=

x yA

yAy x

2

2

22

22

22-e o 0=

x x

AyAx x

2

2

22

22

22- -b l 0= (from equation (i))

xA

yAx x

2

2

2

2

22

22+ 0=

SOL 1.2.41 Option (B) is correct.The line integral (circulation) of force F around the closed path can be divided in four sections as shown below.

For segment 1 we have, y z 0= = dl dxax=- , x0 1< <

So, dF l1

:# x dx2

0

1= -^ h# x

3 313

0

1

= - =-: D




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For segment 2 we have, x z 0= = dl dyay= , y0 1< <

So, dF l2

:# xz dy0

1= -^ ^h h# 0=

For segment (3) we have x z= , y 1= dl dx dza ax z= + , x0 1< < , z0 1< <

So, dF l3

:# x dx y dz2 2

0

1

0

1= + -^ h##

x z33

0

1

01= -: 6D @ 3

1 1 32= - =-

For segment 4 we have, y z= , x 1= dl dy dza ay z=- -

So, dF l4

:# xz dy y dz0

12

0

1= - - + - -^ ^ ^ ^h h h h# #

y z2 3

2

0

1 3

0

1

= +; :E D 21

31= + 6

5=

So, the net circulation of force F around the closed path is

dF l:# dF l1 432

:= + + +c m# ###

31 0 3

265=- + - + 9

1=-

SOL 1.2.42 Option (C) is correct.Given, vector position of , ,P x y z^ h

R x y za a ax y z= + +So, R x y z2 2 2= + +or Rn x y z /n2 2 2 2= + +^ h

R Rn x y z x y za a a/nx y z

2 2 2 2= + + + +^ h 6 @

Now we take the divergence of the vector as

R Rn:d ^ h xx x y z

yy x y z

zz x y z/ / /n n n2 2 2 2 2 2 2 2 2 2 2 2

22

22

22= + + + + + + + +^ ^ ^h h h

x y zxx x n x y z x2 2/ /n n2 2 2 2 2 2 2 2 1

22= + + + + + -

^ ^ ^h h h9 C

x y zyy y n x y z y2 2/ /n n2 2 2 2 2 2 2 2 1

22+ + + + + + -

^ ^ ^h h h9 C

x y zzz z n x y z z2 2/ /n n2 2 2 2 2 2 2 2 1

22+ + + + + + -

^ ^ ^h h h9 C

R n R x y z3 2 2 2 2n n 2 2 2 2= + + +-^ h

R nR3 n n= + n R3 n= +^ h

SOL 1.2.43 Option (B) is correct.The line integral (Circulation) of vector field A around the closed path can be divided into four segments as shown in figure below :



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For segment 1, f 0= , z 0= dl d ar= r , 1 2< <r

So, dA l1

:# sin dr f r= ^ h# 0= 0f =^ h

For segment 2 r 2= , z 0= dl d ar f= f 0 < <f p

So, dA l2

:# d2

0r r f=

p

^ h# 8 0f= p6 @ 8p= 2r =^ h

For segment 3 f p= , z 0= 2 1< <r- -

So, dA l3

:# sin d2

1r f r= -

-

-

^ ^h h# 0= f p=^ h

For segment 4, r 1= , z 0= dl d ar f= - f^ h 0 < <f p

So, dA l4

:# d2

0r r f= -

p

^ h# 0f=- p6 @ p=-

Therefore, the net circulation of the vector is

dA l:# dA l4321

:= + + +c m#### 0 8 0p p= + + + 9p=

***********




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SOLUTIONS 1.3

SOL 1.3.1 Option (D) is correct.Divergence of A in spherical coordinates is given as

A:d ( )r r

r A1r2

2

22= ( )

r rkr1 n

22

22= +

( )rk n r2 n2

1= + +

( )k n r2 0n 1= + =- (Given, A 0:d = )So, n 3+ 0=or, n 3=-

SOL 1.3.2 Option (B) is correct.Given, the vector, A xy xa ax y

2= +Differential displacement along any path in the x -y plane is defined as dl dx dya ax y= + (since, dz 0= )So, the line integral of the vector A along the closed square loop is given as

dA lC

:# ( ) ( )xy x dx dya a a ax y x yC

2:= + +# ( )xydx x dy

C

2= +#

xdx xdx dy dy3 34

31

/

/

/

/

2 3

1 3

1

3

3

1

1 3

2 3= + + +# # ##

[3 1] [1 3]21

34

31

23

31

34

34

31= - + - + - + -; ;E E 1=

SOL 1.3.3 Option (C) is correct.Given, V A#d= ...(1)According to Stoke’s theorem the line integral of a vector along a closed loop is equal to the surface integral of the curl of the vector for the loop.

i.e. dA lC

:# dA SSC

# :d= ^ h## ...(2)

where C is a closed path (contour) and SC is the surface area of the loop. From equation (1) and (2) we get

dA lC

:# dV SSC

:= ##SOL 1.3.4 Option (A) is correct.

The transformation of unit vector af in Cartesian coordinate system gives the result. af sin cosa ax yf f= - +^ ^h h

where f is angle formed with x -axis.at point A, f 90c=



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For View Only Shop Online at www.nodia.co.inSo, af ax=- 3a "

at point B , f 90 45 135c c c= + =

So, af a a2

12

1x y=- - 5b "

at point C , f 45c=-

So, af a a2

12

1x y= + 4c "

at point D , f 0c=So, af az= 2d "

SOL 1.3.5 Option (C) is correct.Given, the solution of a Laplaces equation is V sinh cosx kyepz=i.e. the field V satisfies Laplace’s equation. So, we have V2d 0=

or, xV

yV

zV

2

2

2

2

2

2

22

22

22+ + 0= (1)

Now, xV

2

2

22 sinh cosx kyepz=

yV

2

2

22 sinh cosk x kyepz2=-

zV

2

2

22 sinh cosp x kyepz2=

Putting all the values in equation (1), we get sinh cosx kye k p1pz 2 2- +^ ^h h 0= k p1 2 2- + 0= k2 p1 2 2= +Note :

sinh sin cosdxd x dx

djjx

jj jx= = cos jx= and similarly the 2nd derivative.

SOL 1.3.6 Option (B) is correct.The angle between two vector fields A and B is defined as

a cos ABA B1 := -

b l

Given, electric field intensity at point P is E 10 10 10a a ax y z= + +So, the angle formed between the field E and with x -axis ax^ h is

a cos cos cosE aaE

10 310

31

x

x1 1 1:= = =- - -e c co m m

Similarly, we get

b cos4

11g= = -c m

SOL 1.3.7 Option (B) is correct.Laplace equation is defined as




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V2d 0=Now, we consider the option (C)The scalar field is

V r10=

So, the Laplacian of the field V is given as

V2d r rrrV

rV

zV1 1

2 2

2

2

2

22

22

22

22

f= + +b l

r rrr

1 10 0222= - +cd mn 1

r r r10

22=- b l r

102=

i.e. V2d 0!

So, it doesn’t satisfy Laplace’s equation.

SOL 1.3.8 Option (A) is correct.Laplacian of a scalar function is given as V2d V:d d= ^ h div gradV= ^ h

i.e. The Laplacian of a scalar function is divergence of gradient of V .

SOL 1.3.9 Option (D) is correct.Given, the vector field, A 3 ( 2 )x yz x z x y za a ax y z

2 3 3= + + -So, the divergence of vector A is A:d 6 2xyz= - 0!

Therefore, it is neither divergence less and nor solenoidal Now, we determine the curl of vector as A#d 0=Since, the curl of the vector is zero so, it is irrotational (i.e., not rotational).

SOL 1.3.10 Option (D) is correct.Laplacian of a scalar field V in cylindrical coordinates is given by

V2d V VzV1 1

2 2

2

2

2

22

22

22

22

r r r r r f= + +c m

Since, Laplace equation is defined as V2d 0=So, we get

V VzV1 1

2 2

2

2

2

22

22

22

22

r r r r r f+ +c em o 0=

SOL 1.3.11 Option ( ) is correct.The given curve is divided in three segments AB , BC and CD respectively. So, the total integral is given as

dlc# dl

CDBCAB= + +c m###

Rd dl Rda a a/

/

yR

R

20

2

f f= + - +f fp

pp

-

--^ h ###



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For View Only Shop Online at www.nodia.co.in 2R R Ra a a3 2y

p p= - -f f

2Ray=-

SOL 1.3.12 Option (D) is correct.Given vector field, A cosr a4 rf=

For the given contour we integrate the field in three intervals as

dA l:# dr d drA a A a A ar r1 2 3

: : :f= + +f# # #

cos cosr dr r dr2 0 2

0 /2at at

0

1

1

0f f= +

f f p= =

^ ^

^ ^

h h

h h1 2 3444 444 1 2 3444 444# #

r2 22

0

1

= : D 1=

SOL 1.3.13 Option (C) is correct.For a vector field B to be solenoidal B:d 0=

dvB:d̂ h# 0=

dB Ss

:# 0=

SOL 1.3.14 Option (C) is correct. a a a ax y y x# #+^ ^h h a az z= + -^ h 0=

SOL 1.3.15 Option (D) is correct.(a) Curl F^ h 0=It gives the result that F is irrotational 2a "^ h

(b) div F^ h 0=It gives the result that F is solenoidal. 3b "^ h

(c) div Grad f^ h 0= : dfd ^ h 0= 2d f 0=It is Laplace equation. 1c "^ h

(d) div div f^ h 0= : :d d f^ h 0=As f is a scalar quality so its divergence is not defined. 4d "^ h




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SOL 1.3.16 Option (D) is correct.As by observing the given figure we conclude that the closed circular quadrant is in x -y plane and it’s segments are OP drar= PQ 2d af= f

QO drar=So, the closed loop integral is

dA l:# dr d dr2/

0

2

0

2

2

0f= + +

p

# # # 8 4 2p p= =a k

***********

CHAPTER 2ELECTROSTATIC FIELDS

58 Electrostatic Fields Chap 2



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EXERCISE 2.1

Statement for Linked Question 1-2 :Four equal charges of 2 C+ are being placed at the corners of the square of side

m2 in free space as shown in figure.

MCQ 2.1.1 The net force on a test charge nC2+ at the centre O of the square will be(A) 0 N (B) 18 N

(C) 72 N (D) 36 N

MCQ 2.1.2 If one of the four charges is being removed then the magnitude of the net force on the test charge 1 nC+ placed at the centre will be(A) 0 N (B) 18 N

(C) N9 (D) 36 N

MCQ 2.1.3 Two point charges of 9 C and C12 are located on x -axis at a separation of 3 m. A third point charge q is placed on the x -axis at a distance d from the 36 C charge which makes the entire system in equilibrium. The value of q and d are

(A) 4 C and 1 m

(B) 4 C- and 2 m

(C) 4 C and 2 m

(D) 4 C- and 1 m

Chap 2 Electrostatic Fields 59


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For View Only Shop Online at www.nodia.co.inMCQ 2.1.4 Consider the point charges 5 nC- and nC3+ are located at ( , , )4 0 2- - and

( 5,0,3)- respectively The net electric field intensity at point ( , , )7 3 1- - will be

(A) 1.004 1.284 1.4a a az y x- - +

(B) 1.004 1.284 1.4a a az y x- +

(C) 1.004 1.284 1.4a a ax y z- - +

(D) 1.004 1.284 1.4a a ax y z+ + +

MCQ 2.1.5 The three point charges, each nC4+ , are located on the z -axis at z 1=- , 0, 1 in free space. What will be the electric field intensity at point ( , , )P 0 0 3 ?(A) 13.44az (B) 1 . a9 06 x

(C) . a19 06 z (D) 5.8az

MCQ 2.1.6 Charges Q+ and 2Q+ are separated by a distance 1 m. What will be the distance of point P form Q+ charge such that the net electric field intensity at P is zero. (A) 1 (B) 2.414 m-

(C) 1- (D) 0.414 m

Statement for Linked Question 7 - 8A uniform volume charge density of /C m4 3m is present throughout the spherical shell extending from 2 cmr = to 3 cmr = .

MCQ 2.1.7 The total charge present throughout the spherical shell will be(A) 160 pC (B) 40 pC

(C) 80 pC (D) 72 pC

MCQ 2.1.8 For what value of a half of the total charge will be located in the region cm r a4 < < (A) 2.5 cm (B) 2.6 cm

(C) 2.4 cm (D) 2.7 cm

MCQ 2.1.9 Electrons are moving randomly in a fixed region in free space. During a time interval T the probability of finding an electron in a subregion of volume 10 m12 3-

is %30 .The volume charge density in the subregion for the time interval will be(A) 48 /nC m3- (B) 16 /nC m3

(C) 48 /C m3m (D) 48 /nC m3

MCQ 2.1.10 Total stored charge on the cylindrical surface 2r = , 0 1 mz< < having surface charge density /C mz2 2mr is(A) 25.1 Cm (B) 50.2 Cm

(C) 12.55 Cm (D) 15.7 Cm




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MCQ 2.1.11 Consider a triangular surface in the plane z 0= as shown in the figure.

If the triangular surface has charge density Sr = /C mxy9 2 then the total charge on it will be(A) 6.5 C (B) 13 C

(C) 4.5 C (D) 26 C

MCQ 2.1.12 A circular disk of radius 5 m has surface charge density 3rr =S , where r ( 5 m#

) is the distance of any point on the disk from its centre. The total charge stored on the disk is

(A) 50 Cp (B) 125 Cp

(C) 250 Cp (D) C250p

MCQ 2.1.13 Which of the following charge distribution produces the electric field intensity ?

E 4 6 /V mxy yz xza a a4 x y z= + + ?

(A) infinite line charge of 2 /nC m along x -axis

(B) spherical shell of charge density 3 /nC m3

(C) plane sheet of charge density 3 /nC m2 at x -y plane

(D) field doesn’t exist

MCQ 2.1.14 An infinite line charge of 1 /C mm is located on the z -axis. Electric field due to the line charge at point ( , , )2 1 5- - will be

(A) 2.4 1.8a ax y+ (B) 7.2 14.4a ax y+

(C) 7.2 3.6a ax y- - (D) 2a ax y- -

MCQ 2.1.15 Electric field intensity at any point ( , , )x y z in free space is E 2x xya ax y2= + . The

electric flux density at the point ( , , )1 0 1- will be(A) 0 (B) ax0e

(C) ax0e- (D) 4 ay0pe



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For View Only Shop Online at www.nodia.co.inMCQ 2.1.16 Assertion (A) : Net electric field flux emanating from an arbitrary surface not

enclosing a point charge is zero.Reason (R) : Electric field intensity at any point outside the uniformly charged sphere is always zero.(A) A and R both are true and R is correct explanation of A.

(B) A and R both are true but R is not the correct explanation of A.

(C) A is true but R is false.

(D) A is false but R is true.

MCQ 2.1.17 Consider the electric field intensity in some region is found to be 3 /V mrE ar2= , in spherical coordinate system. The total charge stored in a sphere of radius 2 m, centered at origin will be(A) 4.32 pC (B) 5.3 pC

(C) 4.32 nC (D) 5.3 nC

Statement for Linked Question 18 - 19Volume charge density in the free space in spherical coordinate system is given by

vr 1 / 0 3

0 3

C m m

mr

r

r

< <

>

23

= *

MCQ 2.1.18 Net electric flux crossing the surface mr 2= is(A) 4 Cp (B) Cp-

(C) 2 Cp (D) 0

MCQ 2.1.19 Electric flux density at 1 mr = is

(A) /C ma2 r2p (B) /C mar 2

(C) 4 /C mar 2p (D) /C ma ar2+ q

MCQ 2.1.20 A point charge 8 C is located at the origin. The total electric flux crossing the portion of plane mx y 3+ = lying in the first octant is(A) 1C (B) 4 C

(C) 1 /C m (D) 4 /C m

MCQ 2.1.21 A uniform volume charge density /C mv3r is distributed inside the region defined

by a cylindrical surface of cross sectional radius a . The electric field intensity at a distance r a<^ h from the cylindrical axis is proportional to

(A) r (B) ra

(C) r12 (D) ar2




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Common Data for Question 22 - 24Charge density inside a hollow spherical shell of radius 4mr = centered at origin is defined as

0

/ 2 4

for

c m for

r

rr

24v2

3 1

#

#r = *

MCQ 2.1.22 The Electric field intensity at any point in the region 2r# will be(A) 1 /V m- (B) 4 /V m-

(C) 0 (D) 2 /V m

MCQ 2.1.23 Electric field intensity at r 4= will be

(A) a94

r0e

(B) a4r

0e

(C) a920

r0e

(D) a9r

0e

MCQ 2.1.24 If the region outside the spherical shell is charge free then what will be the electric field intensity at r 5= ?

(A) a31

r0e

(B) 16 ar0ep

(C) a825

r0e (D) a258

r0e

MCQ 2.1.25 Assertion (A) : No charge can be present in a uniform electric field.Reason (R) : According to Gauss’s law volume charge density in a region having electric field intensity E is given by vr Eed=(A) A and R both are true and R is correct explanation of A.




MCQ 2.1.26 In a certain region the electric flux density is D /cos sin C mr ra a

2r3 32q q= + q . Volume

charge density in the region will be

(A) 0 /C m3 (B) /cos C mr

24

3q

(C) /sin C mr3

3q (D) /C mr43

3

MCQ 2.1.27 If electric flux density in a certain region is ( 4 ) 2 4 /C my z xy xD a a a2 x y z2 2= + + +

The total charge enclosed by the cube 0 x 2# # , 0 2y# # , z1 1# #- is(A) 9 C (B) 4 C

(C) 16 C (D) 8 C



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For View Only Shop Online at www.nodia.co.inMCQ 2.1.28 Two point charges 1 Cm+ and 1 Cm- are being located at points (0,0,1) and

( , , )0 0 1- respectively. The net electric potential at point ( , , )P 3 0 4- - due to the two charges will be(A) 578.9 V- (B) 0.64 kV

(C) 2.36 V- (D) 5.78 kV-

Statement for Linked Question 29 - 30 :In the entire free space electric potential is given by

V lnxy z x y z43 2 32 3 2 2 2= + + +^ h

MCQ 2.1.29 Electric field at point ( , , )P 3 2 1- will be(A) 7.1 22.8 71.1a a ax y z+ - (B) 3.6 11.4 35.6a a ax y z+ -

(C) 3.6 11.4 35.6a a ax y z- + (D) 2.2 11.4 35.6a a ax y z- +

MCQ 2.1.30 Electric flux density at point P will be(A) 31.4 101 314.5 /pC ma a ax y z

2+ -

(B) 62.8 202 629 /nC ma a ax y z2+ -

(C) 0.095 0.304 0.948 /nC ma a ax y z2- - +

(D) 7.1 22.8 71.1 /pC ma a ax y z2+ -

MCQ 2.1.31 A potential function V satisfies Laplace’s equation inside a certain region. In this region the potential function will have(A) a maxima only (B) a minima only

(C) a maxima and a minima both (D) neither a maxima nor a minima

MCQ 2.1.32 An electric dipole consists of two point charges of equal and opposite magnitudeQ! is lying along x -axis such that Q+ is at /2x d= and Q- is at /2x d=- .

Electric field due to the dipole at any point ( , , )r q f in spherical coordinate system is given by

E cos sinr

Qd a a2

2 r0

32

peq q= + q6 @ where r >>d

The force applied by the dipole on a charge of 1C+ located at point (0, , )y 0 is

(A) yQd a

4 z0

3pe- (B)

yQd a

4 z0

3pe+

(C) yQd

rQda a

4 42

z y0

30

3pe pe- + (D)

yQd a

4 x0

3pe-

***********




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EXERCISE 2.2

MCQ 2.2.1 Two equal point charges of nC21+ each are located at points ( , , )1 0 0- and ( , , )1 0 0

respectively. What will be the position of third point charge of nC2+ such that the net electric field 0E = at ( , , )0 1 0 ?(A) ( , , )1 0 0- (B) ( , , )0 1 0-

(C) ( , , )3 0 0 (D) ( , , )0 3 0

MCQ 2.2.2 Plane 4 0x y5 + = carries a uniform charge distribution with Sr =2 /nC m2. The electric field intensity at point ( , , )1 0 3 will be(A) 67.8 90.48a ax y- - /V m (B) 67.85 90.48a ax y+ /V m

(C) 3 4a ax y+ /V m (D) 3 4a ax y- - /V m

MCQ 2.2.3 Electric field intensity at a distance 3 m above the center of a circular loop of radius 4 m lying in the xy -plane and carrying a uniform line charge /nC m3+ as shown in the figure is

(A) 21.72 10.86 /V ma ax z+ (B) 10.86 /V maz(C) 10.86 21.72 /V ma ax z+ (D) 72 /V maz

MCQ 2.2.4 Consider a point charge Q is located at the origin. Divergence of the electric flux density produced by the charge is (A) 0, at all points (B) 1+ , at all points

(C) 1+ , at all points except origin (D) 0, at all points except origin



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For View Only Shop Online at www.nodia.co.inCommon Data for Question 5 - 6 :In the region of free space that includes the cubical volume x0 < , y , z 1< , electric flux density is given by D /C mx y y xa a2x y

2 2 2 2= +

MCQ 2.2.5 The total flux leaving the closed surface of the cube is(A) 1/6 C- (B) 1/6 C

(C) 0 C (D) 1/3 C

MCQ 2.2.6 div D at center of the cube is(A) /1 2 (B) /3 4

(C) /1 4 (D) /1 6

Common Data for Question 7 - 8 :In free space, flux charge density is given by

D / .

/ / .

nC m m

nC m m

r r

r r

a

a

3 0 5

2 0 5

<r

r

2 2

2 2 $= *

MCQ 2.2.7 Volume charge density at 0.2 mr = will be(A) 5 /nC m3- (C) 4 /nC m3

(C) 20 /C m3 (D) 5 /nC m3

MCQ 2.2.8 Volume charge density at 1 mr = will be(A) 0 (B) 20 /nC m3

(C) 40 /nC m3- (D) 20 /nC m3-

MCQ 2.2.9 A dipole having a moment C mp a5 z0pe -= is located at origin in free space. If the electric field produce due to the dipole is given by E E EE a a a2 x x y y z z= + + then surface on which E 0z = but Ex , E 0y ! will be

(A) a cone of angle .54 7c (B) a cone of angle .125 3c

(C) (a) and (b) both (D) none of these

MCQ 2.2.10 An infinite line charge /nC m2+ is lying along entire z -axis. If the electric potential at the point (1, /2, )5p due to the line charge is zero then the electric potential at any point ( , , )zr f will be

(A) volt18r (B) ln18 1

rbc lm

(C) ln210 19

r-

b l (D) ln29 10 19#

rb l




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Statement for Linked Question 11 -12 :Electric field at any point ( , , )r q f in free space is given by

E ( )rr a4

2r2 2=

+MCQ 2.2.11 The electric potential will be maximum at

(A) infinity (B) origin

(C) at r 2=- (D) r 2=+

MCQ 2.2.12 Potential difference between the spherical surfaces r 0= and r 2= will be(A) 1/2 volt (B) 1 volt

(C) 1/8 volt (D) 1/4 volt

MCQ 2.2.13 An electric dipole having moment nC mp a a a3x y z53 -= - + is located at point

(0,1, 6)B - . The electric potential due to the dipole at point (1,2,2)A will be(A) 4.23 V (B) 1.91 V

(C) 1.31 V (D) 0.6 V

MCQ 2.2.14 A total charge 20 nC is being split into four equal charges spaced at 90c intervals around a circular loop of radius 5 m. The electric potential at the center of the loop will be(A) 108 V (B) 36 kV

(C) 36 V (D) 135 V

MCQ 2.2.15 The work done in carrying a 2 C charge from point , / ,A 1 1 2 3^ h to the point (4,1,0)B in the field 2 2 /V my xE a ax y= + along the curve /y x 2= will be(A) 28 J- (B) 15.5 J-

(C) 2.3 J (D) 15.5 J+

MCQ 2.2.16 In a certain region, the electric field intensity is given as /V mx yE a ax y= - . The amount of work done in moving a 2 C+ charge along a circular arc centred at origin

from 1 mx = to mx y3

1= = in the region will be

(A) 2 J (B) 1 J-

(C) 1 J+ (D) J21-

Statement for Linked Question 17- 18 :Four equal charges of 1 nC+ is being carried from infinity and placed at different corners of a square. Consider the side of the square is 1 m and the charges are being carried as one at a time.



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For View Only Shop Online at www.nodia.co.inMCQ 2.2.17 How much work does it take to bring in the last charge from infinity and place it

in the fourth corner ?(A) 24.36 J (B) 24.36 nJ

(C) 2.71 J (D) 9 nJ

MCQ 2.2.18 Total work done for assembling the whole configuration of four charges will be(A) 15.36 nJ (B) 48.72 nJ

(C) 9 nJ (D) 24.36 nJ

MCQ 2.2.19 The electric field in a certain region is given by E ( 1) /sin cos sin V mza a a2 zf r f r f= + + +r f

Work done in moving a 2 C charge from A(2,0 ,1)c to B ( , , )2 30 1c in the field is(A) 8 nJ- (B) J8-

(C) 32 J (D) J8

MCQ 2.2.20 Total work done in transferring two point charges 1 Cm+ and 2 mC+ from infinity to the points ( 3,6,0)A - and (2, 4, 1)B - - respectively is(A) 1.604 J (B) 1.604 J-

(C) 9 kJ (D) 85.2 mJ

MCQ 2.2.21 Four point charges of nC4 are placed at the corners of a square of side 1 cm. The total potential energy stored in the system of charges is(A) 3.9 kJ (B) 0.490 mJ

(C) 0.312 mJ (D) 2.7 J

Statement for Linked Question 22 - 23 :The potential field in free space is expressed as

V Vxyz4=

MCQ 2.2.22 The total energy stored within the cube x1 < , y , z 2< will be(A) 4.42 10 J12

#- (B) 9.68 10 J13

#- -

(C) 7.68 10 J3# (D) 9.68 10 J13

#-

MCQ 2.2.23 The energy density at the centre of the cube will be(A) 1.33 10 J11

#- (B) 5.18 10 J13

#-

(C) 2.13 10 J12#

- (D) 4.47 10 J13#

-

Common Data for Question 24 - 25 :A uniformly charged solid sphere of radius R has the total charge Q . Consider the electric potential at a distance r from the centre of the sphere is ( )V r .




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MCQ 2.2.24 For r R2 , plot of ( )V r versus r will be

MCQ 2.2.25 With the increase in r potential ( )V r inside the charged sphere will(A) increase (B) decrease

(C) remain constant (D) be zero always

MCQ 2.2.26 If 1 mR = and CQ 2= then the total stored energy inside the sphere will be(A) 4.34 10 J9

# (B) 6.75 10 J9#

(C) .4 5 109# (D) 5.4 10 J9

#

MCQ 2.2.27 The electric field intensity required to counter act the earth’s gravitational force on an electron is(A) 1.79 10 /V m12

#- (B) 5.57 10 /V m13

#-

(C) 5.57 10 /V m11#

- (D) 1.79 10 /V m11#

-

MCQ 2.2.28 Three point charges Q , kQ and kQ are arranged as shown in figure.

What will be the value of k for which the net electric field intensity at the point , ,P 0 4

131_ i is zero ?



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For View Only Shop Online at www.nodia.co.inMCQ 2.2.29 Three point charges Q , Q4- and Q are located at , ,a 0 0^ h, , ,0 0 0^ h and , ,a 0 0-^ h

respectively. The electric field intensity at any point , ,x 0 0^ h for x a>> is

(A) KxQa6

4

2

c m (B) KxQa3

4

2

c m

(C) KxQa6 4

2

c m (D) zero

MCQ 2.2.30 A volume charge is distributed throughout a sphere of radius R , and centered at the origin, with uniform density /C mv

3r . The electric field intensity at a distance r from the origin is inside the sphere(r R# ) outside the sphere(r R> )

(A) Rr a3

vr

0

2

erb l

rR a3

vr

02

3

erc m

(B) r a3v

r0era k

rR a3

vr

02

3

erc m

(C) Rr a3

vr

0

2

erb l

Rr a

3v

r0

2

3

erc m

(D) r a3v

r0era k

Rr a

3

3v

r0

2erc m

Statement for Linked Question 31 - 32 :Consider a total charge of 2 nC is distributed throughout a spherical volume of radius 3 m. A small hole is drilled through the center of the spherical volume charge as shown in figure. The size of the hole is negligible compared to the size of the sphere.

MCQ 2.2.31 If an electron is placed at one end of the hole and released from rest at t 0= then what will be the distance of the electron from center of sphere at sect 2 m= .(A) 0 (B) 2 m

(C) 1.83 m (D) 2.83 m

MCQ 2.2.32 The frequency of the oscillation of point charge is(A) 54.4 KHz (B) 5.44 KHz

(C) 1.83 KHz (D) 27.2 KHz




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MCQ 2.2.33 An infinite line charge of uniform density Lr is situated along the x -axis. The total electric field flux crossing the portion of plane 1 my z+ = lying in the first octant and bounded by the planes x 0= and 1 mx =

(A) 2L

0er (B) 8

L

0er

(C) 4L

0er (D) 4 L

0er

MCQ 2.2.34 Volume charge of uniform density 5 /nC m3 is distributed in the region between two infinitely long, parallel cylindrical surfaces of radii 5 m and 2 m and with their axes separated by distance of 1 m as shown in the figure.

The electric field intensity in the charge-free region inside the cylindrical surface of radius 2 m is(A) 282.5 /V max (B) 5.65 10 /V m11

#

(C) 3.54 /mV max (D) 1.77 10 /V m12#

-

MCQ 2.2.35 A volume charge is distributed throughout a sphere of radius R and centered at the origin with uniform density /C m3 v

3r . The electric potential at a distance r from the origin is inside the sphere(r R# ) outside the sphere(r R> )

(A) r R23

v

0

22

er -b l r

R3v

0

3

er

(B) R r2 3v

0

22

er -b l r

R3v

0

3

er

(C) R r23

v

0

22

er -b l r

R3v

0

3

er

(D) R r2 3v

0

22

er -b l R

R3 v

0

3

er

MCQ 2.2.36 A total charge of 900 Cmp is uniformly distributed over a circular disk of radius 6 m The applied force on a 150 Cm charge located on the axis of disk and 4 m from it’s center as shown in figure is



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(A) 13.5 N (B) 6.78 N

(C) 9.44 N (D) 18.89 N

Common Data for Question 37 - 38 :Two infinite uniform sheets of charge, each with density /C m2 2, are located at y 1=+ and y 1=- as shown in figure.

MCQ 2.2.37 Electric field intensity at the origin will be

(A) 0 (B) /V ma24

y0e

(C) /V ma25

y0e

- (D) /V ma25

y0e

MCQ 2.2.38 If a test charge of 5 Cm is placed at point , ,2 5 4^ h then the force applied by the sheets on test charge is(A) 2.83 mN (B) 2.5 10 N14

#-

(C) 2.83 N (D) 5.65 10 N2#

MCQ 2.2.39 As we move away from the sheet charge located at y 1=- in the region y 1<- , the electric field intensity will be(A) linearly increasing

(B) linearly decreasing

(C) constant

(D) zero




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MCQ 2.2.40 A charged sphere of radius 1 m carries a uniform charge density of 6 /C m3. A redistribution of the charge results in the density function given by vr /C mk r5 2 3= -^ h

where r is distance of the point from center of the sphere. The value of k will be(A) 2.5 (B) 5

(C) 0.5 (D) 40

MCQ 2.2.41 A 50 Cm point charge is located at the origin. The total electric flux passing through the hemispherical surface defined by 48 mr = , /0 2# #q p is(A) 50 Cm (B) 12.5 Cm

(C) 25 Cm (D) 100 Cm

MCQ 2.2.42 Consider a hollow sphere of radius R centred at origin carries a uniform surface charge density rS . The electric field intensity at distance r from the center of the sphere is insider the sphere(r R# ) outside the sphere(r R> )

(A) 0 rR as

r0

2

erb l

(B) as r0er r

R asr

0

2

erb l

(C) as r0er 0

(D) 0 as r0er

MCQ 2.2.43 An air filled parallel plate capacitor is arranged such that the lower side of upper plate carries surface charge density /C m3 2 and upper side of lower plate carries surface charge density 2 /C m2- as shown in figure. The electric field intensity between the plates will be

(A) a2z

0e- (B) a2

z0e

(C) a4z

0e- (D) a4

z0e

MCQ 2.2.44 In a certain region electric potential distribution is as shown in the figure.



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The corresponding plot of electric field component Ey will be

MCQ 2.2.45 Two electrons are moving with equal velocities in opposite directions. A uniform electric field is applied along the direction of the motion of one of the electrons, so the electron gets accelerated while the electron moving in opposite direction gets decelerated. If the gain in the kinetic energy of accelerating electron is .K EGain and the loss in Kinetic energy of decelerating electron is .K ELoss then the correct relation between them is




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(A) . .K E K EGain Loss=

(B) . .K E K E>Gain Loss

(C) . .K E K E<Gain Loss

(D) Can’t be determined as initial velocities are not given

MCQ 2.2.46 Two identical uniform charges with 80Lr = nC/m are located in free space at 0x = , 3 my != . The force per unit length acting on the line at positive y arising

from the charge at negative y is(A) 9.375ay Nm (B) 37.5ay Nm

(C) 19.17ay Nm (D) 75ay Nm

MCQ 2.2.47 Four 2.2 nC point charge are located in free space at the corners of a square 4 cm on a side. The total potential energy stored is(A) 1.75 mJ (B) 2 mJ

(C) 3.5 mJ (D) 0

***********



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EXERCISE 2.3

MCQ 2.3.1 If the electric field intensity is given by ( )x y zE a a a2 x y z= + + volt/m, the potential difference between (2,0,0)X and (1,2,3)Y is(A) 1+ volt (B) 1- volt

(C) 5+ volt (D) 6+ volt

MCQ 2.3.2 There are three charges, which are given by 1 CQ1 m= , CQ 22 m= and 3 CQ3 m= . The field due to each charge at a point P in free space is 2a a ax y z+ -^ h, 3a ay z+^ h and 2a ax y-^ h newtons/coulomb. The total field at the point P due to all three charges is given by(A) 1.6 2.2 2.5a a ax y z+ + newtons/coulomb

(B) 0.3 0.2 0.2a a ax y z+ + newtons/coulomb

(C) 3 2 2a a ax y z+ + newtons/coulomb

(D) 0.6 .2 0.5a a a2x y z+ + newtons/coulomb

MCQ 2.3.3 Given that the electric flux density D ( ) /cos C mzp az2 2F= . The charge density at point (1, /4, 3)p is(A) 3 (B) 1

(C) 0.5 (D) 0.5 az

MCQ 2.3.4 An electric charge of Q coulombs is located at the origin. Consider electric potentialV and electric field intensity E at any point ( , , )x y z . Then

(A) E and V are both scalars (B) E and V are both vectors

(C) E is a scalar and V is a vector (D) E is a vector and V is a scalar

MCQ 2.3.5 Assertion (A) : Capacitance between two parallel plates of area ‘A’ each and distance of separation ‘d ’ is /A de for large /A d ratio.Reason (R) : Fringing electric field can be neglected for large A/d ratio.(A) Both A and R are individually true and R is the correct explanation of A

(B) Both A and R are individually true but R is not the correct explanation of A

(C) A is true but R is false

(D) A is false but R is true

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MCQ 2.3.6 Assertion (A) : In solving boundary value problems, the method of images is used.Reason (R) : By this technique, conducting surfaces can be removed from the solution domain.(A) Both A and R are individually true and R is the correct explanation of A




MCQ 2.3.7 What will be the equipotential surfaces for a pair of equal and opposite line charges ?(A) Spheres

(B) Concentric cylinders

(C) Non-concentric cylinders

(D) None of the above

MCQ 2.3.8 If the potential functions V1 and V2 satisfy Laplace’s equation within a closed region and assume the same values on its surface, then which of the following is correct ?(A) V1 and V2 are identical

(B) V1 is inversely proportional as V2

(C) V1 has the same direction as V2

(D) V1 has the same magnitude as V2 but has different direction

MCQ 2.3.9 Assertion (A) : The expression VE d=- , where E is the electric field and V is the potential is not valid for time varying fields.Reason (R) : The curl of a gradient is identically zero.(A) Both A and R are individually true and R is the correct explanation of A.

(B) Both A and R are individually true but R is not the correct explanation of A.



MCQ 2.3.10 What is the electric flux density ( / )in C m2m at a point ( , , )6 4 5- caused by a uniform surface charge density of 60 /C m2m at a plane 8x = ?(A) 30ax- (B) 60ax-

(C) 30ax (D) 60ax

MCQ 2.3.11 Of two concentric long conducting cylinders, the inner one is kept at a constant positive potential V0+ and the outer one is grounded. What is the electric field in the space between the cylinders?(A) Uniform and directed radially outwards

(B) Uniform and directed radially inwards

(C) Non-uniform and directed radially outwards

(D) Non-uniform and directed parallel to the axis of the cylinders

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For View Only Shop Online at www.nodia.co.inMCQ 2.3.12 In a charge free space, the Poisson’s equation results in which one of the following?

(A) Continuity equation (B) Maxwell’s equation

(C) Laplace equation (D) None of the above

MCQ 2.3.13 W1 is the electrostatic energy stored in a system of three equal point charges arranged in a line with 0.5 m separation between them. If W2 is the energy stored with 1 m separation between them, then which one of the following is correct ?(A) .W W0 51 2= (B) W W1 2=

(C) W W21 2= (D) W W41 2=

MCQ 2.3.14 Equivalent surface about a point charge are in which one of the following forms ?(A) Spheres (B) Planes

(C) Cylinders (D) Cubes

MCQ 2.3.15 Consider the following statements regarding an electrostatic field :1. It is irrotational

2. It is solensoidal

3. It is static only form a macroscopic view point.

4. Work done in moving a charge in the field form one point to another is independent of the path of movement.

Which of the statements given above are correct ?(A) 1, 2 and 3 (B) 1, 2 and 4

(C) Only 2 and 4 (D) 1, 3 and 4

MCQ 2.3.16 The potential (scalar) distribution is given as 20V y x4 4 3= + . If 0e is the permittivity of free space, what is the volume charge density vr at the point (2, 0) ?(A) 200 0e- (B) 200/ 0e-

(C) 200 0e (D) 240 0e-

MCQ 2.3.17 The x -directed electric field Ex having sinusoidal time variation ej tw and space variation in z -direction satisfies the equation 0Ex2d = under source free condition in a lossless medium. What is the solution representing propagation in positive z-direction ?(A) E E ex

kz0= - (B) E E ex

jkz0= +

(C) E E exjkz

0= - (D) E E exkz

0= +

MCQ 2.3.18 An infinitely long uniform charge of density 30 /nC m is located at ,y z3 5= = . The field intensity at (0, 6, 1) is 6 .7 8 .3 /V ma aE 5 4y z= - . What is the field intensity at ( , , )5 6 1 ?

(A) E (B) E5 6 1

6 12 2 2

2 2

+ ++

c m

(C) E5 6 1

6 1 /

2 2 2

2 2 1 2

+ ++

c m (D) E6 1

5 6 1 /

2 2

2 2 2 1 2

++ +

c m

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MCQ 2.3.19 What is the magnetic dipole moment in Am2 for a square current loop having the vertices at the point (10,0,0)A , (0,10,0)B , ( 10,0,0)C - and (0, 10,0)D - and with current 0.01 A flowing in the sense ABCDA ?(A) 2az (B) 2az-

(C) 4az (D) 4( )a ax y+

MCQ 2.3.20 An electric charge Q is placed in a dielectric medium. Which of the following quantities are independent of the dielectric constant e of the medium ?(A) Electric potential V and Electric field intensity E

(B) Displacement density D and Displacement y

(C) Electric field intensity E and Displacement density D

(D) Electric potential V and Displacement y

MCQ 2.3.21 Two coaxial cylindrical sheets of charge are present in free space, S

5 /C m2r = at 2 mr = and

S/C m4 2r =- at 4 mr = . The displacement flux density D at

3 mr = is(A) 5 /C mD ar 2= (B) 2/3 /C mD ar 2=

(C) 10/3 /C mD ar 2= (D) 18/3 /C mD ar 2=

MCQ 2.3.22 An electric potential field is produced in air by point charge 1 Cm and 4 Cm located at ( , , )1 1 5- and ( , , )1 3 1- respectively. The energy stored in the field is(A) 2.57 mJ (B) 5.14 mJ

(C) 10.28 mJ (D) 12.50 mJ

MCQ 2.3.23 A dipole produces an electric field intensity of 1 /mV m at a distance of 2 km. The field intensity at a distance of 4 km will be(A) 1 mV/m (B) 0.75 mV/m

(C) 0.50 mV/m (D) 0.25 mV/m

MCQ 2.3.24 The energy stored per unit volume in an electric field (with usual notations) is given by(A) 1/2 H2e (B) 1/2 Ee

(C) 1/2 E 2e (D) E 2e

MCQ 2.3.25 A positive charge of Q coulomb is located at point (0,0,3)A and a negative charge of magnitude Q coulombs is located at point (0,0, 3)B - . The electric field intensity at point (4,0,0)C is in the (A) negative x -direction (B) negative z -direction

(C) positive x -direction (D) positive z -direction

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For View Only Shop Online at www.nodia.co.inMCQ 2.3.26 The force between two points charges of 1 nC each with a 1 mm separation in air is

(A) 9 10 N3#

- (B) 9 10 N6#

-

(C) 9 10 N9#

- (D) 9 10 N12#

-

MCQ 2.3.27 Gauss law relates the electric field intensity E with the volume charge density vr at a point as(A) E v0e r#d = (B) E v0: e rd =

(C) /E v 0r e#d = (D) /E v 0: r ed =

MCQ 2.3.28 The electric field strength at any point at a distance r from the point charge q located in a homogeneous isotropic medium with dielectric constant e, is given by

(A) r

qE a4 r2

1

pe=-

(B) cosdE D S q= #

(C) rqE a

4 r2pe= (D)

rqE a

4 r2

2

pe=

MCQ 2.3.29 The vector statement of Gauss’s a law is

(A) d dvD Sv

ss

: r= ## (B) d dvD Sv

vs

: r=# #

(C) d dvD Ss

vv

2: r=## # (D) d dvD S v

vs: r= ##

MCQ 2.3.30 Two charges are placed at a distance apart. Now, if a glass slab is inserted between them, then the force between the charge will(A) reduce to zero (B) increase

(C) decrease (D) not change

MCQ 2.3.31 The following point charges are located in air : 0.008 Cm+ at , m0 0^ h

0.05 Cm+ at , m3 0^ h

0.009 Cm- at , m0 4^ h

The total electric flux over a sphere of 8 m radius with centre ,0 0^ h is(A) 0.058 Cm (B) 0.049 Cm

(C) 0.029 Cm (D) 0.016 Cm

MCQ 2.3.32 Electric flux through a surface area is the integral of the(A) normal component of the electric field over the area

(B) parallel component of the electric field over the area

(C) normal component of the magnetic field over the area

(D) parallel component of the magnetic field over the area

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MCQ 2.3.33 Assertion (A) : The electric field around a positive charge is outward.Reason (R) : Gauss’s law states that the differential of the normal component of the outward electric flux density over a closed surface yields the positive charge enclosed.(A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is

the correct explanation of Assertion (A)

(B) Both Assertion (A) and Reason (R) are individually true but Reason (R) is not the correct explanation of Assertion (A)

(C) Assertion (A) is true but Reason (R) is false

(D) Assertion (A) is false but Reason (R) is true

MCQ 2.3.34 Point charges of 2 3andQ nC Q nC1 2= = are located at a distance apart. With regard to this situation, which one of the following statements is not correct ?(A) The force on the 3 nC charge is repulsive.

(B) A charge of nC5- placed midway between andQ Q1 2 will experience no force.

(C) The forces andQ Q1 2 are same in magnitude.

(D) The forces on andQ Q1 2 will depend on the medium in which they are placed.

MCQ 2.3.35 Which one of the following is the correct statement ?Equi-potential lines and field lines(A) are parallel (B) are anti-parallel

(C) are orthogonal (D) bear no definite relationship

MCQ 2.3.36 Point charges of 3- nC and 10 nC are located in free space at ( , , )1 0 0- m and ( , , )1 0 0 m respectively. What is the energy stored in the field ?(A) Zero (B) 450 nJ

(C) 450- nJ (D) 900- nJ

MCQ 2.3.37 A spherical balloon of radius a is charged. The energy density in the electric field at point P shown in the figure given below is w . If the balloon is inflated to a radius b without altering its charge, what is the energy density at P ?

(A) w ab 3

b l (B) w ab 2

b l

(C) w abb l (D) w

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For View Only Shop Online at www.nodia.co.inMCQ 2.3.38 Which one of the following statements does not state that electrostatic field is

conservative ?(A) The curl of E is identically zero

(B) The potential difference between two points is zero

(C) The electrostatic field is a gradient of a scalar potential

(D) The work done in a closed path inside the field is zero

MCQ 2.3.39 Sphere of radius a with a uniform charge density 4 C/mv3r shall have electric flux

density at r a= , equal to

(A) a i3 C/mv r2r (B) i3

1 C/mv r2r

(C) a i C/mv r2r (D) a i4 C/mv r

2r

MCQ 2.3.40 Equipotential surfaces about a pair of equal and opposite linear charges exist in what form ?(A) Concentric spheres (B) Concentric cylinders

(C) Non-concentric cylinders (D) Planes

MCQ 2.3.41 For electrostatic fields in charge free atmosphere, which one of the following is correct ?(A) 0 0E Eand :d#d = =

(B) 0 0E Eand :d!#d =

(C) 0 0E Eand :d !#d =

(D) 0 0E Eand :! !#d d

MCQ 2.3.42 If the electric field established by three point charge Q , Q2 and Q3 exerts a force 3F on Q3 and 2F on Q2 , then what is the force exerted on the point charge Q ?(A) F (B) F-

(C) 5F (D) 5F-

MCQ 2.3.43 Which one of the following is the Poission’s equation for a linear and isotropic but inhomogeneous medium ?

(A) E2 er

d =- (B) V: de rd =-^ h

(C) V:d d e r=-^ h (D) V2 er

d =-

MCQ 2.3.44 Plane 10z m= carries surface charge density 20 nc/m2. What is the electric field at the origin ?(A) 10 va /mz- (B) 18 a v/mzp-

(C) 72 va /mzp (D) 360 a v/mzp-

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MCQ 2.3.45 Consider the following diagram :

The electric field E at a point P due to the presence of dipole as shown in the above diagram (considering distance r >> distance d ) is proportional to(A) /r1 (B) /r1 2

(C) /r1 3 (D) /r1 4

MCQ 2.3.46 What is the value of total electric flux coming out of a closed surface ?(A) Zero

(B) Equal to volume charge density

(C) Equal to the total charge enclosed by the surface

(D) Equal to the surface charge density

MCQ 2.3.47 A charge is uniformly distributed throughout the sphere of radius a . Taking the potential at infinity as zero, the potential at r b a<= is

(A) a

Qr dr4b

03pe

-3# (B)

rQ dr

4b

02pe

-3#

(C) r

Q dra

Qr dr4 4a

a

b

02

03pe pe

- -3# # (D)

rQ dr

4a

02pe

-3#

MCQ 2.3.48 A potential field is given by V x y yz3 2= - . Which of the following is not true ?(A) At the point ( , , )1 0 1- , V and the electric field E vanish

(B) x y 12 = is an equipotential plane in the xy -plane

(C) The equipotential surface V 8=- passes through the point ( , , )P 2 1 4-

(D) A unit vector normal to the equipotential surface V 8=- at P is

( 0.83 0.55 0.07 )x y z- + +

MCQ 2.3.49 The relation between electric intensity E , voltage applied V and the distance d between the plates of a parallel plate condenser is(A) /E V d= (B) E V d#=

(C) /( )E V d 2= (D) ( )E V d 2#=

***********

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SOLUTIONS 2.1

SOL 2.1.1 Option (D) is correct.Since all the charges are exactly equal and at same distance from the centre. So, the forces get cancelled by the diagonally opposite charges and so the net force on the charge located at centre is Fnet 0 N=

SOL 2.1.2 Option (A) is correct.Since one of the four charges has been removed so, it will be treated as an additional

2 C- charge has been put on the corner, so the force due to the additional charge will be :

F ( )

( ) ( )k

12 1 10

2

9# #= - + -

9 10 2 109 9# # #= - - 18 N=

and so the net force experienced by the charge located at center is Fnet 18 N4 22= + =

SOL 2.1.3 Option (A) is correct.Since the two point charges are positive so the introduced third point charge must be negative as to make the entire system in equilibrium as shown below

as the system must be in equilibrium so the force between all the pair of charges will be equali.e. FAB F FCB AC= =

9dq

3 2-^

^

h

h

dq36

336 9

2 2= =^

^

^ ^h

h

h h

Solving the equation we get, q 4 C=- and d 2 m=

SOL 2.1.4 Option (D) is correct.Electric field intensity at any point P due to the two point charges Q1 and Q2 is defined as

E ( ) ( )

kRQ

RQR R

13

11

23

22= +e o




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where, R1 and R2 is the vector distance of the point P from the two point charges.So the net electric field due to the two given point charges is

E ( ) ( ) ( )

( ) ( ) ( ) ( )a a a7 4 3 0 1 2

9 10 5 10 7 4 3 0 1 2x y z

2 2 2

9 9# # #=

- + + - + - +- - + + - + - +-

6 @

( ) ( ) ( )

( ) ( ) ( )a a a7 5 3 0 1 3

9 10 2 10 7 5 3 0 1 3x y z

2 2 2

9 9# # #+

- + + - + - -- + + - + - --

6 @

3 3 2 3 4a a a a a a

19

45

29

18/ /

x y z x y z

3 2 3 2=- - + +

+- + -6 6@ @

1.4 1.284 1.004a a ax y z= - -

SOL 2.1.5 Option (B) is correct.From the positions of the three point charges as shown in the figure below, we conclude that the electric field intensity due to all the point charges will be directed along az .

So the net electric field intensity produced at the point P due to the three point charges is

E RQ a4 R

0pe= / (where R is the distance of point P from the charge Q )

( ) ( ) ( )

Q a4 3 11

31

3 11

z0

2 2 2pe=+

+ +-; E (a aR z= )

5 10 9 10 a161

91

41

z9 9

# # # #= + +-: D 9.0625a2 z=

SOL 2.1.6 Option (C) is correct.Since, both the point charges are positive, so the point P must be located on the line joining the two charges as shown in figure.



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For View Only Shop Online at www.nodia.co.inGiven the net electric field intensity at point P is zero i.e. ES 0=Since, the direction of electric field intensity due to the two charges will be opposite

So, ( )x

qxq

41

41

12

02

02pe pe-

-: =D G 0=

x2 2 ( )x1 2= - x x2 12 + - 0=

x 22 4 4!- + 1 2!=-

x .0 414= and x = .4143-As discussed above the point P must be located between the two charges, so we have the distance of point P from charge Q+ as: 0.414 mx =

SOL 2.1.7 Option (D) is correct.Given the volume charge density, vr 2 2 10C C6m #= = -

So the total charge present throughout the shell is defined as the volume integral of the charge density inside the region:

i.e. Q dvvr= #

( )( )sinr drd d2 10.

.

r

6 2

0 02

0 03

00

2

# q q f=q

p

f

p-

===###

( ) r4 2 10 3 .

.6

3

0 02

0 03

# #p= -: D

1.6 10 160 pC10#= =-

SOL 2.1.8 Option (A) is correct. The charge located in the region 2 cm r a< < is

q Q2= 2

1 160#= 80 pC=

Similarly as calculated in previous question we have

q dvvr= #

or 80 pc ( )( )sinr drd d2 10.

.

r

6 2

0 02

0 03

00

2

# q q f=q

p

f

p-

===###

or 80 10 12#

- r4 2 10 3 .

a6

3

0 02# # #p= : D

therefore, a ( . )4 2 103 80 10 0 02

/

6

123

1 3

# #

# #p

= +-

-

; E .59 .6cm cm4 4= =

SOL 2.1.9 Option (D) is correct.Charge density in a certain region is defined as the charge per unit volume.Since the net charge in the subregion %30= of the electronic charge

So the argch e density argvolume

net ch e=




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( . )

1010030 1 6 10

12

19# #

=-

-

-

4.8 10 8#=- - 48 /nC m3=-

SOL 2.1.10 Option (D) is correct.Given the surface charge density Sr z2r=So the total charge distributed over the cylindrical surface is

, Q dSSr= #

( )( )z d dzz

2

0

2

0

1r r f=

f

p

==## (dS d dzr f= )

z8 22

0

1

02

# # f= p: 6D @ at 2r =

8 21 2 8# # p p= = 5.1 C3 m=

SOL 2.1.11 Option (D) is correct.Given the surface charge density rS 3 /C mxy 2=So, total stored charge on the triangular surface is

Q dSr= S# xy dxdy3y

x

x 1

2 5

1

2=

=

- +

=^ h## 6.5 C=

SOL 2.1.12 Option (B) is correct. Total stored charge on the disk is evaluated by taking surface integral of the charge density.

i.e. Q dSsr= # ( ) 2r rdr30

5p= ^ h#

r6 33

0

5

p= : D 503 p=

SOL 2.1.13 Option (C) is correct.For an electric field to exist, the its curl must be zero. So, we check the existence of the given field vector first.Given the electric field intensity E 2 4 6 /V mxy yz xza a ax y z= + +

So, E#d xy yz xz

a a a2

2 3

x

x

y

y

z

z= 22

22

22

2 3y z xa a a2 0x y z != - - -6 @

Therefore, as the curl of the given electric field is not equal to zero so, the field does not exist.

SOL 2.1.14 Option (B) is correct.Electric field intensity in free space at a distance R from an infinite line charge with charge density Lr is defined as



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E RR

2L

02pe

r=

Given Lr 1 /C mm= 1 10 /C m6#= -

R 2a ax y=- -

So, E ( ) a a

21 10

52 x y

0

6#pe= - --

b l 7.2 3.6 /kV ma ax y=- -

SOL 2.1.15 Option (A) is correct.Electric flux density in a certain region for the given electric field intensity is defined as D E0e= ( 2 )x xya ax y0

2e= + So at the point ( , , )1 0 1- D ( )ax0e=

SOL 2.1.16 Option (B) is correct.According to Gauss law the total outward electric flux from a closed surface is equal to the charge enclosed by it

i.e. y dD S:= # Qenc=

So when the charge enclosed by the volume is zero then the net outward flux is zero, or in other words, the net electric field flux emanating from an arbitrary surface not enclosing a point charge is zero.Now, the electric field intensity outside a charged sphere having total charge Q is determined by treating the sphere as a point charge

i.e. E r

Q a4 r

02pe

=

where r is distance of the point form center of sphere and ar is it’s radial direction.So the electric field intensity at any point outside the charged sphere is not zero.Therefore, Assertion(A) is true but Reason(R) is false.

SOL 2.1.17 Option (C) is correct. E 3r ar2=According to Gauss’s law the total charge stored in a closed surface is equal to the surface integral of its flux density over the closed surface.

i.e. Qenc dD S:= # dE S0 :e= # (3 )r dSar0

2e= #

( )( )r r3 402 2e p= ( dS# 4 r ar2p= )

3 4 204

# # #e p= 2 mr = .5 3 10 9

#= - .3 nC6=

SOL 2.1.18 Option (D) is correct.According to Gauss law net outward electric flux from any closed surface is equal to the total charge enclosed by the volumei.e. y Qenc=

or, y dvvr= #




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( )sinrr drd d1

r2

2

00

2

0

1q q f=

q

p

f

p

===b l###

1 2 2# # p= 4 Cp=

SOL 2.1.19 Option (A) is correct.As we have already determined the total electric flux crossing the surface 1 mr = So, electric flux density D at 1 mr = is evaluated as below:

Total electric flux y dD S:= #So we have dD S:# 4p= ( 4y p= ) (4 )D r2p 4p=

D 1 /C mr12

2= =

Thus D /C ma4 r2=

SOL 2.1.20 Option (D) is correct.As the point charge is located at origin. So flux due to it will be emanating from all the eight quadrants symmetrically.So the flux through the portion of plane 2 mx y+ = lying in first octant is /1 8 of the total flux emanating from the charge located at origin.and from Gauss law, total flux 8 CQenc= =

So, flux through the surface 2 mx y+ = is Q8enc 1C8

8= =

SOL 2.1.21 Option (D) is correct.We construct a Gaussian surface at rr = as shown in figure.

So, according to Gauss law the total outward flux through the surface rr = will be equal to the charge enclosed by it.i.e. rhD 2p^ h r hv

2r p= ^ h (assume the height of the cylinder is h )

So, D r2vr=

Therefore the electric field intensity at a distance r from the cylindrical axis is

E rD2

v

0 0e er= = a k

Thus E r\

SOL 2.1.22 Option (B) is correct.According to Gauss law the surface integral of the electric flux density over a closed



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For View Only Shop Online at www.nodia.co.insurface is equal to the total charge enclosed inside the region defined by closed surface.

i.e. dD S:# Qenc=

or dE S:# Q1ene

0e= (since E D

0e= )

As we have to evaluate E for 2r# and since the charge density is zero for 2r# so Qenc 0= (for r 2# )

Therefore, dE S:# 1 00#e=

E 0=

SOL 2.1.23 Option (D) is correct.Again from Gauss law, we have the surface integral of electric field intensity over the Gaussian surface at r 3= as

dE S:# Q1enc

0e=

dE S:# dv1v

0er= # /dv r dv1 0 1 4

r r

0

2

0

2

2 3

e e= +

1# #

^ h1 2 344 4 1 2 3444 444

# #

4 (3) )E( 2p# ( )sinrr drd d1 4

r02

2

0

2

02

3

e q q f=pp

=b l###

4 9)E( p# ( )4 4 3 20

#e

p= -

E a94

r0e

=

SOL 2.1.24 Option (C) is correct.As calculated in the previous question, we have the surface integral of the electric field intensity over the Gaussian surface r 5= as

dE S:# dv1v

0er= # /dv r dv dv1 0 1 4 1 0

r r r

0

2

0

2

2 4

0

4 5

e e e= + +

1 1# # #

^ h1 2 344 4 1 2 3444 444 1 2 344 4

# # #

4 ( ) )E( 5 2p# ( )sinrr drd d1 4

r02

2

0

2

02

4

e q q f=pp

=b l###

100 )E( p dr4 40 2

4#e

p= #

E a2512

r0e

=

SOL 2.1.25 Option (D) is correct.According to Gauss’s law vr Eed=So when the field intensity is uniform Ed 0=and vr 0Eed= =So no charge can be present in a uniform electric field.




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SOL 2.1.26 Option (D) is correct.According to Gauss law the volume Charge density in a certain region is equal to the divergence of electric flux density in that regioni.e. vr D:d=

cossin

sin sinr r

rr r r

1 122

23 32

222q

q qq q= +b bl l

cos cosr r1 14 4q q=- +

0=

SOL 2.1.27 Option (B) is correct.According to Gauss law the volume Charge density in a certain region is equal to the divergence of electric flux density in that region.i.e. vr D:d= x2=So total charge enclosed by the cube is

Q dvvr= # ( )( )x dxdydz21

1

0

2

0

2=

-###

4 2 2# #= 1 C4=

SOL 2.1.28 Option (D) is correct.Net electric potential due to two or more point charges is defined as :

V RQ

4 0pe= /So, the electric potential at point P due to the two point charges is

V RQ

RQ

4 40 1

1

0 2

2

pe pe= +

where Q1 1 Cm=+ , Q2 1 Cm=- and ,R R1 2 are the distance of the point P from the two point charges respectively.So, we have R1 ( ) ( ) ( ) .3 0 0 0 4 1 5 832 2 2= - - + - + - - = R2 ( ) ( ) ( ) .3 0 0 0 4 1 4 242 2 2= - - + - + - + =

Thus V . . 578.9 V410

5 831

4 241

0

6

pe= - =--

: D

SOL 2.1.29 Option (A) is correct.Electric field at any point is equal to the negative gradient of potential

i.e. E VxV

yV

zVd

22

22

22=- =- + +c m

y zx y z

x xyzx y z

ya a2 33 2

2 36

x y2 3

2 2 23

2 2 2=- ++ +

+ ++ +c cm m=

xy zx y z

z a32 39

z2 2

2 2 2+ ++ +c m G

So, at the point ( 3, 2, 1)P x y z= = =- E 3.6 11.4 35.6 /V ma a ax y z= + -

SOL 2.1.30 Option (D) is correct.Electric flux density in terms of field intensity is defined as



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For View Only Shop Online at www.nodia.co.in D E0e=So, at point ( , , )P 3 2 1- , D (3.6 11.4 35.6 )a a ax y z0e= + - 31.4 101 314.5 /pC ma a ax y z

2= + -

SOL 2.1.31 Option (C) is correct.Laplace’s equation for a scalar function V is defined as V2d 0=but at the point of maxima V2d must have a negative value while at the point of minima V2d must have a positive value. So the condition of maxima/minima doesn’t satisfy the Laplace’s equation, therefore the potential function will have neither a maxima nor a minima inside the defined region.

SOL 2.1.32 Option (D) is correct.Electric force experienced by a point charge q located in the field E is defined as F qE=So, the force applied at the point charge 1C+ located at ( , , )y0 0 is

F ( ) cos sinr

Qd a a14

2 r0

32

peq q= + q6 @ ( 1Cq =+ )

( )sinr

Qd a4

90 z0

3 cpe

= -6 @ ( 90cq = , a az=-q , r y= )

yQd a

4 z0

3pe= -

***********




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SOLUTIONS 2.2

SOL 2.2.1 Option (A) is correct.For determining the position of the third charge, first of all we evaluate the total electric field at the given point C (0,1,0) due to the two point charges located at points A(1,0,0) and B ( , ,1 0 0- ) respectively as shown in figure.

Electric field due to the charge located at point A is

E1 9 10 10kQACAC a a

21

1 1x y

39 9

3# # #= =++-

b^

^l

h

h a a

4 29

x y= +^ h

and the electric field due to charge at point B is

E2 9 10 10kQBCBC a a

21

1 1x y

39 9

3# # # #= =+

- +-^b

^

^hl

h

h a a

4 29

x y= - +^ h

So, E E1 2+ a a a a4 2

94 2

9x y x y= + + - +^ ^h h a

2 29

y=

As the field is directed in ay direction so for making E 0= the third charge of nC2+ must be placed on y -axis at any point 1y > . Consider the position of the

third charge is ( , , )y0 0 . So, electric field at point C due to the third charge is.

E3 ( )( )

( )y

a1

9 10 2 10y2

9 9# # #=

--

-

( )ya

19 2

y2=--

and since the total electric field must be zeroSo, we have E E E1 2 3+ + 0=

( )y

a a2 2

91

9 2y y2-

- 0=

( )y 1 2- 4= or y 3= , 1- as discussed above y 1> , so the point will be located at y 3=i.e. Point P will have the coordinate ( , , )0 3 0



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For View Only Shop Online at www.nodia.co.inSOL 2.2.2 Option (A) is correct.

Electric field intensity at any point P due to the uniformly charged plane with charge density Sr is defined as

E a2s

n0e

r=

where an is the unit vector normal to the plane directed toward point PSince the unit vector normal to any plane f 0= is defined as

an ff

!4

4=

So, we have the unit vector normal to the given charged plane x y3 4 0+ = as

an 3 4a a

3 4x y

2 2!=

++

3 4a a

5x y

!= + ( f x y3 4= + )

Since at point ( , , )1 0 3 f 0> , so, we take the positive value of an .

Therefore, E a2s

n0e

r= ( / )( ) a a

2 10 362 10

53 4x y

9

9#

p=

+-

-

b l ( Sr =2 /nC m2)

a a436 3 4x yp= +^ h 67.85 90.48a ax y= + /V m

SOL 2.2.3 Option (A) is correct.Horizontal component of the electric field intensity will be cancelled due to the uniform distribution of charge in the circular loop. So the net electric field will have only the component in az direction and defined as below :

E (2 )rr zz a4

12 2 /L z

03 2pe r p=

+^ h

(9 10 ) (2 10 ) (2 4) a4 3

3/ z

9 92 2 3 2p# # # # #=+

-

^ h

9 2 a1252 4 3

z# #p

# #= 1 . /V ma2 56 z=

SOL 2.2.4 Option (C) is correct.Electric flux density produced at a distance r from a point charge Q located at origin is defined as




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D rQ a

4 r2p=

So, the divergence of the electric flux density is

D:d 0r r

rrQ1

422

222

p= =c m

So it is 0 for all the points but at origin ( )r 0= its divergence can’t be defined.

SOL 2.2.5 Option (D) is correct.

The total flux leaving the closed surface is

y dD S:= # ( dS is normal vector to surface)The closed cube has total eight surfaces but as the vector field has no component in az direction so we have the integrals only through the four separate surfaces as shown in the figure

So, y x y dydz x y dydz

, ,at front at backx x

2

0

1

0

1

0

2

0

1

0

1

1

= + -

= =

## ##

x y dxdz x y dxdz

, ,at left at righty y

2 2

0

1

0

1

0

2 2

0

1

0

1

1

+ - +

= =

## ##

ydydz x dxdz0

1

0

12

0

1

0

1=- +## ##

y z x z2 3

2

0

1

01

3

0

1

01=- +; 6 : 6E @ D @ 1 12

151

41

# #=- + =-

SOL 2.2.6 Option (A) is correct.Given the electric flux density D /C mx y y xa ax y

2 2 2 2= +

So, div D D:d=x y z

x y y xa a a a ax y z x y2 2 2

:22

22

22= + + +c ^m h

D:d xy x y2 2 2= +6 @

21

41

43= + = (center of the cube is located at , ,2

121

21

b l)



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From the given data we have the electric flux density at 0.2 mr = as D 5 /nC mr ar2 2=According to Gauss law the volume charge density at any point is equal to the divergence of the flux density at that point, so we have the volume charge density at 0.2 mr = as vr D:d=

( )r r

r r1 522 2

22= ^ h

rr1 5 42

3# #=

r20= 4 /nC m3= ( 0.2 mr = )

SOL 2.2.8 Option (D) is correct.Again from the given data we have the electric flux density at 1 mr = as D 2/ /nC mr ar2 2=So, the volume charge density at 1 mr = is

vr D:d= r r

rr

1 22

222

2= bc lm 0=

SOL 2.2.9 Option (B) is correct.Given the moment p 4 C maz0pe -=The electric field intensity at any point ( , , )r q f produced due to an electric dipole lying along z -axis and having the dipole moment p in az direction is defined as

E (2 )cos sinrp a a

4 r0

3peq q= + q

E cos sinr

a a1 2 r3 q q= + q^ h ( 4 C mp az0pe -= )

Now, given that the z -component of electric field is zeroi.e. Ez E az:= 0=

( ) ( )cos sinr

a a a a1 2 r z z3 : :q q+ q6 @ 0=

cos sinr1 23

2 2q q-6 @ 0=

cos sin2 2 2q q- 0= E 0z =

1 3 2cos21 q+6 @ 0=

Thus .54 7cq = or .125 3cq =Therefore the conical surface of angle .54 7cq = or .125 3c will have the electric field component E 0z = .

SOL 2.2.10 Option (A) is correct.Electric field intensity produced at a distance r from an infinite line charge with charge density Lr is defined as

E 2L

0pe rr=

and since the electric potential at point ( , / , )1 2 2p is zero so, the electric potential at




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point ( , , )zr f will be equal to the integral of the electric field from point ( , / , )1 2 2p to the point to ( , , )zr f .

i.e. V dE l( , / , )

( , , )z

1 2 2:=-

p

r f

# d2L

01 pe rr r=-

r

b l# ( )ln2L

0 1per r= -

r: D

V 2 10 9 10 ln 19 9

r# # #= -b l ln18 1

r= b l ( 1 nCLr =+ )

Note: Since the infinite line charge has the equipotential cylindrical surface so for taking the line integral, f and z has not been considered.

SOL 2.2.11 Option (A) is correct.Electric potential at any point for a given electric field E is defined as

i.e. V d CE l:=- +#Now given the electric field intensity in spherical coordinate system

E ( )rr a4

2r2 2=

+and since the differential displacement in the spherical system is given as dl sindr rdr r da a ar q f= + +q f

So we have the electric potential

, V ( )rr dr C4

22 2=-+

+# r

C4

12=+

+

At maxima, drdV 0=

( )

2r

r41

2 2 #+- 0=

Solving the equation we get, r 0= and r 3=

at r 0= drd V

2

2

ve=-

So the electric potential will be maximum at origin.

SOL 2.2.12 Option (B) is correct.As calculated in the previous question, the electric potential at point ( , , )r q f is

V r

C4

12=+

+

So at r 0= , electric potential is V1 C41= +

and at r 2= electric potential is V2 C81= +

So potential difference between the two surfaces is :

V12 C C41

81= + - +b bl l volt2

1=

SOL 2.2.13 Option (C) is correct.Electric potential at a distance R from a dipole having moment p is defined as

V RRp

4 03

:

pe=

So we have the potential at point A due to the dipole located at point B as:



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V AB

p AB4 0

3:

pe= ( )

2 ( )a a a a a a

4 1 1 8

53 8 10x y z x y z

02 2 2 3

9: #

pe=

+ +

- + + + -b l

0.6 V=

SOL 2.2.14 Option (B) is correct.Since the charge is being split and placed on a circular loop so the distance of all the newly formed point charges from the center of the loop will be equal as shown in the figure.

Therefore, the potential at the center of the loop will be

V 4/r

Q4

40pe= c m (9 10 )

( )5

20 1099

## #=

-

( 20 nCQ = )

36 V=

SOL 2.2.15 Option (A) is correct.The work done in carrying a charge q from point A to point B in the field E is defined as

W q dE lA

B:=- #

Given the electric field intensity in the cartesian system as E 2 2y xa ax y= +and since the differential displacement in cartesian system is given as dl dx dy dza a ax y z= + +So, the work done in carrying charge 2 Cq =+ from point , / ,A 1 1 2 3^ h to the point

(4,1,0)B is

W 2 ydx xdy2 2/yx 1 2

1

1

4=- +

==; E##

The curve along which the charge is being carried is given as

y x2= x& y2 2=




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Therefore, we have W / 2x dx y dy2 2 2 2/

2

1 2

1

1

4=- +_ ^i h; E##

x y4 32

32/

/3 2

1

4 31 21=- +6 6; @ @ E 4 3

7 2127=- +; E

15.5 J=-


The work done in carrying a charge q from one point to other point in the field E is defined as

W q dE l:=- #and since the differential displacement for the defined circular arc is dl d ar f= f as obtained from the figure

So, the work done is W 2 ( ) ( )x y da a a/

x y0

4

: r f=- -f

p

f=

#now we put cosx r f= , siny r f= and sina ax : f=-f , cosa ay : f=f in the expression to get

W sin cos d2 2/

2

0

4

r f f f=- -p

# ( )sin d2 1 2/

0

4

# f f=- -p

# ( 1r = )

1 J=+

SOL 2.2.17 Option (A) is correct.

Consider the last charge is being placed at corner D so the potential at D due to the charges placed at the corners , ,A B C is



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V rq

41

0pe= / 410

11

21

11

0

9

pe= + +-

< F

(9 10 ) 10 22

19 9# # #= +-

c m

4.36 volt3=As the potential at infinity is zero so the work done in carrying the last charge from infinity to the fourth corner is W qV= .10 24 369

#= - ( 1 nCq = ) 24.36 nJ=

SOL 2.2.18 Option (A) is correct.Consider the first charge is being placed at A so the potential at A will be zero as there is no any charge present at any of the corner and therefore the work done in carrying the first charge is W1 0=now consider the second charge is being placed at B so the potential at B will be only due to the charge at corner A

i.e. V2 aq

4 0pe=

and therefore the work done in placing the second charge at B is

W2 qV2= 4q aq

0pe= b l

9 nJ41

110

0

18

pe #= =-

and similarly the potential at the corner C will be due to the charges at corners A and B

i.e. V3 rq

41

0pe= / 410

11

21

0

9

pe= +-

c m

therefore the work done in placing the third charge at C is

W3 qV3= q 41 1

21

0pe= +c m= G

(9 10 ) 102

1 19 18# #= +-

c m

and the work done in placing the last charge at D has already been calculated in previous questioni.e. W4 24.36 nJ= So the total work done in assembling the whole configuration of four charges is W W W W W1 2 3 4= + + + . .0 9 15 36 24 36= + + + 48.72 nJ=

SOL 2.2.19 Option (A) is correct.The work done in carrying a charge q from point A to point B in the field E is defined as

W q dE lA

B:=- #




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Given that q 2 C= E ( 1)sin cos sinza a azf r f r f= + + +r f

and since the given points A and B have 21 2r r= = and z z 11 2= = so the differential displacement in the cylindrical coordinate system from A to B may be given as dl d ar f= f for 0 30< < cf .

Therefore the work done is, W 2 ( )cosz d10

30r f r f=- +

c

c

f=^_ h i#

2 ( ) ( ) sin1 1 2 2

030f# # #=- +c

c6 @

J8 21 4#=- =-

SOL 2.2.20 Option (D) is correct.Consider the 1 Cm+ charge is transferred first, from infinity to the given point

( , , )A 3 6 0- so the work done for transferring the charge will be zero as there is no charge initially present.now the potential at point B due to the charge at A is

V qAB4

10pe= A

9 105 10 1

10126

9 1092 2 2

6 3

##=

+ +=

- ( 1 )Cq m=A

So the work done in transferring the charge 2 mC+ at point B is W q V= B

(2 10 )126

9 1033

##

#= -c m ( 2 mCq =B )

1.604 J=


The total potential energy stored in the system is given by

W q V21

nn

n1

4

==/

where qn is the charges at the four corners and Vn is the total electric potential at the corresponding corners.For the 1st corner :



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For View Only Shop Online at www.nodia.co.inCharge, q1 8 nC=and potential, V1 V V V21 31 41= + +where V21, V31 and V41 are the potential at the 1st corner due to the charges q2, q3and q4 respectively

So, V1 . . .q q q

41

0 01 0 01 2 0 010

2 3 4

pe= + +: D ( 8 nCq q q2 3 4= = = )

. . .48 10

0 011

0 01 21

0 011

0

9#pe= + +

-

< F

1.944 10 V4#=

Since all the charges are equal so the potential will be same at all the corners and therefore the total potential energy stored in the system of the charges is

W 4 q V21

1 1#= ^ h

2 (8 10 ) ( . )1 944 109 4# ## #= - 0.312 mJ=

SOL 2.2.22 Option (C) is correct.Energy density in a certain region in free space having electric field intensity E is defined as

wE E E21

0 :e=

and since the electric field is equal to the negative gradient of the potential so we have E Vd=-

xV

yV

zVa a ax y z2

222

22=- + +; E

/V mx yz xy z xyz

a a a1 1 1x y z2 2 2= + +; E

So the energy density inside the cube will be

wE (E E)21

0 :e= x y z x y z x y z2

1 1 1 10 4 2 2 2 4 2 2 2 4e= + +; E

Therefore the total energy stored in the cube is

WE w dvE= #

WE x y z x y z x y zdxdydz2

1 1 1 10 4 2 2 2 4 2 2 2 4

1

2

1

2

1

2e= + +; E###

x y z xy z xy z

dydz2 31 1 1 10

3 2 2 4 2 2 41

2

1

2

1

2e= - - -b l; E##

2 3 9670

# #e= .68 10 J12 13

#= -

SOL 2.2.23 Option (A) is correct.As calculated in the above question energy density at any point inside the cube is

wE x y z x y z x y z21 1 1 1

0 4 2 2 2 4 2 2 2 4e= + +; E

So, at the centre of the cube (1.5, 1.5, 1.5) the energy density is

wE ( . ) ( . ) ( . )21

1 5 1 5 1 53

0 4 2 2e= ; E 5.18 10 J13#= -




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SOL 2.2.24 Option (A) is correct.The charged sphere will be treated as a point charge for the field at any point outside the sphere. So, the electric field at distance r from the centre of the sphere will be :

E rQ

41

02pe= (For r R2 )

So the electric potential at the point will be :

( )V r dE lr

:=-3# (Taking 3 as a reference point)

rQ dr4

1 r

02pe=-

3# r

Q41 r

0pe=- -3

: D

rQ

41

0pe #=

So, ( )V r r1

\

The graph of ( )V r will be as :


For determining the electric field inside the spherical region at distance ( )r R# from the centre of sphere we construct a Gaussian surface as shown in the figure. So the surface integral of the electric field over the Gaussian surface is given as

( )E r4 2p Q1enc

0e= Q

Rr1

0 34 334 3

e pp

= e o> H

So, the electric field at a distance r from the center is



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E rQRr a

41

r20

3

3

p e= c m QRr a4 r

03pe= (for r R# )

Therefore the electric potential at the point P will be the line integral of the field intensity from infinity to the point P

i.e. ( )V r d dE r E rR

R

r

1 2: :=- +3

; E# #where E1 " electric field outside the sphere as calculated in previous question. E2 " electric field inside the sphere

( )V r rQ dr

RQ r dr4

141

R

rR

02

03pe pe=- +

3c m= G##

QR R

r R4

1 120

3

2 2

pe= - -b l; E

So, ( )V r decreases with increase in r .

SOL 2.2.26 Option (C) is correct.The total stored energy inside a region having charge density vr and potential V is defined as

WE Vdv21

vr= #As calculated in previous question the electric potential at any point inside the sphere is

( )V r QR R

r R4

1 120

3

2 2

pe= - -b l; E

(3 )r41

21

0

2

pe= -: D ( 1 mR = , 1CQ = )

Therefore the total energy stored inside the sphere is

WE ( ) (4 )RQ r r dr2

141

21 3

34 3

0

2 2

0

1

#p pe p= -e o: D#

( )r r dr83

41

24 3

0

2 4

0

1

# #p pep= -# ( 1 mR = , 1CQ = )

r r16

350

35

0

1

pe= -: D

163

54

4 53 9 10 4

0

9

##

# # #pe= = 5

27 109#= .4 10 J24 9

#=

SOL 2.2.27 Option (B) is correct.The electric field to counter act the gravitational force must produce the same force as applied by gravity but in opposite direction.i.e. e E^ h m g ae r= -^ h

where e is the charge of an electron, me is the mass of electron, g is acceleration due to gravity and ar is radial direction of earth.So, taking the magnitude only we have the required field intensity,

E em ge=

.. .

1 6 109 1 10 9 8

19

31

#

# #= -

-^ h

5.57 10 /V m1 11#= -




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Consider the electric field intensity produced at point , ,P 0 41

41

b l due to the charges located at points A, B and C respectively as shown in figure is EA, EB and EC respectively.

So the net electric field at point P is Enet E E EA B C= + +and since the electric field intensity at any distance R from a point charge Q is

defined as E QRR

4 03pe=

So Enet Q kQ kQPA PB

PBPCPCPA

41

03 3 3pe= + += G

Q kQ kQa a a a a a

41

41

41

41

41

43

41

43

41

41

43

41

43

/ / /

y z y z y z

0 2 2 3 2 2 2 3 2 2 2 3 2pe=+

++

+

- ++

+

-

b b

b

b b

b

b b

b

l l

l

l l

l

l l

l

; ; ;E E E

and since E 0net = so we have

k k

241 16

43

41

43

43

41

41

/

/

/ /3 2

3 2

2 2 3 2 2 2 3 2

#

-+

++^

b ^

b b b bh

l h

l l l l; ;E E

0=

Solving the equation we get k 5.591=

SOL 2.2.29 Option (D) is correct.The point charges can be represented as shown below.

So the electric field at point , ,x 0 0^ h will be directed along x -axis. Taking only magnitude we have the net electric field intensity at , ,x 0 0^ h as

E x aQ

xQ

x aQ

4 42

402

02

02pe pe pe

=-

- ++^ ^h h

... ....x

Qxa

xa

xQ

xQ

xa

xa

41 2 3

42

41 2 3

02

2

02

02

2

pe pe pe= + + + - + - + -a ak k: :D D

Since x a>> , neglecting higher powers of xaa k we get



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E x

Qxa

xa

xQ

xQ

xa

xa

41 2 3

42

41 2 3

02

2

02

02

2

pe pe pe= + + - + - +a ak k: :D D

x

Qa46

40

2

pe= K

xQa6

4

2

= c m

SOL 2.2.30 Option (A) is correct.According to Gauss law the surface integral of electric field intensity over a Gaussian surface is defined as

dE S:# Q1enc

0e=

So for the Gaussian surface outside the sphere at a distance ( )r R2 from the centre of the sphere we have

( )E r4 2p ( )Rv

0

34 3

er p

= (there is no charge outside the sphere)

Therefore at any point outside the sphere r R>^ h the electric field intensity will be

E rRa

4v

r0

234 3

per p

= ^ h

rR a3

vr

02

3

er= c m

and for the Gaussian surface inside the sphere at a distance ( )r R# from the center of the sphere we have

( )E r4 2p ( )rv

0

34 3

er p

=

Therefore at any point inside the sphere, the electric field intensity will be

E r

ra1

434

v

r0

2

3

e p

r p=

b l r a3

vr

0er= a k

SOL 2.2.31 Option (C) is correct.As discussed in Q.55. The electric field at any point inside a charged solid sphere is

E r a3v

r0er= a k

where r is the distance from center of the sphere and vr is the volume charge density given as

vr RQ

34 3p

= 3

2 1034 3

9#p

=-

^ h ( 2 nCQ = , 3 mR = )

1.77 10 /C m11 3#= -

So the force acting on electron when it is at a distance r from the center of the sphere is F eE= (e is the charge of an electron)

mdtd re 2

2

e r3

v

0er= a k (me is mass of an electron)

dtd r

2

2

. .. . r

9 1 10 8 85 101 6 10 1 77 10

331 12

19 11

# #

# ##=

-- -

- -

^ ^

^ ^

h h

h h

dtd r

2

2

. r2 17 1011#=-^ h




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.dtd r r1 17 102

211

#+ ^ h 0=

Solving the differential equation we have r . .cos sinA t A t1 17 10 1 17 101

112

11# #= +^ ^h h ....(i)

where A1 and A2 are constants.Now, at t 0= , 3 mr = as the electron is located at one end of the hole.So putting it in equation (i) we get, A1 3=

again at t 0= , drdr 0= as the electron is released from rest.

So putting it in equation (i) we get A2 0=Thus the position of electron at any time t is r .cos t3 1 17 1011

#= ^ h

at t 1 secm= r 2.83 m=

SOL 2.2.32 Option (D) is correct.As calculated in above question the position of the electron at any time t is r .cos t3 1 17 1011

#= ^ h

So, f2p .1 17 1011#=

f .2

1 17 1011#p= 5.44 10 Hz4

#= 54.4 KHz=

SOL 2.2.33 Option (B) is correct.The portion of the plane 1 my z+ = lying in the first octant bounded by the planes x 0= and 1 mx = has been shown in the figure through which we have to determine the total electric field flux.

According to Gauss law the total outward flux through a closed surface is equal to the charge enclosed by it.

i.e. y dD S:= # Qenc=

So the total electric field flux emanating flux from the line charge between x 0= and 1 mx = is

dE S:# Qenc0e

=1L L

0 0er

er= =^ h

and by symmetry, flux through the defined surface will be one fourth of the total electric field flux emanating from the defined portion .



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i.e. the electric flux crossing the surface dE S

4

:=

#4L

0er=

Note: It must be kept in mind that the total electric flux is dD S:# while the

total electric field flux is dE S:#

SOL 2.2.34 Option (D) is correct.Consider a point P inside the cylindrical surface of 2 m as shown in figure.

Now we make the use of superposition to evaluate the electric field at point P by considering the given charge distribution as the sum of two uniformly distributed cylindrical charges, one of radius 5 m and the other of radius 2 m, and such that the total charge in the hole is zero. Thus we obtain the net electric field at point P as Enet E E1 2= +where E1 is the electric field intensity at point P due to the uniformly charged cylinder of radius 5 m that has the charge density /nC m5 3

^ h, while E2 is the electric field intensity at point P due to charged cylinder of radius 2 m that has the charge density 5 /nC m3-^ h

As calculated in MCQ.61 the electric field intensity at a distance r from the cylindrical axes having uniform charge density vr is

E r2v

0er=

So we have E1 R R2 25 10v

01

0

9

1#

er

e= =-

and E2 R R2 25 10v

02

0

9

2#

er

e= = - -

So the net electric field at point P is

Enet R R25 10

0

9

1 2#e= -

-

^ h

By the triangle law of vector R R1 2- C ax= = (separation 1 m= )

So, Enet a25 10

x0

9#e=

-

^ h 282.5 /V max=




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SOL 2.2.35 Option (A) is correct.As we have calculated the electric field for the same distribution in Q.55. So we evaluate the electric potential by taking the line integral of the field intensity.

i.e. V dE l:=- #

E

for

for

r r R

rR r R

a

a

3

3>

vr

vr

0

02

3

#er

er=a

c

k

m

Z

[

\

]]

]]

The electric potential at any point outside the sphere r R>^ h is

V dE lr

:=-3#

lR dl3

vr

02

3

er=-

3c m#

Rl31vr

0

3

er=- -

3; E r

R3v

0

3

er=

and the electric potential at any point inside the sphere r R#^ h is

V d dlE E lR

R

r: :=- +

3; E# #

rR dr l dl3 3v v

R

rR

02

3

0er

er=- -

3b l##

Rr

l3

13 2

v R v

R

r

0

3

0

2

er

er=- - -

3: :D D R

Rr R

31

3 2 2v v

0

3

0

2 2

er

er=- - - -: :D D

R r3 2

32

v

0

2 2

er= -b l R r

2 3v

0

22

er= -b l


Given the total charge on the disk is Q 900 900 10C C6mp p#= = -

radius of the disk is a 6 m=and since the charge has been distributed uniformly over the surface so the small charge element dQ on the disk at a distance r from the center as shown in figure is given as

dQ SQ dS= b l rdrd

6900 10

2

6#

pp f=

-

^^

hh

rdrd25 10 6# f= -

The force applied by the charge element dQ on the 150 Cm charge located at point P is

dF ( )

RdQ150 10

2

6#=

-

( )( )

rdQ

16150 10

2

6#=+

-



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For View Only Shop Online at www.nodia.co.inAs the disk has uniformly distributed charge so the horizontal component of the field is get cancelled and the net force will have the only component in az direction and the net force by projection on z -axis is given as

F 25 10r

rdrdr4 16

150 1016

4

r 02

6 6

20

6

0

2 ##pe

f#=+ +f

p - -

== ^

^ ^d

h

h hn##

F r

27016

12

0

6p= -

+; E 9.44 N=

SOL 2.2.37 Option (D) is correct.Electric field at any point due to infinite surface charge distribution is defined as

E a2s

n0e

r=

where sr " surface charge density an " unit vector normal to the sheet directed toward the point where field is to be determined.At origin electric field intensity due to sheet at y 1=+ is

E1 a a2 25s

y y0 0e

re= - =-^ h a an y=-^ h

and electric field intensity at origin due to sheet at 1y =- is

E 1- a a2 25s

y y0 0e

re= =^ h a an y=^ h

So net field intensity at origin is

E E E1 1= ++ - a a25

25

y y0 0e e=- + 0=

SOL 2.2.38 Option (D) is correct.As the test charge is placed at point , ,2 5 4^ h. So it will be in the region y 1>+ for which electric field is given as E E E1 1= ++ -

a a2 2s

ys

y0 0e

rer= +^ ^h h (for both the sheet a an y= )

a22 5 10

y0

9# #

e=-

^ h a5 10

y0

9#e=

-

Therefore the net force on the charge will be

F qE= a5 10 5 10y

6

0

9

##e= -

-

^ ch m 2.83 10 N3#= -

SOL 2.2.39 Option (B) is correct.Since the electric field intensity due to a sheet charge is defined as

E a2s

n0e

r=

So it doesn’t depend on the distance from the sheet and given as E E E1 1= ++ -

a a2 2s

ys

y0 0e

rer= - + -^ ^h h a2

sy

0er=- a2

5 10y

0

9#e=-

-

So, it will be constant as we move away from the sheet.




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SOL 2.2.40 Option (D) is correct.As the charge is redistributed so the total charge will remain same on the sphere.Total charge before redistribution.

Q1 dvvr= # 6 /C m 34 13 3p= ^ ^bh h l ( 6 / )C mv

2r =

8p= Coulomband total charge after redistribution

Q2 dvvr= # 4k r r dr3r

2 2

0

1p= -

=^ h#

Since Q1 Q2=

So, we have 8p k r r dr4 3 2 4

0

1p= -^ ^h h# k r r4 5

35

0

1

p= -: D k4 1 51p= -: D

or k .53=

SOL 2.2.41 Option (B) is correct.According to Gauss’s law the total electric flux through any closed surface is equal to the total charge enclosed by the volume.Now consider the complete spherical surface defined by 48 mr = through which the total flux is equal to the point charge.So the total flux passing through the hemispherical surface will be half of the point charge.

i.e. y 50 25C CQ2 2

m m= = =

SOL 2.2.42 Option (D) is correct.For any point inside the sphere when we draw a symmetrical spherical surface (Gaussian surface) then the charge enclosed is zero as all the charge is concentrated on the surface of the hollow sphere.So according to Gauss’s law

dE S0 :e # dv 0vr= =#therefore E 0= at any point inside the hollow sphere.now at any point outside the sphere at a distance r from the center when we draw a symmetrical closed surface(Gaussian surface) then the charge enclosed is Qenc R4s 2r p= ^ h

and according to Gauss’s law

dE S0 :e # Qenc=

E R402e p^ h R6s 2r p= ^ h

E rR as

r0

2

2

er= c m

SOL 2.2.43 Option (D) is correct.Electric field intensity at any point due to uniform surface charge distribution is defined as

E a2s

n0e

r=



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For View Only Shop Online at www.nodia.co.inwhere sr " surface charge density an " unit vector normal to the sheet directed toward the point where field is to be determined.The electric field intensity due to the upper plate will be

EU a22

z0e

= -^ h a an z=-^ h

and the field intensity due to lower plate will be

El a22

z0e

=- ^ h a an z=^ h

So the net field between the plates is E E EU l= +

a a22

22

z z0 0e e= - + -^ ^h h: D a2

4z

0e=- a2

z0e

=-

SOL 2.2.44 Option (D) is correct.Electric field intensity at any point is equal to the negative gradient of electric potential at the pointi.e. E Vd=-So, the y -component of the field is

Ey yV22=-

Now, for the interval 3 2y# #- - , V t20 3= +^ h

Ey 20 /V myV22=- =-

For the interval 2 1y# #- - , V 20=

So, Ey yV 022=- =

For the interval 1 1y# #- + , V t20=-

So, Ey 20 /V myV22=- =

For the interval 1 2y# # , V 20=-So Ey 0=For the interval 2 3y# # , V t20 3= -^ h

So, Ey 0 /V myV 322=- =-

Therefore, the plot field component Ey with respect to y for the defined intervals will be same as in option (A).

SOL 2.2.45 Option (A) is correct.Since the electrons are moving with equal but opposite velocities so assume that their velocities are v ax0+ and v ax0- .Now let the electric field is applied in ax directioni.e. E E ax0=So the force applied on the electrons will be F .eE E1 6 10 19

#= =- -^ h




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mdtdv . E1 6 10 19

#=- -^ h

therefore, change in the velocity

dv .m dt

E1 6 10 19#=-

-^

^h

h .

mE dt

a1 6 10

x

190#=-

-^ h

So, the velocity of electron moving in ax+ direction will change to

v1 .

v mE dt

a a1 6 10

x x0

190#= -

-^ h

.

v mE dt

a1 6 10

x0

190#= -

-^ h

; E

Since velocity deceases so loss in K.E. is

.K ELoss mv mv21

21

02

12= -

..

E dt mE dt

1 6 10 21 1 6 1019

0

19 202 2

##= --

-

^^ ^

hh h

...(i)

Again the velocity of electron moving in ax- direction will change to

v2 .

v mE dt

a a1 6 10

x x0

190#=- -

-^ h

.

v mE dt

a1 6 10

x0

190#=- +

-^ h

; E

Since velocity increases, so Gain in K.E. is

.K EGain mv mv21

21

22

02= -

..

E dt mE dt

1 6 10 21 1 6 1019

0

19 202 2

##= +-

-

^^ ^

hh h

...(2)

Comparing eq (1) and eq (2) we get .K EGain .K E> Loss

SOL 2.2.46 Option (B) is correct.The electric field intensity produced at a distance r from a line charge of density Lr is defined as

E a2L

0pe rr= r

where ar is unit vector directed toward point P along r . So, the electric field acting on the line charge at 3 my = due to the line charge located at 3 my =- is

E a2 6

80 10y

0

9#pe

=-

^ h 80 , , 6nC ma aL yr r= = =r^ h

240 /V may=Therefore, the force per unit length exerted on the line charge located at 3 my = is

F dz ELz 0

1r=

=^ ^h h# 240a80 10 y

9#= -

^ ^h h 19.2 Nay m=

SOL 2.2.47 Option (D) is correct.The four charges located at the corners of square 4 cm has been shown in figure below :



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The net potential at the charge located at A due to the other three charges is

VA ABq

ACq

ADq

41 B C D

0pe= + +a k

.9 10 1 2 104 10

14 2 10

14 10

19 92 2 2# # #

# # #= + +-

- - -c m

.4

10 8 10 22

12#= +c m

730.92 Volt=Similarly, the electric potential at all the corners will be VB 730.92 VoltV V VC D A= = = =Therefore, the net potential energy stored in the system is given as

W qV21= / q V q V q V q V2

1A A B B C C D D= + + +^ h

. .21 4 1 2 10 730 929# # # #= -

^ ^h h

.75 J3 m=

***********




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SOLUTIONS 2.3

SOL 2.3.1 Option (B) is correct.Given, the electric field intensity, E x y za a ax y z= + + dl dx dy dza a ax y z= + +So, the potential difference between point X and Y is

VXY dE lX

Y:=- # xdx ydy zdz

2

0

1

2

3

0= + +## #

x y z2 2 22

1

2 2

2

0 2

3

0

=- + += G

2 1 0 2 0 321 2 2 2 2 2 2=- - + - + -6 @ 5=

SOL 2.3.2 Option (B) is correct.Given the electric field vector at point P due to the three charges Q1, Q2 and Q3 are respectively. E1 2a a ax y z= + - E2 3a ay z= + E3 2a ax y= -So, the net field intensity at point P is E E E E1 2 3= + + 3 2a a a5x y z= + +

SOL 2.3.3 Option (B) is correct.Charge density at any point in terms of electric flux density D is defined as vr D:d=Since, D /cos C mz az2 2r f= ^ h

So, we get vr D:d= coszz az2

22 r f= ^ h6 @ /cos C m2 3r f=

At point 1, , 34p

a k, vr cos1 42 p= ^ ah k 2

1= 0.5 /C m3=

SOL 2.3.4 Option (C) is correct.Electric field intensity E is a vector quantity while the electric potential V is a scalar quantity.

SOL 2.3.5 Option (D) is correct.For an ideal capacitance the area of plates, A is assumed very high in comparison to the separation d between the plates.

i.e. dA 3.



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For View Only Shop Online at www.nodia.co.inSo, the fringing effect at the plates edges can be neglected and therefore, we get the capacitance between the parallel plates as

C dA

4e=

So A and R both true and R is correct explanation of A.

SOL 2.3.6 Option (D) is correct.By using method of images, the conducting surfaces are being replaced by the image of charge distribution which gives a system of charge distribution.So, in solving boundary value problems we can avoid solving Laplace’s or Poission’s equation and directly apply the method of images to solve it.Thus both A and R are true and R is correct explanation of A.

SOL 2.3.7 Option (C) is correct.For a pair of line charges equipotential surface exists where the normal distance from both the line charges are same. So, the plane surface between the two line charges will be equipotential.

This is the similar case to method of images.

SOL 2.3.8 Option (D) is correct.According to uniqueness theorem : If a solution to Laplace’s equation (a) be found that satisfies the boundary condition then the solution is unique.Here it is given that the potential functions V1 and V2 satisfy Laplace’s equation within a closed region and has the same value at its boundary so both the functions are identical.

SOL 2.3.9 Option (D) is correct.From Maxwell’s equation we have

E#d tB22=-

E#d t

A#22 d=- ^ h B A#d=^ h

t

E A#d22+b l 0=




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Since, the curl of a gradient of a scalar field is identically zero. So, we get

t

E A22+ Vd=-

i.e. VE! d- in time valeying field therefore A and R both are true and R is the correct explanation of A.

SOL 2.3.10 Option (D) is correct.The surface charge density at plane x 8= is shown in the figure.

The point P is located at , ,6 4 5-^ h. So, the normal vector to the plane x 8= pointing toward P is an ax=-Therefore, the electric flux density produced at point P is

D a2sn

r= 30a a260

x x= - =-^ h

SOL 2.3.11 Option (B) is correct.Consider the coaxial cylinder is located along z -axis. So at any point between the two surfaces the electric field is given as

E Vd=- Va22r=- r (Since all other derivatives will be zero)

Given that the inner surface is at potential V0 while the outer one is grounded so



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For View Only Shop Online at www.nodia.co.inthe region between the two surfaces will have a gradually decreasing potential and so, E will not be uniform and it is radially directed as calculated above (in ar direction).

SOL 2.3.12 Option (B) is correct.The Poisson’s equation is defined as

V2d v

er=-

where V is electric potential and vr is charge density. So, in charge free space (0vr = ) we get the Poisson’s equation as

V2d 0= which is Laplace equation.

SOL 2.3.13 Option (B) is correct.Consider the three equal charges of CQ is placed at a separation of 0.5 m as shown in figure below :

The net stored charge in the system of n charges is defined as

W Q V21

k kk

n

1

==/

where Qk is one point charge and Vk is the net electric potential at the point charge due to the other charges. Now, we have the net electric potential at any of the point charge Q located in the system as

V1 . .Q Q

41

0 5 0 50pe= +b l Q

0pe=

So, total energy stored in the system of charges is given as

W1 3 QV Q21

23

10

2

pe= =b l (1)

Now, when the charges are separated by 1 m then the electric potential at any of the charge Q due to the other two charges is

V2 Q Q Q

41

1 1 20 0pe pe= + =b l

So, the stored energy in the new system is

W2 QV Q3 21

43

20

2

pe= =b l (2)

From equation (1) and (2) we have W2 . W0 5 1= or W1 W2 2=

SOL 2.3.14 Option (D) is correct.Electric potential due to point charge is defined as

V rQ

31

0pe=




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So, for the equal distance r potential will be same i.e. equipotential surface about a point charge is sphere.

SOL 2.3.15 Option (C) is correct.An electrostatic field has its curl always equals to zero. So electric field is irrotational. Statement 1 is correct.Electric field divergence is not zero and so it is not solenoidal. Statement 2 is correct.Electric field is static only from a macroscopic view point. Statement 3 is correct.Work done in moving a charge in the electric field from one point to other is independent of the path. Statement 4 is correct.

SOL 2.3.16 Option (C) is correct.Given electric potential, V 0 0y x2 14 3= +From Poisson’s equation we have

V2d v

0er=-

where, V" Electric potential v "r Charge density

x y z

y x10 202

2

2

2

2

24 3

22

22

22+ + +e ^o h v

0er=-

x y120 120+ v

0er=-

vr 120 2 120 00 # #e= +^ h (x 2= , y 0= ) vr 012 0e=-

SOL 2.3.17 Option (B) is correct.Given, the wave equation in space for a propagating wave in z -direction is E k Ex x

2 2d + 0=Now, from option (C) we have the electric field component as Ex E e jkz0= -

The Lapalacian of electric field is Ex2d jk E e jkz2

0= - -^ h

Ex2d k E e jkz20=- - k Ex2=-

or, E k Ex x2 2d + 0= So, it satisfies the wave equation.

SOL 2.3.18 Option (D) is correct.Consider the infinitely long uniform charge density shown in the figure.The electric field intensity produced at a distance r from an infinite line charge with density r

L is defined as

E 2 0pe rr= L



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Since, the normal distance vector of points P (0, 6, 1) and Q ( , , )5 6 1 from the line charge will be same so, the field intensity produced due to the infinite line at both the points P and Q will be same.Therefore, the field intensity at (5,6,1) is E .

SOL 2.3.19 Option (D) is correct.Consider the square loop ABCD carrying current 0.1 A as shown in figure.

The magnetic dipole moment is m IS=where I is current in the loop and S is the area enclosed by loop.So, m .0 01 10 2

2= ^ ^h h 2 A m2-=

The direction of the magnetic dipole moment is determined by right hand rule.i.e. m 2 A maz 2

-=

SOL 2.3.20 Option (A) is correct.Electric flux density at a distance r from a point charge Q is defined as

D rQ a

4 r2p=




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and the total flux through any defined surface is

y dD S:= #So, both the quantities has not the permittivity e in their expression. Therefore, D and y are independent of permittivity e of the medium.

SOL 2.3.21 Option (B) is correct.According to Gauss’s law, the total outward flux through a closed surface is equal to the charge enclosed inside it.

i.e. dD S:# Qenc=Now, consider the height of cylinder is h . So, the cylindrical surface at 3r = encloses the charge distribution

S5 /C m2r =^ h located at 2 mr = . Therefore, we

get D h2 3p^^ h h h5 2 2# p= ^ h

or, D a320= r

SOL 2.3.22 Option (A) is correct.The electric potential produced by 1 Cm at a distance r is

V r r9 101 10 90009

6

##= =

-^ h

So, the potential energy stored in the field will be the energy of the charges as,i.e. W qV=

r4 10 90006#= -

^ h r36 10 3

#=-

where r is the distance between the charges given as r 2 1 1 3 5 12 2 2= - - + - + +^ ^ ^h h h 7=

So, W 5.15 10 Joule736 10 3

3##= =

--

SOL 2.3.23 Option (C) is correct.Electric field intensity due to a dipole having moment P at a distance r from it is

E r13\

EE

1

2 rr

2313

=

E1

2 42

3

3

=^

^

h

h

E2 /mV m81=

SOL 2.3.24 Option (B) is correct.Energy density (energy stored per unit volume) in an electric field is defined as

we D E21

:= EE E21

21

0 02

:e e= =

SOL 2.3.25 Option (A) is correct.The position of points A, B and C are shown below



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Since, position charge is placed at A and negative charge at B so, their resultant field intensity at C is as shown below :

Since, the forces F F1 2= so the vertical component F1 V and F2 V are get cancelled while F2 H and F1 H are get summed to provide the resultant field in az- direction.

SOL 2.3.26 Option (D) is correct.Given,Charges, Q1 1 10nC CQ2

9= = = -

Separation between charges, r 1 10mm m3= = -

So, the force acting between the charges is

F rkQ Q

109 10 10

21 2

3 2

9 9 2#= = -

-

^

^

h

h

10 N12 3#= -

SOL 2.3.27 Option (B) is correct.According to Gauss’s law, the surface integral of flux density through a closed surface is equal to the charge enclosed inside the closed surface (volume integral of charge density)

i.e. dD S:# dvvr= #In differential form, the Gauss’s law can be written as D#d vr=

E#d v

0er= D E0e=^ h




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SOL 2.3.28 Option (D) is correct.The electric field at a distance r from the point charge q located in a medium with permittivity e is defined as

E rq a

4 r2pe=

rq a4 r2

1

pe=-

SOL 2.3.29 Option (C) is correct.For according to Gauss’s law the total outward electric flux through a closed surface is equal to the charge enclosed by the surface.

i.e. dD Ss

:# Qenc=

or, dD Ss

:# dvvvr= #

SOL 2.3.30 Option (B) is correct.The force between the two charges q1 and q2 placed in a medium with permittivity e located at a distance r apart is defined as

F rq q

41

21 2

pe=

or F 1\ e

i.e. force is inversely proportional to permittivity of the medium.Since, glass has the permittivity greater than 1 (i.e. permittivity of free space) So, the force between the two charges will decreases as the glass is placed between the two charges.

SOL 2.3.31 Option (A) is correct.According to Gauss’s law the total electric flux through a closed surface is equal to the charge enclosed by it. Since, the sphere centred at origin and of radius 5 m encloses all the charges therefore, the total electric flux over the sphere is given as Ey Q Q Q1 2 3= + + . . .0 008 0 05 0 009= + - 0. 49 C0 m=

SOL 2.3.32 Option (D) is correct.Electric flux through a surface area is the integral of the normal component of electric field over the area.

SOL 2.3.33 Option (B) is correct.The electric field due to a positive charge is directed away from it (i.e. outwards.)According to Gauss’s law the surface integral of normal component of flux density over a closed surface is equal to the charge enclosed inside it.So, A is true but R is false.

SOL 2.3.34 Option (A) is correct.Force between the two charges Q1 and Q2 is defined as

F R

Q Q a4 R

02

1 2

pe=



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For View Only Shop Online at www.nodia.co.inWhen the charges are of same polarity then the force between them is repulsive.The electric force on both the charges will have same magnitude. As the expression of Force includes the term e (permittivity of the medium) so it depends on the medium in which the charges are placed.So the statements (a), (c) and (d) are correct while (b) is incorrect.

SOL 2.3.35 Option (B) is correct.Since the electric field is negative gradient of the electric potential so the field lines will be orthogonal to the equipotential lines (surface).

SOL 2.3.36 Option (B) is correct.Electric potential at 10 nC- due to 10 nC charge is

V rQ

41

0pe=

9 102 0 0

10 1092

9

# ##=+ +

-

45 Volt=and so the energy stored is We QV= 10 10 459

# #= - -^ h

50 nJ1=-

SOL 2.3.37 Option (C) is correct.According to Gauss’s the outward electric flux density through any closed surface is equal to the charge enclosed by it. So electric field out side the spherical balloon doesn’t change with the change in its radius and so the energy density at point P is wE for the inflated radius b of the balloon.

SOL 2.3.38 Option (A) is correct.The curl of E is identically zero.i.e. E#d 0=So, it is conservative.The electrostatic field is a gradient of a scalar potential.i.e. E Vd=-So, E#d 0= (Conservative)Work done in a closed path inside the field is zero

i.e. dE l:# 0= E#d 0= (Conservative)So, (a), (c) and (d) satisfies that the field is conservative.As the potential difference between two points is not zero inside a field so, the statement (b) is incorrect.

SOL 2.3.39 Option (D) is correct.Net outward electric flux through the spherical surface, r a= is

dD S:# a34

v3y r p= = b l




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D a4 2p^ h a34v 3r p=

D /C ma a3v

r2r=

SOL 2.3.40 Option (C) is correct.For a pair of equal and opposite linear chargers the electric potential is defined as

V rQ

rQ

4 40 1 0 2pe pe= -

where r1 and r2 are the distances from the charges respectively. For the same value of V (equipotential surface) a plane can be defined exactly at the centre point between them.

SOL 2.3.41 Option (D) is correct.In a charge free region 0vr =^ h electrostatic field has the following characteristic

E:d 0v

er= =

and E#d 0= (for static field)

SOL 2.3.42 Option (C) is correct.Consider the force experienced by Q is F1. Since, there is no any external applied field (or force) so, sum of all the forces in the system of charges will be zero.i.e. FS 0=or, 3 2F F F1+ + 0= F1 5F=-

SOL 2.3.43 Option (A) is correct.Poissions law is derived from Gauss’s law as D:d r=For inhomogeneous medium e is variable and so, E:d e^ h r= V:d de -^ h6 @ r= V:d de^ h r=-This is the Poissions law for inhomogenous medium.

SOL 2.3.44 Option (C) is correct.Electric field intensity due to a infinite charged surface is defined as

E a2 n0e

r= S

where rS is surface charge density and an is the unit vector normal to the surface

directed towards the point of interest.Given that, r

S 20 /nC m2= 20 10 /C m9 2

#= -

and an az=- (Since the surface 10 mz = is above the origin).So we have,

E a220 10

z0

9#e= -

-

^ h



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a220 10 9 10 4

z

9 9# # # # p=

-

360 /V mazp=-

SOL 2.3.45 Option (B) is correct.Electric field intensity due to a short dipole having a very small separation d , a+ a distance R from it is defined as

E 2 cos sinR

Qd a a4 r

03pe

q q= + q^ h (for d R<< )

So, for the given dipole, 90cq =and R r d r2 2 .= - (r d>> )

Therefore, E 0r

Qd a4 0

3pe= + f^ h

i.e. E r13\

SOL 2.3.46 Option (B) is correct.According to Gauss’ law the total outward flux from a closed surface is equal to the total charge enclosed by the surface.

SOL 2.3.47 Option (B) is correct.Electric field intensity at any point r outside the sphere is defined as

E r

Q a4 r

02pe

= for r a>

and the field intensity inside the sphere is

E a

Qr

ra

34 4

34

r3 0

2

3

p pe

p=

b

b

l

l for r a#

a

Qr a4 r

03pe

=

So the electric potential at any point r b a<= is

V dE l:=- # dr E drE a ar r

a

b

r

a: :=- -

3=^ ^h h##

r

Q dra

Qr dr4 4a

ba

02

03pe pe

=- -3

##




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SOL 2.3.48 Option (D) is correct.Given, the electric potential, V x y yz3 2= -Electric field intensity at any point is equal to the negative gradient of the potential.i.e. E Vd=- xy x z ya a a6 3x y z

2=- - - - -^ ^ ^h h h at (x 1= , y 0= , z 1=- ) E 4 0ay !=-So, electric field does not vanish at given point.

SOL 2.3.49 Option (D) is correct.Consider two parallel plates separated by a distance d is connected to a voltage source V . So, the field intensity between the plates is defined as

E dV2=

***********

CHAPTER 3ELECTRIC FIELD IN MATTER

128 Electric Field in Matter Chap 3



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EXERCISE 3.1

MCQ 3.1.1 A certain current density at any point ( , , )zr f in cylindrical coordinates is given by

J ( ) /A me a a5 zz

2 2r= +r .

The total current passing the plane z 0= , 0 2# #r in the az direction is(A) 100p Ampere (B) 4p Ampere

(C) 40p Ampere (D) 0 Ampere

MCQ 3.1.2 In a certain region the current density is given by

J /cos sin A mr r ra a ar2 2 2 2q q= + -q f .

The total current crossing the surface defined by 90cq = , 0 2< <f p , 0 1 mr< < is

(A) A2p (B) A2

p-

(C) A41 (D) 2 A3

p

MCQ 3.1.3 The current density in a cylindrical wire of radius mm8 placed along the z -axis is

/A mJ a50z

2

r= . The total current flowing through the wire is

(A) 80.38 mA (B) 800 mA

(C) 0 A (D) 5.026 A

Common Data for Question 4 - 5 :In a certain region current density is given by

J ( )

/sin A ma a401

20z2

2

r rf= -

+r

MCQ 3.1.4 Total current crossing the plane z 2= in the az direction for 4<r will be(A) 0 A (B) 1.5 mA

(C) 32 A- (D) 20 A

MCQ 3.1.5 Volume charge density in the region at a particular point ( , , )z0 0 0r f will be(A) non uniform (B) linearly increasing with time

(C) linearly decreasing with time (D) constant with respect to time

Chap 3 Electric Field in Matter 129


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For View Only Shop Online at www.nodia.co.inStatement for Linked Question 6 - 9 :In a cylindrical system, two perfectly conducting surfaces of length 2 m are located at 3r = and 5 cm1r = . The total current passing radially outward through the medium between the cylinders is 6 A.

MCQ 3.1.6 If a conducting material having conductivity 0.05 /S ms = is present for 3 5 cmr1 # then the electric field intensity at 4 cmr = will be(A) 238.7 /V mar (B) 150 /V mar(C) 318.3 /V mar (D) 0 /V m

MCQ 3.1.7 The voltage between the cylindrical surfaces will be(A) 4.88 volt (B) 1.45 volt

(C) 2.32 volt (D) 3 volt

MCQ 3.1.8 The resistance between the cylindrical surfaces will be(A) 0.813W (B) 2.44 W

(C) 0.5W (D) 8.13W

MCQ 3.1.9 The total dissipated power in the conducting material will be(A) 175.7 W (B) 18 W

(C) 29.3 W (D) 0.8 W

MCQ 3.1.10 A solid wire of radius r and conductivity 1s has a jacket of material having conductivity 2s . If the inner and outer radius of the jacket are r and R respectively then the ratio of the current densities in the two materials will

(A) depend on r only (B) depend on R only

(C) depend on both r and R (D) independent of both r and R

Statement for Linked Question 11 - 12 :Atomic hydrogen contains 5.5 10 /atom cm19 3

# at a certain temperature and pressure. If an electric field of 40 /kV m is applied, each dipole formed by the electron and positive nucleus has an effective length of 7.1 10 m16

#- .

MCQ 3.1.11 The polarization due to the induced dipole will be(A) 12.5 /nC m2 (B) 8.8 10 /C m6 2

#

(C) 6.25 /nC m2 (D) 3.9 10 /C m9 2#

MCQ 3.1.12 Dielectric constant of the atomic hydrogen will be(A) .2 77 (B) .1 0177

(C) .0 982 (D) .0 0177




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MCQ 3.1.13 The dielectric constant of the material in which the electric flux density is double of the polarization is(A) 2 (B) 1/2

(C) 3 (D) 1

Statement for Linked Question 14- 15 :The potential field in a slab of a dielectric material that has the relative permittivity

/56re = is given by V y500=- .

MCQ 3.1.14 Electric field intensity in the material will be(A) 50 /V may (B) 500 /V may(C) 500 /V may- (D) 0

MCQ 3.1.15 The electric flux density inside the material will be(A) 4.43 /nC m2

(B) 3.54 /C may 2m

(C) 8.85 /nC m2

(D) 7.08 /nC may 2

MCQ 3.1.16 Polarization of the material will be(A) 2.66 /nC may 2 (B) .08 /nC m14 2

(C) 5.31 10 /C may12 2#

- (D) 3 /C may 2

Statement for Linked Question 17 - 18 :Two perfect dielectrics with dielectric constant 2r1e = and 5r2e = are defined in the region 1 ( )y 0$ and region 2 ( )y 0< respectively. Consider the electric field intensity in the 1st region is given by E1 20 10 /kV ma a a25 x y z= + -

MCQ 3.1.17 The Flux charge density in the 2nd region will be(A) 2.21 0.35 0.44 /C ma a ax y z

2m+ -

(B) 2.21 0.35 0.44 /nC ma a ax y z2+ -

(C) 2.21 0.88 0.44 /nC ma a ax y z2+ -

(D) 0.4 0.07 0.08 /nC ma a ax y z2+ -

MCQ 3.1.18 The energy density in the 2nd region will be(A) 66.37 /mJ m3 (B) 118 /mJ m3

(C) 472 10 /J m6 3# (D) 59 /mJ m3



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For View Only Shop Online at www.nodia.co.inMCQ 3.1.19 The electric field in the three regions as shown in the figure are respectively E1, E2

and E3 and all the boundary surfaces are charge free.

If 1 3 2!e e e= , then the correct relation between the electric field is(A) E E E1 2 3! ! (B) E E E1 3 2!=

(C) E E E1 2 3= = (D) E E E1 2 3!=

MCQ 3.1.20 An infinite plane dielectric slab of thickness d and having permittivity 4 0e e= occupies the region z d0 < < . If a uniform electric field EE az0= is applied in the free space then the electric flux density(Din) and electric field intensity(Ein) inside the dielectric slab will be respectively

(A) E a4 z0 and E az0 0e (B) E az0 0e and E a4 z

0

(C) 4E az0 and E a4 z0 (D) E az0 0e and 4E az0

MCQ 3.1.21 The energy stored in an electric field made up of two fields E1 and E2 is Wnet where as the energies stored in individual fields E1 and E2 are W1 and W2 respectively so the correct relation between the energies is(A) W W W1 2= +

(B) W WW1 2=

(C) W W W> 1 2+

(D) W W W< 1 2+

MCQ 3.1.22 An electric dipole is being placed in an electric field intensity E 1.5 /V ma ax z= -If the moment of the dipole be 3 C mp a a4 x y -=- + then energy of the dipole will be(A) 6 J (B) 0 J

(C) 3 J- (D) 3 J+

MCQ 3.1.23 When a neutral dielectric is being polarized in an electric field then the total bound charge of the dielectric will be(A) zero

(B) positive

(C) negative

(D) depends on nature of dielectric




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Statement for Linked Question 24 - 25 :A lead bar of square cross section has a hole of radius .5 cm2 bored along its length as shown in the figure.

(Conductivity of lead 5 10 ( )m6 1W#= - )

MCQ 3.1.24 If the length of the lead bar is 8 m then the resistance between the square ends of the bar will be(A) 1.78 mW (B) . m3 64 W

(C) 1.95 mW (D) 269 mW

MCQ 3.1.25 If the hole in the lead bar is completely filled with copper then the resistance of the composite bar will be (Resistivity of copper

1.72 10 )m8 W#= -

(A) 188 mW (B) 924.6 mW

(C) 3.708 mW (D) 1.76 mW

MCQ 3.1.26 A cylindrical wire of length l and cross sectional radius r is formed of a material with conductivity 10 ( )m6 1W - . If the total conductance of the wire is 10 ( )6 1W - then the correct relation between l and r is

(A) r lp= (B) r l

p=

(C) 2 r lp = (D) r l=

Statement for Linked Question 27 - 28 :A capacitor is formed by two concentric conducting spherical shells of radii 1 cma = and 2 cmb = centered at origin. Interior of the spherical capacitor is a perfect dielectric with 4re = .

MCQ 3.1.27 The capacitance of the capacitor will be(A) 8.9 pF (B) 2.25 pF

(C) 890 pF (D) 225 pF

MCQ 3.1.28 If a portion of dielectric is removed from the capacitor such that 1re = for < <2 f pp and 4re = for the rest of the portion, then the capacitance of the composite capacitor will be



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For View Only Shop Online at www.nodia.co.in(A) 0.56 pF (B) 946 pF

(C) 236.5 pF (D) 6.7 pF

MCQ 3.1.29 Two conducting surfaces are present at x 0= and 5 mmx = and the space between them are filled by dielectrics such that .2 5r1e = for 0 1 mmx< < and 4r2e = for 1 3 mmx< < rest of the region is air filled. The capacitance per square meter of surface area will be(A) 22.1 /nF m2 (B) 3.05 /nF m2

(C) 442.5 /nF m2 (D) 44.25 /nF m2

MCQ 3.1.30 Two coaxial conducting cylinders of radius 4 cm and 8 cm is lying along z -axis. The region between the cylinders contains a layer of dielectric from 4 cmr = to

cm12r = with 4re = . If the length of cylinders is 1 m then the capacitance of the configuration will be(A) 0.55 pF (B) 7 10 F9

#

(C) 1.83 nF (D) 143 pF

MCQ 3.1.31 A parallel plate capacitor is quarter filled with a dielectric ( 3re = ) as shown in the figure. The capacitance of the capacitor will be

(A) 1.38 pF (B) 2.76 pF

(C) 9.95 pF (D) 6 pF

MCQ 3.1.32 Medium between the two conducting parallel sheets of a capacitor has the permittivity e and conductivity s . The time constant of the capacitor will be

(A) se (B) e

s

(C) se (D) /1 se

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EXERCISE 3.2

Common Data for Question 1 -2 :In spherical coordinate system, the current density in a certain region is given by

J /A mr e a2 tr

10 24

= -

MCQ 3.2.1 At 1 mst = , how much current is crossing the surface r 5= ?(A) 75.03 A (B) 27.7 A

(C) 0.37 A (D) 2.77 A

MCQ 3.2.2 At a particular time t , the charge density ( , )r tvr at any point in the region is directly proportional to. (Assume 0v "r as t" 3)

(A) r (B) r1

(C) r12 (D) r2

MCQ 3.2.3 The velocity of charge density at 0.6 mr = will be(A) 6 /m sar

(B) 1000 /m sar(C) 0.6 10 /m sar3

#-

(D) 600 /m sar

Statement for Linked Question 4 - 5 :Two uniform infinite line charges of 5 /pC m each are located at x 0= , y 1= and x 0= , y 2= respectively. Consider the surface y 0= is a perfect conductor that has the zero potential.

MCQ 3.2.4 Electric potential at point ( , , )P 1 2 0- - will be(A) 1.2 volt

(B) 0.2 volt-

(C) 0.2 volt+

(D) 0.04 volt-



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For View Only Shop Online at www.nodia.co.inMCQ 3.2.5 Electric field at the point P will be

(A) 0.12 0.003 /V ma ax y-

(B) 0.12 0.086 /V ma ax y-

(C) 723 18.9 /V ma ax y-

(D) 0.024 0.086 /V ma ax y-

MCQ 3.2.6 A thin rod of certain cross sectional area extends along the y -axis from 0 my = to 5 my = . If the polarization of the rod is along it’s length and is given by

2 3P yy2= + then the total bound charge of the rod will be

(A) 0 (B) 50 C

(C) 48 C (D) can’t be determined

MCQ 3.2.7 A neutral atom of polarizability a is situated at a distance 1 m from a point charge 1/9 nC. The force of attraction between them will be

(A) 2 Na (B) N92a

(C) 9 Na (D) 18 Na

Common Data for Question 8 - 9 :The two dipoles P1, P2 with dipole moment nC m4 - and 9 nC m- respectively are placed at 1 m distance apart as shown in figure.

MCQ 3.2.8 The torque on P2 due to P1 will be(A) 18 10 N m18

# --

(B) 2 nN m-

(C) 8.1 N m-

(D) 0.16 N mm -

MCQ 3.2.9 The torque on P1 due to P2 will be(A) . 10 N m3 24 7

# -- (B) 2 nN m-

(C) . 10 N m1 62 7# -

- (D) . N m3 24 m -

Common Data for Question 10 -11 :A sphere carries a polarization ( )rP ra3 r= where r is the distance from the center of the sphere.




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MCQ 3.2.10 Consider Er is the electric field component in the radial direction inside the sphere. The plot of Er with respect to r will be

MCQ 3.2.11 If the radius of the sphere is a then the electric field outside the sphere will be(A) a4 3p- (B) a8 3p

(C) 0 (D) a8 3p-

Statement for Linked Question 12 - 13 :A thick spherical shell is made of dielectric material with a polarization

( )rP /nC mr a5

r2= where r is the distance from its centre.

MCQ 3.2.12 If the spherical shell is centred at origin and has the inner radius 2 m and outer radius 6 m then the electric field intensity at 1 mr = will be(A) 0 (B) 40 /V mp-

(C) 20 /V mp (D) 20 /V mp-

MCQ 3.2.13 Electric field at 7 mr = will be(A) 100 /V mp- (B) 140 /V mp-

(C) 0 (D) 20 /V mp-



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For View Only Shop Online at www.nodia.co.inMCQ 3.2.14 Electric field intensity at 5 mr = will be

(A) a52

r0e

- (B) a1r

0e

(C) a1r

0e- (D) a5

1r

0e

MCQ 3.2.15 A spherical conductor of radius 1 m carries a charge 3 mC. It is surrounded, out to radius 2 m, by a linear dielectric material of dielectric constant 3re = , as shown in the figure. The energy of this configuration will be

(A) 27 kJ (B) 500 J

(C) 270 J (D) 324 J

MCQ 3.2.16 A sphere of radius / m2 p is made of dielectric material with dielectric constant 2re = . If a uniform free charge density 0.6 /nC m3 is embedded in it then the

potential at the centre of the sphere will be(A) 3 volt (B) 5.4 volt

(C) 0 volt (D) 9 volt

Statement for Linked Question 17 - 19 :A short cylinder of radius r and length L carries a uniform polarization P , parallel to its axis as shown in the figure.




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MCQ 3.2.17 Total bound charge by the cylinder will be(A) 2 coulombP (B) coulombP

(C) 0 coulomb (D) coulombP-

MCQ 3.2.18 If L r2= then the electric field lines of the cylinder will be as

MCQ 3.2.19 The lines of flux charge density will be as



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For View Only Shop Online at www.nodia.co.inMCQ 3.2.20 A parallel plate capacitor is filled with a non uniform dielectric characterized by

(1 0 )a3 5r2e = + where a is the distance from one plate in meter. If the surface area

of the plates is 0.2 m2 and separation between them is 10 cm then the capacitance of the capacitor will be(A) 22.5 pF (B) 90.2 pF

(C) 45.1 pF (D) 4.51 pF

MCQ 3.2.21 A two wire transmission line consists of two perfectly conducting cylinders, each having a radius of 0.2 cm, separated by a centre to centre distance of 2 cm. The medium surrounding the wires has relative permittivity 2re = . If a 100 V source is connected between the wires then the stored charge per unit length on each wire will be(A) 3.64 /nC m (B) 3.64 10 /C m11

#-

(C) 1.82 /nC m (D) 2.5 10 /C m8#

-

MCQ 3.2.22 A tank is filled with dielectric oil of susceptibility 1ec = . Two long coaxial cylindrical metal tubes of radii 1 mm and 3 mm stand vertically in the tank as shown in the figure. The outer tube is grounded and inner one is maintained at 2 kV potential. To what height does the oil rise in the space between the tubes ? (mass density of oil 0.01 /gm cm3= )

(A) 41.1 mm (B) 45.5 mm

(C) 20.5 mm (D) 82.4 mm

MCQ 3.2.23 An infinite plane conducting slab carries uniformly distributed surface charges on both of it’s surface. If the sum of the charge densities on the two surfaces is

/C mso2r then the surface charge densities on the two surfaces will be

(A) /2sor , /2sor (B) 2 sor , sor-

(C) 0, sor (D) None of these




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MCQ 3.2.24 Two infinite plane parallel conducting slabs carry uniformly distributed surface charges S11r , S12r , S21r and S22r on all the four surfaces as shown in the figure.

Which of the following gives the correct relation between the charge densities ?(A) s s11 22r r= , s s12 21r r=

(B) s s11 22r r= , s s12 21r r=-

(C) s s11 12r r= , s s21 22r r=

(D) s s11 22r r=- , s s12 21r r=-

Common data for Question 25 - 26 :The plane surfaces x 0= , x 1= , y 0= and y 1= form the boundaries of conductors extending away from the region between them as shown in the figure.

If the electrostatic potential in the region between the surfaces is given by xy6 volts then the surface charge density on the surface ;

MCQ 3.2.25 x 0= is(A) y5 0e- (B) x5 0e-

(C) x y5 0e- +^ h (D) xy5 0e ^ h

MCQ 3.2.26 y 0= is(A) y5 0e- (B) x5 0e-

(C) x y5 0e- +^ h (D) xy5 0e



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For View Only Shop Online at www.nodia.co.inMCQ 3.2.27 Two infinitely long coaxial, hollow cylindrical conductors of inner radii 2 m and

5 m respectively and outer radii 3 m and 6 m, respectively as shown in the figure, carry uniformly distributed surface charges on all four of their surfaces.

If net surface charge per unit length is 10 /C m and 6 /C m for the inner and outer conductor respectively then the surface charge densities on the four surface will beSurface " 2 mr = 3 mr = 5 mr = 6 mr =(A) 0 /5 3p /1 p- /4 3p(B) /5 3p /1 p- 0 /4 3p(C) /1 p /1 p- /2 p /2 p-(D) 0 /1 p- /1 p 0

MCQ 3.2.28 A conducting spherical shell of inner radii 2 m and outer radii 3 m carry uniformly distributed surface charge on it’s inner and outer surfaces. If the net surface charge is 9 C for the conducting spherical shell then, the surface charge density on inner and outer surfaces are respectively

(A) 0, /C m41 2

p (B) /C m41 2

p , 0

(C) 0, 4 /C m2p (D) 4 /C mp , 0

MCQ 3.2.29 Plane z 0= defines a surface charge layer with the charge density /C mn3 2r =S as shown in figure. If the electric field intensity in the region z 0< is E2 2 3 2 /V ma a ax y z= + -then the field intensity E1 in the region z 0> will be

(A) 220 219 2a a ax y z+ - (B) 2 3 224a a ax y z+ +

(C) 222 221 2a a ax y z+ + (D) 2 3 226a a ax y z+ +




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MCQ 3.2.30 An infinite plane dielectric slab with relative permittivity 5re = occupies the region x 0> . If a uniform electric field 10 /V mE ax= is applied in the region x 0< (free space) then the polarization inside the dielectric will be(A) 8 /C max0

2e (B) 4 /C ma0 x2e

(C) 2 /C max02e (D) 10 /C max0

2e

Statement for Linked Question 31 - 32 :An infinite plane dielectric slab of 1 m thickness is placed in free space such that it occupies the region 0 1 my< < as shown in the figure.

Dielectric slab has the non uniform permittivity defined as

e y1 3

22

0e=+^ h

for y0 1< <

MCQ 3.2.31 If a uniform electric field 4 /V mE ay= is applied in free space then bound surface charge densities on the surface y 0= and y 1= will be at y 0= at y 1=(A) 0 3 0e-(B) 3 0e- 0(C) 3 0e 0(D) 5 0e- 8 0e

MCQ 3.2.32 As we move from the surface y 0= toward the surface y 1= inside the dielectric slab, polarization volume charge density will be(A) linearly increasing (B) linearly decreasing

(C) Constant (D) zero at all points

MCQ 3.2.33 In a spherical coordinate system the region a r b< < is occupied by a dielectric material. A point charge Q is situated at the origin. It is found that the electric field intensity inside the dielectric is given by

E b

Q a4 r

02pe

= for a r b< <



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For View Only Shop Online at www.nodia.co.inThe relative permittivity of the dielectric will be(A) /b r2 2

^ h (B) /a r2 2^ h

(C) /r a2 2^ h (D) /a b2 2

^ h

MCQ 3.2.34 Two perfectly conducting, infinite plane parallel sheets separated by a distance 2 m carry uniformly distributed surface charges of equal and opposite densities

5 /nC m2+ and 5 /nC m2- respectively. If the medium between two plates is a dielectric of uniform permittivity 4 0e e= then the potential difference between the two plates will be(A) 283 KV (B) 1130 KV

(C) 283 V (D) 1.13 KV

MCQ 3.2.35 The medium between two perfectly conducting infinite plane parallel sheets consists of two dielectric slabs of thickness 1 m and 2 m having permittivities 21 0e e= and

42 0e e= respectively as shown in the figure.

If the conducting sheets carry uniformly distributed surface charges of equal and opposite densities 0.6 /nC m2 and 0.6 /nC m2- respectively then the potential difference between the sheets will be(A) 67.8 Volt (B) 6.78 Volt

(C) 33.9 Volt (D) 17.4 Volt

MCQ 3.2.36 Two perfectly conducting, infinite plane parallel sheets separated by a distance d carry uniformly distributed surface charges of equal and opposite densities r

0S and

r-0S respectively. The medium between the sheets is filled by a dielectric of non

uniform permittivity which varies linearly from a value of 1e near one plate to value of 2e near the second plate. The potential difference between the two sheets will be

(A) d2 1e er-0S (B) lnd 2 1 1

2

e er

ee

-0S

^a

hk

(C) lnd2 1 1

2e er

ee

-S0

a k (D) ln1

2r ee

S0 a k




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MCQ 3.2.37 Two perfectly conducting, infinite plane parallel sheet separated by a distance 0.5 cm carry uniformly distributed surface charges of equal and opposite densities. If the potential difference between the two plates is kV7 and the medium between the plates is free space then the charge densities on the plates will be(A) 6.23 Cm (B) 88.5 Cm

(C) 8.85 Cm (D) 17.7 Cm

MCQ 3.2.38 A parallel plate capacitor has two layers of dielectrics with permittivities 31 0e e= and 22 0e e= as shown in the figure.

If the total voltage drop in the capacitor is 9 Volt then the voltage drop in 1st and 2nd dielectric region will be respectively(A) Volt11

18 , Volt1181 (B) 3 Volt, 6 Volt

(C) Volt1181 , Volt11

18 (D) 6 Volt, 3 Volt

MCQ 3.2.39 A dielectric slab is inserted in the medium between two plates of a capacitor as shown in the figure

The capacitance across the capacitor will remain constant(A) if the slab is moved rightward or leftward

(B) if the slab is pulled outward of the capacitor

(C) (A) and (B) both

(D) none of these



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For View Only Shop Online at www.nodia.co.inMCQ 3.2.40 A steel wire has a radius of 2 mm and a conductivity of 2 106

# S/m. The steel wire has an aluminium ( 3.8 107s #= S/m) coating of 2 mm thickness. The total current carried by this hybrid conductor be 80 A. The current density in steel Jst is(A) 1.02 10 /A m6 2

# (B) .2 10 /A m3 5 2#

(C) . 10 /A m2 04 5 2# (D) 1.10 10 /A m5 2#

MCQ 3.2.41 A potential field in free space is given as

40cos sin VVr2

q f=

Point ( 2, /3, /2)P r q p f p= = = lies on a conducting surface. The equation of the conducting surface is(A) 32cos sin r3q f = (B) 16cos sin r3f q =

(C) 16cos sin r3q f = (D) 32cos sin r3f q =

***********




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EXERCISE 3.3

MCQ 3.3.1 A parallel plate air-filled capacitor has plate area of 10 m2 4 2#

- and plate separation of 10 m3- . It is connected to a 0.5 V, 3.6 GHz source. The magnitude of the displacement current is ( 1036

1 9e = p- F/m)

(A) 10 mA (B) 100 mA

(C) 10 A (D) 1.59 mA

MCQ 3.3.2 Medium 1 has the electrical permittivity 1.81 0e e= farad/m and occupies the region to the left of x 0= plane. Medium 2 has the electrical permittivity 2.52 0e e= farad/m and occupies the region to the right of x 0= plane. If E1 in medium 1 is

(2 3 1 )E a a ax y z1 = - + volt/m, then E2 in medium 2 is(A) (2.0 7.5 2.5 )a a ax y z- + volt/m

(B) (2.0 2.0 0.6 )a a ax y z- + volt/m

(C) (2.0 3.0 1.0 )a a ax y z- + volt/m

(D) (2.0 2.0 0.6 )a a ax y z- + volt/m

MCQ 3.3.3 The electric field on the surface of a perfect conductor is 2 V/m. The conductor is immersed in water with 80 oe e= . The surface charge density on the conductor is (

3610 9

e = p-

F/m)(A) 0 C/m2 (B) 2 C/m2

(C) 1.8 10 11#

- C/m2 (D) 1.41 10 9#

- C/m2

MCQ 3.3.4 The space between the plates of a parallel-plate capacitor of capacitance C is filled with three dielectric slabs of identical size as shown in the figure. If dielectric constants are 1e , 2e and 3e , the new capacitance is

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(A) C3 (B) C

31 2 3e e e+ +^ h

(C) C1 2 3e e e+ +^ h (D) 9

1 2 3

1 2 3

e e ee e e+ +^ h

MCQ 3.3.5 If the potential, VV x2 4= + , the electric field is(A) 6 V/m

(B) 2 V/m

(C) 4 V/m

(D) 4 /V max-

MCQ 3.3.6 Two dielectric media with permittivities 3 and 3 are separated by a charge-free boundary as shown in figure below :

The electric field intensity in media 1 at point P1 has magnitude E1 and makes an angle 601 ca = with the normal. The direction of the electric field intensity at point

,P2 2a is

(A) sin E231 1-

c m (B) 45c

(C) cos E231 1-

c m (D) 30c

MCQ 3.3.7 Assertion (A) : Under static conditions, the surface of conductor is an equipotential surface.Reason (R) : The tangential component of electric field on conductor surface is zero.(A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is





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MCQ 3.3.8 A long 1 metre thick dielectric ( )3 0e e= slab occupying the region x0 5< < is placed perpendicularly in a uniform electric field 6E ax0 = . The polarization Pi inside the dielectric is(A) 4 ax0e (B) 8 ax0e

(C) 36 ax0e (D) Zero

MCQ 3.3.9 The flux and potential functions due to a line charge and due to two concentric circular conductors are of the following form :(A) Concentric circular equipotential lines and straight radial flux lines.

(B) Concentric circular flux lines and straight equipotential lines

(C) Equipotentials due to the charge are concentric cylinders and equipotentials due to two conductors are straight lines.

(D) Equipotentials due to line charge are straight flat surfaces and those due to two conductors are concentric cylinders.

MCQ 3.3.10 There are two conducting plates of sizes 1 1 .m m and m m3 3# # Ratio of the capacitance of the second one with respect to that of the first one is(A) 4 (B) 2

(C) 1/2 (D) 1/4

MCQ 3.3.11 Consider the following :In a parallel plate capacitor, let the charge be held constant while the dielectric material is replaced by a different dielectric. Consider 1. Stored energy 2. Electric field intensity.

3. Capacitance

Which of these changes ?(A) 1 only (B) 1 and 2 only

(C) 2 and 3 only (D) 1, 2 and 3

MCQ 3.3.12 By what name is the equation J 0: =d frequently known ?(A) Poisson’s equation

(B) Laplace’s equation

(C) Continuity equation for steady currents

(D) Displacement equation

MCQ 3.3.13 Method of images is applicable to which fields ?(A) Electrostatic fields only

(B) Electrodynamic fields only

(C) Neither electrostatic fields nor electrodynamic fields

(D) Both electrostatic fields and electrodynamic fields

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For View Only Shop Online at www.nodia.co.inMCQ 3.3.14 What is the unit of measurement of surface or sheet resistivity?

(A) Ohm/metre (B) Ohm metre

(C) Ohm/sq. meter (D) Ohm

MCQ 3.3.15 Which one of the following statements is correct ?On a conducting surface boundary, electric field lines are(A) always tangential

(B) always normal

(C) neither tangential nor normal

(D) at an angle depending on the field intensity

MCQ 3.3.16 Which one of the following is correct ? As frequency increases, the surface resistance of a metal(A) decreases

(B) increases

(C) remains unchanged

(D) varies in an unpredictable manner

MCQ 3.3.17 Application of the method of images to a boundary value problem in electrostatics involves which one of the following ?(A) Introduction of an additional distribution of charges and removal of a set of

conducting surfaces

(B) Introduction of an additional distribution of charge and an additional set of conducting surfaces

(C) Removal of a charge distribution and introduction of an additional set of conducting surfaces

(D) Removal of a charge distribution as well as a set of conducting surfaces

MCQ 3.3.18 Assertion (A) : Potential everywhere on a conducting surface of infinite extent is zero.Reason (R) : Displacement density on a conducting surface is normal to the surface.(A) Both A and R are true and R is the correct explanation of A

(B) Both A and R are true but R is NOT the correct explanation of A



MCQ 3.3.19 A parallel plate capacitor of 5 pF capacitor has a charge of 0.1 Cm on its plates. What is the energy stored in the capacitor ?(A) 1 mJ (B) 1 μJ

(C) 1 nJ (D) 1 pJ

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MCQ 3.3.20 A charge of 1 Coulomb is placed near a grounded conducting plate at a distance of 1 m. What is the force between them ?

(A) 41

0pe N (B) 81

0pe N

(C) 161

0pe N (D) 4 0pe N

MCQ 3.3.21 The capacitance of a parallel plate capacitor is given by dAr0e e where A is the area of

each plates. Considering fringing field, under which one of the following conditions is the above expression valid ?

(A) dA is tending towards zero (B) d

A is tending towards infinity

(C) dA is 1 (D) d

A is 1r 0e e

MCQ 3.3.22 What is the expression for capacitance of a solid infinitely conducting solid sphere of radius ‘R ’ in free space ?(A) 2 R0pe (B) 4 R0pe

(C) 8 R0pe (D) 0.5 R0pe

MCQ 3.3.23 A point charge of 10 Cm+ placed at a distance of 5 cm from the centre of a conducting grounded sphere of radius 2 cm is shown in the diagram given below :

What is the total induced charge on the conducting sphere ?(A) 10 μC (B) 4 μC

(C) 5 μC (D) 12.5 μC

MCQ 3.3.24 For an electric field sinE E t0 w= , what is the phase difference between the conduction current and the displacement current ?(A) 0c (B) 45c

(C) 90c (D) 180c

MCQ 3.3.25 An infinitely long line charge of uniform charge density /C mrL

is situated parallel to and at a distance from the grounded infinite plane conductor. This field problem can be solved by which one of the following ?(A) By conformal transformation

(B) By method of images

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For View Only Shop Online at www.nodia.co.in(C) By Laplace’s equation

(D) By Poisson’s equation

MCQ 3.3.26 An air condenser of capacitance of 0.00 F2 m is connected to a d.c. supply of 500 Volts, disconnected and then immersed in oil with a dielectric constant of 2.5. Energy stored in the capacitor before and after immersion, respectively is(A) 500 10 250 10J and J4 4

# #- -

(B) 250 10 500 10J and J4 4# #

- -

(C) 625 10 250 10J and J4 4# #

- -

(D) 250 10 625 10J and J4 4# #

- -

MCQ 3.3.27 A 3 Fm capacitor is charged by a constant current of 2 Am for 6 seconds. The voltage across the capacitor to the end of charging will be(A) 3 V (B) 4 V

(C) 6 V (D) 9 V

MCQ 3.3.28 Consider the following statements :A parallel plane capacitor is filled with a dielectric of relative permittivity r1e and connected to a d.c. voltage of V volts. If the dielectric is changed to another with relative permittivity 2r r1 1e e= , keeping the voltage constant, then1. the electric field intensity E within the capacitor doubles.

2. the displacement flux density D doubles

3. the charge Q on the plates is reduced to half.

4. the energy stored in the capacitor is doubled.

Select the correct answer using the codes given below :(A) 1 and 2 (B) 2 and 3

(C) 2 and 4 (D) 3 and 4

MCQ 3.3.29 A coil of resistance 5W and inductance 0.4 H is connected to a 50 V d.c. supply. The energy stored in the field is(A) 10 joules (B) 20 joules

(C) 40 joules (D) 80 joules

MCQ 3.3.30 The normal components of electric flux density across a dielectric-dielectric boundary(A) are discontinuous

(B) are continuous

(C) depend on the magnitude of the surface charge density

(D) depend on electric field intensity

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MCQ 3.3.31 Consider the following statements in connection with boundary relations of electric field :1. In a single medium electric field is continuous.

2. The tangential components are the same on both sides of a boundary between two dielectrics.

3. The tangential electric field at the boundary of a dielectric and a current carrying conductors with finite conductivity is zero.

4. Normal components of the flux density is continuous across the charge-free boundary between two dielectrics.

Which of these statements is/are correct ?(A) 1 only

(B) 1, 2 and 3

(C) 1, 2 and 4

(D) 3 and 4 only

MCQ 3.3.32 The capacitance of an insulated conducting sphere of radius R in vacuum is(A) 2 R0pe

(B) 4 R0pe

(C) 4 R0 2pe

(D) 4 /R0pe

MCQ 3.3.33 A parallel plate air capacitor carries a charge Q at its maximum withstand voltage V . If the capacitor is half filled with an insulating slab of dielectric constant 4 as shown in the figure given below, what are the maximum withstand voltage and the charge on the capacitor at this voltage, respectively ?

(A) . V2 5 , Q (B) , .V Q4 2 5

(C) , .V Q2 5 (D) / ,V Q4

MCQ 3.3.34 When an infinite charged conducting plate is placed between two infinite conducting grounded surfaces as shown in the figure given below, what would be the ratio of the surface densities 1r and 2r on the two sides of the plate ?

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(A) ( )( )d td t

2

1

++

(B) ( )( )d td t

1

2

++

(C) dd

2

1 (D) dd

1

2

MCQ 3.3.35 The polarization in a solid dielectric is related to the electric field E and the electric flux density D according to which on of the following equations ?(A) E D P0e= +

(B) D E P0e= +^ h

(C) D E P0e= +

(D) E D P0e= +

MCQ 3.3.36 Image theory is applicable to problems involving(A) electrostatic field only

(B) magnetostatic field only

(C) both electrostatic and magnetostatic fields

(D) neither electrostatic nor magnetostatic field

MCQ 3.3.37 Six capacitors of different capacitances , , , ,C C C C C Cand1 2 3 4 5 6 are connected in series. C C C C C C> > > > >1 2 3 4 5 6. What is the total capacitance almost equal to ?(A) C1 (B) C3

(C) C4 (D) C6

MCQ 3.3.38 Two extensive homogeneous isotropic dielectrics meet on a plane z 0= . For z 0$ , 4r1e = and for z 0# , 3r2e = . A uniform electric field exists at z 0$ as

5 2 3E a a a kw/mx y z1 = - + . What is the value of E z2 in the region z 0# ?

(A) 3az (B) 5 2a ax y-

(C) 6az (D) a ax y-

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MCQ 3.3.39 A flat slab of dielectric, 5re = is placed normal to a uniform field with a flux density D 1= Coulomb/m2. The slab is uniformly polarized. What is the polarization P of the slab in Coulomb/m2 ?(A) 0.8 (B) 1.2

(C) 4 (D) 6

MCQ 3.3.40 Which one of the following gives the approximate value of the capacitance between two spheres, whose separation is very much larger than their radii R ?(A) 2 / R0p e (B) 2 R0pe

(C) 2 /R0pe (D) 4 /R0pe

MCQ 3.3.41 Assertion (A) : For steady current in an arbitrary conductor, the current density is solenoidalReason (R) : The reciprocal of the resistance is the conductivity.(A) Both A and R are true and R is the correct explanation of A




MCQ 3.3.42 Assertion (A) : Displacement current can have only a.c components.Reason (R) : It is generated by a change in electric flux.(A) Both A and R are true and R is the correct explanation of A




MCQ 3.3.43 A plane slab of dielectric having dielectric constant 5, placed normal to a uniform field with a flux density of 3 C/m2, is uniformly polarized. The polarization of the slab is(A) 0.4 C/m2 (B) 1.6 C/m2

(C) 2.0 C/m2 (D) 6.4 C/m2

MCQ 3.3.44 Ohm’s law in point form in the field theory can be expressed as(A) V RI= (B) /J E s=

(C) J Es= (D) /R l Ar=

MCQ 3.3.45 A medium behaves like dielectric when the(A) displacement current is just equal to the conduction current

(B) displacement current is less than the conduction current

(C) displacement current is much greater than the conduction current

(D) displacement current is almost negligible

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For View Only Shop Online at www.nodia.co.inMCQ 3.3.46 A copper wire carries a conduction current of 1.0 A at 50 Hz. For copper wire

,0e e= ,0m m= 5.8 10 mho/ms #= . What is the displacement current in the wire ?(A) 2.8 10 A#

(B) 4.8 10 A11#

-

(C) 1 A

(D) It cannot be calculated with the given data

MCQ 3.3.47 Assertion (A) : When there is no charge in the interior of a conductor the electric field intensity is infinite.Reason (R) : As per Gauss’s law the total outward electric flux through any closed surface constituted inside the conductor must vanish.(A) Both A and R are true and R is the correct explanation of A




MCQ 3.3.48 A point charge Q+ is brought near a corner of two right angle conducting planes which are at zero potential as shown in the given figure. Which one of the following configurations describes the total effect of the charges for calculating the actual field in the first quadrant ?

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comMCQ 3.3.49 The electric field across a dielectric-air interface is shown in the given figure. The

surface charge density on the interface is

(A) 4 0e- (B) 3 0e-

(C) 2 0e- (D) 0e-

MCQ 3.3.50 When air pocket is trapped inside a dielectric of relative permittivity ‘5’, for a given applied voltage across the dielectric, the ratio of stress in the air pocket to that in the dielectric is equal to(A) 1/5 (B) 5

(C) 1 5+ (D) 5 1-

***********

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SOLUTIONS 3.1

SOL 3.1.1 Option (A) is correct.For a given current density, the total current that passes through a given surface is defined as

I dJ S:= #where dS is the differential surface area having the direction normal to the surface.So we have dS d d azr r f= for the plane z 0=Therefore, the current passing the plane z 0= , 0 2# #r is

I ( )e d da a a10 zz z

2

0

2

0

2:r r r f= +r

rf

p

==^ h8 B##

10 e dzdz

0

2

0

2r f=

p

##

d d100

2

0

2r r f=

p

## (z 0= )

10 2

2

0

2

02

#r f= p; 6E @

10 23 p# #= 0 A6 p=

SOL 3.1.2 Option (C) is correct.For a given current density, the total current that passes through a given surface is defined as

I dJ S:= #where dS is the differential surface area having the direction normal to the surface.So, we have dS ( )( )sinr d dr aq f= q for the surface 90cq =Therefore, the total current crossing the surface 90cq = ,0 2< <f p ,0 1 mr< < is

I cos sin sinr r r r d dra a a ar2 2 2

:q q q f= + -q f q^ ^h h#

sinr d drr

3 2

0

2

0

1q f=

f

p

==## at 90cq =

r40

24

0

1

#f= p6 :@ D 2 A8

14p p

#= =

SOL 3.1.3 Option (B) is correct.For a given current density, the total current flowing through a cross section is defined as

I dJ S:= #




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where dS is the differential cross sectional area vector having the direction normal to the cross section.So we have dS d d azr r f= (since the cylindrical wire is lying along z -axis)Therefore the total current flowing through the wire (cross section) is

I 50 a d d az z:r r r f= b ^l h

50 d d0

2

0

16 10 3

r r r f=#

f

p

r ==

-

b ^l h##

50016 10

023r f# #= # p-

6 6@ @

0 16 10 26 3 p# # #= -

SOL 3.1.4 Option (C) is correct.For a given current density, the total current that passes through a given surface is defined as

I dJ S:= #where dS is the differential surface area having the direction normal to the surface.So we have dS d d azr r f= for the plane z 2=Therefore the total current crossing the plane z 2= , 4<r is

I 40( 1)20 sin d da a a2 z zr r

f r r f= -+re ^o h

1

20 sin d d20

4

0

2

rf r r f= -

+rf

p

==e ^o h##

sind d1

202

0

4

0

2

0

rr r f f=-

+r f

p

= =; ;E E

1 2 344 44

# #

A4=

SOL 3.1.5 Option (B) is correct.From the equation of continuity we have the relation between the volume charge density, vr and the current density, J as

tv

22r J:d=-

Given the current density,

J ( )

/sin A ma a401

20z2

2

r rf= -

+r

So, we have the components J 40r=r ,J 0=f and

( )sinJ

120

z 2rf=-

+

Therefore, tv

22r J

JzJ1 1 z

22

22

22

r r r r f=- + +rf

^ h= G

( ) sinz

1 401

2022

222

r r rf= +

+-

e o> H

2=So, volume charge density will be constant with respect to time.



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For View Only Shop Online at www.nodia.co.inSOL 3.1.6 Option (C) is correct.

Given the current 6 AI = is flowing radially outward (in ar direction) through the medium between the cylinders. So the current density in the medium between the cylinders is

J lI a2pr= r a2 2

6#pr= r ( 2 ml = )

/A ma25 2

pr= r

For a given current density in a certain medium having conductivity s , the electric field intensity is defined as

E Js= a1

23

s pr= rb l

.2 4 10 0 05

32

# # #p= - ( 4 10 m2r #= - , 0.05 /S ms = )

238.73 /V m=

SOL 3.1.7 Option (C) is correct.Voltage between the cylindrical surfaces is defined as the line integral of the electric field between the two surfaces

i.e. V dE l:=- #Now the electric field intensity in the medium between the two cylindrical surfaces

as calculated in previous question is E a123

s pr= rb l

and the differential displacement between the two cylindrical surfaces is dl d ar= r

So the voltage between the cylindrical surfaces is

V 23 da a

3 10

5 10

2

2

:prs r=-#

#

r rr= -

-

b ^l h# ln23

35

ps=- b l

.88 volt5=-So, the voltage between them will be .88 volt5 .

SOL 3.1.8 Option (C) is correct.As we have already calculated the voltage between the two cylindrical surfaces and the current flowing radially outward in the medium between the surfaces is given in the question. So the resistance between the cylindrical surface can be evaluated directly as

R IV= . 0.8136

4 88 W= = (V 4.88 volt= ,I 6 A= )

SOL 3.1.9 Option (A) is correct.Since voltage between the cylindrical surfaces is V 4.88 volt=and current flowing in the medium is I 6 A=So, Power dissipated in the medium is P VI= ( . )4 88 6#= 29.28 watt=

SOL 3.1.10 Option (B) is correct.Consider a constant voltage is applied across the ends of the wire so, the electric




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field intensity throughout the wire cross section will be constant.

i.e. E J J1

1

2

2s s= =

where J1 is the current density in the material having conductivity 1s . J2 is the current density in the material having conductivity 2s .So,the ratio of the current density is

JJ

2

1 2

1ss=

i.e. it will be independent of both r and R .

SOL 3.1.11 Option (A) is correct.Since hydrogen atom contains a single electron ( ve- charge) and a single proton (

ve+ charge). So, the dipole moment due to one atom of the hydrogen will be p qd= where q is electronic charge and d is effective lengthi.e. 1.6 10 Cq 19

#= - and d 7.1 10 m16#= -

So, p 1.6 10 7.1 1019 16# # #= - -

^ ^h h

and since the polarization in a material is defined as the dipole moment per unit volume.Therefore P np= where n is the number of atoms per unit volume.i.e. n 5.5 10 /atoms cm19 3

#= 5.5 10 /atoms m25 3

#=So, P 5.5 10 1.6 10 7.1 1025 19 16

# # # # #= - -^ ^h h

.25 10 /C m4 9 2#= -

SOL 3.1.12 Option (D) is correct.When an electric field E is applied to a material with dielectric constant re then the polarization of the material is defined as P ( 1)Er0e e= -

So, 1re - EP0e

=.

.8 85 10 40 10

6 25 1012 3

9

# # #

#= -

-.1 7655 10 2

#= -

re .1 0 0177= + .1 0177=

SOL 3.1.13 Option (C) is correct.Given D 2P= & P /2D=If the polarization of a dielectric material placed in an electric field E is P , then the electric flux density in the material is defined as D E P0e= + /2E D0e= +or D 2 E0e= ..........(i)and since the relation between the electric field, E and flux density, D inside a dielectric material with dielectric constant re is defined as D Er0e e=So, comparing the result with equation (i) we get, 2re = .



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Electric field intensity is defined as the negative gradient of the potentiali.e. E Vd=-

xV

yV

zVa a ax y z2

222

22=- + +c m

0 /V ma25 y=

SOL 3.1.15 Option (B) is correct.For a given electric field intensity E in a material having relative permittivity re , the electric flux density is defined as : D Er 0e e=

8.85 10 (500 )a58 12

y# # #= -^ h /8 5re =

.08 /nC ma9 y2=

SOL 3.1.16 Option (C) is correct.For an applied electric field intensity E in a material having relative permittivity re , the polarization of the material is defined as

P 1 Er0e e= -^ h 8.85 10 ( . 1) 500a1 612

y# #= --^ ^h h

8.85 10 0.6 500ay12# # #= -

.66 10 a4 y9

#= -

SOL 3.1.17 Option (D) is correct.Since the two regions is being separated by the plane y 0= , so the tangential and normal component of the electric field to the plane y 0= are given as E t1 50 10a ax z= - E n1 20ay=From the boundary condition, the tangential component of electric field will be uniform.i.e. E t2 E t1= 50 10a ax z= -and the normal component of the field is nonuniform and given as E n2 2e E n1 1e=

E n1 20 8E a a52

n y y2

11e

e= = =^ h

So the electric field intensity in the second region is E2 E Et n2 2= + 50 10 8a a ax z y= - +^ ^h h 50 8 10 /kV ma a ax y z= + -Therefore the electric flux density in the region 2 is D2 Er2 0 2e e= 5 8.85 10 (50 8 10 ) 10a a ax y z

12 3# # #= + --

.21 0. 5 0.44 /C ma a a3 5x y z2m= + -

SOL 3.1.18 Option (B) is correct.Energy density in the region having electric field intensity E2 is defined as




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WE E E21r2 0 2 2:e e= , where the relative permittivity of the medium is r2e

As calculated in previous question the electric field intensity is E2 50 8 10 /kV ma a ax y z= + -So the energy density in the region 2 is

WE ( ) ( ) ( )21 5 50 8 10 100

2 2 2 6# # #e= + +6 @

9 /mJ m7 3=

SOL 3.1.19 Option (D) is correct.According to boundary condition the tangential components of electric field are uniformi.e. E t1 E Et t2 3= = ...(i)but the normal component of electric fields are non uniform and defined as E n1 1e E En n2 2 3 3e e= =Since 1e 3e= (Given)So, Ein E En n3 2!= ...(ii)and as the net electric field is given by E E Et n= + (sum of tangential and normal component)Therefore by combining the results of eq (i) and (ii) we get E1 E E3 2!=

SOL 3.1.20 Option (D) is correct.As the dielectric slab occupies the region z d0 < < and the field intensity in the free space is in az+ direction so, the field will be normal to the boundary of plane dielectric slab.So from the boundary condition the field normal to the surface are related as Eine E0e=

Ein E Ea a4 4z z0

00

0

ee= = ( 4 0e e= )

Therefore, Din 4 E EE a a4in z z00

0 0e e e= = =

SOL 3.1.21 Option (A) is correct.Total energy stored in a region having electric field is given as

W dvE E21

v0 :e= ^ h#

dvE E E E21

v0 1 2 1 2:e= + +^ ^h h# (E E E1 2= + )

2E E dvE E21

v0 1

222

1 2:e= + +^ h#

E dv E dv dvE E21

21

v v0 1

20 2

20 1 2:e e e= + + ^ h# # #

W W dvE Ev

1 2 0 1 2:e= + + ^ h#



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Energy on a dipole with moment p in an electric field E is defined as WE p E:=- ( 2 3 ) (1.5 )a a a ax y x z:=- - + - ( )3 3=- - - 6 J=

SOL 3.1.23 Option (C) is correct.Consider a neutral dielectric is placed in an electric field E , due to which the dielectric gets polarized with polarization P , the bound surface charge density of the dielectric be pr S and the bound volume charge density be pvr .So the total bound charge by the dielectric is given as

Qbound dS dvp pvvS

r r= +S ##Since for a given polarization P of a dielectric material, the bound surface charge density over the surface of material is defined as pr S aP n:= where an is the unit vector normal to the surface directed outward.while the bound volume charge density inside the material is defined as pvr P:d=-

So we have, Qbound d dvP a S PnS v

: :d= -^ h ##

d dvP S Ps v

: :d= - ## (dS da Sn = )

But according to the divergence theorem

dP Ss

:# dvPv

:d= #Therefore, Qbound 0=

SOL 3.1.24 Option (A) is correct.Resistance of a conductor of length l and having uniform cross sectional area S is

R Sls= where s is the conductivity of the conductor

Given the conductivity, s 5 10 ( )m6 1W#= -

the length of the conductor, l 8 m=side of the square cross section, a 3 cm=and radius of the bored hole, r 0.5 cm=So, the net cross sectional area is S = area of square cross section(bar) - area of circular cross section(hole)or S a r2 2p= - (3) (0.5)2 2p= -

cm9 42p= -a k

The total resistance between the square ends is given as

R Sls=

5 10 9 4 108

6 4#

p# #

=- -

^ ah k9 C

.948 102 3 W#= -




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SOL 3.1.25 Option (D) is correct.The two materials of composite bar will behave like two wires of resistance RL(resistance due to lead) and RC (resistance due to copper) connected in parallel.As from the previous question we have the resistance due to the lead is RL 1.948 mW=and since the area of the cross section filled with copper is equal to the area of the cross section defined by hole so we have

cross sectional area SC cm42p=

length of the bar l 8 m=

and conductivity of the copper, Cs resistivity of the copper1=

1.72 101

8#

= -

So the resistance due to copper is

RC SlC Cs

= 10

84

41.72 10

18#

= p#

--^ _h i

1.76 mW=

Therefore the equivalent resistance of the composite bar is

R ||R RC L= . .( . ) ( . )

1 948 1 761 948 1 76#= +

24.62 105 6 W#= -

SOL 3.1.26 Option (D) is correct.Given the conductivity of material, s 10 ( )m6 1W= -

and conductance of the wire, G 10 ( )6 1W= -

Since the conductance of a wire of length l having cross sectional area S is

G lSs=

So we have, 106 lr106 2

# p= (S r2p= )

r lp=

SOL 3.1.27 Option (C) is correct.Given the radii of spherical shell as a 1 0.01cm m= = b 2 0.02cm m= =The capacitance of a spherical capacitor having inner and outer radii a and b respectively is defined as

C 1 1a b

4 r 0pe e=-b l

. .

.

0 011

0 021

4 4 8 85 10 12# # #p=

-

- 8.9 pF=

SOL 3.1.28 Option (B) is correct.Since the dielectric has been removed from the portion defined by < <2 f pp

^ h so the composite capacitor will have the dielectric filled only in th4

3 portion of the total capacitor and so the configuration can be treated as the two capacitors connected in parallel with each other.



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For View Only Shop Online at www.nodia.co.inThe capacitance of the portion carrying air( 1re = ) as the medium between the spherical shells

C1

a b41

1 14 0

#pe=-

. .

.41

0 011

0 021

4 8 85 10 12

## #p=

-

-

.0 56 10 12#= - 0.56 pF=

The capacitance of the portion carrying dielectric( 4re = ) as the medium between the spherical shells

C2 C43#=

Where C is the capacitance if no any portion of dielectric was removed as already calculated in previous question.

So we have C2 .43 8 9 10 12# #= - .6 7 10 12

#= - 6.7 pF=

Therefore the equivalent capacitance of the composite capacitor is, Ceq C C1 2= + 0. 6 6.7 .6 2 12= + =


Capacitance of a parallel plate capacitor is defined as

C dSe=

where S is the surface area of the parallel plates d is the separation between the platesHere, the three different regions will be treated as the three capacitors connected in series as shown below

So the capacitance of the region 1 is C1 .S

0 001r1 0e e= 2500 S0e=




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the capacitance of the region 2 is C2 .S

0 002r2 0e e= 2000 S0e=

the capacitance of the region 3 is C3 .S

0 0020e= 500 S0e=

Therefore the equivalent capacitance of the whole configuration is calculated as

C1eq

C C C1 1 1

1 2 3= + + S

12500

12000

15001

0e= + +b l

So, Ceq 3.45 10 S20e#=

The capacitance per square meter of surface area will be

Ceql 3.45 10 .05 /nF mSC

4eq 20

2e#= = =

SOL 3.1.30 Option (B) is correct.Capacitance between the two cylindrical surfaces is defined as

C /ln b al2pe=

^ h

Where l " length of the cylinder a " inner radius of the cylinder b " outer radius of the cylinderSince, the medium between the conducting cylinders includes the dielectric layer

4re =^ h from 4 cmr = to 6 cmr = and air 1re =^ h from 6 cmr = to 8 cmr = , so the configuration can be treated as the two capacitance connected in series.Now for the dielectric layer ( 4re = ) from 4 cmr = to 6 cmr = , capacitance is

C1 ( / )ln 6 42 1r0pe e=

(1.5)ln8 0pe= ( 1 ml = )

and for the air medium 1re =^ h from 6 cmr = to 8 cmr = , capacitance is

C2 ( / )ln 8 62 10 #pe=

( / )ln 4 32 0pe=

So, the equivalent capacitance of the configuration is evaluated as

C1eq

C C1 1

1 2= +

( . ) ( / )ln ln8

1 52

4 30 0pe pe= +

Ceq 143 pF=

SOL 3.1.31 Option (D) is correct.The equivalent arrangement of the capacitor can be drawn in form of circuit as below

For which the capacitances are calculated as below



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C1 //dS

dS

220 0e e= =

4 1010 10

30

4

#

# #e= -

-

40e=

C2 //

dS

dS

22r r0 0e e e e= = 4

3 0e= 3re =

C3 /dS 20e=

2 4 1010 10

30

4

# #

# #e= -

-

80e=

Therefore the equivalent capacitance of the capacitor is

Ceq C C CC C

31 2

1 2= + + 84 4

34 4

30

0 0

0 0e

e e

e e= +

+ 2.76 pF4

0e= =

SOL 3.1.32 Option (C) is correct.As the medium between capacitor plates is conducting so it carries the resistive as well as capacitive property.Consider the plates are separated by a distance d and the surface area of plates is S as shown in the figure.

So the total resistance of the medium between plates is

R Sds=

and capacitance of the capacitor is

C dSe=

Therefore the time constant of the capacitor will be

t RC se= =

***********




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SOLUTIONS 3.2

SOL 3.2.1 Option (D) is correct.For a given current density, the total current that passes through a given surface is defined as

I dJ S:= #where dS is the differential surface area having the direction normal to the surface.Since the current density is independent of q and f so we can have directly the current I J S:= (4 )rJ ar2p=

r e r1 4t10 23

p= - ( ) e4 6 612 10 103 3

# # #p= #- -

4 3 e e121 1p p# #= =- -

SOL 3.2.2 Option (A) is correct.From the equation of continuity we have the relation between the volume charge density, vr and the current density, J as

tv

22r J:d=-

and since the current density have only the component in ar direction so we have,

tv

22r

r rr J1

r22

22=- ^ h

tv

22r

r rr r e

1 1 t2

2 103

22=- -

b l

Integrating both sides we get,

( , )r tvr ( )re dt f r1 t

2103

=- +-#where ( )f r is the function independent of time.

( , )r tvr ( )re f r10 t

2

3103

= +-

-

Now for t" 3 ( , )r tvr 0=So, we put the given condition in the equation to get ( )f r 0=

therefore ( , )r tvr re10 t

2

3103

=-

-

i.e. ( , )r tvr r12\



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For View Only Shop Online at www.nodia.co.inSOL 3.2.3 Option (B) is correct.

The velocity of charge density can be defined as the ratio of current density to the charge density in the region

i.e. v Jvr

= /

(1/ )r e

r e a10 t

tr

3 2 10

10

3

3

=- -

-

^ h 10 r ar3=

So, at 0.6 mr = , v 10 0.6ar3#= 00 /m sa3 r=

SOL 3.2.4 Option (D) is correct.The given problem can be solved easily by using image theory as the conducting surface y 0= can be replaced by the equipotential surface in the same plane y 0= and image of line charges ( L 5 /pC mr =-l at x 0= , y 1=- and x 0= , y 2=- ) as shown in the figure

.The work done to carry a unit positive charge from a point located at a distance a from the line charge with charge density rL to another point located at a distance b from the line charge is defined as

Vab ln ab

2 0per=- L

b l

and since the surface y 0= has zero potential, so the potential at point P will be equal to the work done in moving a unit positive charge from the plane y 0= to the point P . So the potential at point P will be

VP ln ab

2 0per=- L

b l/where a is the distance of the surface y 0= from the line charges while b is the distance of point P from the line charges.

So, VP ln ln ln ln25 10

21

12

110

217

0

12#pe=- - - + +

-

b c c cl m m m= G

0. volt3=-




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SOL 3.2.5 Option (C) is correct.Electric field at a distance R from a line charge having uniform charge density rL is defined as

E RR

2 02pe

r= L

So the net electric field intensity produced at the point P due to the four line charges discussed in previous question is given as

E RR

2 02pe

r= L/where R is the distance of point P from the line charges

Therefore, E ( , , )

( , , ) ( , , )( , , )

( , , ) ( , , )2 1 3 0

1 2 0 0 1 01 4 0

1 2 0 0 2 0l

02 2pe

r=- -

- - - +- -

- - ->

( , , )

( , , ) ( , , )( , , )

( , , ) ( , , )1 1 0

1 2 0 0 1 01 0 0

1 2 0 0 2 02 2-

- -- - - - -

-- - - -

H

( , , ) ( , , ) ( , , ) ( , , )

25 10

101 3 0

171 4 0

21 1 0

11 0 0

0

12#pe= - - + +

-

; E

0.12 0.0032a ax y= - 0.12 0.003 /V ma ax y= -

SOL 3.2.6 Option (C) is correct.For a given polarization P inside a material, the bound surface charge density over the surface of material is defined as pr S aP n:= where an is the unit vector normal to the surface directed outward.while the bound volume charge density inside the material is defined as pvr P:d=-Since the component of polarization of rod along y -axis is 2 3P yy

2= + . So, the polarization of the material is P (2 3)y ay2= + . and the charge density on the surface of the rod is pr S aP n:= At y 0= (top surface) S1r (2 3)y a ay y

2:= + -^ h 3=-

At y 5= (bottom surface) S2r (2 3) ( )y a ay y2

:= + 53=and since the polarization has no radial component so no charge will be stored on its curvilinear surface and so the total bound surface charge on the surface of the rod is

QpS dspr= S# S SS S1 2r r= + (S is the cross sectional area) 3 53S S=- + 50 S=Now, the bound volume charge density inside the material is pvr P:d=- (2 3)y ay2

:d=- + 4y=-So the total bound volume charge stored inside the material will be

Qpv dvpvr= # ( )y Sdy40

5= -# 4S y2

2

0

5

=- ; E S50-

So, Total bound charge Qbound Q QS v= + 50 50 0S S= - =



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Electric field produced by the point charge at a distance r is

E rq a4

1r

02pe=

So, the induced dipole moment in the neutral atom due to the electric field E produced by the point charge will be

p Ea= rq a

4 r0

2pea=

and since the electric field intensity produced due to a dipole having moment p at a distance r from the dipole is defined as

Edip 2 cos sinrp a a

4 r0

3peq q= + q^ h

where q is the angle formed between the distance vector r and dipole moment pSo the field produced by the induced dipole at the point charge is

Edip rp

42

03pe

= rrq

4

24

03

02

pepea

=b l

rq

42

02 5pe

a=^ h

(q p= as shown in the figure)

Therefore the force experienced by the point charge due to the field applied by induced dipole is

F qEdip= qr

2 41

0

2

5a pe= b l

( )

2 91 10 9 10

119 9

2

5# # # #a= -b l

SOL 3.2.8 Option (B) is correct.Electric field intensity produced due to a dipole having moment p , at a distance r from the dipole is defined as


4 r0

3peq q= + q^ h

where q is the angle formed between the distance vector r and dipole moment pSo the electric field intensity produced due to dipole P1 at P2 is

E1 rp a

4 03

1

pe= q ( )

a4 12 10

03

9

#

#pe

= q

- ( /2q p= )

Therefore the torque on P2 due to P1 is T p E2 1#= Taking the magnitude only we have the torque on P2 is

T sinp E2 1 q= 9 10 42 10 90sin9

0

9

cpe##= -

-

^ ch m

1.62 10 N m8# -= - 0.16 N mm -=




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SOL 3.2.9 Option (C) is correct.Electric field intensity produced due to a dipole having moment p , at a distance r from the dipole is defined as


4 r0

3peq q= + q^ h

where q is the angle formed between the distance vector r and dipole moment pSo the electric field intensity produce due to dipole P2 at P1 is

E2 2rp a

4 r0

32

pe=-

( )2a

4 19 10

r0

3

9#pe #=-

- (q p= )

Therefore the torque on P1 due to P2 is T p E1 2#=Considering the magnitude only we have the torque on P1 is

T sinp E1 2 q= 2 10 49 10 29

0

9

# ##

#pe= ---

c m ( /2q p= )

3. 4 10 N m6 4# -= - 0.32 N mm -=

SOL 3.2.10 Option (A) is correct.For a given polarization P inside a material, the bound volume charge density inside the material is defined as pvr P:d=-Since the polarization of the sphere is ( )rP 2rar=So the bound volume charge density inside the sphere is

pvr rP4:=- ^ h ( )r r

r r1 222

22=-

rr1 62

2#=- 6=-

Therefore the electric field intensity inside the sphere at a distance r from the center is given by

E rQ a4

1 encr

02pe=

rr a4

1 pvr

0234 3

#pe

r p=

r ra a3 36v

r r0 0e

re= =- ra2

r0e

=-b l

So the radial component of the electric field inside the sphere is

Er r20e

=-

which is linearly decreasing with a slope 20e

-b l with respect to r as shown below :



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For a given polarization P of a material, the bound surface charge density over the surface of material is defined as pr S aP n:=So the bound surface charge density over the spherical surface is pr S rP ar:= ^ h (a an r= ) r2= a2= (at the spherical surface r a= )So, total bound surface charge over the sphere is QpS a a2 4 2

# p= a8 3p=and the bound volume charge density inside the sphere as calculated above is pvr 6=-So, total bound volume charge inside the sphere is

Qpv ( 6)a a34

34

pv3 3r p p#= = -b bl l a8 3p=-

Therefore the total bound charge in the sphere is Qbound Q Qp pv= +S a a12 123 3p p= - 0=According to Gauss law the outward electric field flux through a closed surface is equal to the charge enclosed by the surface and since the total bound charge for any point outside the sphere is zero So, the electric field intensity at any point outside the sphere is 0E = .

SOL 3.2.12 Option (C) is correct.Since the spherical shell is of inner radius 2 mr = so region inside the sphere will have no polarization and therefore the total charge enclosed inside the shell for

2 mr < will be zero.i.e. Qenc 0=According to Gauss law the total outward electric flux from a closed surface is equal to the charge enclosed by the surface and since the total enclosed charge for

2 mr < is zero so the electric field intensity at 1 mr = will be zero.

SOL 3.2.13 Option (A) is correct.Since the total bound charge by a polarized neutral dielectric is zero as discussed in MCQ. 33. So for any point outside the spherical shell the total enclosed charge(bound charge) will be zero and as discussed in the previous question, according to Gauss law the electric field intensity at any point outside the spherical shell will be zero.So, for the surface r 7= Qenc 0=Therefore the electric field intensity is E 0=

SOL 3.2.14 Option (A) is correct.As we have to find electric field at 5 mr = so we determine first the charge enclosed by the surface 5 mr = which will be equal to the sum of the volume charge stored in the region 2 5 mr# # and the surface charge stored at 2 mr = .Since for a given polarization P of a dielectric material, the bound volume charge density inside the material is defined as




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pvr P:d=-So the bound volume charge density inside the dielectric defined in the region 2 mr 6# # will be

pvr rP:d=- ^ h r rr r

1 52

2

22=- b l

5r2

=-

so the total bound volume charge in the region 2 5 mr# # is

Qpv dvpvr= # 4r r dr5r

2

2

5p#= -

=#

r2025p=- 6 @ 60p=-

Now for a given polarization P inside a dielectric material, the bound surface charge density over the surface of dielectric is defined as pr S P an:=where an is the unit vector normal to the surface pointing outward of the material.So the bound surface charge density at 2 mr = is pr S ( )rP ar:= -^ h (a an r=- )Therefore the total bound surface charge over the surface 2 mr = is

QpS r r5 4 2# p=- (for spherical surface S r4 2p= )

25 4 22# #p=- 40p=- 2 mr =

So, the total enclosed charge by the surface 5 mr = is Qenc Q Qpv ps= + 60 40p p=- - 100p=-So the electric field intensity at 5 mr = will be,

E rQ a4

1 encr

02pe #= a4

15100

r0

2pep

#= - a1r

0e=-

SOL 3.2.15 Option (C) is correct.Since the electric field intensity at any point inside a conductor is always zero, so the electric flux density at a distance r from the center of the spherical conductor can be given as

D ,

, m

r

rQ ra

0 1

41

<

>r2p= *

where 3 mCQ = is the total charge carried by the conductor.and since the dielectric material surrounding the spherical conductor has permittivity

3re = , so the electric field intensity at a distance r from the center of the sphere is

E

, m

m

m

r

rQ r

rQ r

a

a

0 1

41 2

42

<

< <

>

rr

r

02

02

pe e

pe

=

Z

[

\

]]]

]]]

So, the total energy of the configuration is

WE dvD E21

:= #



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4 4

44 4

4drrQ

rQ r dr

rQ

rQ r dr2

1 0 2 22

20

22

r0

1

0 21

2

p pe ep

p pep= + +

3

c c ^ c c ^m m h m m h= G# ##

( )Q

rdr

rdr2

14

4 1 1 1 12

2

20

221

2

pp e e= +

3

' 1##

Qr r8

1 1 1 1r

2

0 1

2

0 2p e e e= - + -3

: :D D' 1 Q8 3

121

21

0

2

#pe= +& 0

( )

23 10 9 10

1283 2 9

# # ##=

-

.7 10 J3 4#=

SOL 3.2.16 Option (B) is correct.The electric potential at the centre of sphere will be equal to the work done to carry a unit charge from infinity to the centre of the sphere (the line integral of the electric field intensity from infinity to the center of the sphere)

i.e. V dE l0

:=-3#

Since the sphere has uniform charge density vr =0.6 /nC m3 embedded in it, so the electric field intensity at a distance r from the center of the sphere can be given as

E ,

,

r r R

rR r R

a

a

3

3

<

>

r

vr

vr

0

02

3

e er

er

=

Z

[

\

]]

]]

where R is the radius of the sphere i.e. mR 1p

=

So, the potential at the centre of sphere will be

V dE l0

:=-3# (where differential displacement is d drl ar= )

r

dr r dr3

13

/

/

v v

02

31

1

0

er

p er=- -

3

p

pc m# #

3rr

31 1

2/

/

v

r

v

0

3 1

0

2

1

0

er

p e er

#=- - -3

p

pc m : :D D

331

21v

r

v

0

3

0er

pp e e

rp# #= +c m 3 12

v v

0 0e pr

e pr= +

125

3 45v v

0 0#per

per= = .

35 0 6 10 9 109 9# # # #=

- ( 2re = )

volt5= vr =0.6 /nC m3

SOL 3.2.17 Option (A) is correct.For a given polarization P of a material, the surface charge density over the surface of material is defined as Sr aP n:= where an is the unit vector normal to the surface directed outward of the material.




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while the volume charge density inside the material is defined as vr P:d=-Since the the cylinder has uniform polarization P .So, volume charge density inside the sphere is vr P:d=- 0=and the surface charge density over the top and bottom surface of the cylinder is rS P an:= P!= ( P+ at top surface and P- at bottom surface)So the total bound charge by the cylinder is Qbound Q QS v= +

dS dvS

vv

r r= +S# # ( ) ( )P r P r 02 2p p= + - +6 @ 0=

SOL 3.2.18 Option (C) is correct.As calculated above the volume charge density inside the cylinder is zero while the surface charge density at top and bottom surfaces are respectively P+ and P- , so the cylinder can be considered as the two circular plates (top and bottom surface) separated by a distance L . Since the separation between the plates is larger than the cross sectional radius (L r2= ) so the fringing field(electric field) will exist directed from the upper plate towards the lower plate.

SOL 3.2.19 Option (A) is correct.The electric flux lines will be the same as the electric field intensity outside the cylinder but as the volume charge density is zero 0vr = inside the cylinder so

dD S 0: =# and therefore the flux lines will be continuous.

SOL 3.2.20 Option (A) is correct.Consider the surface charge density on the parallel plates is !rS so the electric flux density between the plates is defined as D anr= S

where an is the unit vector normal to the surface of plates directed from one plate toward the other plate.Since permittivity changes from layer to layer, but the field is normal to the surface so electric flux density D will be uniform throughout the plate separation as from boundary condition.So the electric field intensity at any point between the parallel plates is

E Dr0e e=

( )aa

2 1 100n

02e

r=+S ( )a2 1 100r

2e = +

Therefore the voltage between the plates can be evaluated by taking the line integral of electric field from one plate to the other plate

i.e. V dE l:=- # ( )a

da a2 1 100

.n

a 02

0

0 1:

er=-+=

Sc ^m h# (d dl a= )

S

( . ) ada

2 1001

0 1

.

02 2

0

0 1

er=

+# (the direction of a is along an)

S. .tan a

2 1001

0 11

0 1.

0

1

0

0 1

er

# #= -a k9 C



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S2 10

14 0

0er p

#= -9 C S80 0er p=

Now charge stored at the parallel plates is Q ( )( )Sr= S where S is surface area of the plates ( . )0 2s#r= 0.2 mS 2=So, the capacitance of the capacitor is evaluated as

C VQ=

( )/( . )800 2 16

s

s

0

0#

r p er

pe= =

.51 105 11#= -

SOL 3.2.21 Option (C) is correct.For the two wire transmission line consists of the cylinders of radius b and separated by a distance h2 (centre to centre), the capacitance per unit length between them is defined as

Cl ( / )cosh h b1

pe= -

Here, 2 2 cmh = and 0.2 cmb =

So, Cl / .

.cosh 1 0 22 8 85 10

1

12# # #p= -

-

^ h 3.64 10 /F m11

#= - ( 2re = )

So the charge per unit length on each wire will be, Q C V0= l .3 64 10 10011

# #= - ( 100 VVo = ) .64 10 /C m5 9

#= -

SOL 3.2.22 Option (C) is correct.Consider the oil rises to a height h in the space between the tubes.So, the capacitance of the tube carrying oil partially will be treated as the two capacitors connected in parallel.Since the capacitance between the two cylindrical surfaces is defined as

C /ln b al2pe=

^ h

Where l " length of the cylinder a " inner radius of the cylinder b " outer radius of the cylinderSo the capacitance of the portion carrying oil ( 1ec = ) as the medium between the cylindrical surfaces is

Coil /lnh

3 12 r 0pe e=

^ h ln

h3

4 0pe=^ h

1 2r ee c= + =^ h

and the capacitance of the portion carrying air( 1re = ) as the medium between the cylindrical surfaces is

Cair /1

lnh

3 12 0pe

=-

^

^

h

h

Therefore the equivalent capacitance of the tube carrying oil to the height h is

C C Coil air= + 2( )

( )lnh

31

0pe= +




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Since the energy stored in a capacitor is defined as

WE CV21 2= where V is the applied voltage to the capacitor

So the net upward force due to the capacitance is given by

F dhdWE= V dh

dC21 2=

( )lnV2

13

22 0pe=

and net downward force on the oil due to gravity will be F mg= (0.01 / ) ( )gm cm b a h g3 2 2

# #p= - mass density 0.01 /gm cm3=

. ( ) h g100 01 9 1 106

6# # # #p= --

- . hg0 08p=

Since in equilibrium both the upward and downward forces are equal

So, 0.08 hgp ( )ln

V21

322 0pe=

0.08 (9.8)hp # ( )( )

.ln2

1 2 103

2 8 85 103 212

# # ## #p=

-

h . . ( )

( ) .ln2

10 08 9 8 3

2 10 2 8 85 103 2 12

## #

# # # #=-

4.11 10 m5#= - 4 .1 m5 m=

SOL 3.2.23 Option (C) is correct.Consider the charge densities of the two surface of the slab is /C ms1

2r and /C ms22r

as shown in the figure.As the sum of the charge densities is /C mso

2r so we have s s1 2r r+ s0r= ...(1)and since the electric field intensity inside the conducting slab must be zero so, E E1 2+ 0= ...(2)where E1 is field inside slab due to charge density s1r and E2 is field inside slab due to s2r

As the electric field intensity at any point P due to the uniformly charged plane with charge density Sr is defined as

E a2 n0e

r= S

where an is the unit vector normal to the plane directed toward point P

So we have, E1 ( )a2 z0e

r= -S1 (a an z=- )



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E2 a2 z0e

r= S2 (a an z= )

From equation (ii) a a2 2s

zs

z0

1

0

2

er

er- +^ h 0=

s1r s2r=Putting the result in equation (i) we get

s1r 4ss

20r r= =

SOL 3.2.24 Option (D) is correct.As the slabs are conducting so net electric field inside the slab must be zero.and since the electric field intensity at any point P due to the uniformly charged plane with charge density Sr is defined as

E a2 n0e

r= S

where an is the unit vector normal to the plane directed toward point P

So, the net electric field intensity inside slab 1 is

a a a a2 2 2 2S

zS

zS

zS

z0

11

0

12

0

21

0

22

er

er

er

er- + + +^ h 0=

(a an z=- for S11r while a an z= for rest of the charge densities) s s s s11 12 21 22r r r r- + + + 0= ...(1)and the net electric field intensity inside slab 2 is

( ) ( ) ( )a a a a2 2 2 2S

zS

zS

zS

z0

11

0

12

0

21

0

22

er

er

er

er- + - + - + 0=

(a an z= for S22r while a an z=- for rest of the charge densities)

s s s s11 12 21 22r r r r- - - + 0= ...(2)Solving eq. (i) and eq (ii) we get, s11r s22r=and s12r s21r=-

SOL 3.2.25 Option (C) is correct.As all the four surfaces form the boundaries of the conductors extending away from




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the region between them so, the medium outside the defined region is conductor and so the field intensity outside the region will be zero.Now the electric potential in the non conducting region is given as V xy5=So the electric field intensity in the region is E Vd=- 5 5y xa ax y=- -From the conductor-free space boundary condition we have the surface charge density on the boundary surface defined as sr En0e= where En is the normal component of the electric field intensity in the free space.So, the surface charge density on the surface x 0= is sr y50e= -^ h (the normal component E y5n =- for the surface x 0= ) y5 0e=-

SOL 3.2.26 Option (D) is correct.Again as discussed in above question, the surface charge density on the surface y 0= will be given by sr En0e= and since the field component normal to surface y 0= is En x5=-So, the surface charge density on the surface y 0= is sr x5 0e=-

SOL 3.2.27 Option (C) is correct.From the symmetry associated with the charge distribution the electric field must be radially directed. Then choosing Gaussian surfaces which are cylinders having the same axis ( 0r = ) as the conductors and of length l , we get l E2pr r^ h 0= for 2 m<r(since there is no charge enclosed by the Gaussian surface)Thus Er 0= for 2 m<rNow, since the field inside the conductor 2 3 m< <r is zero; there cannot be any charge on the surface 2 mr = .i.e. rS 0= at 2 mr =and all the charge associated with the inner conductor resides on the surface

3 mr = .

i.e. rS 10 /C m2 3p

=^ h

/C m35 2

p= at 3 mr =

Proceeding further we have

2 lEpr r 10 /C m l10e

= ^ h for 5 6 m< <r

where l is length of the cylinder.

So E a210

0pe r= r for 5 6 m< <r



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For View Only Shop Online at www.nodia.co.inThis the field produced by the inner conductor but the fact is that the field inside the conductor 5 6 m< <r is zero that gives

at mS 5r

r=6 @ E aat0 5:e= -r r= ^ h6 @

( )

( )a a2 5

100

0:e pe= -r rc m /C m

2 510 1 2

p p=- =-^ h

and at m6

rr=S6 @ /C m

2 61 6 2 5

at 5pr p= -

r=S^^

hh6 @$ .

121 6 10p= +^ h 3

4p=

SOL 3.2.28 Option (C) is correct.From the symmetry associated with the charge distribution the electric field must be radially directed. As, there is no charge enclosed by the surface 2 mr = so we get Er 0= for 2 mr <Now from the conductor-free space boundary condition we have the surface charge density on the boundary surface defined as sr En0e= where En is the normal component of the electric field intensity in the free space.So the charge density at 2 mr = is S1r Er= 0=Therefore the total charge will be concentrated over the outer surface which is given as

S2r /C mrQ

4 2 39

81

2 22

p p p= = =^ h

SOL 3.2.29 Option (D) is correct.From the boundary condition for the charge carrying interface, the tangential component of electric field on either side of the surface will be same.i.e. E t1 E t2=while the normal components are related as

E En n1 2- s

0er=

now as the field intensity in the region z 0< is E2 2 3 2a a ax y z= + -So the tangential component, E t2 2 3a ax y= +and the normal component, E n2 2az=-Therefore the field components in region z 0>^ h are E t1 2 3E a at x y2= = +

and E n1 E ns

20er= +

.a2

8 85 102 10

z12

9

#

#= - + -

-

; E 224az=

So the net field intensity in the region z 0> is E1 E Et n1 1= + 2 3 224a a ax y z= + +




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SOL 3.2.30 Option (C) is correct.As the dielectric slab occupies the region x 0> and the electric field in the free space is directed along ax so, the field will be normal to the boundary surface, x 0= of the dielectric slab.So from the boundary condition the field normal to the interface of dielectrics are related as Er i0e e E0e= (where Ei is the field inside the dielectric)

Ei 10 2aE a5r

xxe= = = ( )5re =

So, the polarization inside the dielectric is P Ei0e e= -^ h E5 i0 0e e= -^ h 8 ax0e=

SOL 3.2.31 Option (D) is correct.As the dielectric slab occupies the region 0 1 my< < and the electric field in the free space is directed along ay so, the field will be normal to both the boundary surfaces y 0= and y 1= .So from the boundary condition the field normal to the interface of dielectrics are related as Eie E0e= (where Ei is the field inside the dielectric)

Ei y E4 10

0 2

ee= +^ h 4

ya4

1y

2

=+^

^h

h since ey1

42

0e=+^ h

1 y ay2= +^ h

So the polarization inside the dielectric is P E Ei i0e e= -

yE E

14

i i20

0e e=+

-^ h

y

y a14 1 y2

00

2e e=+

- +^

e ^h

o h

y a4 1 y2

0e= - +^ h8 B

Now for a given polarization P inside a dielectric material, the surface charge density over the surface of dielectric is defined as pr S aP n:= where an is the unit vector normal to the surface directed outward of the dielectric.So, the bound surface charge density at y 0= is

0atps yr

=6 @ P ay:= -^ h (an ay=- ) y4 1 12

0e= - + -^ ^h h6 @ 4 1 0e=- -^ h 3 0e=- y 0=and the surface charge density at 1 my = is

1atps yr

=6 @ P ay:= ^ h (an ay= ) y4 1 12

0e= - +^ ^h h8 B 4 4 00e= - =^ h

SOL 3.2.32 Option (C) is correct.As calculated in previous question, polarization inside the dielectric is P y a4 1 y0e= - +^ h6 @

Since for a given polarization P inside a material, the bound volume charge density inside the material is defined as



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For View Only Shop Online at www.nodia.co.in pvr P:d=-So the volume charge density inside the dielectric is

vr y

y4 1 202

2 e=- - +^ h8 B

y2 1 0e= +^ h

So when we move from y 0= to 1 my = , the volume charge density will be linearly increasing.

SOL 3.2.33 Option (C) is correct.As the charge is being located at origin so the field intensity due to it will be in radial direction and normal to the surface of the dielectric material.Therefore the flux density will be uniform(as from boundary condition) and at any point r inside the dielectric flux density will be

D rQ a

4 r2p=

Now it is given that electric field intensity at any point inside the dielectric is

E b

Q a4 r

02pe

=

and since in a medium of permittivity r 0e e e= the flux density is defined as D Er 0e e=So for the given field we have

rQ a

4 r2p

bQ a

4r r00

2e epe

= c m

re rb

2

2

=

SOL 3.2.34 Option (C) is correct.Consider the parallel sheets arrangement as shown in the figure.

Electric field intensity at any point P due to the uniformly charged plane with charge density Sr is defined as

E a2S

ner=

where an is the unit vector normal to the plane directed toward point P and e is the permittivity of the medium.




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So the field intensity inside the dielectric due to the left sheet will be

E1 a25 10

y

9#

e=-

^ h a an y=^ h

and again the field intensity inside the dielectric due to right sheet will be

E2 a25 10

y

9#e= - -

-

^ h a25 10

y

9#

e=-

a an y=-^ h

so the net field intensity inside the dielectric will be

E E E a5 10y1 2

9#e= + =

-

Since the field intensity is uniform inside the dielectric So potential difference between the plates will be directly given as V E#= (distance between the plates)

45 10 2

0

9#

#e=-

2.824 10 Volt2#= 283 kV= 4 0e e=^ h

Option (C) is correct.As calculated in previous question the electric field between the two dielectrics having surface charge densities Sr and Sr- is

E Ser=

where e is the permittivity of the medium between the sheets.

So electric field in slab 1 is E1 2S S

0er

er= =

and electric field in slab 2 is E2 4S S

0er

er= =

Since the electric field between the sheets is uniform so the potential difference between the plates will be V tandis ceE#= ^ h/ 1 2m mE E1 2= +^ ^h h

( )2 1 4 20 0e

rer= +S S ^ h s

0er=

..

8 85 100 6 10

12

9

#

#= -

-

7.8 Volt5=

SOL 3.2.35 Option (C) is correct.As calculated in previous question the electric field between the two dielectrics having surface charge densities Sr and Sr- is

E Ser=

where e is the permittivity of the medium between the sheets.

So electric field in slab 1 is E1 2S S

0er

er= =

and electric field in slab 2 is E2 4S S

0er

er= =

Since the electric field between the sheets is uniform so the potential difference



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For View Only Shop Online at www.nodia.co.inbetween the plates will be V tandis ceE#= ^ h/ 1 2m mE E1 2= +^ ^h h

( )2 1 4 20 0e

rer= +S S ^ h s

0er=

..

8 85 100 6 10

12

9

#

#= -

- 6 .8 Volt4=

SOL 3.2.36 Option (A) is correct.The electric field between the plates carrying charge densities r+ S0 and r- S0 is defined as

E er= S0

where e is the permittivity of the medium between the plates.Now consider that near the plate 1 permittivity is 1e and near the plate 2, permittivity is 2e . So at any distance x from plate 1 permittivity is given by

e d x12 1e e e= + -

a k (Since the permittivity is linearly increasing)

So the field intensity at any point in the medium will be

E

d x12 1e e er=

+ -S0

a k

Therefore the potential difference between the plates will be

V xdx

2

d

12 10 e e er=

+ -S0

a k# ln

dd x

d

2 11

2 1

0

e er e e e= - + -S0

aa

kk:

>DH

lnd2 1 1

2e er

ee= -

S0a k

SOL 3.2.37 Option (A) is correct.Assume that the surface charge densities on the plates is !r 0S so the electric field intensity between the plates will be

E 0e

r= S0

and the potential difference between the plates will be given by V E#= (Distance between plates)

5 103# .0 5 10

0

2#e

r#= -S0

b ^l h

Therefore the surface charge density is

rS0 ..

8.85 C0 5 10

8 85 10 5 102

12 3

#

# # # m= =-

-

^

^ ^

h

h h

SOL 3.2.38 Option (C) is correct.The capacitor of a parallel plate capacitor is defined as

C dSe=

So, the capacitance in 1st dielectric region will be

C1 3S S

1 101e e= =

and the capacitance in 2nd dielectric region




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C2 S S3 3

22 0e e= =

Therefore the voltage drop in 1st dielectric region is

V1 VC CC

1 2

2= + (where V is total voltage drop)

3

/Volt

S SS2 3

90 3

20

0

e ee

=+

^^

hh Volt11

18=

and similarly, V2 VC CC

1 2

1= + SS

3 23 9S

0 0 3

0

e ee=+

^ h Volt1182=

SOL 3.2.39 Option (C) is correct.Consider the dielectric slab is of thickness t and d1, d2 are the remaining width in the medium as shown in the figure.

Now the capacitance of the whole configuration will be considered as the three capacitors (capacitance in the three regions) connected in series as shown in the figure

So, C1 dS1

1e= , C2 tS2e= and C3 d

S2

1e=

The equivalent capacitance, is defined as

C1eq

C C C1 1 1

1 2 3= + + S

tS

d d1 2

e e= ++^ h

Since t ; d d1 2+^ h will be constant although if the dielectric slab is moved leftward or rightward so the equivalent capacitance will be constant. But if the slab is pulled outward then the capacitance will change as the effective surface area of the capacitance due to dielectric slab changes.

SOL 3.2.40 Option (B) is correct.Since, the wire is coated with aluminum So,the configuration can be treated as the



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For View Only Shop Online at www.nodia.co.intwo resistance connected in parallel and therefore, the field potential will be same across both the material or we can say that the field intensity will be same inside both material.i.e. Est Eal=where Est " Field intensity insteel Eal " Field intensity in aluminum.

or, Jst

st

s Jal

al

s=

where Jst " current density in steel Jal " current density in aluminum sts " conductivity of steel als " conductivity of aluminum

So, we get, JJal

st .3 8 102 10

191

al

st7

6

#

#ss= = =

Jal J19 st= ...(1)Now, the total current through the wire is given as, I J a J b ast al

2 2 2p p p= + -^ ^h h

where a " cross sectional radius of inner surface (steel wire) b " cross sectional radius of outer surface (with coating)Since, thickness of coating is t 2 10 3

#= -

So, b a t 2 10 2 10 4 103 3 3# # #= + = + =- - -

^ ^ ^h h h

Therefore, we get, 80 J J4 10 16 10 4 10st al

6 6 6# # #p p p= + -- - -

^ ^ ^h h h6 @

or, 80 J J4 10 19 12 10st st6 6

# #p p= +- -^ ^h h6 @ (from eq.(1))

So, Jst 232 1080

6#p

= - 1.10 10 /A m5 2#=

SOL 3.2.41 Option (A) is correct.Given, the potential field in free space

V cos sinr40

3 q f=

So, the potential at point P (r 2= , 3q p= , 2f p= ) is given as

VP cos sin240

3 23p p=

^a a

hk k 2.5 Volt=

Now, as the conducting surface is equipotential, so, the potential at any point on the conducting surface will be equal to the potential at point P .i.e. V 2.5 VoltVP= =

or cos sinr40

3 q f .2 5=

cos sin16 q f r3= This is the equation of the conducting surface.

***********




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SOLUTIONS 3.3

SOL 3.3.1 Option (B) is correct.The capacitance of a parallel plate capacitor is defined as

C .dA

108 85 10 10o

3

12 4# #e= = -

- - 8.85 10 13

#= -

The charge stored on the capacitor is Q CV= 8.85 10 4.427 1013 13

# #= =- -

Therefore, the displacement current in one cycle

I TQ fQ= = 4.427 10 3.6 10 .59213 9

# # #= =- mA ( 3.6 GHzf = )

SOL 3.3.2 Option (A) is correct.The electric field of the EM wave in medium 1 is given as E1 2 3 1a a ax y z= - +Since the interface lies in the x 0= plane so, the tangential and normal components of the field intensity in medium 1 are E t1 3a ay y=- + and 2E an x1 =From the boundary condition, tangential component of electric field is uniform. So, we get the tangential component of the field intensity in medium 2 as E t2 3E a at y y1= =- +Again from the boundary condition the for normal component of electric flux density are uniformi.e. D n1 D n2=or E n1 1e E n2 2e=So, we get 1.5 2ao xe 2.5 Eo n2e=

or E n2 . 1.2a a2 53

x x= =

Thus, the net electric field intensity in the medium 2 is E2 E Et n2 2= + 3 1.2a a ay z x=- + +

SOL 3.3.3 Option (B) is correct.The surface charge density on a conductor is equal to the electric flux density at its boundary.i.e. s D= E E80 0e e= = ( 80 oe e= ) 80 8.854 10 212

# # #= - 1.41 10 9#= - C/m2



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The configuration shown in the figure can be considered as the three capacitors connected in parallel as shown below

Now, consider the distance between the two plate is d and the total surface area of the plates is S . So, for the three individual capacitors the surface area is /S 3 and the separation is d . Therefore, we get,

C1 /dS 30 1e e

= ^ h

C2 /dS 30 2e e

= ^ h

C3 /dS 30 3e e

= ^ h

Since, the three capacitance are in parallel So, the equivalent capacitance is Ceq C C C1 2 3= + +

/ / /dS

dS

dS3 3 30 1 0 2 0 3e e e e e e

= + +^ ^ ^h h h

dS

30 1 2 3e e e e= + +

b cl m C35 21 2 3e e e= + +

b l C dS0e=b l

SOL 3.3.5 Option (B) is correct.The electric field is equal to the negative gradient of electric potential at the point.i.e. E Vd=-Given, electric potential V x4 2= +So, E 4 /V max=-

SOL 3.3.6 Option (D) is correct.The angle formed by the electric field vector in two mediums are related as

tantan

2

1aa

2

1ee=

So, for the given field vectors we have,

tantan60

2

ca

33=

tan 2a 1=or 2a 1tan 1= -

^ h 45c=




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SOL 3.3.7 Option (C) is correct.The tangential component of electric field on conducting surface is zero (since the surface conducts current) So, under static condition we have, Vd- 0E= =or V = constanti.e. the conducting surface is equipotential.So, (A) and (R) both true and R is correct explanation of A.

SOL 3.3.8 Option (C) is correct.Since, the electric field is incident normal to the slab. So, the electric field intensityEi^ h inside the slab is given as

Eie E0 0e=

Ei 6a

3x

0

0

ee

= ^ h 2ax=

Therefore, the polarization inside the slab is given as Pi EXe i0e=where Xe is electric susceptibility defined as 1Xe re= - . So, we have Pi E3 1 i0e= -^ h 4 ax0e=

SOL 3.3.9 Option (C) is correct.Due to both the line charge and concentric circular conductors, the equipotential surfaces are circular (cylinder) i.e. concentric equipotential lines.The flux lines due to both the configurations (line charge and concentric circular conductors) are in straight radial direction.

SOL 3.3.10 Option (C) is correct.Capacitance of 1st plate is given as

C1 dS

d d1 11 #e e e= = =^ h

The capacitance of 2nd plate is

C2 dS

d d3 2 62 #e e e= = =^ h

So, the ratio of capacitances is

CC

1

2 4=

SOL 3.3.11 Option (B) is correct.Consider the dielectric material with permittivity 1e is replaced by a dielectric material with permittivity 2e .The capacitance of parallel plate capacitor is defined as

C dSe=

i.e. the capacitance depends on the permittivity of the medium and so, due to the replacement of the material between the plates the capacitance changes.Now, the charge is kept constant



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For View Only Shop Online at www.nodia.co.ini.e. Q1 Q2=or, C V1 1 C V2 2=So, due to the change in capacitance voltage on the capacitor changes and therefore the electric field intensity between the plates changes.The stored energy in the capacitance is defined as

W CQ2

2

=

As total stored charge Q is kept constant while capacitance changes so, the stored energy in the capacitance also changes.Thus, all the three given quantities changes due to the replacement of material between the plates.

SOL 3.3.12 Option (A) is correct.According to continuity equation we have

J:d tv

22r=-

As for electrostatic field t

0v

22r = so, we get

J:d 0=

SOL 3.3.13 Option (C) is correct.Electrostatic fields only.

SOL 3.3.14 Option (A) is correct.Surface or sheet resistivity is defined as resistance per unit surface area. So, the unit of surface resistivity is Ohm/sq. meter.

SOL 3.3.15 Option (D) is correct.Since a conducting surface is equipotential so no electric field component exists tangential to the surface and therefore the electric field lines are normal to a conducting surface boundary.

SOL 3.3.16 Option (D) is correct.Surface resistance of a metal is defined as

Rs f

2 22

. swm

sp m=

So, as frequency ( f ) increases the surface resistance increases.

SOL 3.3.17 Option (C) is correct.When we determine force using method of images then in this method, the conducting surface is being removed and an additional distribution of charge is being introduced symmetrical to the existing charge distribution.

SOL 3.3.18 Option (D) is correct.The conducting surface is equipotential and since the potential at infinity is zero so, the potential every where on a conducting surface of infinite extent is zero.Since the conducting surface is equipotential so displacement density on a conducting




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surface is normal to the surface.So A and R both true but R is not correct explanation of A.

SOL 3.3.19 Option (C) is correct.Capacitance, C 5 5 10pF F2

#= = -

Charge on capacitance, Q 0.1 0.1 10C C6m #= = -

The energy stored in the capacitor is defined as

W CQ2

2

= .

1 mJ2 5 100 1 10

12

6 2

# #

#= =-

-^ h

SOL 3.3.20 Option (A) is correct.Consider the charge of 1C is placed near a grounded conducting plate at a distance of 1 m as shown in figure.

Using image of the charge we have one negative charge opposite side of the plate at the same distance as shown in the figure and the force between them is

F Nr4

1 14 2

116

10

20

20pe pe pe=

-= - = -^ ^

^

h h

h

Negative sign indicates that the direction of force is attractive.

SOL 3.3.21 Option (D) is correct.Fringing field has been shown below in the figure

The capacitance of a parallel plate capacitor is given as



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For View Only Shop Online at www.nodia.co.in C d

Ar0e e=

It is valid only when the fringing is not taken into account. Now, the fringing field can be ignored only when the separation d between the plates is much less than the plate dimensions. So, for the fringing field taken under consideration, /A d is tending towards infinity.

SOL 3.3.22 Option (D) is correct.The capacitance of a solid infinitely conducting sphere is defined as C 4 R0pe= .where R is radius of the solid sphere.

SOL 3.3.23 Option (C) is correct.The electric potential produced by a point charge Q at the a distance r from it is defined as

V rQ

4pe=

where e is permittivity of the medium. So, the electric potential produced by the point charge 10 Cm+ at the centre of the sphere is

V rQ

4 4 5 1010 10

o 02

6

#

#pe pe

= = -

-

^ h (Given 5 cmr = )

As the surface of sphere is grounded so, the total voltage on the spherical capacitor will be equal to the potential at its centre as calculated above.Now, the capacitance of the isolated sphere is defined as C a4pe=where a is the radius of the sphere. Therefore, the induced charge stored on the sphere is given as

Qind CV= a44 5 10

10 100

02

6

#

#pepe

= -

-

^^

^h

h

h

5 10

2 10 10 102

2 6

#

# # #= -

- -

^

^ ^

h

h h (Given 2 cma = )

5 5C C10 6# m= =-

SOL 3.3.24 Option (A) is correct.Given electric field E sin tE0 w=The conduction current is defined as Jc sin tE E0s s w= =^ h

where s is conductivity and E is electric field intensity.and the displacement current density is

Jd tD

tE

22

22e= =

cos tE0e w w= ^ h sin tE 20ew p w= -a k

So the phase difference between Jc and Jd is 90c.




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SOL 3.3.25 Option (D) is correct.Method of images are used for the charge distribution at a distance from the grounded plane conductor.

SOL 3.3.26 Option (A) is correct.GivenCapacitance of condenser, C 0.005 5 10F F9m #= = -

Supply voltage, V 500 V=Permittivity of oil, re .2 5= (immersed oil)So, the energy stored in condenser before immersion is

W CV21 2= 2

1 5 10 5009 2# # #= -

^ h

6.25 10 J4#= -

After immersing the condenser in oil the capacitance changes while the total charge remains same.i.e. Qafter immersion Qbefore immersion= 5 10 5009

#= -^ ^h h

2.5 10 Coulomb6#= -

The capacitance of the condenser after immersion is Cafter immersion Cre= .2 5 5 10 9

#= -^ ^h h . F1 25 10 8

#= -

Therefore, the stored energy in the condenser immersed in oil is

, W CQ

2 ( )after immersion

2

= ..

2.5 10 J2 1 25 10

2 5 108

64

#

##= =-

--

^

^

h

h

SOL 3.3.27 Option (D) is correct.Given,Capacitance, C 3 3 10F F6m #= = -

Current, I 2 2 10A A6m #= = -

Charging time, t sec6=So, the total charge stored on capacitor is Q = Charge transferred It= 2 10 66

#= -^ ^h h

Therefore, the voltage across the charged capacitor is

V CQ

4 102 10 8

6

6

#

#= = -

-^ ^h h

Volt5=

SOL 3.3.28 Option (A) is correct.Given, the total charge on capacitor V=(1) Electric field between the plates will be given as

E Vd=- which is independent of permittivity of the material filled in capacitor so E

will be constant.

(2) The displacement flux density inside the capacitor is given as



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For View Only Shop Online at www.nodia.co.in D Ee= As E is constant while permittivity is doubled so D will also be doubled.

(3) The charge stored on the plates is given as

Q CV= where V is constant but capacitance C will be doubled as it is directly

proportional to the permittivity given as

C dSe=

So, the charge on plates will be get doubled.

(4) As discussed already, the capacitance will get doubled.

Therefore, the statements 2 and 4 are correct.

SOL 3.3.29 Option (D) is correct.Since, resistance doesn’t store any energy. So, the energy stored in the coil is only due to inductance and given as

W LI21 2=

where L is the inductance and I is the current flowing in the circuit. At the fully charged condition, inductor is short circuit and therefore, current through the circuit is

I 10 ARV

550= = =

So, the energy stored in the field (in the inductor) is

W . 0 Joules21 0 6 10 32= =^ ^h h

SOL 3.3.30 Option (A) is correct.The normal component of electric flux density D^ h across a dielectric-dielectric boundary is given as D Dn n1 2- sr=where sr is the surface charge density at the interface.So, the normal components of electric flux density across a dielectric-dielectric boundary is dependent on the magnitude of surface charge density.

SOL 3.3.31 Option (A) is correct.Statement 1, 2 and 4 are correct while statement 3 is incorrect.

SOL 3.3.32 Option (D) is correct.The capacitance of an insulated conducting sphere of radius R in vacuum is C R4 0pe=

SOL 3.3.33 Option (A) is correct.Maximum withstand voltage is the value that the dielectric between capacitor plates can toterate without any electrical breakdown. Maximum withstand voltage is larger for any dielectric material than that for free space (air).Since the maximum withstand voltage across the capacitor filled with air is V so




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the maximum withstand withstand voltage for the composite capacitor will be also V as the capacitors are connected in parallel. Now, the capacitance before filling the dielectric is

C dA0e=

and after filling the dielectric Ceq C C1 2= +

/ /dA

dA

42 2

0 0e e= + dA

25 0e=

So, the stored charge Q1 after filling dielectric is determined as below

QQ

1 CCeq

= (Since voltage is constant)

or, Q1 .

dA

Q dA

Q25

2 50

0

e

e= =

Therefore, the maximum withstand voltage of the capacitor is V and charge is 2. Q5 .

SOL 3.3.34 Option (B) is correct.Since, the potential on both sides of plate will be same (Consider the potential is V ). So, the charge densities on the two sides is determined as below : 2r C V2= l

and 1r C V1= l

where C 2l and C 1l are the capacitance per unit area of the capacitance formed by the region d1 and d2.

Therefore, 2

1

rr

C VC V

2

1=ll

d V

d V

2

0

1

0

e

e= d

d1

2=

SOL 3.3.35 Option (A) is correct.Electric flux density in a polarized dielectric is defined as D P E0e= +

SOL 3.3.36 Option (C) is correct.Image theory is applicable only for static charge distribution (electrostatic field).

SOL 3.3.37 Option (B) is correct.The equivalent capacitance of series connected capacitance has the value less than the smallest capacitance here the smallest capacitance is C6 so the total capacitance is less then C6

i.e. Ceq C< 6

or Ceq C6.



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Given, the electric field intensity in medium 1. E1 5 2 3a a ax y z= - +Since, the medium interface lies in plane z 0= .So, we get the field components as E t1 5 2a ax y= -and E n1 3az=Now, From the boundary condition for electric field we have E t1 E t2= E n1 1e E n2 2e=So, the field components in medium 2 are E t2 5 2E a at x y1= = -

E n2 6E an z2

11e

e= =

Therefore, the net electric field intensity in medium 2 is given as E2 E Et n2 2= + 5 2 6a a ax y z= - +So, the z -component of the field intensity in medium 2 is E z2 6az=

SOL 3.3.39 Option (C) is correct.Electric flux density, D 1 /C m2=Relative permittivity, re 5=Since, the normal component of flux density is uniform at the boundary surface of two medium so, the flux density inside the slab is D 1 /C m2=Therefore, the polarization of the slab is given as

P D1r

r

ee= -

b l 184#= .0 8=

SOL 3.3.40 Option (D) is correct.The capacitance of a isolated spherical capacitor of radius R is defined as C R4 0pe=Since the two spheres are identical and separated by a distance very much larger then R . So, it can be assumed as the series combination of capacitances. Therefore, the net capacitance between two spheres is given as

i.e. C C CC C

R RR R

4 44 4

1 2

1 2

0 0

0 0

pe pepe pe

= + = +^ ^h h

R2 0pe=

SOL 3.3.41 Option (A) is correct.For steady current in an arbitrary conductor the current density is given as

J AI=

and since I is constant So, J is constant and therefore 0J#d =So, the current density is solenoidal. i.e. Assertion (A) is true.The reciprocal of resistivity is conductivity. i.e. Reason (R) is false.




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SOL 3.3.42 Option (C) is correct.Since, the displacement current density is defined as

Jd tD22=

So, it is generated by a change in electric flux and therefore the displacement current has only A.C. components as derivative of D.C. components is zero.i.e. A and R both are true and R is correct explanation of A.

SOL 3.3.43 Option (D) is correct.Dielectric constant, re 5=Flux density, D 2 /C m2=So, the polarization of the medium is given as

P D1r

r

ee= -

b l 54 2#= .6 /C m2 2=

SOL 3.3.44 Option (A) is correct.The ohm’s law in point form in field theory is expressed as below V RI= (For constant voltage)

lE Al AJr=

where l is length integral and A is the cross sectional area. So, we get E Jr=

E Js=

i.e. J Es=

SOL 3.3.45 Option (A) is correct.Displacement current density is defined as

Jd tE

22e=

and the conduction current density is defined as Jc Es=for a dielectric e must be larger while conductivity must tend to zero.So, we get Jd J>> c

i.e. displacement current is much greater than conduction current.

SOL 3.3.46 Option (D) is correct.Conduction current, Ic 1 A=Operating frequency, f 50 Hz=Medium permittivity, e 0e=Permeability m 0m=Conductivity, s 5.8 10 /mho m#=The ratio of conduction current density to the displacement current density is

JJd

c wes=

or, //I AI Ad

c wes= (A is cross sectional area)



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Id . (1)I 5 8 102 50

c0

## #

swe p e= = . 10 A3 3 11

#= -

SOL 3.3.47 Option (B) is correct.When there is no charge in the interior of a conductor, the electric field intensity is zero according to Gauss’s law the total outward flux through a closed surface is equal to the charge enclosed.Now if any charge is introduced inside a closed conducting surface then an electric field will be setup and the field exerting a force on the charges and making them move to the conducting surface. So all the charges inside a conductor is distributed over its surface. Therefore the outward flux through any closed surface constructed inside the conductor must vanish.A is false but R is true.

SOL 3.3.48 Option (B) is correct.When the method of images is used for a system consisting of a point charge between two semi infinite conducting planes inclined at an angle f, the no. of images is given by

N 360 1cf= -c m

Here the angle between conducting planes is 90cf = .So, N 3=and since all the images lie an a circle so we have the image charges as shown in figure.

SOL 3.3.49 Option (D) is correct.Consider the two dielectric regions as shown below.




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Since the field is normal to the interface So, the normal components of the fields are, E n1 1= and E 2n2 =From boundary condition we have E En n1 1 2 2e e- sr=(where sr is surface charge density on the interface). 1 2 20 0e e-^ ^ ^ ^h h h h sr= sr 3 0e=-

SOL 3.3.50 Option (D) is correct.The stress is called the force per unit area which is directly proportional to the electric field intensity and electric field intensity is inversely proportional to the permittivity of dielectric material.

i.e. E 1\ e

So, ratio of stress is EE

2

1 //

11 0

ee=

//

1 51

0

0

ee= 5=

***********

CHAPTER 4MAGNESTOSTATIC FIELDS

202 Magnestostatic Fields Chap 4



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EXERCISE 4.1

MCQ 4.1.1 Assertion (A) : For a static magnetic field the total number of flux lines entering a given region is equal to the total no. of flux lines leaving the region.Reason (R) : An isolated magnetic charge doesn’t exist.(A) Both A and R one true and R is the correct explanation of A.

(B) Both A and R are true but R is not the correct explanation of A.



MCQ 4.1.2 Match List-I with List-II and select the correct answer using the codes given below. (Notations have their usual meaning).

List-I List-II

a. Ampere’s law 1. D v: rd =

b. Conservative nature of magnetic field

2. d dJ S H lLS

: := ##

c. Gauss’s law 3. 0B:d =

d. Non existence of magnetic monopole

4. dE l 0: =#

Codes : a b c d(A) 3 1 4 2(B) 2 1 4 3(C) 2 4 1 3(D) 3 4 2 1

MCQ 4.1.3 Magnetic field intensity H exists inside a certain closed spherical surface. The value of H:d will be(A) 0 at each point inside the sphere.

(B) 0 at the center of the sphere only.

(C) 0 at the outer surface of the sphere only.

(D) Can’t be determined as H is not given.

Chap 4 Magnestostatic Fields 203


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For View Only Shop Online at www.nodia.co.inMCQ 4.1.4 The source which doesn’t cause a magnetic field is

(A) A charged disk rotating at uniform speed

(B) An accelerated charge

(C) A charged sphere spinning along it’s axis

(D) A permanent magnet

MCQ 4.1.5 A circular loop of radius a , centered at origin and lying in the xy plane, carries current I as shown in the figure.

The magnetic field intensity a the centre of the loop will be

(A) aI a2 z (B) a

I a2 z-

(C) aI a4 z (D) a

I a2z

MCQ 4.1.6 A conducting filament carries a current 5 A from origin to a point , ,3 0 4^ h. Magnetic field intensity at point (3, 4, 0) due to the filament current will be(A) 0.23 /wb ma 2

f (B) 0.095 /wb ma 2f

(C) 0.074 /wb ma 2f (D) 0.074 /wb maz 2

MCQ 4.1.7 A circular conducting loop of radius 2 m, centered at origin in the plane z 0= carries a current of 4 A in the af direction. What will be the magnetic field intensity at origin ?

(A) /A ma21

zp (B) /A maz

(C) 2 /A maz (D) /A maz-

MCQ 4.1.8 The correct configuration that represents magnetic flux lines of a magnetic dipole is




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MCQ 4.1.9 The correct configuration that represents current I and magnetic field intensity H is

MCQ 4.1.10 A long straight wire placed along z -axis carries a current of AI 5= in the az+ direction. The magnetic flux density at a distance 5 cmr = from the wire will be

(A) 4 10 /wb m5 2#

- (B) 2 10 /wb m5 2#

-

(C) /wb m100 2

p (D) 2 10 /wb m6 2#

-

MCQ 4.1.11 For the currents and the closed path shown in the figure what will be the value of

dH l:# ?



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For View Only Shop Online at www.nodia.co.in(A) 30 A (B) 20 A

(C) 20 A- (D) 10 A

MCQ 4.1.12 Two infinitely long wires separated by a distance m2 , carry currents I in opposite direction as shown in the figure.

If 8 AI = , then the magnetic field intensity at point P is

(A) a5yp (B) a5

yp-

(C) a81

yp (D) a81

yp-

MCQ 4.1.13 In the free space a semicircular loop of radius a carries a current I . What will be the magnitude of magnetic field intensity at the centre of the loop ?

(A) aI (B) a

I2

(C) aI4 (D) a

I4

Common Data for Question 14 - 15 :A long cylindrical wire of cross sectional radius R carries a steady current I distributed over its outer surface.

MCQ 4.1.14 Magnetic field intensity inside the wire at a distance ( )r R< from it’s center axes will be(A) non uniform

(B) zero

(C) uniform and depends on r only

(D) uniform and depends on both r and R

MCQ 4.1.15 The magnetic flux density outside the wire at a distance ( )r R< from it’s center axes will be proportional to(A) r (B) /r1

(C) /r R (D) /R1




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MCQ 4.1.16 Two point charges Q1 and Q2 are located at , ,0 0 0^ h and , ,1 1 1^ h respectively. A current of 16 A flows from the point charge Q1 to Q2 along a straight wire connected

between them. What will be the value of dB l:# around the closed path formed by the triangle having the vertices , ,1 0 0^ h, , ,0 1 0^ h and , ,0 0 1^ h ?

(A) 22 /Wb m0m (B) 6 /Wb m02m

(C) 14 /Wb m0m (D) 6 /Wb m0m

Common Data for Question 17 - 18An infinite current sheet with uniform current density /A mK a15 x= is located in the plane z 2= .

MCQ 4.1.17 Magnetic field intensity at origin will be(A) 1 /A ma1 y (B) 10 /A may-

(C) 0 /A m (D) 20 /A may

MCQ 4.1.18 Magnetic field intensity at point (2, 1- , 5) will be(A) 10 /A may (B) 10 /A may-

(C) 0 /A m (D) 20 /A may

MCQ 4.1.19 Two infinite current carrying sheets are placed parallel to each other in free space such that they carry current in the opposite direction with the same surface current density. The magnetic flux density in the space between the sheets will be(A) zero

(B) constant

(C) linearly increasing from one sheet to other

(D) none of these

MCQ 4.1.20 In a spherical co-ordinate system magnetic vector potential at point ( , , )r q f is given as 12cosA aq= q . The magnetic flux density at point ( , , )3 0 p will be(A) 4af (B) 0

(C) 4aq (D) 36af

MCQ 4.1.21 An infinite plane current sheet lying in the plane y 0= carries a linear current density /A mKK az= . The magnetic field intensity above (y 0> ) and below y 0<^ h the plane will be

y 0> y 0<

(A) K a2 x K a2 x-

(B) K a2 x- K a2 x



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For View Only Shop Online at www.nodia.co.in(C) 2Kax- 2Kax

(D) K a2 y- K a2 y

MCQ 4.1.22 In the free space two cylindrical surfaces 0. cm3r = and 0.25 cmr = carries the uniform surface current densities 2 /A maz and 0.8 /A maz- respectively and a current filament on the entire z -axis carries a current of 14 mA in the az+ direction. What will be the surface current density on the cylindrical surface at 8 cmr = . which will make the net magnetic field 0H = for 8 cm>r will be

(A) 0.13 /A maz- (B) 0.13 /A maz+

(C) 64.3 /mA maz (D) 0.10 /A maz-

Statement for Linked Question 23 - 24 :In a cartesian system, vector magnetic potential at a point ( , , )x y z is defined as A 2 /wb mx y y x xyza a a4 3x y z

2 2= + -

MCQ 4.1.23 The magnetic flux density at point ( , , )1 2 5- - will be

(A) 40 6 /wb ma ax z2+ (B) 40 80 6 /wb ma a ax y z

2+ +

(C) 40 80 6 /wb ma a ax y z2- - - (D) 80 6 /wb ma ay z

2-

MCQ 4.1.24 The total magnetic flux through the surface z 4= , x0 1# # , y1 4# #- will be

(A) 20 wb (B) 10/3 wb-

(C) 40 wb (D) 130/3 wb

Statement for Linked Question 25 - 26 :An infinite current sheet with uniform surface current density /A mK a8 x= is located at z 0= as shown in figure.




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MCQ 4.1.25 Magnetic flux density at any point above the current sheet ( )z 0> will be(A) 2 /wb may0

2m- (B) 2 /wb may02m

(C) /wb ma2 y0 2m (D) /wb may0

2m-

MCQ 4.1.26 The vector magnetic potential at z 2=- will be(A) 4 /wb max0m

(B) 4 /wb may0m-

(C) 2 /wb max0m

(D) 4 /wb max0m-

MCQ 4.1.27 In the free space, magnetic field intensity at any point ( , , )zr f is given by 2 /A maH 2r= f . The current density at 2 mr = will be

(A) 12 /A maz (B) 24 /A maz(C) 4 /A maz (D) 0

MCQ 4.1.28 The current density that would produce the magnetic vector potential 2A a= f in cylindrical coordinates is

(A) a10

2m r f (B) a20

2

mr

f

(C) a20m r f (D) a2

02m r f

MCQ 4.1.29 Magnetic field intensity produced due to a current source is given as

H cosz ay z ea a2 4yy

x= + +^ ^h h

The current density over the xz plane will be(A) /A ma a ax y z

2- -^ h

(B) a a ax y z- + -

(C) 2 2a a ax y z- + -

(D) a a ax y z+ +

MCQ 4.1.30 Assertion (A) : In a source free region, magnetic field intensity can be expressed as a gradient of scalar function.Reason (R) : Current density for a given magnetic field intensity is defined as J H#d=(A) A and R both are true and R is correct explanation of A.

(B) A and R both are true and R is not the correct explanation of A.


(D) R is true but A is false.



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For View Only Shop Online at www.nodia.co.inMCQ 4.1.31 Magnetic field intensity produced at a distance r from an infinite cylindrical wire

located along entire z -axis is 3 /A mar f . The current density within the conductor will be(A) 6 /A maz 2r

(B) 3 /A maz 2

(C) 6 /A maz 2

(D) 3 /A maz 2r

MCQ 4.1.32 An electron beam of radius a travelling in az direction, the current density is given as

J 2 a a1 zr= -a k For a<r

The magnetic field intensity at the surface of the beam will be

(A) a a3 f (B) a a6 f

(C) a a32 2p

f (D) a a32

f

MCQ 4.1.33 In a certain region consider the magnetic vector potential is A and the current density is J . Which of the following is the correct relation between J and A ?(A) A Jd = (B) A J2

0md =

(C) A Jm#d =- (D) A J20md =-

MCQ 4.1.34 A circular loop of wire with radius .5 mR 1= is located in plane x 0= , centered at origin. If the loop carries a current 7 AI = flowing in clockwise as viewed from negative x -axis then, its magnetic dipole moment will be(A) 0.12 A max 2

-

(B) 5.5 A max 2--

(C) 5.5 A max 2-

(D) 22 A max 2-

***********




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EXERCISE 4.2

MCQ 4.2.1 In the free space, the positive z -axis carries a filamentary current of 10 A in the az- direction. Magnetic field intensity at a point (0, , )3 2 due to the filamentary

current will be(A) 0.73 /A max- (B) 1.46 /A max(C) 0.40 /A max (D) 0.73 /A max

MCQ 4.2.2 If there is a current filament on the x -axis carrying 4.4 A in ax direction then what will be the magnetic field intensity at point (4, 2, 3) ?(A) 0.1( 2 ) /A ma az y- (B) 1.76 1.62 /A ma az y-

(C) ( 1.077 1.62 ) /A ma az y- + (D) 0.1(2 )a az y- -

MCQ 4.2.3 A filamentary conductor is formed into an equilateral triangle of side 2 m that carries a current of 4 A as shown in figure. The magnetic field intensity at the center of the triangle will be

(A) /A ma9zp (B) /A ma3

zp

(C) /A ma6zp (D) 0

MCQ 4.2.4 A current sheet 4 /A mK ay= flows in the region z2 2< <- in the plane x 0= . Magnetic field intensity at point ( , , )P 3 0 0 due to the current sheet will be(A) 1.5 /A maz- (B) 0.75 /A maz-

(C) 0.75 /A maz+ (D) 2.1 /A maz-



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For View Only Shop Online at www.nodia.co.inMCQ 4.2.5 In the plane z 0= a disk of radius m3 , centered at origin carries a uniform

surface charge density 2 /C m2r =S . If the disk rotates about the z -axis at an angular velocity 2 /rad sw = then the magnetic field intensity at the point ( , , )P 0 0 1 will be(A) /A maz (B) 2 /A maz(C) /A may (D) 2 /A may

MCQ 4.2.6 A square conducting loop of side 1 m carries a steady current of 2 A. Magnetic flux density at the center of the square loop will be. (A) 17.78 10 /Wb m7 2

#- (B) 0.45 /Wb m2

(C) 2.26 10 /Wb m6 2#

- (D) 4 10 /Wb m7 2#

-

MCQ 4.2.7 A filamentary conductor is formed into a loop ABCD as shown in figure. If it carries a current of .2 A5 then the magnetic field intensity at point P will be

(A) 0.2 /A m (B) 0.8 /A m

(C) 0.26 /A m (D) 1.01 /A m

MCQ 4.2.8 The magnetic field intensity at point P due to the steady current configurations shown in figure will be

(A) 0.82 A m (B)0.32 A m

(C) 0.5 A m (D) 0.18 A m

MCQ 4.2.9 In the plane 5 mz = a thin ring of radius, 3 ma = is placed such that z -axis passes through it’s center. If the ring carries a current of 50 mA in af direction then the magnetic field intensity at point (0, 0, 1) will be(A) 0.9 /mA maz (B) 1.8 /mA maz(C) 0.6 /mA maz (D) 0.5 /A maz




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MCQ 4.2.10 An infinite solenoid (infinite in both direction) consists of 1000 turns per unit length wrapped around a cylindrical tube. If the solenoid carries a current of 4 mA then the magnetic field intensity at its axis will be(A) 4 /A m (B) 0

(C) 2000 /A m (D) 0.2 /A m

Common Data for Question 11 - 13 :The two long coaxial solenoids of radius a and b carry current mAI 6= but in opposite directions. Solenoids are placed along y -axis as shown in figure. The inner solenoid has 2000 turns per unit length and outer solenoid has 1000 turns per unit length.

MCQ 4.2.11 Magnetic field intensity inside the inner solenoid will be

(A) 3 /A may- (B) 3 /A may+

(C) 6 /A may- (D) 6 /A may+

MCQ 4.2.12 The magnetic field intensity in the region between the two solenoids will be(A) 3 /A may (B) 0

(C) 6 /A may (D) 3 /A may-

MCQ 4.2.13 The magnetic field outside the outer solenoid will be(A) 6 /A may (B) 3 /A may-

(C) 0 (D) 3 /A may+

Statement for Linked Question 14 - 15 :In a cartesian system two parallel current sheets of surface current density

3 /A mK az1 = and 3 /A mK az2 =- are located at 2 mx = and 2 mx =- respectively. The net vector and scalar potential due to the sheets are zero at a point ( , , )P 1 2 5 .

MCQ 4.2.14 Consider the scalar potential at any point ( , , )x y z in the region between the two



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For View Only Shop Online at www.nodia.co.inplanar sheets is Vm . The plot of Vm versus y will be

MCQ 4.2.15 The vector potential at origin will be(A) 3 /Wb maz0m (B) 3 /Wb maz0m-

(C) 0 (D) 3 /Wb m-

MCQ 4.2.16 A long cylindrical wire lying along z -axis carries a total current 5 mAI 10 = as shown in the figure. The current density inside the wire at a distance r from it’s axis is given by rJ\ .

If the cross sectional radius of the wire is 2 cm then the magnetic flux density at 1 cmr = will be

(A) 25 /nWb m2 (B) 6.25 10 /Wb m4 2#

-

(C) 1.25 /nWb m2 (D) 12.5 /nWb m2




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MCQ 4.2.17 Magnetic field intensity is given in a certain region as

H 3 /A mxx yz x z y

xyza a a1 1x y z

22 2

2

= + + - +The total current passes through the surface 2 mx = , 1 4 my# # , 3 4 mz# # in ax direction will be(A) 259 A- (B) 259 A

(C) 18.2 A (D) 1.98 10 A3#

MCQ 4.2.18 A phonograph record of radius 1 m carries a uniform surface charge density

S 20 /C m2r = . If it is rotating with an angular velocity 0.1 /rad sw = ; then the magnetic dipole moment will be(A) 4 A m2p - (B) /2 A m2p -

(C) 2 A m2p - (D) 2 A m32p

-

Statement for Linked Question 19 - 20 :A uniformly charged solid sphere of radius r is spinning with angular velocity

/rad s6w = about the z -axis. The sphere is centered at origin and carries a total charge 5 C which is uniformly distributed over it’s volume.

MCQ 4.2.19 The plot of magnetic dipole moment of the sphere, ( )m r versus the radius of the sphere, r will be

MCQ 4.2.20 The average magnetic field intensity within the sphere will be

(A) r a2p q (B) r a

2zp

(C) r a2

q (D) r a21

zp



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For View Only Shop Online at www.nodia.co.inMCQ 4.2.21 A rectangular coil, lying in the plane . .x y z3 1 5 3 5+ - = carries a current 7 A such

that the magnetic moment of the coil is directed away from the origin. If the area of the rectangular coil is 0.1 m2 then the magnetic moment of the coil will be(A) 0.2 0.2 0.3 A ma a ax y z

2-- - +

(B) 2 6 3 A ma a ax y z2

-+ -

(C) 1.4 4.2 2.1 A ma a ax y z2

-+ -

(D) 0.2 0.6 0.3 A ma a ax y z2

-+ -

MCQ 4.2.22 Vector magnetic potential in a certain region of free space is A (6 2 ) 4y z xza ax y= - +The electric current density at any point ( , , )x y z will be(A) ( 8 2 6 ) /A ma a ax y z

2- + + (B) (3 ) /A ma ay z2+

(C) 0 (D) (8 2 6 ) /A ma a ax y z1 2

0 + -m

Statement for Linked Question 23 - 24 :A circular toroid with a rectangular cross section of height mh 5= , carries a current 10 AI = flowing in 105 turns of closely wound wire around it as shown in figure. The inner and outer radii of toroid are 1 ma = and 2 mb = respectively.

MCQ 4.2.23 The total magnetic flux across the circular toroid will be(A) 1.39 Wb (B) 0.14 Wb

(C) 15.1 Wb (D) 0 Wb

MCQ 4.2.24 If the magnetic flux is found by multiplying the cross sectional area by the flux density at the mean radius then what will be the percentage of error ?(A) . %4 31- (B) . %3 14-

(C) . %4 61- (D) . %6 14-

MCQ 4.2.25 Magnetizing force at any point P on z -axis due to a semi infinite current element placed along positive x -axis is H . If one more similar current element is placed along positive y -axis then the resultant magnetizing force at the point P will be(A) /H 2 (B) H2

(C) 2H (D) H2-




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MCQ 4.2.26 An infinitely long straight wire carrying current 5 A and a square loop of side 2 m are coplanar as shown in the figure. The distance between side AB of square loop and the straight wire is 4 m. What will be the total magnetic flux crossing through the rectangular loop ?

(A) 2.55 Wbm (B) 81.1 Wbm

(C) 0.81 Wbm (D) 8.11 Wbm

MCQ 4.2.27 A .5 m1 square loop is lying in x -y plane such that one of it’s side is parallel to y-axis and the centre of the loop is 0.3 m away from the y -axis. How much current must flow through the entire y -axis for which the magnetic flux through the loop is 5 10 Tesla m5 2

#- ?

(A) A417 (B) A834

(C) 208.5 A (D) 280 A

MCQ 4.2.28 A L-shaped filamentary wire with semi infinite long legs making an angle 90c at origin and lying in y -z plane as shown in the figure.

If the current flowing in the wire is 4 AI = then the magnetic flux density at 2 ,0,0m^ h will be

(A) 2 10 /Wb ma ay z7 2

#- +-^ h (B) 2 10 /Wb ma ay z

7 2# +-

^ h

(C) 4 10 /Wb ma ay z7 2

#- +-^ h (D) 4 10 /Wb ma ay z

7 2# +-

^ h

Common Data for Question 29 - 30 :An infinitely long straight conductor of cylindrical cross section and of radius R carries a current I , which is uniformly distributed over the conductor cross section.



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For View Only Shop Online at www.nodia.co.inMCQ 4.2.29 If the conductor is located along z -axis then the magnetic flux density at a distance

( )R>r from the cylindrical axis will be

(A) RI a

2 20

pm r

f (B) I a2 2

0

prm

f

(C) I a20

pm

f (D) RI a2

02m r

f

MCQ 4.2.30 Magnetic flux density at a distance ( )R>r from the cylindrical axis will be proportional to

(A) 1r (B) 1

2r(C) r (D) 2r

MCQ 4.2.31 Consider a filamentary wire is bent to form a square loop of side m3 lying in the x -y plane as shown in the figure. If the current flowing in the wire is 1 AI = then the magnetic flux density at the center of the loop will be

(A) /Wb ma2 2 10 z7 2

#- (B) 4 10 /Wb ma2 z

7 2#

-

(C) 2 10 /Wb maz7 2#

- (D) 4 10 /Wb ma2 z7 2

#- -

MCQ 4.2.32 An infinitely long straight wire carrying a current 20 A and a circular loop of wire carrying a current I are coplanar as shown in the figure.

The radius of the circular loop is 10 cm and the distance of the centre of the loop from the straight wire is 1 m. If the net magnetic field intensity at the centre of the loop is zero then the current I is(A) A2

p (B) A20p

(C) A2p (D) 2 Ap




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MCQ 4.2.33 The magnitude of the magnetic field intensity produced at center of a square loop of side a carrying current I is

(A) aI2 2

p (B) aI2

p

(C) a

I2 p

(D) aI8

p

MCQ 4.2.34 For the single turn loop of current shown in the figure the magnetic field intensity at the center point P of the semi circular portion will be

(A) 5.8 /A m outward (B) 5.8 /A m inward

(C) 3.8 /A m outward (D) 3.8 /A m inward

MCQ 4.2.35 Two perfect conducting infinite parallel sheets separated by a distance 2 m carry uniformly distributed surface currents with equal and opposite densities 4ax and

4ax- respectively as shown in figure.

The medium between the two sheets is free space. What will be the magnetic flux between the sheets per unit length along the direction of current ?(A) 0 (B) 8 /Wb may0m

(C) 8 /Wb may0m- (D) 4 /Wb may0m-

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EXERCISE 4.3

Statement for Linked Question 1 - 2 :An infinitely long uniform solid wire of radius a carries a uniform dc current of density J

MCQ 4.3.1 The magnetic field at a distance r from the center of the wire is proportional to(A) 1/for and forr r a r r a< >2

(B) 1/for and forr a r r a0 < >

(C) 1/for and forr r a r r a< >

(D) 0 1/for and forr a r r a< >2

MCQ 4.3.2 A hole of radius ( )b b a< is now drilled along the length of the wire at a distance d from the center of the wire as shown below.

The magnetic field inside the hole is(A) uniform and depends only on d

(B) uniform and depends only on b

(C) uniform and depends on both andb d

(D) non uniform

MCQ 4.3.3 Two infinitely long wires carrying current are as shown in the figure below. One wire is in the y z- plane and parallel to the y- axis. The other wire is in the x y- plane and parallel to the x- axis. Which components of the resulting magnetic field are non-zero at the origin ?

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(A) , ,x y z components (B) ,x y components

(C) ,y z components (D) ,x z components

MCQ 4.3.4 A flux of .2 mWb2 exerts in a magnet having a cross-section of 30 cm2. The flux density in tesla is(A) 4 (B) 0.4

(C) 2.5 (D) 40

MCQ 4.3.5 The magnetic flux density B and the vector magnetic potential A are related as(A) B A4#= (B) A B4#=

(C) B A4:= (D) A B4:=

MCQ 4.3.6 Consider the following statements relating to the electrostatic and magnetostatic field :1. The relative distribution of charges on an isolated conducting body is dependent

on the total charge of the body.

2. The magnetic flux through any closed surface is zero.

Which of the above statements is/are correct ?(A) Neither 1 nor 2 (B) 1 only

(C) 2 only (D) Both 1 and 2

MCQ 4.3.7 The line integral of the vector potential A around the boundary of a surface S represents which one of the following?(A) Flux through the surface S

(B) Flux density in the surface S

(C) Magnetic field intensity

(D) Current density

MCQ 4.3.8 An infinitely long straight conductor located along z-axis carries a current I in the +ve z -direction. The magnetic field at any point P in the x y- plane is in which

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For View Only Shop Online at www.nodia.co.indirection?(A) In the positive z -direction

(B) In the negative z -direction

(C) In the direction perpendicular to the radial line OP (in x y- plane) joining the origin O to the point P

(D) Along the radial line OP

MCQ 4.3.9 A A13 current enters a right circular cylinder of 5 cm radius. What is the linear surface current density at the end surface?(A) (50/ ) /A mp (B) (100/ ) /A mp

(C) (1000/ ) /A mp (D) (2000/ ) /A mp

MCQ 4.3.10 What is the value of the magnetic vector potential due to an infinitesimally small current element, evaluated at infinite distance from it ?(A) Infinity

(B) Unity

(C) Zero

(D) Any number between zero and infinity depending on the strength of the current element

MCQ 4.3.11 What is the magnetic field intensity vector H between two parallel sheets with separation ‘d ’ along z-axis both sheets carrying surface current KK ay y= ?(A) k ay y- (B) k ay y+

(C) k ay x- (D) Zero

MCQ 4.3.12 Current density ( )J , in cylindrical coordinate system is given as :

( , , )zJ r f ( / )forforJ a

aa ba

0 0 < << <z0

2rrr= *

where az is the unit vector along z -coordinate axis. In the region, a r b< < , what is the expression for the magnitude of magnetic field intensity ( )H ?

(A) ( )J a20 3 3

rr - (B) ( )J a2

0 3 3

rr +

(C)( )a

J a3 2

03 3

rr -

(D) ( )J a20 3 3

pr r -

MCQ 4.3.13 Which one of the following concepts is used to find the expression of radiated E and H field due to a magnetic current element ?(A) Concept of vector magnetic potential

(B) Concept of scalar electric potential

(C) Concept of scalar magnetic potential

(D) Concept of vector electric potential

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MCQ 4.3.14 The circulation of H around the closed contour C , shown in the figure is

(A) 0 (B) 2l

(C) 4l (D) 6l

MCQ 4.3.15 The unit of magnetic flux density is(A) gauss (B) tesla

(C) bohr (D) weber/sec

MCQ 4.3.16 The magnetic flux density created by an infinitely long conductor carrying a current I at a radial distance R is

(A) RI

20

pm (B) R2

1p

(C) RI

2 30

pm (D) R I

34 2p

MCQ 4.3.17 Match List I with List II and select the correct answer using the codes given below the lists.

List-I List-II

a. Work 1. Ampere/metre

b. Electric field strength 2. Weber

c. Magnetic flux 3. Volt/metre

d. Magnetic field strength 4. Joule

Codes : a b c d(A) 4 3 2 1(B) 1 3 2 4(C) 4 2 3 1(D) 1 2 3 4

MCQ 4.3.18 A long straight wire carries a current AI 1= . At what distance is the magnetic field 1 Am 1- ?(A) 1.59 m (B) 0.159 m

(C) 0.0159 m (D) 0.00159 m

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For View Only Shop Online at www.nodia.co.inMCQ 4.3.19 How much current must flow in a loop radius 1 m to produce a magnetic field

1 mAm 1- ?(A) 1.0 mA (B) 1.5 mA

(C) 2.0 mA (D) 2.5 mA

MCQ 4.3.20 Assertion (A) : Knowing magnetic vector potential A at a point, the flux density B at the point can be obtained.Reason (R) : 0A:d = .(A) Both A and R are true and R is the correct explanation of A

(B) Both A and R are true and R is not the correct explanation of A



MCQ 4.3.21 The magnetic vector potential A obeys which equations ?1. B Ad#= 2. A J2

0md =-

3. RIdA l

40

pm= #

Select the correct answer using the code given below :(A) 1 and 2 (B) 2 and 3

(C) 1 and 3 (D) 1, 2 and 3

MCQ 4.3.22 A long straight wire carries a current 10I A= . At what distance is the magnetic field 1H Am 1= - ?(A) 1.19 m (B) 1.39 m

(C) 1.59 m (D) 1.79 m

MCQ 4.3.23 What is the magnetic field due to an infinite linear current carrying conductor ?

(A) H rI

2 A/mpm= (B) H r

I2 A/mp=

(C) H rI

2 A/mm= (D) H rI A/m=

MCQ 4.3.24 Equation 0B:d = is based on(A) Gauss’s Law (B) Lenz’s Law

(C) Ampere’s Law (D) Continuity Equation

MCQ 4.3.25 Plane y 0= carries a uniform current density 30a mA/mz . At ( , , )1 20 2- , what is the magnetic field intensity ?

(A) 15a mA/mx- (B) 15a mA/mx

(C) 18.85a mA/my (D) 25a mA/mx

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MCQ 4.3.26 Which one of the following is not the valid expression for magnetostatic field vector B ?(A) B A4:= (B) B Ad#=

(C) 0B:d = (D) B J04 m# =

MCQ 4.3.27 Which one of the following statements is correct ? Superconductors are popularly used for(A) generating very strong magnetic field

(B) reducing i R2 losses

(C) generating electrostatic field

(D) generating regions free from magnetic field

MCQ 4.3.28 Assertion (A) : 0dSBs

: =# where, B = magnetic flux density, dS = vector with direction normal to surface elements dS .Reason (R) : Tubes of magnetic flux have no sources or sinks.(A) Both A and R are true and R is the correct explanation of A




MCQ 4.3.29 Plane defined by z 0= carry surface current density 2a A/mx . The magnetic intensity ‘Hy ’ in the two regions 0z< <a- and 0 z< < a are respectively

(A) ay and ay- (B) ay- and ay(C) ax and ax- (D) ax- and ax

***********

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SOLUTIONS 4.1

SOL 4.1.1 Option (D) is correct.It is not possible to have an isolated magnetic poles (or magnetic charges). If we desire to have an isolated magnetic dipole by dividing a magnetic bar successively into two, we end up with pieces each having north and south poles. So an isolated magnetic charge doesn’t exist.That’s why the total flux through a closed surface in a magnetic field must be zero.

i.e. dB S:# 0=or more clear, we can write that for a static magnetic field the total number of flux lines entering a given region is equal to the total number of flux lines leaving the region.So, (A) and (R) are both true and R is correct explanation of A.


SOL 4.1.3 Option (D) is correct.Since the field intensity exists in a closed surface and lines of field intensity makes a closed curve so the flux lines leaving the spherical surface equal to the total flux entering the surface and So the net flux

F dB S 0:= =#According to divergence theorem

dB S:# dvB:d= ^ h# 0 dvB:d= #Since volume of the sphere will have certain finite value so, B:d 0=or H:d 0= at all points inside the sphere

SOL 4.1.4 Option (C) is correct.The Magnetic field are caused only by current carrying elements and given as

B R

Idl R4 3

0

pm #=

Since an accelerated electron doesn’t form any current element(Idl ) so it is not a source of magnetic field.

SOL 4.1.5 Option (D) is correct.The magnetic field intensity produced due to a small current element Idl is defined




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as

dH R

Idl a4

R2p

#=

where dl is the differential line vector and aR is the unit vector directed towards the point where field is to be determined. So for the circular current carrying loop we have dl ad af= f

aR a=- r

Therefore the magnetic field intensity produced at the centre of the circular loop is

H a

Iad a a4 2

0

2 #

pf

=-f r

f

p

=

^ h# aIa a

4 z2 02

pf= p

^ h6 @ /A maI a4 z=


Magnetic field intensity at any point P due to a filamentary current I is defined as

H cos cosI a4 2 1pr a a= - f6 @

where r " distance of point P from the current filament. 1a " angle subtended by the lower end of the element at P . 2a " angle subtended by the upper end of the element at P .From the figure we have r 3 4 52 2= + = 1a /2r= & cos 1a 0=

and cos 2a 5 12

121312

2 2=

+=

Now we Put these values to get,

H a4 55

1312 0

#p= - fb l ( 5 )AI =

a133p= f 0. /wb ma15 2= f

SOL 4.1.7 Option (C) is correct.According to Biot-savart law, magnetic field intensity at any point P due to the current element Idl is defined as



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For View Only Shop Online at www.nodia.co.in H

RIdl R

4 3#p

= #where R is the vector distance of point P from the current element.Here current is flowing in af directionSo the small current element Idl I d ar f= f 4 2d af#= f 8d af= f

and since the magnetic field to be determined at center of the loop so we have R 2 m= (radius 2 m= )and aR a=- r (pointing towards origin)Therefore the magnetic field intensity at origin is

H ( )

(8 ) ( )d a a4 2 2

0

2

pf #= -f rp

# d a168

z0

2

p f=p

# a4z

02

p f= p6 @

/A ma2 z=

SOL 4.1.8 Option (B) is correct.According to right hand rule if the thumb points in the direction of outward or inward current then rest of the fingers will curl along the direction of magnetic flux lines, This condition is satisfied by the configuration shown in option (C).

SOL 4.1.9 Option (B) is correct.According to right hand rule if the thumb points in the direction of current then rest of the fingers will curl along the direction of magnetic field lines. This condition is satisfied by the configuration shown in option (C).


According to Ampere’s circuital law, the line integral of magnetic field intensity H around a closed path is equal to the net current enclosed by the path. Since we have to determine the magnetic field intensity due to the infinite line current at 5 cmt = so we construct a circular loop around the line current as shown in the figure.Now from Ampere’s circuital law we have

dB lL

:# I0 encm=

or (2 )B pr 100 #m= ( 10 )AIenc =

Therefore we have the magnetic flux density at 5 cmt = as

B 2 5 104 10 10

2

7

# #

# #pp= -

- 10 /wb m5 5 2

#= -




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SOL 4.1.11 Option (B) is correct.According to Ampere’s circuital law the contour integral of magnetic field intensity in a closed path is equal to the current enclosed by the path.

i.e. dH l:# Ienc=Now using right hand rule, we obtain the direction of the magnetic field intensity in the loop as it will be opposite to the direction of L.

So, dH l:# Ienc=- 20 A=-(10 A is not inside the loop. So it won’t be considered.)

SOL 4.1.12 Option (C) is correct.The magnetic field intensity produced at a distance r from an infinitely long straight wire carrying current I is defined as

H I2pr=

As determined by right hand rule, the direction of magnetic field intensity will be same(in ay- direction) due to both the current source. So, at point P the net magnetic field intensity due to both the current carrying wires will be H H H1 2= +

I Ia a2 4 2 1y yp p

= - + -^

^^

^h

hh

h

a85 8

yp=- ^ h a4

yp=- ( 8 AI = )

SOL 4.1.13 Option (B) is correct.As calculated in previous question the magnetic field intensity produced at the centre of the current carrying circular loop is

H aI2=

So by symmetry the semicircular loop will produce the field intensity half to the field intensity produced by complete circular loop.

i.e. Field intensity at the centre of semicircular loop H21= a

I4=

SOL 4.1.14 Option (C) is correct.Since current in the wire is distributed over the outer surface so net enclosed current, Ienc for any Amperian loop inside the wire will be zero.and as from Ampere’s circuital law we have

dH l:# Ienc=

So dH l:# 0= (I 0enc = )

or H 0= for r R<

SOL 4.1.15 Option (C) is correct.Consider the cylindrical wire is lying along z-axis as shown in the figure. As the current I is distributed over the outer surface of the cylinder so for an Amperian



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For View Only Shop Online at www.nodia.co.inloop at a distance ( )r R> from the centre axis, enclosed current is equal to the total current flowing in the wire.

Now from Ampere’s circuital law we have,

dB l:# Ienc0m=

(2 )rB p I0m= (I Ienc = )

or B rI a2

0

pm= f

or B r1

\

SOL 4.1.16 Option (A) is correct.Since the current flows from Q1 and terminates at Q2 and the charge Q2 is located at the surface of the contour so the actual current is not enclosed by the closed path and the circulation of the field is given as

dB l:# Id enc0m= 6 @ I 0c enc =^ h6 @

and Id enc6 @ dtd d dE S E S0 1 0 2: :e e= +: D# #

where E1 is the electric field intensity produced by charge Q1 while E2 is the field intensity produced by charge Q2.

So, Id enc6 @ dtd Q Q

8 200

10

0

2e e e e= +c cm m= G dtdQ

dtdQ

81

211 2= + (1)

As the current flows from Q1 and terminates at Q2 so the rate of change in the net charges is given as

dtdQ1- 16 Adt

dQ2= =

Therefore from equation (1) we have the enclosed displacement current as

Id enc6 @ ( ) ( )81 16 2

1 16= - + 6 A=

Thus, the circulation of magnetic flux density around the closed loop is

dB l:# 60m= ^ h /Wb m8 0m=




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Magnetic field intensity at any point P due to an infinite current carrying sheet is defined as

H K a21

n#=

where K is the current density and an is the unit vector normal to the current sheet directed toward the point P .Since we have to determine the magnetic field intensity at origin so from the figure we have an az=-Therefore the magnetic field intensity at the origin is

H (20 ) ( )a a21

x z#= - /A ma5 y= ( 20K ax= )

SOL 4.1.18 Option (C) is correct.

Magnetic field intensity at any point P due to an infinite current carrying sheet is defined as

H K a21

n#=



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For View Only Shop Online at www.nodia.co.inwhere K is the current density of the infinite sheet and an is the unit vector normal to the current sheet directed toward the point P .Since we have to determine the magnetic field intensity at point (2, 1- , 5) which is above the plane sheet as shown in figure, so we have, an az=+Therefore the magnetic field intensity at the point (2, 1- , 5) is

H (20 )a a21

x z#= a220

y=- /A ma5 y=- ( 20K ax= )

SOL 4.1.19 Option (C) is correct.Consider one of the sheet carries the current density K1. So, the other sheet will have the current density K1- .Magnetic flux density produced at any point P due to a current sheet is defined as

B K a2 n0m

#=

where K is current density of the sheet and an is the unit vector normal to the sheet directed towards point P .So for any point in the space between the sheets normal vector will be opposite in direction for the two sheets as shown in figurei.e. an2 an1=-

Therefore, the resultant magnetic flux density at any point in the space between the two sheets will be

B ( ) ( )K a K a2 n n0

1 11 1# #m= + - -6 @ K an0 1 1m #=

Since an1 is unit vector normal to the surface, and K1 is given current density. So the cross product will be a constant.

SOL 4.1.20 Option (D) is correct.The magnetic flux density at any point is equal to the curl of magnetic vector potential A at the point.i.e. B A#d= (12 )cos aq#d= q

(12 )cosr rr a1

22 q= f

cosr a12 q= f




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or, B cos a312 0= f a2= f (at ( , , )3 0 p )

SOL 4.1.21 Option (C) is correct.Consider the current sheet shown in the figure.

Magnetic field intensity produced at any point P due to a current sheet is defined as

H K a21

n#=

where K is current density of the sheet and an is the unit vector normal to the sheet directed towards point P .

So, for y 0> H Ka a21

z y#= ^ ^h h Ka21

x=- ( /A mKK az= , an ay= )

and for y 0< H Ka a21

z y#= -^ ^h h K a2 x= ( /A mKK az= , an ay=- )

SOL 4.1.22 Option (D) is correct.Consider the current density at 8 cmr = is J directed along az+ .Now the magnetic field for 8 cm>r must be zero.i.e. H 0= (for 8 cm>r )So from Ampere’s circuital law we have

dH l:# I 0enc= =

Since for the region 8 cm>r the Amperian loop will have all the current distributions enclosed inside it.i.e. Ienc 14 10 2 (2 0.5 10 ) . ( . )0 8 2 0 25 103 2 2

# # #p p# # # #= + -- - -

(2 8 10 )J 2p# #+ -

6.43 10 (16 10 )J2 2p# #= +- -

So we have 6.43 10 (16 10 )J2 2p# #+- - 0= ( 0Ienc = )

or J .16 106 43 10

2

2

#

#p

=- -

-

or J 0. 3 /A ma2 z=-




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For View Only Shop Online at www.nodia.co.inMagnetic flux density is defined as the curl of vector magnetic potentiali.e. B A#d=

x y y xyz

a a a

2 2 8

x

x

y

y

z

z2 2

=-

22

22

22

R

T

SSSS

V

X

WWWW

( 8 0) (0 8 ) (2 2 )xz yz y xa a ax y z2 2= - - + + + -

So the net magnetic flux density at ( , , )1 2 5- - is B 0 0 /wb ma a a2 6 12x y z

2= + +

SOL 4.1.24 Option (B) is correct.Total magnetic flux through a given surface S is defined as

F dB SS

:= #where dS is the differential surface vector having direction normal to the surfaceSo, for the given surface z 4= , x0 1# # , y1 4# #- we have dS ( )dxdy az=and as calculated in previous question we have B ( 8 0) (0 8 ) (2 2 )xz yz y xa a ax y z

2 2= - - + + + -Therefore, the total magnetic flux through the given surface is

F ( )( )y x dxdy2 2xy

2 2

0

1

1

4= -

==-## y dy x dx2 1 2 52

1

42

0

1

# #= --# #

y x2 3 10 3

3

1

4 3

0

1

= --

; :E D 2 365

310

#= -

40 wb

SOL 4.1.25 Option (D) is correct.For determining the magnetic field at any point above the plane z 0= , we draw a rectangular Amperian loop parallel to the y -z plane and extending an equal distance above and below the surface as shown in the figure.From Ampere’s circuital law,

dB l:# Ienc0m=




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Since the infinite current sheet is located in the plane z 0= so, the z -component of the magnetic flux density will be cancelled due to symmetry and in the closed Amperian loop the integral will be only along y -axis. Thus we have (2 )lB Ienc0m= 2 lB Kl0m= ( )I Klenc =As determined by right hand rule, the magnetic flux density above the plane z 0= will be in ay- direction. So we have the flux density above the current sheet as

B a24

y0 #m=- 2 /wb may0

2m=- ( 4 /A mK = )

Alternate Method :The magnetic flux density produced at any point P due to an infinite sheet carrying uniform current density K is defined as

B ( )K a21

n0m #=

where an is the unit vector normal to the sheet directed toward the point P .So, magnetic flux density at any point above the current sheet 4K ax= is

B a a21 4 x z0m #= ^ ^h h /wb ma4 y0

2m=- (a an z= )

SOL 4.1.26 Option (A) is correct.Magnetic flux density at a certain point is equal to the curl of magnetic vector potential at the point.i.e. B A#d=So from the above determined value of magnetic flux density B we have, A#d 2 /wb may0

2m=- (1)Since A is parallel to K so the vector potential K will depend only on z . Hence, we have A ( )A z ax=From equation (1) we have,

2 ay0m- ( )A z

a a a

0 0

x

x

y

y

z

z= 22

22

22

2 ay0m- ( )zA z ay2

2=-

or ( )A z 2 z0m=So, A 2 zax0m=Therefore the vector magnetic potential at z 2=- is A 4 /wb max0m=-

SOL 4.1.27 Option (D) is correct.Current density at any point in a magnetic field is defined as the curl of magnetic field intensity at the point. i.e. J H#d=Since the magnetic field intensity in the free space is given as



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For View Only Shop Online at www.nodia.co.in H 2 a2r= f

Therefore the current density is

J ( )(2 )a1

z

2

22

r rr r= (2 )a1

z3

22

r r r=

6 azr= /A ma2 z2= ( 2 )mr =

SOL 4.1.28 Option (A) is correct.The magnetic flux density at any point is equal to the curl of the vector magnetic potential at the pointi.e., B A#d=

( )A a1z2

2r r r= f ( )a1 2 z2

2r r r= a2

zr=

The current density J in terms of magnetic flux density B is defined as

J ( )B10m #d= a1 2

0 22

m r r= - fb l; E a22

0m r= f

This current density would produce the required vector potential.

SOL 4.1.29 Option (C) is correct.The current density for a given magnetic field intensity H is defined as J H#d=Given H cosz ay z ea ay

yx= + +^ ^h h

So H#d cosz e z ay

a a a

0

x

xy

y

y

z

z=+22

22

22

^ h

cos coszz ay

zz e

xz ay

yz ea a ax

yy

yz2

222

22

22= - - - + + - +^ ^b ^ch hl hm= G

cosay ea a ax yyz=- + -

or, J H#d= cosay ea a ax yyz=- + -

Therefore the current density in the x -z plane is J /A ma a ax y z

2=- + - (y 0= in x -z plane)

SOL 4.1.30 Option (D) is correct.In a source free region current density, 0J =The current density at any point is equal to the curl of magnetic field intensity H .i.e. J H#d=or H#d 0= ( 0J = )and since the curl of a given vector field is zero so it can be expressed as the gradient of a scalar fieldi.e. H fd=So A and R both are true and R is correct explanation of A.

SOL 4.1.31 Option (B) is correct.Given that the cylindrical wire located along z -axis produces a magnetic field intensity, 3H ar= f .




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So, applying the differential form of Ampere’s circuital low we have the current density with in the conductor as J H#d=

A A A

a a a1

z

z

z

r

r

r= 2

222

22

r

r

r

f

f

f

a a a

1

0 3 0

z

z2

r

r

r= 2

222

22

r

r

f

f

6 /A ma a1 3 z z2 2

22

r r r= =^ h

SOL 4.1.32 Option (D) is correct.As the beam is travelling in az direction so the field intensity produced by it will be in af direction and using Ampere’s circuital law at it’s surface we have H a2pf ^ h Ienc=

H a2pf ^ h a d2 1 2a

0

r pr r= -a k#

H a2pf ^ h a4 2 3

a2 3

0p r r= -; E

H a2pf ^ h a3

2 2p=

or H a a4= f

SOL 4.1.33 Option (A) is correct.Since the magnetic flux density is defined as B A#d=and B#d J0m=Now using the vector identity, we have A#d #d̂ h A A2:d d d= -^ h

or, B#d A A2:d d d= -^ h

or, J0m A A2:d d d= -^ h

As the vector potential is always divergence free so we get, A2d J0m=-

SOL 4.1.34 Option (B) is correct.Magnetic dipole moment of a conducting loop carrying current I is defined as : m IS=where S is the area enclosed by the conducting loop. So we have m ( . )7 0 52

# #p= .5 5= ( 7 , 0.5 )A mI R= =The direction of the moment is determined by right hand rule as when the curl of fingers lies along the direction of current,then the thumb indicates the direction of moment.So, m 5.5 A max 2

-=

***********



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SOLUTIONS 4.2

SOL 4.2.1 Option (D) is correct.Since the current is flowing in the az- directionSo, Idl 10 ( )dz az= -

Magnetic field intensity at any point P due to a filamentary current I is defined as


where r " distance of point P from the current filament. 1a " angle subtended by the lower end of the element at P . 2a " angle subtended by the upper end of the element at P .Now from the figure we have, r 2= 2a p q= -or cos 2a ( )cos p q= -

cosq=-2 3

32 2

=-+ 13

3=-

and 1a 0= (angle subtended by end z 3= )or cos 1a cos0 1= =

So, H cos cosI a4 2 1pr a a= - f6 @

I a4 2 1133

#p= - - fc m= G ( ) a810

2133

p #= + fc m

Now the direction of magnetic field intensity is defined as




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af a al#= r

where al is unit vector along the line current and ar is the unit vector normal to the line current directed toward the point P .So we have af ( ) ( )a az y#= - ax=

Therefore, H ( )a810 1

133

xp= +c m

.73 /A ma1 x=

SOL 4.2.2 Option (D) is correct.According to Biot-savart law, magnetic field intensity at any point P due to the current element Idl is defined as

H R

Idl R4 3#p

= #where R is the vector distance of point P from the current element.

Now the current element carries a current of 4.4 A in ax+ direction.So we have, R (4 2 3 ) ( )xa a a ax y z x= + + - (Since on x -axis y - and z -component will be zero) R (4 ) 2 3x a a ax y z= - + +or R ( )x4 2 32 2 2= - + + x x8 292= - +and Idl 4.4dxax= (filament lies from x 3=- to x 3= )Therefore the magnetic field intensity is

H ( )

( . ) ( )x x

xdx

a a a a4 8 29

4 4 4 2 3/

x x y z2 3 2

#

p=

- +- + +

3

3

-

+ 6 @#

. (2 3 )( )x x

dxa a44 4

8 29 /z y 2 3 2p= -- +3

3

-

+

#

. (2 3 )( )

( )x xxa a4

4 426 8 29

2 8/z y 2 1 2p= -

- +-

3

3

-

+

= G

. (2 3 )a a264 4

z yp= - 0.1077 0.162a az y= - 0.1 0.2 /A ma az y= -



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For View Only Shop Online at www.nodia.co.inAlternate Method :

According to Ampere’s circuital law, the line integral of magnetic field intensity H around a closed path is equal to the net current enclosed by the path.Since we have to determine the magnetic field intensity at point (4, 2, 3) so we construct a circular loop around the infinite current element that passes though the point (4, 2, 3) as shown in the figure.Now from Ampere’s circuital law we have,

dH l:# Ienc=

( )r H2p .4 4= ( 4.4 )AIenc =

or H .2 13

4 4#p

= r 13= from figure.

Now direction of the magnetic field intensity is defined as af a al#= r

where al " unit vector in the direction of flow of current ar " unit vector normal to the line current directed toward the point.

So we have, af ( )

( ) ( )a

a a a a4 4 2 3

4 2 3 4x

x y z x

2 2 2#=- + +

+ + -6 @

(2 3 )a a a

13x

y z#= +

2 3a a

13z y= -

Therefore the magnetic field intensity at the point ( , , )4 2 3 is

H . ( )a a2 13

4 413

2 3z y

p= -

. (2 3 )a a264 4

z yp= -

. . /A ma a1 5 2 5z y= -

SOL 4.2.3 Option (D) is correct.As the magnetic field intensity at the center of the triangle produced by all the




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three sides will be exactly equal so we consider only one side lying along x -axis that carries 4 A current flowing in ax+ direction as shown in the figure.Now the magnetic field intensity at any point P due to a filamentary current I is defined as


where r " distance of point P from the current filament. 1a " angle subtended by the lower end of the element at P . 2a " angle subtended by the upper end of the element at P .

From the figure we have

tan30c 1r= & r

31=

1a /6 65p p p= - = & 5cos cos 61a p= 2

3=-

and 2a 30c= & cos cos302 ca = 23=

So the magnetic field intensity produced by one side of the triangle at centre of the triangle is

H1 cos cos a4

31

42 1

#pa a= - f6 @

a323

23

p= + f; E a5p= f

Now the direction of magnetic field intensity is determined as af a al#= r

where al is unit vector along the line current and ar is the unit vector normal to the line current directed toward the point P .and since the line current is along x -axis so we have af a a ax y z#= = (al ax= , a ay=r )Therefore the net magnetic field intensity due to all the three sides of triangle is

H 3H1= 3 a3zp#= b l /A ma9

zp= ( )a az=f



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According to Biot-savart law, magnetic field intensity at any point P due to the current sheet element dSK is defined as

H R

dSK a4

R

s2p

#= #where R is the vector distance of point P from the current element.Now we consider a point , ,y z0^ h on the current carrying sheet, from which we have the vector distance of point ( , , )3 0 0 R 3 0 0 0 y za a a a a ax y z x y z= + + - + +^ ^h h 3 y za a ax y z= - -^ h

or aR 3

y zy za a a

3x y z

2 2 2=

+ +- -

3

y zy za a a

9x y z

2 2=

+ +- -

Therefore the magnetic field intensity due to the current sheet is

H ( )( )y zy z

dydza a a a

4 94 3

/y x y y

yz2 2 3 2

2

2 #

p=

+ +- -

3

3

=-=-## ( 4K ay= )

( )( )

y zz

dydza a

4 94 3

/x z

yz2 2 3 2

2

2

p=

+ +- -

3

3

=-=-##

We note that the x component is anti symmetric in z about the origin (odd parity). Since the limits are symmetric, the integral of the x component over z is zero. So we are left with

H ( )

12y z

dydza4 9 /

z2 2 3 2

2

2

p=

+ +-

3

3

--##

( )z y z

y dza39 9

z 2 2 22

2

p=-+ + + 3

3

-

+

-; E#

z

dza39

2z 2

2

2

p=-+-

# tan za631

3z1

2

2

p=- -

-a k: D

(2) (0.59)a2zp # #=- /A ma2 z=-


Since the uniformly charged disk is rotating with an angular velocity 2 /rad sw =




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about the z-axis so we have the current density K ( )angular velocitysr #= ( ) 2 2sr wr r# #= =or K 4 ar= f

According to Biot-savart law, magnetic field intensity at any point P due to the current sheet element dSK is defined as

H R

dSK a4

R

s2p

#= #where R is the vector distance of point P from the current element.Now from the figure we have R a az r= - r

or R 1 2r= +

and aR a a

1z

2rr=+

- r

So the magnetic field intensity due to a small current element dSK at point P is

dH R

dSK a4

R2

#p

= ( )

( ) ( )a a a4 1

4/

z2 3 2

#

p rr r=

+-f r

( )( )a a

4 14

/z

2 3 2p rr r=

++r

On integrating the above over f around the complete circle, the ar components get cancelled by symmetry, leaving us with

( )zH ( )

d da4 1

4/

z2 3 2

2

0

3

0

2

p rr r r f=+

p

^ h##

2( )

d a1 / z2 3 2

3

0

3

rr r=+

# 2 a11

1z

22

0

3r

r= + +

+= G

2( ) a

1 33 2 1 1 3

z=+

+ - +> H /A ma5 z=


As all the four sides of current carrying square loop produces the same magnetic field at the center so we consider only the line current AB for which we determine the magnetic field intensity at the center.Now the magnetic field intensity at any point P due to a filamentary current I is defined as


where r " distance of point P from the current filament.



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For View Only Shop Online at www.nodia.co.in 1a " angle subtended by the lower end of the filament at P . 2a " angle subtended by the upper end of the filament at P .From the figure, we have

r m21= , 2a 45c= and 1a 180 45c c= -

So the magnetic field intensity at the centre O due to the line current AB is

H1 cos cosRI

2 2 1p a a= -6 @

( / )

( )cos cos2 1 2

1 45 180 45#

c c cp= - -6 @

12

2#p= /A m2

p=

and the magnetic flux density produced by the line current AB is

B1 H0 1m= 4 10 27# #p p= -

5.66 10 /wb m7 2#= -

Therefore the net magnetic flux density due to the complete square loop will be four times of B1

i.e. B 4 4 (5.66 10 )B17

# #= = -

.26 10 /wb m3 6 2#= -

SOL 4.2.7 Option (D) is correct.According to Biot-savart law, magnetic field intensity at any point P due to the current element Idl is defined as

H R

Id al4

R2p

#= #where R is the vector distance of point P from the current element.As the cross product of two parallel lines is always zero so the straight segments will produce no field at P . Therefore the net magnetic field produced at point P will be only due to the two circular section.i.e. H H HCD AB= +

or H ( ) ( ) ( ) ( )I d I da a a a

4 4

/ /

at m at m2

0

2

12

0

2

2

#

prr f

prr f#= - + - -f r

p

r

f rp

r= == =G G# #

( ) ( )I d I da a

4 1 4 2

//z z

0

2

0

2

p f p f= -pp

##

. a43 2 1 2

12 zpp

# #= - a k: D 0.2 /A m=

Alternate Method :The magnetic field intensity produced at the center of a circular loop of radius R carrying current I is defined as

H RI

2=




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and since the straight line will not produce any field at point P so due to the two quarter circles having current in opposite direction, magnetic field at the center will be

H aI

bI

41

2 2= -: D

where a " inner radius b " outer radius

H ( ) ( )

41

2 13 2

2 23 2

# #= -; E 0.2 /A m=

SOL 4.2.8 Option (D) is correct.The magnetic field intensity at any point P due to an infinite filamentary current I is defined as

H I2pr=

where r is the distance of point P from the infinite current filament.Now the two semi infinite lines will be in combination treated as a single infinite line for which magnetic field intensity at point P will be

H1 RI

2p= (R is the length of point P from line current)

2 24#p= 1

p= ( 4 , 2 )A mI R= =

As the magnetic field intensity produced at the center of a circular loop of radius R carrying current I is defined as

H RI

2=

So magnetic field produced at point P due to the semi circular segment is

H2 RI

21

2#= 21=

Therefore net magnetic field intensity produced at point P is

H H H1 2= + 121

p= +

0.82 /A m=




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Magnetic field intensity produced at any point P on the axis of the circular loop carrying current I is defined as

H ( )hI

2 /2 2 3 2

2

rr=+

where h is the distance of point P from the centre of circular loop and r is the radius of the circular loop.From the figure we have r 3 m= and h 5 1 4 m= - =and using right hand rule we conclude that the magnetic field intensity is directed along az+ . So the magnetic field intensity produced at point P is

H ( )

( ) a2 3 4

50 10 3/ z2 2 3 2

3 2#=

+

-

a2 1259 50 10

z

3

## #=

- .8 /mA ma2 z=


Let the cylindrical tube is of radius a for which we have to determine the magnetic field intensity at the axis of solenoid.

Now we consider a small ring (small section of solenoid) of the width dz at a distance z from point P lying on the axis of the solenoid as shown in the figure.

The total current flowing in the loop of the ring will be dI nIdz= where n is the no of turns per unit length




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Since magnetic field intensity produced at any point P on the axis of the circular loop carrying current I is defined as

H ( )hI

2 /2 2 3 2

2

rr=+

where h is the distance of point P from the centre of circular loop and r is the radius of the circular loop.So we have the magnetic field intensity due to the ring as

dH ( )( )a znIdz a

2 2 /

2

2 3 2=+

( ,a h zr = = )

From the figure we have

z cota q= & dzsina d2q

q=-

and sinq ra

a za2 2

= =+

& a z

1/2 2 3 2+^ h

sina3

3q=

The total magnetic field intensity produced at point P due to the solenoid is

H ( )( )a znIdz a

2 /z

2 2 3 2

2

=+3

3

=-# ( )

sinsinnIaa ad2 3 2

2 3

0 qq q= -

q

p

=#

sinnI d20

q q=-q p=# ( )cos cosnI

2 0 p= -

nI= 1000 106 3# #= - /A m6= ( 1000, 4 mAn I= = )

SOL 4.2.11 Option (D) is correct.As calculated in the previous question the magnetic field intensity inside a long solenoid carrying current I is defined as H nI= where n is no. of turns per unit lengthand since using right hand rule we conclude that the direction of magnetic field intensity will be right wards ( ay+ ) due to outer solenoid and left wards (( )ay- ) due to inner solenoid. So the resultant magnetic field intensity produced inside the inner solenoid will be H H H1 2= + ( )n I n Ia ay y1 2= - +where n1 and n2 are the no. of turns per unit length of the inner and outer solenoids respectively.So H (3 10 )( 000) (3 10 )( 000)a a2 1y y

3 3# #=- +- -

3 10 ( 1000)ay3#= -- 3 /A may=-

SOL 4.2.12 Option (D) is correct.Since no any magnetic field is produced at any point out side a solenoid so in the region between the two solenoids field will be produced only due to the outer solenoid.i.e. H n Iay2= 100 10 a4 y

4# #= - /A ma4 y=

SOL 4.2.13 Option (B) is correct.Since no any magnetic field is produced at any point out side a solenoid so, at any point outside the outer solenoid, the net magnetic field intensity produced due to



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For View Only Shop Online at www.nodia.co.inthe two solenoids will be zero.

SOL 4.2.14 Option (B) is correct.The magnetic field intensity produced at any point P due to an infinite sheet carrying uniform current density K is defined as

H ( )K a21

n#=

where an is the unit vector normal to the sheet directed toward the point P .So the field intensity produced between the two sheets due to the sheet 3K az1 = located at 2 mx = is

H1 (3 ) ( )a a21

z x#= - /A ma23

y=- (a an x=- )

and the field intensity produced between the two sheets due to the sheet 3K az2 =- located at 2 mx =- is

H2 ( 3 ) ( )a a21

z x#= - /A ma23

y=- ( )a an x=

Therefore the net magnetic field intensity produced at any point between the two sheets is H H H1 2= + 3ay=-Since the magnetic field intensity at any point is the equal to the negative gradient of scalar potential at the point i.e. H Vmd=-

So for the field H 3ay=- in the region between the two current carrying sheets, we have

3ay- dydV am y=- (the field has a single component in ay direction)

or Vm y C3 1= + where C1 is constantPutting V 0m = for point ( , , )P 1 2 5 (given), we have 0 ( ) C3 2 1#= +or C1 6=-Thus, Vm (3 6)Ay= -and the graph of Vm versus y will be as plotted below




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SOL 4.2.15 Option (C) is correct.The magnetic flux density at any point is equal to the curl of the vector magnetic potential at the pointi.e., B A#d= (1)Since B H0m= 3 /Wb may0

2m=- (calculated in previous question)As the magnetic flux density is in ay direction so A is expected to be z -directed. Therefore from eq (1) we have

xAz22- 3 0m=-

or Az x C3 0 2m= +Putting 0Az = at point ( , , )P 1 2 5 (given), we have 0 C3 0 2m= +or C2 3 0m=-So, Az ( )x3 10m= - 3 0m=- at origin ( , , )0 0 0Thus, the magnetic vector potential at origin is A /Wb ma8 z0m=-

SOL 4.2.16 Option (A) is correct.Since the current density inside the wire is given by J r\

So we have, J kr= where k is a constant.and the total current flowing in the wire is given by

I0 dJ Ss

:= #

or 5 10 3#

- ( )k d20

2 10 2

r pr r=#

-

# ( 5 mAI0 = )

5 10 3#

- ( )k

32 2 10 2 3

#p=-

So we have k 2 8 103 5 10

1615 106

33

# #

# ##p p= =-

-

Now for the Amperian loop at 1 cmr = enclosed current is

Ienc dJ Ss

:= # ( )k d20

1 10 2

r pr r=#

r=

-

# 1615 10 2 3

33

0

1 10 2

# #p p r=#

-

b l ; E

815

31 10 3

# #= - 2415 10 3

#= -

So from Ampere’s circuital law we have

dB lL

:# Ienc0m=

(2 )B pr 2415 100 3

#m= -

Therefore the magnetic flux density at 1 cmr = is

B ( )24

152 1 10

10 4 102

37

##

# #pp= -

--

.1 25 10 8#= - 12.5 /nWb m2=



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Current density at any point in a magnetic field is defined as the curl of magnetic field intensity at the point.i.e. J H#d=So the current density component in ax direction is

Jx ( )H x#d= yH

zHz y

22

22= -

( )yxz x z a

16 x2

22=-

++e o

Therefore the total current passing through the surface 2 mx = , 1 4 my# # , 3 4 mz# # is

I dJ SxS

:= # ( )yxz x z dydz

16

yz2

22

1

4

3

4=-

++

==e o## (d dydzS ax= )

( )yz z dydz1

2 242

2

1

4

3

4=-

++e o## ( 2 mx = )

yz zy dz1

2 242

3

4

1

4

=- +- +; E# z z dz5

3 722

3

4=- +b l#

A145=-


Magnetic dipole moment of a conducting loop carrying current I is defined as : m IS=where S is the area enclosed by the conducting loop. So for a ring of radius r , magnetic dipole moment m ( )I r2p= .Now as the charged disk(charge density, S 20 /C m2r = ) is rotating with angular velocity 0.1 /rad sw = so, the current in the loop is given as dI S rdrr w=Therefore the magnetic dipole moment is

m ( )dI r2p= # S( )( )rdr rr

2

0

1r w p=

=# S r dr3

0

1r wp= #

. r20 0 1 44

0

1

# # p= : D A m32p

-=




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SOL 4.2.19 Option (B) is correct.Magnetic dipole moment of a spherical shell of radius r having surface charge density Sr is given as

m S r34 4p r w= where w is angular velocity.

Since the total charge of 5 C is distributed over the volume of the sphere so, the magnetic dipole moment of the sphere is given as

( )m r ( )dr r34

v4p r w= # (r =

Sdrvr )

where vr is uniformly distributed volume charge density of the sphere. Therefore, we have

( )m r r34

5v

5p r w= Q r51 2w=

rQ

34 3vr p

=e o

(5) (4) r51 2# # #= 4 A mr2 2

-= ( 5 CQ = , 4 / sradw = )

SOL 4.2.20 Option (C) is correct.The average magnetic field intensity over a sphere of radius r , due to steady currents within the sphere is defined as

Have rm

41 2

3p= rr

41 2 4

3

2#

p= r2p= (m r4 2= )

As the sphere is spinning about the z -axis so, the produced magnetic field will be in az direction as determined by right hand rule. Thus, we have

Have r a3

zp=

SOL 4.2.21 Option (A) is correct.Magnetic dipole moment of a conducting loop carrying current I is defined as : m ISan=where S is the area enclosed by the conducting loop and an is normal vector to the surface. So we have m (7)(0.1)an= ( 7 , 0.1A mI S 2= = )Now the given plane is .x y z3 1 5+ - .3 5=For which we have the function f 1.5x y z2= + -



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For View Only Shop Online at www.nodia.co.inand the normal unit vector to the plane is,

an ff

4

4= ( . )

.a a a1 3 1 5

3 1 5x y z2 2 2

=+ + -+ -

So the magnetic dipole moment of the coil is

m (0.7) .( . )a a a

3 53 1 5x y z= + -

.2 0.6 0.3 A ma a a1 x y z2

-= + -

SOL 4.2.22 Option (B) is correct.The magnetic field intensity, (H ) in the terms of magnetic vector potential, ( )A is defined as

H ( )A10m #d= ( )y z xza a1 6 2 4x y

0#dm= - +6 @

a a a1 3 2 4x y z0m

= - - +6 @

Since the electric current density at any point is equal to the curl of magnetic field intensity at that point.i.e. J H#d=So, we have the electric current density in the free space as

J a a a1 8 2 6x y z0m#d= - - +6 @ 0=

SOL 4.2.23 Option (D) is correct.Magnetic flux density across the toroid at a distance r from it’s center is defined as

B rNI a2

0

pm= f

where N " Total no. of turns I " Current flowing in the toroidSo, the total magnetic flux across the toroid is given by the surface integral of the flux density

i.e. f dB SS

:= #where dS is differential surface area vector.Consider a width dr of toroid at a distance r from it’s center as shown in figure

So we have the total magnetic flux across the toroid as

f ( )rNI hdra a2mr

0

1

2

pm= f f

=b l# (dS hdra= f)




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ln24 10 10 10 10

127 5

# # # #p

p=-

b l (N 105= , 10 AI = )

1.39 Wb=

SOL 4.2.24 Option (B) is correct.As determined in previous question the magnetic flux density across the toroid at a distance r from it’s center is

B rNI a2

0

pm= f

So at the mean radius,

r 1.5 ma b2= + =

we have, B NI a30

pm= f ( 1.5 mr = )

Therefore the total magnetic flux is

'f dB S:= # 3 ( )NI hdra a0

r 1

2

pm= f f

=b l# (dS hdra= f)

r34 10 10 10 107 5

12# # # #

pp=

-

6 @ (N 105= , 10 AI = )

1.33 Wb=Thus, the percentage of error is

% error ' %100#ff f= - ( 1.39 wbf = as calculated above)

.1.33 . %1 39

1 39 100#= - .31%12=-


Consider the point P on z -axis is ( , , )h0 0 and current flowing in the current element is I in ax direction. Since the magnetic field intensity at any point P due to a current element I is defined as




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For View Only Shop Online at www.nodia.co.inwhere r " distance of point P from the current element. 1a " angle subtended by the lower end of the element at P . 2a " angle subtended by the upper end of the element at P .So for the given current element along positive x -axis we have 1a 90c= 2a 0c=

Therefore, H hI a4p= f ( hr = )

Now the direction of magnetic field intensity is defined as af a al#= r

where al is unit vector along the line current and ar is the unit vector normal to the line current directed toward the point P .So, af a a ax z y#= =-Therefore magnetizing force is

H ( )hI a4 yp= -

or H hI

4p= ...(1)

Now consider the current flowing in the current element introduced along the positive y -axis is I in ay direction. So, the magnetic field intensity produced at point P due to the current element along the positive y -axis is


cos coshI a4 0 90c cp= - f6 @ ( , ,h 90 01 2c cr a a= = = )

hI a4 xp= (af aa ay z x#= = )

Therefore the resultant magnetic field intensity produced at point P due to both the current elements will be

Hnet ( )hI a a4 y xp= - +

or, Hnet 2hI

4p=

Thus, from equation (1) we have Hnet H3=

SOL 4.2.26 Option (B) is correct.The magnetic flux density produced at a distance r from a straight wire carrying current I is defined as

B I2

0

prm=

Now consider a strip of width dr of the square loop at a distance r from the straight wire as shown in the figure.




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Total magnetic flux crossing the strip is d my B d2 r= ^ h (area of strip d2 r= )

( )I d2 20

prm r=

So, the flux crossing the complete square loop is

my d my= # I d2 20

4

6

prm r=

r=^ h#

lnI0 46

pm r= 6 @ ln4 10 5

467

# #p

p=-

b l

8.11 10 Weber7#= -

. Wb2 42 m=

SOL 4.2.27 Option (B) is correct.As calculated in previous question the total flux crossing through the square loop due to the straight conducting element is

my I Ld2a

b0

prm r=

r=^ h#

where I is the current carried by the conductor, L is the side of the square loop and ,a b are the distance of the two sides of square loop from the conductor.

So we have L 0.5 m=

a 0.3 . 0.05 m20 5= - =

b 0.3 . 0.55 m20 5= + =

Thus, my .I d2 0 5.

.0

0 05

0 55

prm r=

r=^ h# lnI4 .

.00 050 55

pm r= 6 @ lnI4 110

pm= ^ h

Therefore the current that produces the net flux 5 10 Tmm5 2y #= - is

I ln 14 5 10

0

5# #m

p= -

^^ hh

2 8.5 A3=

SOL 4.2.28 Option (D) is correct.Consider the flux density at the given point due to semi infinite wire along y -axis is B1 and the flux density due to wire along z -axis is B2.The magnetic flux density B produced at any point P due to a straight wire carrying current I is defined as

B cos cosI a40

2 1prm a a= - f6 @



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where r " distance of point P from the straight wire. 1a " angle subtended by the lower end of the wire at P . 2a " angle subtended by the upper end of the wire at P .and the direction of the magnetic flux density is given as af a al#= r

where al is unit vector along the line current and ar is the unit vector normal to the line current directed toward the point P . So, we have r 2 m= af a ay x#= ^ h az=- ( ,a a a al y x= =r )

1a 2p= , 02a = (as y tends to 3)

Therefore the magnetic flux density produced at point P due to the semi infinite wire along y -axis is

B1 cos cos a4 2

40 2 z

0

pm p= - -

^

^a ^h

hk h a2 z

0

pm=-

Similarly we have the magnetic flux density produced at point P due to semi infinite wire along z -axis as

B2 a2 y0

pm=-

Thus, the net magnetic flux density produced at point P due to the L-shaped filamentary wire is

B a a2 2y z0 0

pm

pm=- -

3 3 /Wb ma a 10 4y z

2#=- + -

^ h

SOL 4.2.29 Option (D) is correct.Using ampere’s circuital law we have

dB l:# Ienc0m=As the conductor carries current I which is uniformly distribute over the conductor cross section so, the current density inside the conductor is

J RI

2p=

We construct an Amperian loop of radius r( )R< inside the cylindrical wire for




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which the enclosed current is

Ienc RI

22

ppr= b l I

R2

2r= c m

and since the current is flowing along z -axis so using right hand rule we get the direction of magnetic flux density along a+ f .Thus, from Amperes circuital law, we have ( )( )B 2prf Ienc=

or Bf RI

20

2

2

prm r= c m

or B RI a

6 20

pm r= f

SOL 4.2.30 Option (D) is correct.Similarly as calculated above we construct an Amperian loop of radius ( )R>r outside the cylinder for which the entire current flowing in the wire will be enclosed.i.e. Ienc I=and from Ampere’s circuital law we get, B 2prf ^ h I0m=

Bf I

20

prm=

So B 1\ r

SOL 4.2.31 Option (C) is correct.We consider only the half side of the loop to determine the flux density at the center as shown in the figure.

The magnetic flux density B produced at any point P due to a straight wire carrying current I is defined as

B cos cosI a40

2 1prm a a= - f6 @

where r " distance of point P from the straight wire. 1a " angle subtended by the lower end of the wire at P . 2a " angle subtended by the upper end of the wire at P .



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For View Only Shop Online at www.nodia.co.inand the direction of the magnetic flux density is given as af a al#= r

where al is unit vector along the line current and ar is the unit vector normal to the line current directed toward the point P .Therefore, the magnetic flux density produced at centre O due to the half side of the square loop is

B1 cos cosI a40

2 1prm a a= - f^ h

where 1 mr = 1a 2p= and cos

1 11

21

2 2 2a =

+=

Thus, B1 a4 1

4 10 12

1 0 z

7#

pp

= --

^

^ ^ch

h hm (af a ay x#= -^ h az= )

/Wb ma2

10z

42=

-

As all the half sides of the loop will produce the same magnetic flux density at the centre so, the net magnetic flux density produced at the centre due to whole square loop will be B 8B1= 4 10 /Wb ma2 z

7 2#= -

SOL 4.2.32 Option (D) is correct.Using right hand rule we conclude that the field intensity produced at centre of the loop by the loop wire and the straight wire are opposing each other, so, the field intensity at the centre of the loop will be zero if Hwire Hloop= ...(1)where Hwire is the field intensity produced at the center of loop due to the straight wire and Hloop is the field intensity produced at the center of loop due to the current in the circular loop.Since the magnetic field intensity produced at a distance r from an infinitely long straight wire carrying current I is defined as

H I2pr=

So we have Hwire I

2 1 220 10

p p p= = =^ h

,A mI 20 1r= =^ h

and as calculated in Q.59 the field intensity produced by circular loop at its center is

Hloop aI2= where a is the radius of the loop

or, Hloop I I I

2 10 10 210 52

#= = =-

^ h ( 10 cma = )

So putting the values in eq. (1) we get

10p I5=

Thus, I A6p=




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SOL 4.2.33 Option (D) is correct.Consider one half side of the square loop to determine the magnetic field intensity at the centre O as shown in the figure.

The magnetic field intensity H produced at any point P due to a straight wire carrying current I is defined as

H cos cosI4 2 1pr a a= -6 @

where r " distance of point P from the straight wire. 1a " angle subtended by the lower end of the wire at P . 2a " angle subtended by the upper end of the wire at P .So we have r /2a= 1a /2p=and 2a /4p=Therefore the magnetic field intensity produced at centre O due to the half side of the square loop is

H1 ( / )cos cos

aI

2 2 4 2pp p= -a k

aI2 p

=

As all the eight half sides produces same field intensity at the centre of the loop so, net field intensity produced at the center due to the complete square loop is

Hnet 8a

I2 p

= c m aI2 2

p=

SOL 4.2.34 Option (A) is correct.For the shown current loop we divide the loop in two segments as shown in figure

Now the field intensity due to segment (1) (Semicircular loop) at point P can be given directly as calculated in Que.60



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For View Only Shop Online at www.nodia.co.ini.e. H1 a

I4= where a is radius of semicircular loop

or H1 2 /A m4 1

8= =^ h

( 1 ma = )

again for determining the field intensity due to segment (2) we consider it as the half portion of a complete square loop of side 2 m and since the field intensity due to a completer square loop of side a carrying current I can be directly given from previous question.

i.e. H aI2 2

p=

so the field intensity due to the half portion of square loop is

H2 H aI

21 2

p= =

or H2 228 4 2

p p= =^

^

h

h ( 8 AI = , 2 ma = )

As determined by right hand rule the direction of field intensity produced at point P due to the two segments will be same (inward) therefore, the net magnetic field intensity produced at point P will be

Hnet H H1 2= + 2 4 2p= + 3.8 /A m= inward.

SOL 4.2.35 Option (B) is correct.The flux density due to infinite current carrying sheet is defined as

B K a2 n0m

#=

where K is surface current density and an is unit vector normal to the surface directed toward the point where flux density is to be determinedSo, for the sheet in z 0= plane,

B1 4a a2 x z0

#m= ^ ^h h 2 ay0m=- a an z=^ h

and for the sheet in 2 mz = plane

B2 4a a2 x z0

#m= - -^ ^h h 2 ay0m=- a an z=-^ h

Therefore, the net flux density between the sheets is B 4B B ay1 2 0m= + =-Thus the magnetic flux per unit length along the direction of current is /lmy tanDis ce between the plates( )B#= 8 /Wb may0m=-

***********




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SOLUTIONS 4.3

SOL 4.3.1 Option (B) is correct.For r a> , Ienclosed ( )a J2p=

dH l:# Ienclosed=

2H rp^ h ( )a J2p=

H rI

2o

p= ( )I a Jo2p=

i.e. H r1

\ , for r a>

For r a< , Ienclosed ( )a

J raJr

2

2

2

2

pp= =

So, dH l:# Ienclosed=

2H rp^ h aJr

2

2

=

H aJr

2 2p= & H r\ , for r a<

SOL 4.3.2 Option (C) is correct.Assume that the cross section of the wire lies in the x -y plane as shown in figure below :

Since, the hole is drilled along the length of wire. So, it can be assumed that the



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For View Only Shop Online at www.nodia.co.indrilled portion carriers current density of J- .Now, for the wire without hole, magnetic field intensity at point P will be given as

( )H R21 pf ( )J R2p= & H 1f JR2=

Since, point O is at origin and the cross section of the wire located in x -y plane. So, in vector form the field intensity due to the current carrying wire without considering the hole is given as

H1 ( )J x ya a2 x y= + (1)

Again, only due to the hole magnetic field intensity at point P will be given as ( )( )H r22 pf ( )J r 2p=-

H 2f Jr2= -

Again, if we take Ol at origin then in vector form

H2 ( )J x ya a2 x y= - +l l (2)

where xl and yl denotes point ‘P ’ in the new co-ordinate system.Now the relation between two co-ordinate system will be x x d= +l and y y= l

So, putting it into equation (2) we have

H2 [( ) ]J x d ya a2 x y= - - +

Therefore, the net magnetic field intensity at point P is

Hnet J dH H a3 x1 2= + =

i.e. the magnetic field inside the hole will depend only on d .

SOL 4.3.3 Option (A) is correct.Due to 1 A current wire in x -y plane, magnetic field be at origin will be in x direction as determined by right hand rule.Due to 1 A current wire in y -z plane, magnetic field be at origin will be in z direction as determined by right hand rule.Thus, x and z -component is non-zero at origin.

SOL 4.3.4 Option (D) is correct.The total flux, F 1.2 mWb= 1.2 10 Wb3

#= -

Cross sectional area, A 30 cm2= 30 10 m4 2#= -

So, the flux density is given as

B AF= .

30 101 2 10

4

3

#

#= -

- 0.4 Tesla=

SOL 4.3.5 Option (D) is correct.The relation between magnetic flux density B and vector potential A is given as B A#d=





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For an isolated body the charge is distributed over its region which depends on the total change and the curvature of the body. Thus Statement 1 is correctSince the magnetic flux lines form loop so the total magnetic flux through any closed surface is zero. Thus Statement 2 is correct.

SOL 4.3.7 Option (D) is correct.The magnetic flux density in terms of vector potential is defined as B A#d=

dB S:# dSA#d= ^ h# F dA l:= #i.e. the line integral of vector potential A around the boundary of a surface S is equal to the flux through the surface S .

SOL 4.3.8 Option (B) is correct.Consider the current element along z -axis as shown in the figure.

Using right hand rule we get the direction of magnetic field directing normal to radial line OP .

SOL 4.3.9 Option (D) is correct.For the given circular cylinder, consider the surface current density is K . So, the total current I through the cylinder is given as K r2p^ h I=where r is radius of circular cylinder.

So, K rI

2p= 2 5 10

52

#p= -

^ h /A m50

p=

SOL 4.3.10 Option (B) is correct.Magnetic vector potential of an infinitesimally small current element is defined as

A RIdl

40

pm= #

where R is the distance from current element. Given that R" 3

So A 0=

SOL 4.3.11 Option (A) is correct.Consider the two parallel sheets are separated by a distance ‘d ’ as shown in the figure below



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The two sheets carries surface currents K K ay y=At any point between them the magnetic field intensity is given as

H K a a21

nu nl#= +^ h

where anu is the normal vector to the upper plate and anl is normal vector to the lower plate both directs toward the point between themi.e. anu az=- and anl az=

So, H K a a a21

y y z z#= - +^ h 0=

SOL 4.3.12 Option (B) is correct.For the given current distribution, the current enclosed inside the cylindrical surface of radius r for a b< <r is

Ienc Ja

d2a

0 2r pr r=

r

b ^l h# J a32

20 3 3

rp r= -^ h

and the magnetic field intensity is given as

dH l:# Ienc=

H 2pr^ h aJ a

32

20 3 3p r= -^ h

H a

J a4 2

03 3

rr

=-^ h

SOL 4.3.13 Option (D) is correct.The radiated E and H field are determined by following steps(1) Determine magnetic field intensity H from the expression B H Am #d= =(2) then determine E from the expression

H#d tE

22e=

So, the concept of vector magnetic potential is used to find the expression of radiated E and H field.




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SOL 4.3.14 Option (B) is correct.Using right hand rule, we conclude that the direction of field intensity is same as determined for the two correct elements I3 and I2 while it is opposite for the current element I . Therefore, from the ampere’s circuital law, we get the circulation of H around the closed contour as

dH l:# Ienclosed= I I I2 3= + - I4=

SOL 4.3.15 Option (C) is correct.The unit of magnetic flux density (B) is Tesla or /Wb m2

SOL 4.3.16 Option (D) is correct.From Ampere’s circuital law, the circulation of magnetic field intensity in a closed path is equal to the current enclosed by the path

i.e. dH l:# Ienc=So, for the current I the circulation at a radial distance R is given as H R2p^ h I=

or H RI

2p=

Therefore, the magnetic flux density at the radial distance R is

B H0m= RI

50

pm=

SOL 4.3.17 Option (D) is correct.Unit of work is Joule.Unit of electric field strength E^ h is volt/meter.Unit of magnetic flux is Weber.Unit of magnetic field strength is Ampere/meter.So, in the match list we get, A 4" , B 3" , C 2" ,D 1" .

SOL 4.3.18 Option (D) is correct.Magnetic field intensity at a distance r from a long straight wire carrying current I is defined as

H rI

2p=

1 r21p=

r 21p= 1.59 m=

SOL 4.3.19 Option (B) is correct.Consider the current flowing in the loop is I and since the magnetic field intensity is maximum at the centre of the loop given as

H rI2=

where r is radius of the loop. So, the current that must flow in the loop to produce the magnetic field 1 /mA mH = is



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For View Only Shop Online at www.nodia.co.in I rH2= 2 1 1 2 mA# #= =

SOL 4.3.20 Option (C) is correct.Magnetic flux density B in terms of vector potential A is defined as B A#d=So, B can be easily obtained from A also we know 0A:d = but it is not the explanation of Assertion (A).i.e. A and R both are true but R is not the correct explanation of A.

SOL 4.3.21 Option (A) is correct.(1) Magnetic flux density in terms of vector potential is given as

B A#d=(2) Poission’s equation for magnetic vector potential is

A2d J0m=-(3) Magnetic vector potential for a line current is defined as

A RIdl

20

pm= #

So, all the statements are correct.

SOL 4.3.22 Option (B) is correct.Magnetic field intensity due to a long straight wire carrying current I at a distance r from it is defined as

H rI

2p=

1 r210p=

r 1.59 m210p= =

SOL 4.3.23 Option (C) is correct.Magnetic field intensity due to an infinite linear current carrying conductor is defined as

dH l:# Ienc=

H r2p^ h I= & H rI

2p=

SOL 4.3.24 Option (D) is correct.The net outward magnetic flux through a closed surface is always zero as magnetic flux lines has no source or sinks.

i.e. dB S:# 0= (1)

Now, from Gauss’s law we have

dvB:d̂ h# 0dB S:= =# (2)

So, comparing the equation (1) and (2) we get B:d 0=




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SOL 4.3.25 Option (D) is correct.Given, the plane y 0= carries a uniform current density 30 /mA maz and since the point A is located at (1, 20, 2- ) so, unit vector normal to the current sheet is an ay=Therefore, the magnetic field intensity

H K a21

n#= 30a a21

z y#= ^ ^h h 15 /mA max=- ( 30K a mA/mz= )

SOL 4.3.26 Option (D) is correct.The magnetic flux density at any point is curl of the magnetic vector potential at that point.i.e. B A#d=From the Maxwell’s equation, the divergence of magnetic flux density is zero.i.e. B:d 0=Again from the Maxwell’s equation, the curl of the magnetic field intensity is equal to the current density.i.e. H4# J=or, B#d J0m= (B H0m= )The expression given in option (A) is incorrecti.e. B A:! d

SOL 4.3.27 Option (D) is correct.Superconductors are popularly used for generating very strong magnetic field.

SOL 4.3.28 Option (D) is correct.As the magnetic flux lines have no source or sinks i.e. it forms a loop. So the total outward flux through a closed surface is zero.

i.e. dB S:# 0=

SOL 4.3.29 Option (D) is correct.The magnetic field intensity due a surface current density K is defined as

H K a21

n#=

Where an is unit normal vector to the current carrying surface directed toward the point of interest.Given that, K 2ax= .and since the surface carrying current is in plane z 0= .So, for z 0< <a- an az=-

and H1 2a a21

x z#= -^ ^h h ay=

For z x0 < < , an az=

and H2 a a41 2 y z#= ^ ^h h az2

1=-

***********

CHAPTER 5MAGNESTOSTATIC FIELDS IN MATTER

268 Magnestostatic Fields in Matter Chap 5



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EXERCISE 5.1

MCQ 5.1.1 Path of a charged particle A that enters in a uniform magnetic field B (pointing into the page) is shown in the figure.

The deflection in the path of the particle shows that the particle is(A) positive charged (B) negatively charged

(C) uncharged (D) can’t be determined

MCQ 5.1.2 Unit of a magnetic point charge is(A) Ampere meter (B) coulomb meter

(C) Ampere meter square (D) doesn’t exit

MCQ 5.1.3 Assertion (A) : Both the electric force and magnetic force are produced when a charged particle moves at a constant velocity.Reason (R) : Electric force is an accelerating force where as magnetic force is a purely deflecting force.(A) Both A and R are true and R is correct explanation of A.




MCQ 5.1.4 An electron beam is passed through a uniform crossed electric and magnetic fields 15 /V mE ay= and /wb mB a23 z

2= (E and B are mutually perpendicular and both of them perpendicular to the beam). If the beam passes the field without any deflection then the velocity of the beam will be (A) 5 /m s (B) 45 /m s

(C) 30 /m s (D) 18 /m s

Chap 5 Magnestostatic Fields in Matter 269


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For View Only Shop Online at www.nodia.co.inMCQ 5.1.5 An electron is moving in the combined fields E 0.1 0.2 0.3 /kV ma a ax y z= - +

and B Teslaa a a2 3x y z=- + - . If the velocity of the electron at t 0= is (0) (200 300 400 ) /m sV a a ax y z= - - then the acceleration of the electron at t 0=

will be (charge on electron, 1.6 10 Ce 19

#= - ; mass of electron, 9.1 10 kgme 31#= - )

(A) 1.75 10 (1.1 1.4 0.5 ) /m sa a ax y z13 2

# + -

(B) 2.1 10 ( ) /m sa a ax y z4 2

# + -

(C) 3.5 10 (6 6 ) /m sa a ax y z13 2

# + -

(D) 3.19 10 (6 6 ) /m sa a ax y z17 2

# + --

MCQ 5.1.6 A current element of 2 m length placed along z -axis carries a current of 3 mAI = in the az+ direction. If a uniform magnetic flux density of /wb mB a a7x y

2= + is present in the space then what will be the force on the current element in the presence of the magnetic flux density ?(A) 6 18 mNa ax y- (B) 18 6 mNa ax y- +

(C) 18 6 mNa ax y- (D) 1.8 6 mNa ax y- +

MCQ 5.1.7 Consider two current loops C1 and C2 carrying current I1 and I2, separated by a distance R . If the force experienced by the current loop C2 due to the current loop C1 is F , then the force experienced by current loop C1 due to the current loop C2 will be(A) F- (B) F

(C) IIF2

1- b l (D) IIF

1

2b l

MCQ 5.1.8 Which of the following statements is correct for a current free interface between two different magnetic media ?(A) Normal component of magnetic field intensity will be continuous.

(B) Tangential component of magnetic flux density will be continuous.

(C) Magnetic scalar potential will be same in both the medium.

(D) None of these

Statement for Linked Question 9 - 10 :In the free space three uniform current sheets with surface current densities

4K ax1 = , K a4 x2 =- , 2K ax3 =- are located in the plane z 0= , z 1= and z 1=- respectively.

MCQ 5.1.9 Net magnetic field intensity produced between the sheets located at z 0= and z 1= will be(A) 2 /A may (B) 4 /A may(C) 2 /A may- (D) 0




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MCQ 5.1.10 If a conducting filament located along the line y 0= , 0.2 mz = carries 7 A current in ax+ direction then what will be the force per unit length exerted on it ?(A) 14 /N maz- (B) 14 /N max0m

(C) 14 /N maz0m (D) 14 /N maz0m-

MCQ 5.1.11 A rectangular coil of area 1 m2 carrying a current of 5 A lies in the plane x y z2 6 3 4+ - = . Such that magnetic moment is directed away from origin. If the

coil is surrounded by a uniform magnetic field 4 5B a a a2 x y z= + + /wb m2 then the torque on the coil will be(A) 3 20 20 N ma a ax y z -- - (B) 30 20 20 N ma a ax y z -- -

(C) 21 14 14 N ma a ax y z -- - (D) 6 4 4 N ma a ax y z -- -

MCQ 5.1.12 A circular current loop of radius 1 m is located in the plane z 0= and centered at origin. What will be the torque acting on the loop in presence of magnetic field

4 4 2 /wb mB a a ax y z2= - - , if a uniform current of 10 A is flowing in the loop ?

(A) 20 (2 )a ax zp - (B) 40 ( )a ax yp +

(C) 4 ( )a ax yp + (D) 40 ( )a ax yp -

MCQ 5.1.13 List I shows the type of magnetic materials and List-II shows their criterions. Match List I with List II and select the correct answer using the codes given below : (Notations have their usual meaning)

List-I List-II

a. Ferromagnetic 1. 0mc = , 1rm =

b. Diamagnetic 2. 0>mc , 1r Lm

c. Non-magnetic 3. 0<mc , 1r Km

d. Paramagnetic 4. 0>>mc , 1>>rm

Codes : a b c d(A) 2 3 1 4(B) 4 3 1 2(C) 4 1 3 2(D) 1 3 4 2

MCQ 5.1.14 Which of the following is a diamagnetic material ?(A) copper (B) sodium

(C) carbon (D) aluminum

MCQ 5.1.15 Assertion (A) : Aluminium is a paramagnetic material.Reason (R) : A paramagnetic material have an odd no. of electrons.(A) Both A and R are true and R is correct explanation of A.




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For View Only Shop Online at www.nodia.co.in(C) A is true but R is false.


MCQ 5.1.16 Magnetic flux density inside a medium is /mwb ma6 z2. If the relative permeability

of the medium is .2 3 then the magnetization inside the medium will be(A) 3979 /A m (B) 2249 /A m

(C) 9151 /A m (D) 8650 /A m

MCQ 5.1.17 Magnetic flux density inside a magnetic material is B . If the the permeability of the material is 3 0m m= then the vector magnetization of the material will be

(A) B3 0m (B) 3B2 0m

(C) B2 0m (D) 2B3 0m

MCQ 5.1.18 A portion of B -H curve for a ferromagnetic material can be approximated by the analytical expression kB H0m= . The magnetization vector M inside the material is(A) k H10m -^ h (B) kH

(C) k H1+^ h (D) k H1-^ h

MCQ 5.1.19 A magnetic material of relative permeability /4rm p= is placed in a magnetic field having strength 2 /A mH a2r= f . The magnetization of the material at 2r = will be(A) 8.19 /A maf (B) 1.10 /A maf(C) 0.546 /A maf (D) 2.19 /A maf

MCQ 5.1.20 A metallic bar of cross sectional area 2 m2 is placed in a magnetizing field /A mH 8=. If the field causes a total magnetic flux of 4.2 mWbF = in the bar then the susceptibility of the bar will be(A) .22 87 (B) .23 87

(C) .46 74 (D) 3 10 5#

-

MCQ 5.1.21 A large piece of magnetic material carries a uniform magnetization M and magnetic field intensity H0 inside it. The magnetic flux density inside the material is given by B0 ( )H M20 0m= +If a small spherical cavity is hollowed out of the material then the magnetic field intensity H at the center of the cavity will be

(A) 2H0 (B) H M30 +

(C) 2H M30 - (D) H M

30 -




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Statement for Linked Question 22 - 23 :A nonuniform magnetic field B inside a medium with magnetic susceptibility

2mc = is given as B Teslaza3 x=

MCQ 5.1.22 Bound current density inside the medium will be

(A) /A ma83

y0 2m (B) /A ma3

8y

0

2

m

(C) /A ma34

y0

2

m (D) /A ma316

y0

2

m

MCQ 5.1.23 Total current density inside the medium will be

(A) /A ma4t

0

2

m (B) /A ma34

y0

2

m

(C) /A ma38

y0

2

m (D) 4 /A may02m

MCQ 5.1.24 A uniformly magnetized circular cylinder of infinite length has magnetization M along it’s axis. The magnetic field intensity outside the cylinder will be(A) non uniform

(B) uniform and depend on the radius of circular cylinder

(C) zero

(D) none of these

MCQ 5.1.25 An infinite circular cylinder is located along z -axis that carries a uniform magnetization . /A mM a1 2 z= . The magnetic flux density due to it inside the cylinder will be(A) 2. 10 a2 7

# f- (B) 0.7af

(C) 8.8 10 a7# f- (D) 4.4af

MCQ 5.1.26 Magnetic flux lines are passing from a nickel material to the free space. If the incident of the flux line makes an angle 751 ca = to the normal of the boundary in the nickel side as shown in figure then what will be the angle 2a with normal of the flux when it comes out in free space ? (relative permeability of Nickel 600= )

(A) 15c (B) .1 23c

(C) .2 7c (D) .0 356c



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For View Only Shop Online at www.nodia.co.inStatement for Linked Question 27 - 28 :The two homogenous, linear and isotropic medium is defined in a Cartesian system such that medium 1 with relative permeability 8r1m = is located in the region y 0# and medium 2 with relative permeability 6r2m = is in the region y 0> .

MCQ 5.1.27 The magnetic field intensity in the 1st medium is 16 10H a a a13 x y z1 = + - . What will be the magnetic field intensity in the 2nd medium ?(A) 9 18.67 10 /A ma a ax y z- +

(B) 9 2.63 10 /A ma a ax y z+ -

(C) 9 18.67 10 /A ma a ax y z+ -

(D) 18.67 9 10 /A ma a ax y z- +

MCQ 5.1.28 Magnetic flux density in medium 2 will be(A) (6.8 14.1 7.5 ) 10 /wb ma a ax y z

5 2#- + -

(B) (6.8 14.1 7.5 ) 10 /wb ma a ax y z5 2

#+ - -

(C) (14.1 6.8 7.5 ) 10 /wb ma a ax y z5 2

#- + -

(D) (54 117 60 ) /wb ma a ax y z2+ -

MCQ 5.1.29 The magnetic flux density in the region z 0< is given as 4 /Wb mB a a8x z2= + .If

the plane z 0= carries a surface current density 4 /A mK ay= ; then the magnetic flux density in the region z 0> will be(A) 4 3 /Wb ma a1x z0

2m+ +^ h

(B) 4 3a a ax y z0m+ +

(C) /Wb ma a4 3 1x z 02m+ +^ ^h h

(D) 4 4 3 /Wb ma a ax y z02m+ +

MCQ 5.1.30 An infinite plane magnetic material slab of thickness d and relative permeability rm occupies the region x d0 < < . An uniform magnetic field BB az0= is applied in free space (outside the magnetic material). The field intensity Hin and flux density Bin inside the material will be respectively

(A) B0r 0m m and Br 0m (B) Br 0

0

m m and B0

(C) Br 0m and B0

0

m (D) B0

0

m and Br 0m

MCQ 5.1.31 Two infinite plane conducting sheets are located in the plane z 0= and 2 mz =.The medium between the plates is a magnetic material of uniform permeability

4 0m m= . If in the region between the plates a uniform magnetic flux density is defined as 10 /Wb mB a a12 4x y

3 2#= + -

^ h then the magnetic energy stored per unit area of the plates will be(A) 2.5 /J m2 (B) 4.1 /J m2

(C) 5 /J m2 (D) 7.8 /J m2




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MCQ 5.1.32 In the two different mediums of permeability 1m and 2m , the magnetic fields are (B1, H1) and (B2, H2) respectively as shown in the figure.

If the interface carries no current then the correct relation for the angle 1q and 2q is(A) cos cosB B1 1 2 2q q=

(B) cos cosH H1 1 2 2q q=

(C) sin sinB B1 1 1 2q q=

(D) Both (B) and (C)

MCQ 5.1.33 In a three layer medium shown in the figure below, Magnetic flux impinges at an angle 1q on the interface between regions 1 and 2. The permeability of three regions are 1m , 2m and 3m . So the angle of emergence 4q will be independent of

(A) 1m and 2m both (B) 2m only

(C) All 1m , 2m and 3m (D) 1m only

MCQ 5.1.34 A conducting wire is bent to form a circular loop of mean radius 0 cm2 . If cross sectional radius of the wire is a , such that 0 cma 2<< then the internal inductance of the loop will be(A) 125 H (B) 785 nH

(C) 157.1 nH (D) 250 nH



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For View Only Shop Online at www.nodia.co.inMCQ 5.1.35 The magnetic circuit shown in the figure has N turns of coil. Electrical analog for

the magnetic circuit shown in the figure is

MCQ 5.1.36 The coil of the magnetic circuit shown in figure has 100 turns.

Which of the following is correct electrical analog for the magnetic circuit ?




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MCQ 5.1.37 A 200 turns of a coil is wound over a magnetic core of length 15 cm that has the relative permeability of 150. The current that must flow through the coil to produce 0.4 Tesla of flux density in the core is(A) 320 A (B) 1.6 A

(C) 20.1 A (D) 0.63 A

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EXERCISE 5.2

MCQ 5.2.1 In the free space the magnetic flux density B points in the az direction and electric field E points in the ay direction as shown in the figure. If a charged particle at rest is released from the origin, then what path will it follow ?

MCQ 5.2.2 A point charge 2 C+ of mass kgm 2= is injected with a velocity 2 /m sv ay0 = into the region y 0> , where the magnetic field is given by 3 /wb mB ax 2= . If the point charge is located at origin at the time of injection then in the region y 0> the point charge will follow




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(A) a circular path centered at ( , , )0 0 2-

(B) an elliptical path centered at origin

(C) a circular path centered at ( , , )1 2 0

(C) a parabolic path passing through origin

Statement for Linked Question 3 - 4 :A filamentary conductor is formed into a rectangle such that it’s corners lies on points ( , , )P 1 1 0 , ( , , )Q 1 3 0 , ( , ,0)R 3 4 , ( , , )S 4 1 0 . An infinite straight wire lying on entire x -axis carries a current of 5 A in ax direction.

MCQ 5.2.3 If the filamentary conductor carries a current of 3 A flowing in ax+ direction from Q to R then the force exerted by wire on the side QR of rectangle will be(A) 3 10 Nay6

#- (B) 2 10 Nay6#- -

(C) 6 10 Nay6#- - (D) 3 10 Nay6

#-

MCQ 5.2.4 The total force exerted on the conducting loop by the straight wire will be(A) 6 10 Nay6

#- - (B) 12 10 Nay6#

-

(C) 6 10 Nay6#

- (D) 12 10 Nay6#- -

MCQ 5.2.5 Two filamentary currents of 5ax- and 5 Aax are located along the lines y 0=, 1 mz =- and y 0= , 1 mz = respectively. If the vector force per unit length exerted on the third filamentary current of 10 Aax located at y k= , z 0= be F then the plot of F versus k will be



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For View Only Shop Online at www.nodia.co.inMCQ 5.2.6 A current filament placed on x -axis carries a current 10 AI = in ax+ direction. If

a conducting current strip having surface current density 3 /A mK ax= is located in the plane y 0= between .z 0 5= and 1.5 mz = then what will be the force per unit meter on the filament exerted by the strip ?(A) 6.6 /N max m (B) 6.6 /N maz m

(C) 6 /N maz m (D) 0

MCQ 5.2.7 A conducting current strip of 5 m length is located in the plane x 0= between y 1= and y 3= . If surface current density of the strip is 6 /A mK az= then the force exerted on it by a current filament placed on z -axis that carries a current

10 AI = in az+ direction will be(A) 16.4 Nay m- (B) 4.8 Nay m-

(C) 26.4 Nay m- (D)26.4 Nay m

Statement for Linked Question 8 - 9 :A thick slab extending from y a=- to y a=+ carries a uniform current density JJ ax0=

MCQ 5.2.8 Plot of magnetizing factor H at any point in the space (inside or outside slab) versus y will be




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MCQ 5.2.9 If a magnetic dipole of moment mm ax0= is placed at the origin then the force exerted on it due to the slab will be(A) 0 N (B) m J yaz0 0 0m

(C) m J az0 0 0m (D) m J yaz0 0 0m-

Statement for Linked Question 10 - 11 :A long circular cylinder placed along z -axis carries a magnetization M a2 2r= f .

MCQ 5.2.10 The volume current density J at any point inside the cylinder is proportional to(A) r (B) /1 r

(C) sinr f (D) 2r

MCQ 5.2.11 The plot of the magnetic flux density B inside the cylinder versus r will be

MCQ 5.2.12 A short cylinder placed along z -axis carries a “frozen-in” uniform magnetization M in az+ direction. If length of the cylinder is equal to its cross sectional diameter then pattern of its surface current density K will be as



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MCQ 5.2.13 Magnetization of a long circular cylinder is M along it’s axis. Which of the following gives the correct pattern of magnetic field lines (B).

Statement for Linked Question 14 - 15 :A conducting rod of square cross section of side 2 cm carries a uniform magnetization

4 /A mM = along it’s axis. Length of the rod is cmL 4>> .

MCQ 5.2.14 If the rod is bent around it into a complete circular ring then magnetic flux density inside the circular ring will be(A) 4 /wb m2 (B) 4 /wb m0

2m

(C) 2 /wb m02pm (D) /wb m0

2m

MCQ 5.2.15 Assume that there remains a narrow gap of width 0.1 mm between the ends of the rod when it is formed into a circular ring. The net magnetic flux density at the center of the gap will be(A) 50.04 10 /wb m7 2

#- (B) 49.88 10 /wb m6 2

#-

(C) 51.23 10 /wb m6 2#

- (D) 34.66 10 /wb m6 2#

-




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MCQ 5.2.16 Magnetic flux density B inside a sphere that carries a uniform magnetization M will be

(A) 0 (B) M31

0m

(C) M2

0m (D) M32

0m

MCQ 5.2.17 A short cylinder of length equals to it’s diameter carries a uniform magnetization M as shown in the figure.

The correct sketch for the magnetic field intensity H inside the cylinder is

MCQ 5.2.18 Mutual inductance between an infinite current filament placed along y -axis and rectangular coil of 1500 turns placed in x -y plane as shown in figure will be



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For View Only Shop Online at www.nodia.co.in(A) 41.4 mH (B) 0.33 mH

(C) 2.38 mH (D) 33 Hm

MCQ 5.2.19 An infinitely long straight wire of radius a , carries a uniform current I . The energy stored per unit length in the internal magnetic field will be(A) uniform and depends on I only

(B) non uniform

(C) uniform and depends on a only

(D) uniform and depends on both I and a

MCQ 5.2.20 A planar transmission line consists of two conducting plates of 2 m width placed along x -z plane such that the current in one plate is flowing in az+ direction. While in the other it is flowing in az- direction. If both the plate carries 4 A current and there is a very small separation between them then what will be force of repulsion per meter between the two plates ?(A) 16 0m (B) 4 0m

(C) 8 0m (D) /40m

MCQ 5.2.21 A very long solenoid having ,20 000 turns per meter. The core of solenoid is formed of iron. If the cross sectional area of solenoid is 0.04 m2 and it carries a current

100 mAI = then what will be the energy stored per meter in it’s field ? (relative Permeability of iron, 100rm = )(A) 8.1 /J m (B) 20.11 /J m

(C) 100.5 /J m (D) 10.05 /J m

MCQ 5.2.22 A mass spectrograph is a device for separating charged particles having different masses. Consider two particles of same charges Q but different masses m and 2m injected into the region of a uniform field B with a velocity v normal to the magnetic field as shown in the figure. When the particles will be releasing out of the spectrograph the separation between them will be

(A) 2Bqmv (B) Bq

mv2

(C) Bqmv (D) 0




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Statement for Linked Question 23 - 24 :Consider a conducting filamentary wire of length 1 meter and mass 0.3 kg oriented in east-west direction, situated in the earth’s magnetic field at the magnetic equator.(Assume the magnetic field at equator has a value of 0. 10 /Wb m2 4 2

#- and

directed northward)

MCQ 5.2.23 The current that required to counteract the earth’s gravitational force on the wire must flow from(A) west to east (B) east to west

(C) any of (A) and (B) (D) none of these

MCQ 5.2.24 What will be the magnitude of the current flowing in the wire as to counteract the gravitational force ?(A) 49 kA (B) 24.5 kA

(C) 98 kA (D) 4.9 kA

MCQ 5.2.25 A rigid loop of wire in the form of a square is hung by pivoting one of it’s side along the x -axis as shown in the figure. The loop is free to swing about it’s pivoted side without friction. The mass of the wire is 0.2 /kg m and carries a current 2 A. If the wire is situated in a uniform magnetic field .96 /Wb mB 2 2= then the angle by which the loop swings from the vertical is

(A) /3 4p (B) /4p

(C) /6p (D) /2p

MCQ 5.2.26 Electron beams are injected normally to the plane edge of a uniform magnetic field HH ax0= as shown in figure.



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For View Only Shop Online at www.nodia.co.inThe path of the electrons ejected out of the field will be in(A) ay+ direction (B) ay- direction

(C) a ay z+^ h direction (D) a ay z-^ h direction

MCQ 5.2.27 The medium between the two infinite plane parallel sheets carrying current densities 4ax and /A ma8 x- , consists of two magnetic material slabs of thickness 1 m and 2 m having permeabilities 21 0m m= and 42 0m m= respectively as shown in the figure.

What will be the magnetic flux per unit length between the current sheets along the direction of flow of current ?(A) 24 /Wb may0m- (B) 16 /Wb may0m-

(C) 32 /Wb may0m- (D) 40 /Wb may0m-

MCQ 5.2.28 Two perfectly conducting, infinite plane parallel sheets separated by a distance d carry uniformly distributed surface currents with equal and opposite densities K and K- respectively. The medium between the two plates is a magnetic material of non uniform permeability which varies linearly from a value of 1m near one plate to a value of 2m near the second plate. What will be the magnetic flux between the current sheets per unit length along the direction of flow of the current ?

(A) Kd21 2m m+

b l (B) Kd1 2m m+^ h

(C) Kd1 11 2m m+b l (D) Kd2

2 1m m-a k

Statement for Linked Question 29 - 30 :The magnetic field intensity inside an infinite plane magnetic material slab is given as H a a12 24x y= + . The permeability of the magnetic material is m 2 0m=

MCQ 5.2.29 If the magnetic material slab occupies the region 0 2 mz< < then the magnetization surface current densities at the surfaces z 0= and 2 mz = will be respectively(A) 4 2a ay x- +^ h and 4 2a ay x-^ h (B) 2 4a ax y- +^ h and 2 4a ax y-^ h

(C) 4 4a ax y+^ h and 2 4a ax y-^ h (D) 2 4a ax y+^ h and 2 4a ax y- +^ h




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MCQ 5.2.30 The magnetization volume current density Jm will be(A) 0 (B) 4 2a ax y+

(C) 8 4a ax y+ (D) 4 2a ax y- -

MCQ 5.2.31 B -H curve for a ferromagnetic material is given as B 2 HH0m= . What will be the work done per unit volume in magnetizing the material from zero to a certain value B H20 0 0

2m= ?

(A) H4 3002

m (B) H4 3003

m

(C) H4 0 02m (D) H2 3

05

MCQ 5.2.32 Two infinitely long straight wire and a third wire of length l are parallel to each other located as shown in the figure.

Infinitely long wire carries a current I while the wire of length l shown at the top carries a current I2 . The magnitude of the force experienced by the top wire is

(A) I2

pm (B) I 2mp

(C) lI

2

2

pm (D) l

I2pm

MCQ 5.2.33 Two infinite plane conducting sheets lying in the plane x 0= and 5 cmx = carry surface current densities 0 /mA ma1 y+ and 20 /mA may- respectively. If the medium between the plates is a magnetic material of uniform permeability 2 0m m= then what will be the energy stored per unit area of the plates ?(A) 800 /J m2p

(B) 160 /J m2p

(C) 8 /J m2p

(D) 1.6 /J m2p

MCQ 5.2.34 Medium 1 comprising the region z 0> is a magnetic material with permeability 41 0m m= where as the medium 2, comprising the region z 0< is a magnetic material

with permeability 22 0m m= . Magnetic flux density in medium 1 is given by B1 0.4 0.8 /Wb ma a ax y z

2= + +^ h



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For View Only Shop Online at www.nodia.co.inIf the boundary z 0= between the two media carries a surface current of density K given by

K 0.2 0.4 /A ma a1x y

0m= -^ h

then the magnetic flux density in medium 2 will be(A) 0.8 /Wb ma a ax y z

2+ +^ h

(B) 0.8 /Wb ma a ax y z2- + -^ h

(C) 0.8 /Wb ma a ax y z2+ +^ h

(D) 0.8 /Wb ma ax y2+^ h

MCQ 5.2.35 A square loop of a conductor lying in the yz plane is bisected by an infinitely long straight wire carrying current 2 A as shown in the figure. If the current in the square loop is 4 A then the force experienced by the loop will be

(A) 1.6 Nay m (B) 6.4 Nay m

(C) 3.2 Nay m (D) 0.64 Nay m

MCQ 5.2.36 A certain region z 0< comprises a magnetic medium with permeability 52 0m m= . The magnetic flux density in free space z 0>^ h makes an angle q , with the interface whereas in medium 2 flux density makes an angle 2q as shown in the figure.

If 1.2 0.8B a ay z2 = + then what will be the angular deflection 1 2q q-^ h ?(A) .50 6c

(B) .19 47c

(C) .31 15c

(D) .12 06c




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Statement for Linked Question 37 - 38 :Consider the magnetic circuit shown in figure

The cross sectional area of the section on which coil is wound is S1 where as all the rest of the section has the cross sectional area S2. Magnetic core has the permeability 005 0m m= .

MCQ 5.2.37 If 5 cmS12= and 10 cmS2

2= then the total reluctance of the circuit will be(A) /41 100 0m (B) /9 20 0m

(C) /20 90m (D) /39 100 0m

MCQ 5.2.38 If the no. of turn of the coil is 100 then the equivalent self inductance of the coil is(A) 22.22 kH (B) 1.41 mH

(C) 27.9 mH (D) 4.5 kH

MCQ 5.2.39 The coil of a magnetic circuit has 50 turns. The core of the circuit has a relative permeability of 600 and length of the core is 0.6 m. What must be the core cross section of the magnetic circuit so that the coil may have a 0.2 mH inductance ?(A) 6.4 cm2 (B) 0.64 cm2

(C) 11.94 cm2 (D) 1.56 cm2

Statement for Linked Question 40 - 41 :A System of three coils on an ideal core is shown in figure



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For View Only Shop Online at www.nodia.co.inThe cross sectional area of all the segments of the core is 0 cmS 45 2= .

MCQ 5.2.40 If N 5001 = then what will be the self inductance of the coil having N1 turns ?(A) 0.125 mH (B) 62.8 mH

(C) 31.41 mH (D) 6.28 mH

MCQ 5.2.41 If N 2502 = then the self inductance of the coil N2 will be(A) 2.6 mH (B) 23.6 mH

(C) 70.7 mH (D) 2.36 mH

MCQ 5.2.42 A system of three coils on an ideal core that has two air gaps is shown in the figure.

All the segments of core has the uniform cross sectional area 2 00 mm5 2.What will be the mutual inductance between N1 and N2 ?(A) 39.27 mH (B) 52.36 mH

(C) 26.18 mH (D) 78.54 mH

MCQ 5.2.43 The mutual inductance between N2 and N3 will be(A) 0 (B) 78.54 mH

(C) 52.36 mH (D) 39.27 mH

MCQ 5.2.44 The magnetization curve for an iron alloy is approximately given by

B /Wb mH H31 2 2m= +

If H increases from 0 to 210 A/m, the energy stored per unit volume in the alloy is(A) 6.2 /MJ m3 (B) 1.3 /MJ m3

(C) 2.3 /kJ m3 (D) 2.9 /kJ m3

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EXERCISE 5.3

MCQ 5.3.1 A current sheet /A mJ a5 y= lies on the dielectric interface x 0= between two dielectric media with 5, 1r r1 1e m= = in Region-1 ( )x 0< and 2, 2r r2 2e m= = in Region-2 ( 0)x2 . If the magnetic field in Region-1 at x 0= - is 3 30 /A mH a ax y1 = + the magnetic field in Region-2 at x 0= + is

(A) 1.5 30 10 /A mH a a ax y z2 = + - (B) 3 30 10 /A mH a a ax y z2 = + -

(C) 1.5 40 /A mH a ax y2 = + (D) 3 30 10 /A mH a a ax y z2 = + +

MCQ 5.3.2 A bar magnet made of steel has a magnetic moment of 2.5 A m2- and a mass

of 6.6 10 kg3#

- . If the density of steel is 7.9 10 /kg m3 3# , the intensity of

magnetization is

(A) 8.3 10 /A m7#

- (B) 3 10 /A m6#

(C) 6.3 10 /A m7#

- (D) 8.2 10 /A m6#

MCQ 5.3.3 If the current element represented by 10 a2 y4

#- Amp-m is placed in a magnetic

field of 5 /H ax m= A/m, the force on the current element is(A) 2.0 mNaz- (B) 2.0 mNaz(C) 2.0 Naz- (D) 2.0 Naz

MCQ 5.3.4 Match List I with List II and select the correct answer using the code given below the lists :

List I List II

a. MMF 1. Conductivity

b. Magnetic flux 2. Electric current

c. Reluctance 3. EMF

d. Permeability 4. Resistance

GATE 2011

IES EC 2012

IES EC 2011

IES EC 2011



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For View Only Shop Online at www.nodia.co.inCodes : a b c d(A) 3 4 2 1(B) 1 2 4 3(C) 3 2 4 1(D) 1 4 2 3

MCQ 5.3.5 Consider the following statements associated with boundary conditions between two media:1. Normal component of B is continuous at the surface of discontinuity.

2. Normal component of D may or may not be continuous.

Which of the statement(s) given above is/are correct?(A) 1 only (B) 2 only

(C) Both 1 and 2 (D) Neither 1 nor 2

MCQ 5.3.6 Magnetic current is composed of which of the following ?(A) Only conduction component

(B) Only displacement component

(C) Both conduction and displacement components

(D) Neither conduction component nor displacement component

MCQ 5.3.7 Which one of the following is the correct expression for torque on a loop in magnetic field B ? (Here M is the loop moment)(A) T B4:= (B) T M B:=

(C) T M B#= (D) T B M#=

MCQ 5.3.8 Match List I with List II and select the correct answer using the code given below the lists :

List-I List-II

a. Line charge 1. Maxwell

b. Magnetic flux density 2. Poynting vector

c. Displacement current 3. Biot-Savart’s law

d. Power flow 4. Gauss’s lawCodes : a b c d(A) 1 2 4 3(B) 4 3 1 2(C) 1 3 4 2(D) 4 2 1 3

IES EC 2008

IES EC 2007

IES EC 2006

IES EC 2006




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MCQ 5.3.9 What does the expression J A21 : represent ?

(A) Electric energy density (B) Magnetic energy density

(C) Power density (D) Radiation resistance

MCQ 5.3.10 Two thin parallel wires are carrying current along the same direction. The force experienced by one due to the other is(A) Parallel to the lines

(B) Perpendicular to the lines and attractive

(C) Perpendicular to the lines and repulsive

(D) Zero

MCQ 5.3.11 A boundary separates two magnetic materials of permeability 1m and 2m . The magnetic field vector in 1m is H1 with a normal component Hn1 and tangential component Ht1 while that in 2m is H2 with a normal component Hn2 and a tangential component Ht2. Then the derived conditions would be(A) andH H H Ht t1 2 1 2= =

(B) andH H H Ht t n n1 2 1 1 2 2m m= =

(C) andH H H Hn n1 2 1 1 2 2m m= =

(D) , andH H H H H Ht t n n1 2 1 2 1 1 2 2m m= = =

MCQ 5.3.12 The dependence of B (flux density) on H (magnetic field intensity) for different types of material is

MCQ 5.3.13 Statement I : Polarization is due to the application of an electric field to dielectric materials.

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For View Only Shop Online at www.nodia.co.inStatement II : When the dipoles are created, the dielectric is said to be polarized or in a state of polarization.(A) Both Statement (I) and Statement (II) are individually true and Statement

(II) is the correct explanation of Statement (I)

(B) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I)

(C) Statement (I) is true but Statement (II) is false

(D) Statement (I) is false but Statement (II) is true

MCQ 5.3.14 The following equation is not valid for magneto-static field in inhomogenous magnetic materials(A) 0B:d = (B) 0H:d =

(C) A B#d = (D) H J#d =

MCQ 5.3.15 Assertion (A) : Superconductors cannot be used as coils for production of strong magnetic fields.Reason (R) : Superconductivity in a wire may be destroyed if the current in the wire exceeds a critical value.(A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is





MCQ 5.3.16 A conductor 2 metre long lies along the z -axis with a current of 10 A in az direction. If the magnetic field is 0. 5 TB a2 x= , the force on the conductor is(A) 4.0 Nay (B) 1.0 Naz(C) 1.0 Nay (D) 3.0 Naz

MCQ 5.3.17 The force on a charge moving with velocity v under the influence of electric and magnetic fields is given by which one of the following ?(A) q vE B#+^ h

(B) q vE H#+^ h

(C) q vH E#+^ h

(D) q vE B#+^ h

MCQ 5.3.18 If a very flexible wire is laid out in the shape of a hairpin with its two ends secured, what shape will the wire tend to assume if a current is passed through it ?(A) Parabolic (B) Straight line

(C) Circle (D) Ellipse

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MCQ 5.3.19 Consider the following :Lorentz force e vF B#= ^ h where ,e v and B are respectively the charge of the particle, velocity of the particle and flux density of uniform magnetic field. Which one of the following statements is not correct ?(A) Acceleration is normal to the plane containing the particle path and B

(B) If the direction of the particle path is normal to B , the acceleration is maximum

(C) If the particle is at rest, the field will deflect the particle

(D) If the particle path is in the same direction of B , there will be no acceleration

MCQ 5.3.20 What is the force on a unit charge moving with velocity v in presence of electric field E and magnetic field B ?(A) E v B:- (B) E v B:+

(C) E B v#+ (D) E v B#+

MCQ 5.3.21 What is the force experienced per unit length by a conductor carrying 5 A current in positive z -direction and placed in a magnetic field B a a3 4x y= +^ h ?(A) 15 20a a N/mx y+ (B) 20 15a a N/mx y- +

(C) 20 15a a N/mx y- (D) 20 20a a N/mx y- -

MCQ 5.3.22 Which one of the following formulae is not correct for the boundary between two magnetic materials ?(A) B Bn n1 2=

(B) B B Bn t2 2 2= +

(C) H H Hn t1 1 1= +

(D) ( )a H H Kn21 1 2# - = where an21 is a unit vector normal to the interface and directed from region 2 to region 1.

MCQ 5.3.23 Interface of two regions of two magnetic materials is current-free. The region 1, for which relative permeability 2r1m = is defined by z 0< , and region 2, z 0> has

1r2m = . If 1. . .4 TB a a a4 2 2 1x y x1 = + + ; then H2 is(A) 1/ 0.6 0.4 0.4a a a A/mx y z0m + +6 @

(B) 1/ 1.2 0.8 0.8a a a A/mx y z0m + +6 @

(C) 1/ 1.2 0.4 0.4a a a A/mx y z0m + +6 @

(D) 1/ 0.6 0.4 0.8a a a A/mx y z0m + +6 @

MCQ 5.3.24 If A and J are the vector potential and current density vectors associated with a

coil, then dvA J:# has the units of(A) flux-linkage (B) power

(C) energy (D) inductance

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SOLUTIONS 5.1

SOL 5.1.1 Option (A) is correct.Force applied by a magnetic field B on a moving charge with velocity v is defined as F v B#=Since the direction of velocity v and B are perpendicular to each other as obtainde from the shown figure so the resultant force will be perpendicular to both of them.i.e. the force on the moving charged particle will be in upward direction.and as the particle is also deflected in upward direction with the applied force so it gives the conclusion that the particle will be positively charged.

SOL 5.1.2 Option (D) is correct.Since a magnet bar must have south and north pole i.e. a single pole charge can’t exist. So, a magnetic point charge doesn’t exit.

SOL 5.1.3 Option (B) is correct.Force applied on a moving charge in the presence of electric and magnetic field is defined as F ( )qF F E v Be m #= + = +where Fe and Fm are the electric and magnetic forces applied on the charge so it is clear that the moving charge experiences both the electric and magnetic forces.The electric force is applied in a uniform direction (in direction of electric field) i.e. it is an accelerating force while, the magnetic force is applied in the normal direction of both the magnetic field and velocity of the charged particle i.e. it is a deflecting force.Therefore, both the options are correct but R is not the correct explanation of A.

SOL 5.1.4 Option (A) is correct.For a moving charge Q in the presence of both electric and magnetic fields, the total force on the charge is given by F ( )Q vE B#= +6 @

where E " electric field v " velocity of the charge B " magnetic flux densitySince the electron beam follows its path without any deflection so the net force applied by the field will be zero




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i.e. Q E v B#+ ^ h6 @ 0= 15 3va ay z#+ 0=As the electric field is directed along az and magnetic field is directed along ay so the velocity of beam will be in ax direction (perpendicular to both of the field).Consider the velocity of the beam is V kax=So we have 15 3ka a ay x z#+ 0= 15 3ka ay y- 0=

k 315= 5 /m s=

So, the velocity of the beam will be 5 /m s along the x -axis.

SOL 5.1.5 Option (C) is correct.For a moving charge Q in the presence of both electric and magnetic fields, the total force on the charge is given by F ( )Q vE B#= +6 @

where E " electric field v " velocity of the charged particle B " magnetic flux density So, at time t 0= total force applied on the electron is (0)F (0)e E V B#= + ^ h6 @

Now we have (0)V B# (200 300 400 ) ( 3 2 )a a a a a ax y z x y z#= - - - + - 1100 1400 500a a ax y z= + -therefore the applied force on the electron is (0)F ( . ) (0.1 0.2 0.3 )a a a1 6 10 10x y z

19 3# #= - +-

6 1100 1400 500a a ax y z+ + - @

(0)m ae 1.6 10 ( ) ( ) ( )a a a100 1100 1400 200 300 500x y z19

#= + + - + --6 @

( (0) (0)mF ae= , where (0)a is acceleration of electron at t 0= )

(0)a .. 200(6 6 )a a a

9 1 101 6 10

x y z31

19

#

##= + --

-

.5 10 ( ) /m sa a a6 7 2x y z13 2

#= + -

SOL 5.1.6 Option (B) is correct.Force F applied on a current element in the presence of magnetic flux density B is defined as F I L B#= ^ h

where I " current flowing in the element L " vector length of current element in the direction of current flowingSo, F 3 10 ( )a a a2 3z x y

3##= +-

6 @

6 10 a a3y x3

#= --6 @ 1 6 mNa a4 1x y=- +

SOL 5.1.7 Option (A) is correct.The magnitude of the force experienced by either of the loops will be same but the direction will be opposite.So the force experienced by C1 due to C2 will be F- .




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For View Only Shop Online at www.nodia.co.inThe boundary condition for the current interface holds the following results.(i) normal component of magnetic flux density is continuous.

i.e. B n1 B n2=(ii) Tangential component of magnetic field intensity is continuous.

i.e. H t1 H t2=So, (A) and (B) are wrong statement. Now, we check the statement (C).

Consider the magnetic field intensity in 1st medium is H1 and magnetic field intensity in 2nd medium is H2. So, it’s tangential component will be equali.e. H t1 H t2= (tangential component)Since scalar magnetic potential difference is defined as the line integral of magnetic field intensity

i.e. V V1 2- d IH l:= =#and since there is no current density at boundary.So, we have V V1 2- 0= V1 V2=i.e. magnetic scalar potential will be same in both medium.

SOL 5.1.9 Option (C) is correct.The magnetic field intensity produced at any point in the free space will be the vector sum of the field intensity produced by all the current sheets.Since, the magnetic field intensity produced at any point P due to an infinite sheet carrying uniform current density K is defined as

H ( )K a21

n#=

where an is the unit vector normal to the sheet directed toward the point P . So in the region 1z0 < < magnetic field intensity due to K2 and K3 will be cancelled as the unit normal vector to the two sheets will be opposite to each other.Therefore in this region magnetic field intensity will be produced only due to the current density 4K ax1 = which is given as

H K a21

n1#= (4 ) ( )a a21

x z#= (an az= )

/A ma4 z=-

SOL 5.1.10 Option (D) is correct.As the conducting filament is located along the line y 0= , 0.2 mz = which is in the region 0 1 mz< < , so, the net magnetic field intensity produced on the conducting filament by the current sheets is H 2 /A may=- (as determined in previous question)or, B H0m= 2 ay0m=-Now the force experienced by a current element Idl in the presence of magnetic flux density B is defined as dF Idl B#=where I is the current flowing in the element and dl is the differential vector length




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of the current element in the direction of flow of current.So force per unit length experienced by the conducting filament is

dldF 7 ( 2 )a ax y0m#= - ( 7 AI = , d dll ax= )

14 /N maz0m=-

SOL 5.1.11 Option (B) is correct.Magnetic dipole moment of a coil carrying current I and having area S is given by m ISan=where an is normal vector to the surface of the loop.Since the coil is lying in the plane x y z2 6 3 4+ - = so the unit vector normal to the plane of the coil is given as.

So, an ff

4

4 ( )

2 6 3a a a2 6 3x y z2 2 2

=+ + -+ -

( f x y z2 6 3= + - )

Therefore the magnetic dipole moment of the coil is

m (5)(1)( )a a a

72 6 3x y z= + -

( 5 , 1A mI S 2= = )

( )a a a

75 2 6 3x y z= + -

As the torque a magnetic field B on the loop having magnetic moment m is defined as T m B#=So the torque on the given coil is

T ( )

(6 4 5 )a a a a a a7

5 2 6 3x y zx y z#= + - + +; E

3 N ma a a5 8x y z -= - -

SOL 5.1.12 Option (B) is correct.Magnetic dipole moment of a coil of area S carrying current I is defined as m ISan=where an is the unit vector normal to the surface of the loop.and since from the given data we have I 10 A= S ( )r 12 2

#p p= = p= an az= (normal vector to the surface z 0= )So the magnetic moment of the circular current loop lying in the plane z 0= is m 10 azp=Now the torque on an element having magnetic moment m in the presence of magnetic flux density B is defined as T m B#=Therefore, the torque acting on the circular loop is T (10 ) (4 4 2 )a a a az x y zp #= - - (B 4 4 2a a ax y z= - - ) 10 (4 4 )a ay xp= + 0 ( )a a3 x yp= +



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SOL 5.1.14 Option (C) is correct.A diamagnetic material carries even no. of electrons inside it’s atom.Number of electron in carbon atom is six.Which is even so it is a diamagnetic material rest of the material having odd no. of electrons.

SOL 5.1.15 Option (A) is correct.A paramagnetic material have an odd no. of electrons and since atomic no. of Al is 13, which is odd. So, it is a paramagnetic material.So, A and R both true and A is correct explanation of R.

SOL 5.1.16 Option (B) is correct.In a magnetic medium the magnetization in terms of magnetic field intensity is defined as M Hmc=where mc is magnetic susceptibility given as mc 1rm= - .1 3= (relative permeability, 2.3rm = )and since the magnetic field intensity in terms of magnetic flux density is given as

H Bm= B

r0m m= (B 5 /mwb maz 2= )

.a

4 10 2 35 10

z7

3

# #

#p

= -

- 1730 /A maz= ( rm .2 3= )

So the magnetization inside the medium is M 2249 /A mHmc= =

SOL 5.1.17 Option (D) is correct.Given the permeability, 3 0m m= and magnetic flux density B=So the field intensity inside the material will be

H B B3 0m m= =

Since the magnetization of a magnetic material is defined as

M B H0m

= -

where B and H are the flux density and field intensity inside the material. So we get

M 2B B B3 30 0 0m m m= - =

SOL 5.1.18 Option (D) is correct.As the magnetic flux density and magnetic field intensity inside a magnetic material are related as B Hr 0m m=So, comparing it with given expression for magnetic flux density we get the relative permeability as rm k= 1k= -




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Therefore, the magnetization vector inside the material is given as M ( 1)Hrm= - k H1= -^ h

SOL 5.1.19 Option (D) is correct.Magnetic flux density in a medium in terms of magnetic filed intensity is defined as B Hm= Hr 0m m= 4/ (4 10 )(2 )a7 2p p r#= f

-^ h ( /4rm p= , 2 /A mH a2r= f )

10 a16 7 2r#= f-

Again the magnetic flux density inside a magnetizing material is defined as B H M0m= +^ h

where M is the magnetization of the material. So, we have

M B H0m

= - a4 10

32 10 27

7 22

#

#

pr r= - f-

-

; E 2 a4 12r p= - f: D

At 2r = M . /A ma4 14= f

SOL 5.1.20 Option (A) is correct.GivenMagnetic field intensity, H 70 /A m=Total magnetic flux in the bar, F 4.2 mWb=Cross sectional area of bar, S 2 m2=So we have the magnetic flux density in the bar

B SF= .

24 2 10 3

#=-

2.1 /mwb m2=Since the magnetic field intensity and magnetic flux density are related as B (1 )Hm0m c= +So, we have 2.1 10 3

#- (4 10 )(1 )(70)m

7p c#= +-

( )1 mc+ ( )

( . )70 4 10

2 1 107

3

#

#

p= -

-

mc 4 103 10 17

5

#

#p

= --

-

c m

( . )23 87 1= - 2 . 72 9=

SOL 5.1.21 Option (B) is correct.For the spherical cavity of magnetization M , the flux density is given by

Bcavity M32

0m=

Since the cavity is hollowed. So not magnetic flux density at the center of cavity is

Bnet B Bcavity0= - B M32

0 0m= -

and so the net magnetic field intensity at the center of cavity is

Hnet B1net

0m= B M1

32

00 0m m= -: D

H M M132

00 0 0 0m m m m= + -: D H M

30= +: D (B H M0 0 0m= +^ h)



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In a magnetic medium the magnetic field intensity and magnetic flux density are related as B (1 )Hm0m c= +So the magnetic flux density inside the medium is

H ( )B

1 m 0c m= + z a34

x0m

= (B 4 Tzax= , mc 2= )

Now the magnetization of a magnetic medium having magnetic field intensity H is given as M Hmc=

M 2 z a34

x0m

= b l z a38

x0m

=

The bound current density inside a medium having magnetization M is given as Jb M#d=

38z a

0x4 m#= b l /A ma3

12y

0

2

m=

SOL 5.1.23 Option (A) is correct.Total current density inside a medium having magnetic flux density B is given as

JT B0m

#d= ( )zz a1 4

y0 22

m= ; E (B 4 Tzax= )

/A ma43

y0

2

m=


Volume current density inside a material is equal to the curl of magnetization M i.e. J M#d=and the surface current density in terms of magnetization is defined as K M an#=where an is unit vector normal to the surface. Consider the cylinder is placed along z -axisSo, an a= r

and M Maz=Therefore the volume current density inside the cylinder is J ( )Ma 0z4#= = (M is not the function of z )




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and the surface current density of the cylinder is K Ma az#= r Ma= f

So the current flowing in cylinder is just similar to a solenoid and the field intensity produced due to a solenoid at any point outside it is zero. Thus we have the magnetic field intensity outside the cylinder as Houtside 0=


Volume current density inside a material is equal to the curl of magnetization M i.e. J M#d=So the volume current density inside the cylinder is J (0.7 )az#d= 0= ( 0.7 /A mM az= )and since the surface current density in terms of magnetization is defined as K M an#= where an is unit vector normal to the surface.So the surface current density of the cylinder is K (0.7 )a az #= r 0.7a= f ( 0.7 /A mM az= , a an = r)Therefore the current flowing in cylinder is just similar to a solenoid and the field intensity produced due to a solenoid at any point inside it is given as B K0m= nI0m=where n is the no. of turns per unit length of the solenoid and I is the current flowing in the solenoid.Thus, the magnetic flux density inside the cylinder is (direction is determined by right hand rule) B 0.7 az0m= .8 10 a2 z

7#= - ( 0.7K = )

SOL 5.1.26 Option (D) is correct.From Snells law we have the relation between the incidence and refracted angle of magnetic flux lines as

tantan

2

1aa

r

r

2

1

mm=

where r1m and r2m are relative permeability of the two medium.



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For View Only Shop Online at www.nodia.co.in tan

tan752

ca 1

600= (relative permeability of air 1= )

tan 2a tan600

75c=

2a tan tan600

751 c= -: D .0 356c=

SOL 5.1.27 Option (C) is correct.Magnetic field intensity in 1st medium is given H1 9 16 10a a ax y z= + - H Ht n1 1= +where H1t and H n1 are respectively the tangential and normal components of the magnetic field intensity to the boundary interface in medium 1.From boundary condition we have H t1 H t2=and H n2 2m H1n1m=where H t2 and H n2 are respectively the tangential and normal component of magnetic field intensity in medium 2. So we get the components in medium 2 as H t2 9 10a ax z= -

and H2n H n2

11m

m= Hr

rn

2 0

1 01m m

m m=

(16 ) 18.67a a67

y y= =

Therefore, the net magnetic field intensity in medium 2 is H2 H Ht n2 2= + 1 .67 1 /A ma a a12 3 5x y z= + -

SOL 5.1.28 Option (B) is correct.Magnetic flux density in any medium in terms of magnetic field intensity is defined as B Hm=where m is the permeability of the medium. So, the magnetic flux density in medium 2 is given as B2 H2 2m= Hr2 0 2m m= 6 (4 10 ) (9 18.67 10 )a a ax y z

7p# # #= + -- ( 6r2m = ) (6.8 14.1 7.5 ) 10 /wb ma a ax y z

5 2#= + - -

SOL 5.1.29 Option (D) is correct.The magnetic flux density in region z 0< is given as B 4 3 /Wb ma ax z

2= +




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B1 4 3a ax z= +Therefore the tangential component B t1 and normal component B n1 of the magnetic flux density in region 1 are B t1 4ax=and B n1 3az=From the boundary condition the tangential and normal components of magnetic flux density in two mediums are related as B n1 B n2= B Bt t2 1- K0m=where B t2 and B n2 are respectively the tangential and normal components of the magnetic flux density in region 2 and K is the current density at the boundary interface.So, we get B n2 B n1= 3az= (B n1 3az= )and B t2 4 4a ax y0m= + ^ h (B t1 4ax= , 4 /A mK ay= ) 4 4a ax y0m= +Therefore the net flux density in region 2 z 0>^ h is B2 B Bt n2 2= + 4 a a a2 5x y z0m= + +

SOL 5.1.30 Option (D) is correct.As the surface boundary of the slab is parallel to yz -plane so the given magnetic flux density will be tangential to the surface.i.e. Bto B0=

and Hto B Bto

0 0

0

m m= =

Since the tangential component of magnetic field intensity is uniform at the boundary of the magnetic material So, magnetic field intensity inside the material is

Hin H Bto

0

0

m= =

Therefore, the flux density inside the material is

Bin Hinm= Br 0

0

0m m m= Br 0m=



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The magnetic stored energy per unit volume of the plate for a given uniform flux density (uniform permeability) is defined as

wm H B21

:=

Given B 3 4 10 /Wb ma ax y3 2

#= + -^ h

So we have, H 3 4

10 /A mB a a4x y

0

3

m m #= = + -c m

and therefore wm H B21

:= 21

49 16 10

0

6#m= + -

: D 8 4 10

25 107

6

# #

#p

= -

-

2.49 /J m3=now the separation between the plates is given as 2 md =Thus magnetic energy stored per unit area of the plate is /W Am w dm#= .2 49 2#= ^ h 5 /J m2=

SOL 5.1.32 Option (D) is correct.From boundary condition the normal component of flux density is uniform at boundaryi.e. B n1 B n2= sinB1 1q sinB2 2q=and the tangential component of field intensity is uniformi.e. H t1 H t2= cosH1 1q cosH2 2q=

SOL 5.1.33 Option (B) is correct.Relation between 1q and 2q at boundary of region (1) and region (2) as tan1 1m q tan2 2m q=and at the interface between region (2) and region (3) is tan2 3m q tan3 4m q= ; since 2 3q q=So, combining the two eq. we get, tan1 1m q tan3 4m q=Thus, 4q will be independent of 2m only.

SOL 5.1.34 Option (C) is correct.Internal inductance of a loop of radius r is defined as

Lin ( )r8 20

pm p= x

84 10 2 50 107 2

# # # #p

p p=- -

( 50 cmr = )

157.1 nH=

SOL 5.1.35 Option (A) is correct.From the analogy between electrical and magnetic circuits, we have the following relations, F (magnetomotive force) " V (voltage) f (magnetic flux) " I (current)




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R (Reluctance) " R (Resistance)now, magnetomotive force, F NI0=and so, the electrical analog of the magnetic circuit is

SOL 5.1.36 Option (A) is correct.For drawing the electrical analog replace the coil by a source (magnetomotive force) and each section of the core by a reluctance. In the shown magnetic material there are 9 sections so we draw the reluctance for each of them and we get the magnetomotive force as F I1000= (N 1000= )So the equivalent circuit is

SOL 5.1.37 Option (B) is correct.Given that,the magnetic flux density, B 0.4 T=no. of turns of coil, N 200=length of magnetic core, l 15 cm 15 10 2

#= = -

permeability of the core, rm 150=So, current required to produce the given magnetic field is

i NBlm=

.150 200

0 4 15 100

2#

m=

-

^ ^

^ ^

h h

h h

.6 A2=

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SOLUTIONS 5.2

SOL 5.2.1 Option (A) is correct.Consider the particle carries a total charge Q . Since for a moving charge Q in the presence of both electric and magnetic fields, the total force on the charge is given by F ( )Q vE B#= +6 @

where E " electric field v " velocity of the charged particle B " magnetic flux density So initially the magnetic force on the particle will be zero as the particle is released at rest (v 0= ). Therefore the electric field will accelerate the particle in y -direction and as it picks up speed (consider the velocity is kv ay= , k is very small) a magnetic force develops which will be given by F v B#=since the magnetic field is in az direction while the beam has the velocity in ay direction so the magnetic force will be in ax (a ay z# ) direction.Therefore the magnetic force will pull the charged particle around to the right and as the magnetic force will be always perpendicular to both the velocity of particle and electric field. So the particle will initially goes up in the y -direction and then following a curve path lowers down towards the x -axis.

SOL 5.2.2 Option (A) is correct.For a moving charge Q in the presence of both electric and magnetic fields, the total force on the charge is given by F ( )Q vE B#= +6 @

where E " electric field v " velocity of the charged particle




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B " magnetic flux density Since initially the velocity of the charge (at the time of injection) is 2 /m sv ay0 =and for the region y 0> magnetic flux density is B 3 /wb max 2= .so there will be no any velocity component in ax+ direction caused by the field (since the magnetic field is in ax direction).i.e. vx 0=So we consider the velocity of the point charge in the region y 0> at a particular time t as v v va ay y z z= +Therefore we have the force applied by the field on the charge particle at time t as F Q v va a a3y y z z x#= +^ ^h h6 @

m dtdv

dtdva ay

yzz+; E 3 3Q v va ay z z y= - +6 @

So, we get dtdvy m

Q v3z=

and dtdvz m

Q v3y=-

From the two relations we have

dtd v

mQ v3z

z2

2 2

+ b l 0=

vz cos sinA mQ t B m

Q t3 31 1= +b bl l

where A1 and B1 are constants.and since at t 0= , 0vz = (since charge was injected with a velocity in ay direction)Putting the condition in the expression we get A1 0=

and so we have vz sin sinB mQ t B t3

1 1= =b l 2 CQ = , 6 kgm =

Again, dtdvz m

Q v3y=-

so vy Qmdtdv

3z=-b l cos cosB m

q t B t31 1=- =-b l

and since at t 0= , 2 /m svy =Putting the condition in the expression we get, 2 cosB 01=- B1 2=-

So, we have, vz sin t2=- " sindtdz t2=-

vy cos t2= " cosdtdy t2=

Solving the equations we get, z cos t C2 2= +and y sin t C2 3= +and since at t 0= , y z 0= = (charge is located at origin at the time of injection)



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For View Only Shop Online at www.nodia.co.inPutting the condition in the expression we get C2 2=- and C 03 =So we have z cos t2 2= - &z 2+ cos t2= and y 2 sin t=Therefore the equation of the path that the charged particle will follow is ( )y z 22 2+ + 4=This is the equation of a circle centred at (0,0, )5- .


The magnetic flux density produced at a distance r from an infinitely long straight wire carrying current I is defined as

B I2

0

prm=

So the magnetic flux density produced by the straight wire at side QR of the loop is (direction of magnetic flux density is determined by right hand rule)

BQR ( )I a

2 3 z0 1

pm= ( 3r = )

a65

z0

pm= ( 5 AI1 = )

Force experienced by a current element Idl in the presence of magnetic flux density B is defined as dF Idl B#=where I is the current flowing in the element and dl is the differential vector length of the current element in the direction of flow of current.So the force exerted by wire on the side QR of the square loop is

FQR I dl BQRQ

R

2 #= #where I2 is the current flowing in the square loop as shown in the figure. So, we get

FQR ( )dxa a3 65

xz

x

0

1

4

# pm=

=b l# ( 3 ,AI d dxl ax2 = = )

[4 1]( )a25

y0

pm= - - a2

5 4 10 3y

7# # #

pp= - -

3 10 Nay6#=- -




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SOL 5.2.4 Option (C) is correct.Total force on the loop will be the vector sum of the forces applied by the straight wire on all the sides of the loop. The forces on sides PQ and RS will be equal and opposite due to symmetry and so we have F FPQ RS+ 0=Therefore the total force exerted on the conducting loop by the straight wire is Ftotal F FQR SP= + (1)where FQR and FSP are the forces exerted by the straight wire on the sides QR and SP of the conducting loop respectively.As calculated in previous question we have FQR 3 10 Nay6

#=- -

Similarly we get the force exerted by the wire on the side SP of the loop as

FSP I dl BSPS

P

2 #= #where BSP is the magnetic flux density produced by the wire on the side SP . So, we get

BSP ( )I a

2 1 z0 1

pm= ( 1r = )

a25

z0

pm= ( 5 AI1 = )

FSP ( )dxa a3 25

x z0

1

4

# pm= -# ( 3 ,AI d dxl ax2 = =- )

9 10 Nay6#= -

Thus, from equation (1), the total force exerted by the straight wire on the conducting loop is Ftotal 3 10 9 10a ay y

6 6# #=- +- -

10 Na12 y6

#= -


Net magnetic flux density arising from the two current filaments 5ax- and 5 Aax



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For View Only Shop Online at www.nodia.co.inat the location of third filament is given by B B B1 2= + (1)where B1 and B2 are the magnetic flux density produced by the current filaments 5ax and 5ax- respectively. Since the magnetic flux density produced at a distance r from a straight wire carrying current I is defined as

B I a20

prm= f

and the direction of the magnetic flux density is given as af a al#= r

where al is unit vector along the line current and ar is the unit vector normal to the line current directed toward the point P . So, the magnetic flux density produced by the current filament 5ax is

B1 ( )k k

ka a a2 1

51

xy z

20

2#p

m=+ +

-e o> H

( )

( )kka a

2 15

z y20

pm=+

+

Similarly the magnetic flux density produced by the current filament ( 5 )axis

B2 ( )

( )( )

k kka a a

2 15

1x

y z2

02

##

pm=

+-

++

e o> H

( )

( )k

ka a2 1

5z y2

0

pm=+

- +

Therefore from equation (1), we get the net magnetic flux density experienced by the third filamentary current of 10 Aaz as

B ( )k

k ka a a a2 1

5z y z y2

0

pm=+

+ - +6 @

( )

(2 )k

a2 1

5y2

0

pm=+

( )k

a15

y20

pm=+

As the force experienced by a current element Idl in the presence of magnetic flux density B is defined as dF Idl B#=where I is the current flowing in the element and dl is the differential vector length of the current element in the direction of flow of current.Force per unit meter length experienced by the third filament is

F ( )( )

dxk

a a1015

x yx

20

0

1

# pm=+=

# (d dxl ax= )

( )k

a1

10 5 4 10z2

7# # #

pp=

+

-

( )

Nka

120 z

2 m=+

or, F ( )

Nk1

122 m=

+Thus, the graph between F and k will be as shown in the figure below :




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Consider the strip is formed of many adjacent strips of width dz each carrying current Kdz .

Since the magnetic flux density produced at a distance r from a straight wire carrying current I is defined as

B I2

0

prm=

So the magnetic flux density produced by each differential strip is

dB ( )zKdz a2 y

0

pm= (I Kdz= )

(using right hand rule we get the direction of the magnetic flux density along ay)Therefore the net magnetic flux density produced by the strip on the current filament is

B z dza

23

.

.y

z

0

0 5

1 5

pm=

=# .

.ln a23

0 51 5

y0

pm= b l ( 3 /A mK = )

6.6 10 /wb may7 2#= -

As the force experienced by a current element Idl in the presence of magnetic flux density B is defined as dF Idl B#=where I is the current flowing in the element and dl is the differential vector length



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For View Only Shop Online at www.nodia.co.inof the current element in the direction of flow of current. So the force exerted on the filament per unit length is

F Idl B#= # ( ) ( . )dxa a10 6 6 10x yx

7

0

1

# #= -

=#

. /N ma2 4 z m=

SOL 5.2.7 Option (C) is correct.Consider the strip as made up of many adjacent strips of width dy , each carrying current Kdy

Since the magnetic flux density produced at a distance r from a straight wire carrying current I is defined as

B I2

0

prm=

So the magnetic flux density produced at distance y from the current filament located along z -axis as shown in the figure will be

B ( )yI a2 x

0

pm= - (Direction is determined using right hand rule)

y a210

x0

pm=-

As the force experienced by a current element Idl in the presence of magnetic flux density B is defined as dF Idl B#=and since the length of strip is 2 ml = so, the force exerted on the width dy of strip is given by dF ( )l dyK B#=Therefore the net force exerted on the strip is

F (2)(6 ) y dya a210

z xy

0

1

3

pm

#= -=

c m# ( 2 , 6ml K az= = )

lnya60y

013

pm=- 6 @

.4 Na13 y m=-




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SOL 5.2.8 Option (A) is correct.Consider a rectangular Amperian loop of dimension ( ) ( )l y2# inside the slab as shown in the figure below.

As from the Amperes circuital law, we have

dH l:# Ienc=So for the Amperian loop inside the slab we get 2 )l(H ( )( )y l J2 0#= for a y a# #-(Net magnetic field intensity along the edge y2 will be cancelled due to symmetry)Therefore the magnetic field intensity (magnetizing factor) at any point inside the slab is H J yaz0=or H J y0= (for y a# )and the magnetic field intensity (magnetizing factor) at any point outside the slab is H J a0= (for y a> )Thus, the plot of H versus y will be as shown below

SOL 5.2.9 Option (A) is correct.Force on any dipole having moment m due to a magnetic flux density B is defined as F ( )m B:d=Since the magnetic moment of the dipole is given as



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For View Only Shop Online at www.nodia.co.in m m ax0= (1)and as calculated in previous question the magnetic field intensity produced due to the slab is H J yaz0=So we get the magnetic flux density produced due to the slab as B H0m= J yaz0 0m= (2)Therefore from equation (1) and (2) we get m B: 0=Thus the force acting on the dipole is F 0=

SOL 5.2.10 Option (A) is correct.Volume current density inside a magnetic material is equal to the curl of its magnetization M i.e. J M#d=So volume current density due inside the circular cylinder is

J (5 )a1z

2

22

r r r r= 15 azr=

or J \ r

SOL 5.2.11 Option (B) is correct.As calculated above the volume current density inside the cylinder is J 15 azr=So, we can get the flux density by Ampere’s circuital law as

dB l:# Ienc0m=

( )(2 )B pr Ienc0m=

B I2 enc0

prm=

Now the enclosed current in the loop is

Ienc dJ SS

:= # ( )( )d d150

2

0r r r f=

pr## 2 15 3

3

0#p r=

r

; E

10 3pr=So, the magnetic flux density inside the cylinder is

B I2 enc0

prm= 5 0

2m r= ( 10I 3pr= )

Thus the plot of magnetic flux density B versus r is as shown below




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SOL 5.2.12 Option (B) is correct.The surface current density of a material in terms of its magnetization is defined as K M an#= where an is unit vector normal to the surface.So, the surface current density of the cylinder is K ( ) ( )Ma az #= r Ma= f ( ,MM a a az n= = r)Therefore the surface current density is directed along af as shown in option (B).

SOL 5.2.13 Option (A) is correct.Magnetic flux density inside a magnetic material is defined as B ( )H M0m= + So, B and M will be in same direction inside the cylinder.Now as the magnetic field lines are circular so outside the cylinder it will make a loop. Thus, the magnetic field lines will be as shown below




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For View Only Shop Online at www.nodia.co.inLet the circular ring being placed such that magnetization M is in af direction and the ring is centered at origin.So, we have M 4a= f

As the surface current density of a material in terms of its magnetization is defined as K M an#= where an is unit vector normal to the surface.So the surface current density of the ring is K 4 ( )a a#= -f r 4az= (M 4a= f , a an =- r)and since the volume current density inside a material is equal to the curl of magnetization M i.e. J M#d=So the volume current density inside the ring is J (4 ) 0a#d= =f ( 4M a= f)Now from Ampere’s circuital law we have

dB lL

:# Ienc0m=

and for determining the field inside the circular ring, the current present on the inner surface of ring will be considered only. So we get ( )(2 )B pr ( )(2 )K0m pr=Therefore the magnetic flux density inside the circular ring is B ( )(4)0m= 4 /wb m0

2m= ( 4 /A mK = )Alternate Method :Magnetic flux density inside a magnetic material is defined as B M0m=and since the magnetization of the rod is 4 /A mM = so, we can have directly the magnetic flux density inside the ring as B /Wb m5 0

2m=

SOL 5.2.15 Option (A) is correct.As calculated above for the complete circular ring, magnetic flux density inside the ring is B 4 /wb ma0 2m= f

(magnetic flux density will be directed along the assumed direction of magnetization)




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Now we calculate the flux density contributed by the gap at its centre when it was the complete ring. The gap has its cross section in form of a square loop as shown in figure below

As calculated in previous question the surface current density of the ring is K 4 /A m=and since the width of the gap(square loop) is w so, net current in the loop is I Kw= w4=Now the magnetic flux density at any point P due to a filamentary current I is defined as


where r " distance of point P from the current filament. 1a " angle subtended by the lower end of the filament at P . 2a " angle subtended by the upper end of the filament at P .So the flux density at center of the square loop produced due to one side of the loop is

Bsq1 ( )I

4 10 22

20

##p

m= - c m ( 1 , 135 , 45cm 1 2c cr a a= = = )

Summing the flux density produced due to all the four sides of loop, we get total magnetic flux density produced by the square loop as

Bsq 44 (10 )

2I2

0# p

m= -f p ( )

10w2 40 2

pm

#= (I Kw= )

( )( . )2 4 0 1 10 100

3 2# # #

pm=

-

( .1 mmw 0= )

a4 2 10 2

0#p m= f

-

Therefore at the centre of the gap the net magnetic flux density will reduce by this amount of the flux density. Thus at the centre of the gap the net magnetic flux density at the centre of the loop will be Bnet B Bs= -

4 4 2 100

2

0#m p m= -

-

4 4 2 100

2#m p= -

-

c m

50.04 10 /wb m7 2#= -



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Let the magnetized sphere be of radius r , centered at origin and the magnetization be M in az direction as shown in figure.Volume current density inside a material is equal to the curl of magnetization M i.e. J M#d=So the volume current density inside the cylinder is J ( ) 0aM z#d= =and since the surface current density in terms of magnetization is defined as K M an#= where an is unit vector normal to the surface.So the surface current density on the sphere is K ( )Ma az r#= ^ h (an ar= ) sinM aq= f ...(i)Now, consider a rotating spherical shell of uniform surface charge density s , that corresponds to a surface current density at any point ( , , )r q f . So we have K sinR asw q= f ...(ii)where w " angular velocity of spherical shell across z -axis R " radius of the sphere.and the magnetic flux density produced inside the rotating spherical shell is defined as

B R32

0m sw= ...(iii)

Comparing the eq.(i) and eq.(ii) we get M Rsw=

Putting this value in eq.(iii) we get the magnetic flux density for the magnetized sphere as

B M32

0m= ( )M Rsw=




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SOL 5.2.17 Option (B) is correct.Since the magnetic flux density inside a magnetic material is defined as B ( )H M0m= +

So, we have the magnetic field intensity inside the material as

H B M10m

= -

and outside the material the magnetic field intensity is

H B10m

=

So the field lines outside the material will be same as for the case of magnetic flux density shown in the MCQ. 1.31. Whereas inside the material the direction of magnetic field intensity will be opposite to the direction of magnetization. Thus the sketch of the field intensity will be same as shown in the option (B).

SOL 5.2.18 Option (B) is correct.The magnetic flux density produced at any point P due to an infinite filamentary current I is defined as

B I2

0

prm=

where r is the distance of point P from the infinite current filament.Now consider a small area dS of the coil located at a distance x from the current filament. The magnetic flux density produced on it due to the current filament along y -axis is

B xI

20

pm= ( xr = )

Since the flux density will be normal to the surface of the coil as determined by right hand rule therefore, the total magnetic flux passing through the coil is

my dB S:= # xI dxdy2xy

0

2

6

0

1

pm=

==b ^l h## lnI2 30

pm=

As the mutual inductance in terms of total magnetic flux my is defined as

M IN my=

where I " current flowing in the element that produces the magnetic flux. N " Total no. of turns of the coil that experiences the magnetic flux.Thus the mutual inductance between the current filament and the loop is

M 2 3lnII1500 0

pm= b l 0.33 mH= 1500N =

SOL 5.2.19 Option (A) is correct.Consider the wire is lying along z -axis. So at any point inside the wire (at a distance a<r from it’s axis) magnetic field intensity will be determined as

dH l:# Ienc= (Ampere’s circuital law)

(2 )H pr Ia2

2

ppr= c m (for Amperian loop of radius r)



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or, H aI a

2 2pr= f

The direction of the magnetic field intensity is determined using right hand rule.Now the stored energy in the magnetic field H is defined as

Wm H dv21v

02m= #

So the stored energy in the internal magnetic filed per unit length (over the unit length in z -direction) will be

Wm ( )aI d d dz

2 2a

z2 2

02 2

00

2

0

1

pm r r r f=

rf

p

===### I

160

2

pm=

Therefore, the energy per unit length depends only on I and is uniform for the uniform current.

SOL 5.2.20 Option (B) is correct.Since the two conducting plates of width 2 mw = carry a uniform current of 4 AI = each so, the surface current density of each plate is

K 2 /A mwI

24= = =

Now consider the first plate carrying current in az+ direction is located at y 0= and the second plate carrying current in az- direction is located at y d= , where d is a very small separation between the plates.Since the magnetic field intensity produced at any point P due to an infinite sheet carrying uniform current density K is defined as

H ( )K a21

n#=

where an is the unit vector normal to the sheet directed toward the point P . So, the magnetic field intensity produced at the second plate due to the first plate is

H12 ( )a a21 2 z y#= ax=- ( 2 ,K a a az n y1 = = )

Now the force per meter, exerted on the 2nd plate due to the 1st plate will be

F12 ( )K dB S2 120

2

0

1

#= ##where K2 " current density of the 2nd plate B12 " magnetic flux density produced at the 2nd plate due to 1st plate

So, F12 ( 2 ) ( )dydzHaz 0 120

2

0

1m#= -## ( 2 ,K a B Hz2 12 0 12m=- = )

( ) ( )dydza a2 z x00

2

0

1

# m= - -## 4 ay0m=

As the force applied by first plate on the 2nd plate is in ay direction so it is a repulsive force. Therefore the repulsive force between the plates is 4 0m .

SOL 5.2.21 Option (D) is correct.No, of turns, n 20,000 /turns meter=Relative permeability, rm 100=




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cross sectional area, S 0.04 m2=Current in the solenoid, I 100 10 A3

#= -

So, its self inductance will be, Ll n Sr0

2m m= ( ) ( ) ( , ) ( . )4 10 100 20 000 0 047 2# # # #p= -

.2 011 103#=

Therefore the energy stored per unit length in the field is

W ml L I21 2= l .2

1 2 011 10 103 2# # #= - 10.05 /J m=

SOL 5.2.22 Option (A) is correct.Consider the path followed by the two particles are the curvatures having radii r , and r2 as shown in figure. So at balanced condition centrifugal force will be equal to magnetic force.Therefore for the first charged particles

rmv

1

2

Bqv= & r Bqmv

1 =

and rm v2

2

2^ h

Bqv= & r Bqmv

2 =

So the distance between the two particles at releasing end is

d r r2 22 1= - Bqmv

Bqmv2 2 2= -b bl l Bq

mv2=

SOL 5.2.23 Option (A) is correct.The wire is oriented in east-west direction and magnetic field is directed northward as shown in the figure.

Since the direction of gravitational force will be into the paper(toward the earth) so for counteracting the gravitational force, applied force must be outward.Now the force experienced by a current element Idl in a magnetic field B is



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For View Only Shop Online at www.nodia.co.in F Idl B

L#= ^ h#

As the magnetic field B is directed toward north therefore, using right hand rule for cross vector we conclude that for producing the outward force current must flow from west to east as shown in the figure below.

Since

SOL 5.2.24 Option (A) is correct.Consider the current flowing in the wire is I . So the magnetic force applied by the field B0 on the wire is Fm ILB0= where L is length of the wireAt balanced condition the magnetic force will be equal to the gravitational force : Fm mg=where m is the mass of the wire and g is acceleration due to gravity.So comparing the two results we get the current flowing in the wire as

I LBmg

0=

Since B0 0.6 10 /Wb m4 2#= - , m 0.3 kg= and L 1 m=

Therefore I .

. .1 0 6 10

0 3 9 84

# #

#= -^ ^

^

h h

h 49 kA= ( 9.8 /m sg = )

SOL 5.2.25 Option (B) is correct.Consider the square loop has side a . Now, when the loop is situated in the field

1.96 /Wb mB 2= . Suppose it swings with an angle a. So in the new position the torque must be zero. Gravitational forces acting on all the sides of loop will be down wards and the force due to magnetic field will be in horizontal direction as shown in the figure.




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So, in balanced condition,from the shown figure we have

sin sina mg a a mg a2 2a a+^ ah k I aB=

tana 2 . ..

mgIB

2 0 2 9 82 1 96

= =^ ^ ^

^ ^

h h h

h h

a /tan 1 41 p= =-^ h

SOL 5.2.26 Option (B) is correct.As discussed in Que 51 the path of electron will be parallel to the input beam but in opposite direction. So the ejected electrons will be flowing in the ay- direction.

SOL 5.2.27 Option (D) is correct.At any point in between the two parallel shuts the net magnetic flux density produced by the two sheets is given as B B B1 2= + where B1 is the flux density produced by the lower sheet and B2 is the flux density produced by upper sheet.Now the magnetic flux density produced at point P due to a plane sheet having current density K is defined as

B K a2 nm

#=

where an is the unit vector normal to the sheet and directed toward point P . So, the flux density produced by lower sheet is

B1 aa2 4 x zm

#= ^ h ( 4K ax= , a an z= )

and the flux density produced by the lower sheet is

B2 ( )aa2 4 x zm

#= - -^ h ( 4K ax=- , a an z=- )

So the net magnetic flux density produced in the region between the two sheets is

B 4 4aa a a2 2x z x z# #m m= + - -^ ^ ^h h h

4 aym=-where m is the permeability of the medium.Therefore the flux density in region 1 is Bregion 1 4 ay1m=- 8 ay0m=- ( 21 0m m= )and the flux density in region 2 is Bregion 2 4 ay2m=- 16 ay0m=- ( 42 0m m= )So the net flux per unit length in the region between the two sheets is

lf Bregion 1= ^ h (width of region 1) + Bregion 2^ h (width of region 2)

8 16a a1 2y y0 0m m= - + -^ ^ ^ ^h h h h

0 /Wb ma5 y0m=-

SOL 5.2.28 Option (A) is correct.As the permeability of the medium varies from 1m to 2m linearly. So at any distance



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For View Only Shop Online at www.nodia.co.inz from one of the plate near to which permiability is 1m , the permeability is given as

m d z12 1

mm m

= +-^ h

(1)

The magnetic flux density between the two parallel sheets carrying equal and opposite current densities is defined as B Km=where K is the magnitude of the current density of the sheets.Therefore the flux per unit length between the two sheets is

lf Bdz

d

0= # where d is the separation between the two sheets.

Kdzd

0m= # K d z dz

d

12 1

0m

m m= +

-^ h; E# (from equation (1))

K z dz2

d

12 1

2

0

m m m= + -b l; E K d2

1 2m m= +b l

SOL 5.2.29 Option (B) is correct.Given the field intensity inside the slab is H 4 2a ax y= +So the magnetic flux density inside the slab is given as B Hm= where m is the permeability of the material. 4 2a a2 x y0m= +^ h (m 2 0m= )Therefore the magnetization of the material is

M B H0m

= -

8 4 4 2a a a ax y x y= + - +^ h 4 2a ax y= +Now the magnetization surface current density at the surfaces of a magnetic material is defined as Km M an#=where an is the unit vector normal to the surface directed outward of the material So, at z 0= magnetization surface current density is K 0atm z=6 @ M az#= -^ h ( )a an z=- 4 2a ay x= -and at z d= , the magnetization surface current density isSo, K atm z d=6 @ M az#= ^ h (a an z= ) 4 2a a ax y z#= +^ ^h h 4 2a ay x=- +

SOL 5.2.30 Option (A) is correct.As calculated in the previous question the magnetization vector of the material is M 4 2a ax y= +The magnetization volume current density inside a magnetic material is equal to the curl of magnetization.,i.e. Jm M#d=Therefore the magnetization volume current density inside the slab is




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Jm a a a

4 2 0

x

x

y

y

z

z= 22

22

22 0=

SOL 5.2.31 Option (B) is correct.Given the B -H curve for the material, B 2 HH0m=The work done per unit volume in magnetizing a material from 0 to B0 that has non uniform permeability is defined as

wm dH BB

0

0

:= #Now for determining dB . we can express B 2 H aH0

2m=where aH is the unit vector in direction of H .

So, dHdB 4 HaH0m=

4 H0m=

and wm 4H HH

00

0

: m= ^ h# 4 H3

H

0

3

0

0

m= ; E H4

3 0 03m=

SOL 5.2.32 Option (A) is correct.As 7 shows the direction into the paper while 9 shows the direction out of the paper. So the wire of length l carries current I2 that flows out of the paper. The Magnetic field intensity produced at a distance r from an infinite straight wire carrying current I is defined as

H I2pr=

So the magnetic field intensity produced at the top wire due to the infinite wire carrying current inward is

H 1f l

I2 2p

=^ h

( l2r = )

and the magnetic field intensity at top wire due to the infinite wire carrying current outward is

H 2f l

I2 2p

=^ h

( l2r = )



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For View Only Shop Online at www.nodia.co.inTherefore the resultant field intensity at the wire of length l is HT cosH H1 2 q= +f f^ h

l

I2 2

22

1#

p=

^ h lI

2p=

Since the force exerted on a current element Idl by a magnetic field H is defined as dF ( )( )H Idlm=So the force experienced by the wire of length l is

F ( )H I l2Tm= ^ h lI I l2 2m p= b ^l h I

2

2

pm=

SOL 5.2.33 Option (C) is correct.Consider the sheets as shown in figure that having the surface current densities

20 /mA may+ and 20 /mA may-

So the field intensity between the plates will be given as H H H1 2= +where H1 is the field intensity produced by the sheet located at x 0= and H2 is the field intensity produced by the sheet located at 5 cmx = .Now the magnetic field intensity produced at point P due to a plane sheet having current density K is defined as

H K a21

n#=

where an is the unit vector normal to the sheet directed toward point P . So, the magnetic field intensity in the region between the plates is

H K a K a21

21

at at cm

n

x

n

x

1 1

0

2 2

5

# #= += =

1 2 344 44 1 2 344 44

20 10 20 10a a a a21

21

y x y x3 3

# ## #= + - -- -^ ^ ^ ^h h h h

20 10 az3#=- -

and magnetic flux density in the region between the sheets is B Hm= 40 10 az0

3m #=- - 2 0m m=^ hTherefore the stored magnetic energy per unit volume in the region is




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wm H B21

:= 21 800 100

6#m= -

^ h

400 4 10 107 6# # #p= - - 160 /J m3p=

Since the separation between plates is 5 cmd = . So, stored energy per unit area between the plates is /W Am w dm#= .160 0 05p #= ^ ^h h /J m4 2p=

SOL 5.2.34 Option (C) is correct.Since the boundary surface of the two medium is z 0= , so the normal component B n1 and tangential component B t1 of magnetic flux density in medium 1 are B n1 az=and B t1 0.4 0.8a ax y= +As the normal component of magnetic flux density is uniform at the boundary of two medium So, the normal component of magnetic flux density in the medium 2 is B n2 B an z1= = (1)Now for determining tangential component of field in medium 2, we first calculate tangential component of magnetic field intensity in medium 1 which is given as

H1t B t

1

1

m= where 1m is the permeability of medium 1.

0.4 0.8a a41

x y0m

= +^ h 0.1 0.2a ax y

0m= +

( 41 0m m= )

Again from the boundary condition the tangential component of magnetic field intensity in the two mediums are related as a H Hn t t1 2# -^ h K=where H t2 and H t1 are the tangential components of magnetic field intensity in medium 2 and medium 1 respectively, K is the surface current density at the boundary interface of the two mediums and an is the unit vector normal to the boundary interface. So we have

. .

H Ha a a a a0 1 0 2z

x ytx x ty y

02 2m#

+ - +^ h< F 0.2 0.4a a1x y

0m= -^ h

. .H Ha a0 1 0 2tx y ty x

02

02m m- - -b bl l . .a a1 0 2 0 4x y

0m= -^ h

Comparing the x and y -components we get

H tx2 . . .0 1 0 4 0 50 0 0m m m= + =

and H ty2 . . .0 2 0 2 0 40 0 0m m m= + =

Therefore the tangential component of magnetic field intensity in medium 2 is

H t2 . .a a0 5 0 4x y

0 0m m= +

and the tangential component of magnetic flux density in medium 2 is B t2 H t2 2m= 0.8a ax y= +Thus the net magnetic flux density in medium 2 is B2 B Bt n2 2= + 0.8a a ax y z= + +



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Consider the square loop is of side a2 as shown in the figure

Since the sides BC and AD crosses the straight wire so no force will be experienced by the sides, while the flux density produced by the straight wire at sides AB and CD will be equal in magnitude.Now the magnetic flux density produced at a distance r from a straight wire carrying current I is defined as

B I2

0

prm=

So the magnetic flux density produced by the straight wire at the two sides of the loop is

B a220

pm

=^

^

h

h a

0

pm= 2 ,AI ar= =^ h

Since the force exerted on a current element Idl by a magnetic field B is defined as dF ( )Idl B#=Therefore the force experienced by side AB of length a2 is

F1 a aa a4 2 z x0

pm

#= ^ h6 9@ C a8y

0

pm= -^ h ( 4 AI = )

Similarly force experienced by side CD is

F2 a aa a4 2 z x0

# pm= - -^ ^ ^h h h6 9@ C a8

y0

pm= ^ h

Thus the net force experienced by the loop is

F F F1 2= + a16y

0

pm= ^ h

16 4 10 ay7# #= - .4 Na2 y m=

SOL 5.2.36 Option (A) is correct.According to Snell’s law the permeability of two mediums are related as tan0 1m q tan 2m q=

tantan

2

1

qq 15

0

0

mm=

tan 1q tan15 2q= ...(i)Now, the given flux density in medium 2 is B2 1.2 0.8a ay z= +




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So the normal and tangential component of the magnetic flux density in medium 2 is B n2 0.8az=and B t2 1.2ay=From the figure we have

tan 2q ..

BB

1 20 8

32

t

n

2

2= = =

or 2q /tan 2 31= -^ h

from equation (1) tan 1q tan15 2q= tan 1q 10= q tan 101= -

^ h

Thus the angular deflection is 1 2q q- /tan tan10 2 31 1= -- -

^ ^h h

5 .64 c=

SOL 5.2.37 Option (B) is correct.For calculating total reluctance of the circuit, we have to draw the electrical analog of the circuit. In the given magnetic circuit, there are total six section for which six reluctance has been drawn below.

For a given cross sectional area S and length of the core l reluctance is defined as

R Slm=

Where m is permeability of the medium in core

So, we have R1 1000 5 105 10

101

04

2

0#

#m m= =-

-

^ ^h h

R2 1000 10 105 10

201

04

2

0#

#m m= =-

-

^ ^h h

R3 1000 10 106 10

503

04

2

0#

#m m= =-

-

^ ^h h

R4 1000 10 1014 10

507

04

2

0#

#m m= =-

-

^ ^h h

R5 503R3

0m= =

R6 1000 10 104 10

251

04

2

0#

#m m= =-

-

^ ^h h



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For View Only Shop Online at www.nodia.co.inSince all the reluctance are connected in series so total reluctance of the magnetic circuit is RT R R R R R R1 2 3 4 5 6= + + + + +

101

201

503

507

503

251

0 0 0 0 0 0m m m m m m= + + + + +

209

0m=

SOL 5.2.38 Option (C) is correct.For a given reluctance R of a magnetic circuit, the self, inductance is defined as

L NR

2

= Where N is no. of turns of coil

Then, L /9 20100

0

2

m=

^

^

h

h 20

9RT0m

=c m

.2 79 10 2#= - 27.9 mH=

SOL 5.2.39 Option (B) is correct.Give thatno. of turns of coil, N 50=length of the core, l 0.6 m=relative permeability, rm 600=inductance of the coil, L 0.2 0.2 10mH H3

#= = -

So, the cross sectional area of core is

SNLl

2m=

. .600 50

0 2 10 0 6

02

3#

m=

-

^ ^

^ ^

h h

h h

6.366 10 m5 2#= - .64 cm1 2=

SOL 5.2.40 Option (B) is correct.Since the core is ideal so it’s reluctance will be zero and so the electrical analog for the magnetic circuit will be as shown below

The reluctance R1, R2 and R3 is produced by the air gap.

R1 Sl

100 104 10 4

0 1

1

04

2

0#

#m m m= = =-

-

^ h

R2 100 102 10 2

04

2

0#

#m m= =-

-

^ h

R3 100 102 10 2

04

2

0#

#m m= =-

-

^ h




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So, the total reluctance seen by coil N1 is RT ||R R R1 2 3= +

4 1 50 0 0m m m= + =

and the self inductance of coil will be

L1 62.8 mHNRT

12

= =

SOL 5.2.41 Option (B) is correct.The total reluctance of the magnetic circuit as seen from the coil N2 is RT ||R R R1 2 3= +^ h

||4 2 20 0 0m m m= +b l (as calculated above)

34 2

0 0m m= + 310

0m=

Therefore the self inductance of the coil N2 is

L NRT

222

= /10 3

2500

2

m=

^

^

h

h 2 .6 mH8=

SOL 5.2.42 Option (D) is correct.Since the coil N1 and N2 are directly connected through ideal core so entire flux produced by N2 will link with N1.The electrical analog of the magnetic circuit is shown below where the reluctance R1 and R2 are the reluctance due to air gap.

So, the reluctance seen by coil N2 is

SlR10m

= 2000 104 10

06

3

#

#

m= -

-

^

^

h

h 2

0m=

Consider the current flowing in coil N2 is i2. So, the total flux produced by N i2 2 is

2f /

N i i i2500 250

R1

2 2

0

20 2m m= = =

^ h

Since the entire flux will link with N1 So mutual induction between N1 and N2 is

M L12= iN

ii250 250

2

1 2

2

0 2f m= =

^ ^h h 78.54 mH=

SOL 5.2.43 Option (A) is correct.As the coil N1 and N2 are directly connected through an ideal core so entire flux will produced by N2 will link with N1 and so flux linked with N3 will be zero.



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For View Only Shop Online at www.nodia.co.inTherefore the mutual inductance between N3 and N2 is zero.

SOL 5.2.44 Option (A) is correct.Given, the expression for magnetization curve,

B /Wb mH H31 2 2m= +

The energy stored per unit volume in a magnetic material having linear magnetic flux density is defined as

wm dH BH

H

0

0

:==

#Since, magnetic field intensity varied from 0 to 210 /A m So, we have

wm HdBH 0

210=

=#

Since, dHdB H3

1 2= +

So, putting it in equation we get,

wm H H dH31 2

H 0

210= +

=b l#

H H6 3

22 3

0

210

= +: D

6.18 10 /J m6 3#= 6.2 /MJ m3=

***********




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SOLUTIONS 5.3

SOL 5.3.1 Option (A) is correct.From boundary condition we have the following relation between the magnetic field intensity in the two mediums : H n1 1m H n2 2m= (1)and ( )H H an1 2 12#- K= (2)where H1 and H2 are the magnetic field intensity in the two mediums, an12 is the unit vector normal to the interface of the mediums directed from medium 1 to medium 2 and K is the surface current density at the interface of the two mediums. Now, the magnetic field intensity in medium 1 is H1 3 30 /A ma ax y= +As the interface lies in the plane x 0= so, we have H n1 3ax=From equation (1), the normal component of the field intensity in medium 2 is given as

H n2 1.5H a2n

x1= =

Therefore, the net magnetic field intensity in medium 2 can be considered as H2 1.5 A Ba a ax y z= + + (3)where A and B are the constants. So, from equation (2) we have (3 30 ) (1.5 )A Ba a a a a ax y x y z x#+ - + +6 @ 10ay= 1.5 (30 )A Ba a a ax y z x#+ - -6 @ 10ay= 0 (30 )A Ba az y- - - 10ay=Comparing the components in the two sides we get A30 - A0 30&= =and B- B10 10&= =-Putting these values in equation (3) we get the magnetic field intensity in medium 2 as H2 1.5 30 10 /A ma a ax y z= + -

SOL 5.3.2 Option (B) is correct.Given,the magnetic moment m 2.5 A m2

-=Mass of magnet, mass 6.6 10 kg3

#= -

density of steel, density 7.9 10 /kg m3 3#=



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For View Only Shop Online at www.nodia.co.inSo, the net volume of the magnet bar is

v ..

densitymass

7 9 106 6 10

3

3

#

#= =-

0.835 10 m6 3#= -

Now, the magnetization of the magnet is defined as the magnetic moment per unit volume so, we get magnetization of the magnet bar as

M .

.vm

0 835 102 5

6#

= = -

3 10 /A m6#=

SOL 5.3.3 Option (A) is correct.Given,

Magnetic field intensity, H 5 /A maxm=

Current element, Idl 4 10 A may4# -= -

So, the magnetic flux density is given as B Hm= 5 /A max=Since, the force exerted on a current element Idl placed in a magnetic field B is defined as F Idl B#= ^ h

So, putting all the values we get, F 4 10 5a ay x

4##= -

^ ^h h

2 10 Naz3#=- - 2 mNaz=-

SOL 5.3.4 Option (C) is correct.The electrical analogy of the magnetic field are listed below :

Electrical field Magnetic field

EMF (electromotive force) " MMF (magneto motive force)

Electric current " Magnetic flux

Resistance " Reluctance

Conductivity " PermeabilitySo, for the given match list we get, A 3" , B 2" , C 4" , D 1" .

SOL 5.3.5 Option (C) is correct.At the surface of discontinuity (interface between two medium) the normal component of magnetic flux density are related as B n1 B n2=i.e. normal component of magnetic flux density is uniform at the surface of discontinuity.Statement 1 is correctAt the boundary interface between two mediums, the normal component of the electric flux density is related as D Dn n2 1- sr= i.e. discontinuous




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where rS is surface charge density at the interface. If the interface is charge free (

0r =S

) then, the equation changes to D n2 D n1= i.e. continuousSo, the normal component of flux density at the surface of discontinuity may or may not be continuos.

SOL 5.3.6 Option (C) is correct.Magnetic current is composed of both displacement and conduction components.

SOL 5.3.7 Option (C) is correct.Torque exerted on a loop with dipole moment M in a magnetic field B is defined as T M B#=

SOL 5.3.8 Option (B) is correct.Biot savart’s law gives the magnetic flux density as defined below

B R

Idl R

4 2

0

p

m #=

# 3b "^ h

Displacement current is determined by using maxwell’s equation as H#d J Jc d= + where Jd is displacement current density 1c "^ h

Time average power flow in a field wave is determined by poynting vector as

Pave E H21

s s#= 2d "^ h

Using Gauss’s law line charge distribution can be determined. 4a "^ h

SOL 5.3.9 Option (B) is correct.Magnetic energy density in a magnetic field is defined as

wm J A21

:=

SOL 5.3.10 Option ( ) is correct.Consider the two wires carrying current as shown below :

The force exerted due to the wire 2 at wire 1 is given as F Idl B#= ^ ^h h

where Idl is the small current element of the wire 1 and B is magnetic flux density produced by wire 2 at wire 1. As determined by right hand rule the magnetic flux



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For View Only Shop Online at www.nodia.co.indensity produced due to wire 2 at wire 1 is out of the paper. Which will be towards wire 2. In the similar way the force due to wire 1 at wire 2 will be toward wire 1 i.e. attractive and perpendicular to the wire.

SOL 5.3.11 Option (B) is correct.From the boundary condition for magnetic field, we have the following derived condition as Hn1 1m Hn2 2m=and Ht1 Ht2=

SOL 5.3.12 Option (C) is correct.The magnetic flux density B and magnetic field intensity H in a medium with permeability m are related as B H Hr 0m m m= =Now, for the different magnetic material relative permeability rm are listed below :Free space (vacuum) rm 1=Diamagnetic rm 1K

Paramagnetic rm 1L

Ferromagnetic rm 1>>So, the B -H curve for the respective material has been shown below (depending on their slopes m).

SOL 5.3.13 Option (A) is correct.When a dielectric material is placed in an electric field then the electric dipoles are created in it. This phenomenon is called polarization of the dielectric material.So, we conclude that both the statement are correct and statement (II) is correct explanation of (I).

SOL 5.3.14 Option (B) is correct.For an inhomogenous magnetic material, magnetic permeability is a variable and so, it has some finite gradient. Now, from maxwell’s equation we know B:d 0=Since, B Hm=So, B:d H:d m= ^ h

0 H: :md d= +In the above equation :d m have some finite value therefore,




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H:d 0! (in inhomogenous medium)

SOL 5.3.15 Option (C) is correct.Force on a current element Idl kept in a magnetic field B is defined as

F Idl BL

#= # 0.05a a10 2 z x#= ^ ^h h6 6@ @ 1.0 Nay=

SOL 5.3.16 Option (C) is correct.Magnetic energy density in a magnetic field is defined as

wm J A21

:=

where J is the current density and A is the magnetic vector potential.

SOL 5.3.17 Option (D) is correct.The force on a moving charge q with the velocity v in a region having magnetic field B and electric field E is defined as F q E v B#= +^ h

SOL 5.3.18 Option (B) is correct.The currents in the hairpin shaped wire flows as shown in the figure.

As the direction of current are opposite so the force acting between them is repulsive, and So it tend to a straight line.

SOL 5.3.19 Option (C) is correct.Given, the Lorentz force equation, F e v B#= ^ h

If the particle is at rest then 0v = and so there will be no any deflection in particle due to the magnetic field.

SOL 5.3.20 Option (D) is correct.Force acting on a small point charge q moving in an EM wave is defined as F q qE v B= + +^ h

So, for q 1= F E v B#= +

SOL 5.3.21 Option (B) is correct.Given,Current flowing in the conductor, I 5 A=Magnetic flux density, B 3 4a ax y= +



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For View Only Shop Online at www.nodia.co.inSince, the force experienced by a current carrying element Idl placed in a magnetic field B is defined as dF Idl B#= ^ h

As the current flowing in az direction so, we have dl dlaz=and the force experienced by the conductor is dF dla a a5 3 4z x y#= +^ ^h h

Therefore, the force per unit length experienced by the conductor is

dldF 15 20a ay x= -

20 15 /N ma ax y=- +

SOL 5.3.22 Option (B) is correct.From the boundary condition for magnetic field we have the following relation :Normal component of magnetic flux density is continuousi.e. Bn1 Bn2=Any field vector is the sum of its normal and tangential component to any surfacei.e. H1 H Hn t1 1= +When the interface between two medium carries a uniform current K then the tangential component of magnetic field intensity is not uniform.i.e. H Ht t1 2- K=or, a H Hn21 1 2# -^ h K=

But, B2 B Bn t2 2! +

SOL 5.3.23 Option (A) is correct.Given, the magnetic flux density in medium 1 is B1 1.2 0.8 0.4a a ax y z= + +and the interface lies in the plane z 0= .So, the tangential and normal components of magnetic flux density in the two mediums are respectively : B t1 1.2 0.8a ax y= +and B n1 0.4az=Now, from the boundary condition of current free interface, we have the following relations between the components of field in two mediums. B n1 B n2=

and B t

1

1

m B t

2

2

m=

Therefore, we get the field components in medium 2 as B n2 B n1= 0.4az=

and B t2 B t11

2

mm= b l 1.2 0.8 0.6 0.4a a a a2

1x y x y= + = +^ ^h h

Thus, the net magnetic flux density in region 2 is




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B2 B Bn t2 2= + 0.6 0.4 0.4a a ax y z= + +So, the magnetic field intensity in region 2 is

H2 0.6 0.4 0.4 /A mB a a a1x y z

2

2

0m m= = + +^ h

SOL 5.3.24 Option (C) is correct.Energy stored in a magnetic field is defined as

Wm dvA J21

:= #

So, dvA J:# has the units of energy.

***********

CHAPTER 6TIME VARRYING FIELD AND MAXWELL EQUATION

342 Time Varrying Field and Maxwell Equation Chap 6



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EXERCISE 6.1

MCQ 6.1.1 Match List I with List II and select the correct answer using the codes given below (Notations have their usual meaning)

List-I List-II

a Ampere’s circuital law 1. D v: rd =

b Faraday’s law 2. 0B:d =

c Gauss’s law 3.t

E B22

#d =-

d Non existence of isolated magneticharge

4.t

H J D22

#d = +

Codes : a b c d(A) 4 3 2 1(B) 4 1 3 2(C) 2 3 1 4(D) 4 3 1 2

MCQ 6.1.2 Magneto static fields is caused by(A) stationary charges (B) steady currents

(C) time varying currents (D) none of these

MCQ 6.1.3 Let A be magnetic vector potential and E be electric field intensity at certain time in a time varying EM field. The correct relation between E and A is

(A) t

E A22=- (B)

tA E

22=-

(C) t

E A22= (D)

tA E

22=

MCQ 6.1.4 A closed surface S defines the boundary line of magnetic medium such that the field intensity inside it is B . Total outward magnetic flux through the closed surface will be(A) B S: (B) 0

(C) B S# (D) none of these

Chap 6 Time Varrying Field and Maxwell Equation 343


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For View Only Shop Online at www.nodia.co.inMCQ 6.1.5 A perfect conducting sphere of radius r is such that it’s net charge resides on the

surface. At any time t , magnetic field ( , )r tB inside the sphere will be(A) 0

(B) uniform, independent of r

(C) uniform, independent of t

(D) uniform, independent of both r and t

MCQ 6.1.6 The total magnetic flux through a conducting loop having electric field E 0= inside it will be(A) 0

(B) constant

(C) varying with time only

(D) varying with time and area of the surface both

MCQ 6.1.7 A cylindrical wire of a large cross section made of super conductor carries a current I . The current in the superconductor will be confined.(A) inside the wire (B) to the axis of cylindrical wire

(C) to the surface of the wire (D) none of these

MCQ 6.1.8 If Bi denotes the magnetic flux density increasing with time and Bd denotes the magnetic flux density decreasing with time then which of the configuration is correct for the induced current I in the stationary loop ?

MCQ 6.1.9 A circular loop is rotating about z -axis in a magnetic field cosB tB ay0 w= . The total induced voltage in the loop is caused by(A) Transformer emf (B) motion emf.

(C) Combination of (A) and (B) (D) none of these




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MCQ 6.1.10 A small conducting loop is released from rest with in a vertical evacuated cylinder voltage induced in the falling loop is(Assume earth magnetic field 10 T2 6

#= - at a constant angle of 10c below the horizontal)(A) zero (B) 1 mV

(C) 17.34 mV (D) 9.8 mV

MCQ 6.1.11 A square loop of side 2 m is located in the plane x 0= as shown in figure. A non-uniform magnetic flux density through it is given as

B 4z t ax3 2= ,

The emf induced in the loop at time sect 2= will be

(A) 16 volt (B) 4 volt-

(C) 4 volt (D) 2 volt-

MCQ 6.1.12 A very long straight wire carrying a current AI 3= is placed at a distance of 4 m from a square loop as shown in figure. If the side of the square loop is 4 m then the total flux passing through the square loop will be

(A) 0.81 10 wb7#

- (B) 10 wb6-

(C) 4.05 10 wb7#

- (D) 2.0 10 wb7#

-



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For View Only Shop Online at www.nodia.co.inMCQ 6.1.13 A straight conductor ab of length l lying in the xy plane is rotating about the

centre a at an angular velocity w as shown in the figure.

If a magnetic field B is present in the space directed along az then which of the following statement is correct ?(A) Vab is positive (B) Vab is negative

(C) Vba is positive (D) Vba is zero

MCQ 6.1.14 In a certain region magnetic flux density is given as 0. /Wb mtB a2 z2= . An electric

loop with resistance 2W and 4 W is lying in x -y plane as shown in the figure.

If the area of the loop is m2 2 than, the voltage drop V1 and V2 across the two resistances is respectively (A) 66.7 mV and 33.3 mV (B) 33.3 mV and 66.7 mV

(C) 50 mV and 100 mV (D) 100 mV and 50 mV

MCQ 6.1.15 Assertion (A) : A small piece of bar magnet takes several seconds to emerge at bottom when it is dropped down a vertical aluminum pipe where as an identical unmagnetized piece takes a fraction of second to reach the bottom.Reason (R) : When the bar magnet is dropped inside a conducting pipe, force exerted on the magnet by induced eddy current is in upward direction.(A) Both A and R are true and R is correct explanation of A.







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MCQ 6.1.16 A magnetic core of uniform cross section having two coils (Primary and secondary) wound on it as shown in figure. The no. of turns of primary coil is 5000 and no. of turns of secondary coil is 3000. If a voltage source of 4 Volts is connected across the primary coil then what will be the voltage across the secondary coil ?

(A) 72 volt (B) 7.2 volt

(C) 20 volt (D) 7.2 volt-

MCQ 6.1.17 Self inductance of a long solenoid having n turns per unit length will be proportional to(A) n (B) /n1

(C) n2 (D) /n1 2

MCQ 6.1.18 A wire with resistance R is looped on a solenoid as shown in figure.

If a constant current is flowing in the solenoid then the induced current flowing in the loop with resistance R will be(A) non uniform (B) constant

(C) zero (D) none of these

MCQ 6.1.19 A long straight wire carries a current ( )cosI I t0 w= . If the current returns along a coaxial conducting tube of radius r as shown in figure then magnetic field and electric field inside the tube will be respectively.



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For View Only Shop Online at www.nodia.co.in(A) radial, longitudinal (B) circumferential, longitudinal

(C) circumferential, radial (D) longitudinal, circumferential

MCQ 6.1.20 Assertion (A) : Two coils are wound around a cylindrical core such that the primary coil has N1 turns and the secondary coils has N2 turns as shown in figure. If the same flux passes through every turn of both coils then the ratio of emf induced in the two coils is

VV

emf

emf

1

2 NN

1

2=

Reason (R) : In a primitive transformer, by choosing the appropriate no. of turns, any desired secondary emf can be obtained.(A) Both A and R are true and R is correct explanation of A.




MCQ 6.1.21 In a non magnetic medium electric field cosE E t0 w= is applied. If the permittivity of medium is e and the conductivity is s then the ratio of the amplitudes of the conduction current density and displacement current density will be(A) /0m we (B) /s we

(C) /0sm we (D) /we s

MCQ 6.1.22 In a medium where no D.C. field is present, the conduction current density at any point is given as Jd . /cos A mt a5 1 5 10 y

8 2#= ^ h . Electric flux density in the

medium will be(A) 133.3 . /sin nC mt a1 5 10 y

8 2#^ h (B) 13.3 . /sin nC mt a1 5 10 y

8 2#^ h

(C) 1.33 . /sin nC mt a1 5 10 y8 2

#^ h (D) 1.33 . /sin nC mt a1 5 10 y8 2

#- ^ h

MCQ 6.1.23 In a medium, the permittivity is a function of position such that 0d .ee . If the

volume charge density inside the medium is zero then E:d is roughly equal to(A) Ee (B) Ee-

(C) 0 (D) E:ed-




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MCQ 6.1.24 A conducting medium has permittivity, 4 0e e= and conductivity, 1.14 10 /s m8s #=. What will be the ratio of magnitude of displacement current and conduction current in the medium at 50 GHz ?(A) .1 10 104

# (B) .1 025 107#

(C) .9 75 10 17#

- (D) .9 75 10 8#

-

MCQ 6.1.25 In free space, the electric field intensity at any point ( , , )r q f in spherical coordinate system is given by

E sin cos

rt kr

aq w

=-

q^ h

The phasor form of magnetic field intensity in the free space will be

(A) sinr

k e ajkr0wmq

f- (B) sin

rk e ajkr

0wmq- f

-

(C) rk e ajkr0wm

f- (D) sin

rk e ajkrq

f-

MCQ 6.1.26 Magnetic field intensity in free space is given as

H 0. /cos sin A my t bx a2 15 6 10 z9

#p p= -^ ^h h

It satisfies Maxwell’s equation when b equals to(A) 46.5 /rad m! (B) 41.6 /rad m!

(C) 77.5 /rad m! (D) 60.28 /rad m!

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EXERCISE 6.2

MCQ 6.2.1 Two parallel conducting rails are being placed at a separation of 3 m as shown in figure. One end of the rail is being connected through a resistor 10R W= and the other end is kept open. A metal bar slides frictionlessly on the rails at a speed of 5 /m s away from the resistor. If the magnetic flux density 0. /Wb mB 2 2= pointing out of the page fills entire region then the current I flowing in the resistor will be(A) 0.01 A (B) 0.01 A-

(C) 1 A (D) 0.1 A-

Common Data for Question 2 - 3 :A conducting wire is formed into a square loop of side 4 m. A very long straight wire carrying a current 30 AI = is located at a distance 2 m from the square loop as shown in figure.

MCQ 6.2.2 If the loop is pulled away from the straight wire at a velocity of 5 /m s then the induced e.m.f. in the loop after . sec0 6 will be(A) 5 voltm (B) 2.5 voltm

(C) 25 voltm (D) 5 mvolt




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MCQ 6.2.3 If the loop is pulled downward in the parallel direction to the straight wire, such that distance between the loop and wire is always 3 m then the induced e.m.f. in the loop at any time t will be(A) linearly increasing with t (B) always 0

(C) linearly decreasing with t (D) always constant but not zero.

MCQ 6.2.4 An infinitely long straight wire with a closed switch S carries a uniform current 4 AI = as shown in figure. A square loop of side 2 ma = and resistance 4R W= is

located at a distance 4 m from the wire. Now at any time t t0= the switch is open so the current I drops to zero. What will be the total charge that passes through a corner of the square loop after t t0= ?(A) 277 nC (B) 693 nC

(C) 237 nC- (D) 139 nC

MCQ 6.2.5 A circular loop of radius 5 m carries a current 2 AI = . If another small circular loop of radius 1 mm lies a distance 8 m above the large circular loop such that the planes of the two loops are parallel and perpendicular to the common axis as shown in figure then total flux through the small loop will be(A) 1.62 fWb (B) 25.3 nWb

(C) 44.9 fWb (D) 45.4 pWb



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For View Only Shop Online at www.nodia.co.inMCQ 6.2.6 A non magnetic medium at frequency 1.6 10 Hzf 8

#= has permittivity 54 0e e= and resistivity 0.77 mr W -= . What will be the ratio of amplitudes of conduction current to the displacement current ?(A) 0.43 (B) 0.37

(C) 1.16 (D) 2.70

MCQ 6.2.7 Two voltmeters A and B with internal resistances RA and RB respectively is connected to the diametrically opposite points of a long solenoid as shown in figure. Current in the solenoid is increasing linearly with time. The correct relation between the voltmeter’s reading VA and VB will be

(A) V VA B= (B) V VA B=-

(C) VV

RR

B

A

B

A= (D) VV

RR

B

A

B

A=-

Statement for Linked Question 8 - 9 :Two parallel conducting rails are being placed at a separation of 6 m with a resistance

10R W= connected across it’s one end. A conducting bar slides frictionlessly on the rails with a velocity of 4 /m s away from the resistance as shown in the figure.

MCQ 6.2.8 If a uniform magnetic field TeslaB 4= pointing out of the page fills entire region then the current I flowing in the bar will be(A) 0 A (B) 40 A-

(C) 4 A (D) 4 A-




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MCQ 6.2.9 The force exerted by magnetic field on the sliding bar will be(A) 4 N, opposes it’s motion

(B) 40 N, opposes it’s motion

(C) 40 N, in the direction of it’s motion

(D) 0

MCQ 6.2.10 Two small resistor of 225W each is connected through a perfectly conducting filament such that it forms a square loop lying in x -y plane as shown in the figure. Magnetic flux density passing through the loop is given as B 7.5 (120 30 )cos t azcp=- -The induced current ( )I t in the loop will be

(A) . ( )sin t0 02 120 30cp - (B) . ( )sin t2 8 10 120 303# cp -

(C) . ( )sin t5 7 120 30cp- - (D) . ( )sin t5 7 120 30cp -

Common Data for Question 11 - 12 :In a non uniform magnetic field 8 TeslaxB az2= , two parallel rails with a separation of 10 m and connected with a voltmeter at it’s one end is located in x -y plane as shown in figure. The Position of the bar which is sliding on the rails is given as x 1 0.4t t2= +^ h

MCQ 6.2.11 Voltmeter reading at . sect 0 4= will be



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For View Only Shop Online at www.nodia.co.in(A) 0.35 volt- (B) 0.35 volt

(C) 0.29 volt- (D) 1.6 volt-

MCQ 6.2.12 Voltmeter reading at 12 cmx = will be(A) 12.27 mvolt (B) 14.64 mvolt-

(C) 23.4 mvolt (D) 23.4 mvolt-

MCQ 6.2.13 A rectangular loop of self inductance L is placed near a very long wire carrying current i1 as shown in figure (a). If i1 be the rectangular pulse of current as shown in figure (b) then the plot of the induced current i2 in the loop versus time t will be (assume the time constant of the loop, /L R&t )

MCQ 6.2.14 Two parallel conducting rails is placed in a varying magnetic field 0.2 cos tB axw=. A conducting bar oscillates on the rails such that it’s position is given by y 0. cos mt25 1 w= -^ h




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If one end of the rails are terminated in a resistance 5R W= , then the current i flowing in the rails will be

(A) . sin cost t0 01 1 2w w w+^ h

(B) . sin cost t0 01 1 2w w w- +^ h

(C) . cos sint t0 01 1 2w w w+^ h

(D) . sin sint t0 05 1 2w w w+^ h

MCQ 6.2.15 Electric flux density in a medium ( 10re = , 2rm = ) is given as

D .33 . /sin C mt x a2 3 10 0 2 y8 2

# m= -^ h

Magnetic field intensity in the medium will be

(A) 10 . /sin A mt x a3 10 0 2 y5 8

# --^ h

(B) 2 . /sin A mt x a3 10 0 2 y8

# -^ h

(C) 4 . /sin A mt x a3 10 0 2 y8

#- -^ h

(D) 4 . /sin A mt x a3 10 0 2 y8

# -^ h

MCQ 6.2.16 In a non conducting medium ( )0s = magnetic field intensity at any point is given by H /cos A mt bx a10 z

10= -^ h . The permittivity of the medium is e 0.12 /nF m= and permeability of the medium is m 3 10 /H m5

#= - . If no D.C. field is present in medium, then value of b for which the field satisfies Maxwell’s equation is(A) 600 /rad s-

(B) 600 /rad m

(C) 3.6 10 /rad m5#

(D) (A) and (B) both

MCQ 6.2.17 A current filament located on the x -axis in free space with in the interval 0.1 0.1 mx< <- carries current ( ) 8 AI t t= in ax direction. If the retarded vector

potential at point ( , , )P 0 0 2 be ( )tA then the plot of ( )tA versus time will be



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MCQ 6.2.18 In a non-conducting medium ( 0s = , 2r rm e= = ), the retarded potentials are given as V volty x ct= -^ h and /Wb my tA ac

xx= -^ h where c is velocity of waves in

free space. The field (electric and magnetic) inside the medium satisfies Maxwell’s equation if(A) 0J = only (B) 0vr = only

(C) 0J vr= = (D) Can’t be possible

MCQ 6.2.19 Electric field in free space in given as

E 5 sin cosy bx a10 6 10 z9

#p p= -^ ^h h

It satisfies Maxwell’s equation for ?b =(A) 20 /rad m! p (B) /rad m300! p

(C) 10 /rad mp (D) 30 /rad mp

Statement for Linked Question 20 - 21 :In a region of electric and magnetic fields E and B , respectively, the force experienced by a test charge qC are given as follows for three different velocities. Velocity m/sec Force, N ax q a a2 x y+^ h ay qay az 2q a ay z+^ h




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MCQ 6.2.20 What will be the magnetic field B in the region ?(A) ax (B) a ax y-

(C) az (D) a ay z-

MCQ 6.2.21 What will be electric field E in the region ?(A) a ax z- (B) a ay z-

(C) a ay z+ (D) a a ay z x+ -

MCQ 6.2.22 In Cartesian coordinates magnetic field is given by 2/xB az=- . A square loop of side 2 m is lying in xy plane and parallel to the y -axis. Now, the loop is moving in that plane with a velocity v a4 x= as shown in the figure.

What will be the circulation of the induced electric field around the loop ?

(A) x x 2

16+^ h

(B) x8

(C) x x 2

8+^ h

(D) x x

162+^ h

Common Data for Question 23 - 24 :In a cylindrical coordinate system, magnetic field is given by

B sin

for m

for m

for m

ta0 5

2 5 6

0 6

<< <>

z

rw r

r=

Z

[

\

]]

]]

MCQ 6.2.23 The induced electric field in the region 4 m<r will be

(A) 0 (B) cos t a2r

w wf

(C) 2 cos taw- f (D) sin t a21w f

MCQ 6.2.24 The induced electric field at 4.5 mr = is

(A) 0 (B) 17 cos t18

w w-

(C) 4 cos t9

w w (D) 17 cos t4

w w-



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For View Only Shop Online at www.nodia.co.inMCQ 6.2.25 The induced electric field in the region 5 m>r is

(A) cos ta18r w w- f (B) cos t a9

rw w-

f

(C) 9 cos tar w- f (D) cos t a9r

w wf

MCQ 6.2.26 In a certain region a test charge is moving with an angular velocity 2 / secrad along a circular path of radius 4 m centred at origin in the x -y plane. If the magnetic flux density in the region is 2 /Wb mB az 2= then the electric field viewed by an observer moving with the test charge is(A) 8 /V mar (B) 4 /V mar(C) 0 (D) 8 /V ma- r

MCQ 6.2.27 A 8 A current is flowing along a straight wire from a point charge situated at the origin to infinity and passing through the point (1, 1, 1). The circulation of the magnetic field intensity around the closed path formed by the triangle having the vertices , ,2 0 0^ h, , ,0 2 0^ h and , ,0 0 2^ h is equal to

(A) A87 (B) 3 A

(C) 7 A (D) 1 A

MCQ 6.2.28 Magnetic flux density, 0.1tB az= Tesla threads only the loop abcd lying in the plane xy as shown in the figure.

Consider the three voltmeters V1, V2 and V3, connected across the resistance in the same xy plane. If the area of the loop abcd is 1 m2 then the voltmeter readings are

V1 V2 V3

(A) 66.7 mV 33.3 mV 66.7 mV

(B) 33.3 mV 66.7 mV 33.3 mV

(C) 66.7 mV 66.7 mV 33.3 mV

(D) 33.3 mV 66.7 mV 66.7 mV




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Statement for Linked Question 29 - 30 :A square wire loop of resistance R rotated at an angular velocity w in the uniform magnetic field /mWb mB a2 y

2= as shown in the figure.

MCQ 6.2.29 If the angular velocity, 2 / secradw = then the induced e.m.f. in the loop will be(A) 2 /sin V mmq (B) 2 /cos V mmq

(C) 4 /cos V mmq (D) 4 /sin V mmq

MCQ 6.2.30 If resistance, 40 mR W= then the current flowing in the square loop will be(A) 0.2 sin mAq (B) 0.1 sin mAq

(C) 0.1 cos mAq (D) 0.5 sin mAq

MCQ 6.2.31 In a certain region magnetic flux density is given as sinB tB ay0 w= . A rectangular loop of wire is defined in the region with it’s one corner at origin and one side along z -axis as shown in the figure.

If the loop rotates at an angular velocity w (same as the angular frequency of magnetic field) then the maximum value of induced e.m.f in the loop will be(A) B S2

10 w (B) B S2 0 w

(C) B S0 w (D) B S4 0 w



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For View Only Shop Online at www.nodia.co.inMCQ 6.2.32 A 50 turn rectangular loop of area 64 cm2 rotates at 60 revolution per seconds in

a magnetic field 0.25 377 /sin Wb mtB 2= directed normal to the axis of rotation. The rms value of the induced voltage is(A) 2.13 volt (B) 21.33 volt

(C) 4.26 volt (D) 42.66 volt

Statement for Linked Question 33 - 34 :Consider the figure shown below. Let 120 /cos Wb mB t5 2p= and assume that the magnetic field produced by ( )i t is negligible

MCQ 6.2.33 The value of vab is(A) 118.43 120cos tp- V (B) 118.43 120cos tp V

(C) 118.43 120sin tp- V (D) 118.43 120sin tp V

MCQ 6.2.34 The value of ( )i t is(A) 0.47 120cos tp- A (B) 0.47 120cos tp A

(C) 0.47 120sin tp- A (D) 0.47 120sin tp A

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EXERCISE 6.3

MCQ 6.3.1 A magnetic field in air is measured to be Bx yx

x yya aB y x0 2 2 2 2=

+-

+c m What current distribution leads to this field ?[Hint : The algebra is trivial in cylindrical coordinates.]

(A) , 0Bx y

rJ z 10

02 2 !m=+c m (B) , 0B

x yrJ z 2

0

02 2 !m=-+c m

(C) 0, 0rJ != (D) , 0Bx y

rJ z 10

02 2 !m=+c m

MCQ 6.3.2 For static electric and magnetic fields in an inhomogeneous source-free medium, which of the following represents the correct form of Maxwell’s equations ?(A) 0E:d = , 0Bd# = (B) 0E:d = , 0B:d =

(C) 0Ed# = , 0Bd# = (D) 0Ed# = , 0B:d =

MCQ 6.3.3 If C is closed curve enclosing a surface S , then magnetic field intensity H , the current density J and the electric flux density D are related by

(A) dt

dlH S J DS C

$ :22= +b l## ## (B) d

tdH l J D S

S S: :

22= +b l# ##

(C) dt

dlH S J DS C

: :22= +b l### (D) d

tdH l J D S

C S: :

22= +b l###

MCQ 6.3.4 The unit of H#d is(A) Ampere (B) Ampere/meter

(C) Ampere/meter2 (D) Ampere-meter

MCQ 6.3.5 The Maxwell equation t

H J D22

#d = + is based on(A) Ampere’s law (B) Gauss’ law

(C) Faraday’s law (D) Coulomb’s law

MCQ 6.3.6 A loop is rotating about they y -axis in a magnetic field ( )cosB tB ax0 w f= + T. The voltage in the loop is(A) zero

(B) due to rotation only

(C) due to transformer action only

(D) due to both rotation and transformer action

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GATE 1998



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For View Only Shop Online at www.nodia.co.inMCQ 6.3.7 The credit of defining the following current is due to Maxwell

(A) Conduction current (B) Drift current

(C) Displacement current (D) Diffusion current

MCQ 6.3.8 A varying magnetic flux linking a coil is given by /3 t2 3lF = . If at time 3 st = , the emf induced is 9 V, then the value of l is.(A) zero (B) 1 /Wb s2

(C) 1 /Wb s2- (D) 9 /Wb s2

MCQ 6.3.9 Assuming that each loop is stationary and time varying magnetic field B , induces current I , which of the configurations in the figures are correct ?

(A) 1, 2, 3 and 4 (B) 1 and 3 only

(C) 2 and 4 only (D) 3 and 4 only

MCQ 6.3.10 Assertion (A) : For time varying field the relation VE d=- is inadequate.Reason (R) : Faraday’s law states that for time varying field 0Ed# =(A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is





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IES EC 2011

IES EC 2011




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MCQ 6.3.11 Who developed the concept of time varying electric field producing a magnetic field ?(A) Gauss (B) Faraday

(C) Hertz (D) Maxwell

MCQ 6.3.12 A single turn loop is situated in air, with a uniform magnetic field normal to its plane. The area of the loop is 5 m2 and the rate of charge of flux density is 2 / /Wb m s2 . What is the emf appearing at the terminals of the loop ?(A) 5 V- (B) 2 V-

(C) 0.4 V- (D) 10 V-

MCQ 6.3.13 Which of the following equations results from the circuital form of Ampere’s law ?

(A) t

E B22

#d =- (B) 0B:d =

(C) D: rd = (D) t

H J D22

#d = +

MCQ 6.3.14 Assertion (A) : Capacitance of a solid conducting spherical body of radius a is given by 4 a0pe in free space.Reason (R) : jH E Jwe#d = +

(A) Both A and R are individually true and R is the correct explanation of A.




MCQ 6.3.15 Two conducting thin coils X and Y (identical except for a thin cut in coil Y) are placed in a uniform magnetic field which is decreasing at a constant rate. If the plane of the coils is perpendicular to the field lines, which of the following statement is correct ? As a result, emf is induced in(A) both the coils

(B) coil Y only

(C) coil X only

(D) none of the two coils

MCQ 6.3.16 Assertion (A) : Time varying electric field produces magnetic fields.Reason (R) : Time varying magnetic field produces electric fields.(A) Both A and R are true and R is the correct explanation of A




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For View Only Shop Online at www.nodia.co.inMCQ 6.3.17 Match List I (Electromagnetic Law) with List II (Different Form) and select the

correct answer using the code given below the lists :

List-I List-II

a. Ampere’s law 1. D v4: r=

b. Faraday’s law 2. Jth

4:22=-

c. Gauss law 3.t

H J D4

22

# = +

d. Current 4.t

E B4

22

# =-

Codes : a b c d(A) 1 2 3 4(B) 3 4 1 2(C) 1 4 3 2(D) 3 2 1 4

MCQ 6.3.18 Two metal rings 1 and 2 are placed in a uniform magnetic field which is decreasing with time with their planes perpendicular to the field. If the rings are identical except that ring 2 has a thin air gap in it, which one of the following statements is correct ?(A) No e.m.f is induced in ring 1

(B) An e.m.f is induced in both the rings

(C) Equal Joule heating occurs in both the rings

(D) Joule heating does not occur in either ring.

MCQ 6.3.19 Which one of the following Maxwell’s equations gives the basic idea of radiation ?

(A) /

/

t

t

H DE B

2 2

2 2

#

#

d

d

==

4 (B) /

/

t

t

E BD B:

2 2

2 2

#d

d

=-=-

4

(C) 0

DD

:

:

rd

d

==

3 (D) / t

B

H D

:

2 2

r

#

d

d

=

= ^ h4

MCQ 6.3.20 Which one of the following is NOT a correct Maxwell equation ?

(A) t

H D J22

#d = + (B) t

E H22

#d =

(C) D:d r= (D) 0B:d =

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MCQ 6.3.21 Match List I (Maxwell equation) with List II (Description) and select the correct answer :

List I

a. 0dSB: =#b. d dvSD v

v: r= ##

c. dtdl B SE: $

22=- ##

d. ( )d

tdl D J SH: :

22= +##

List II

1. The mmf around a closed path is equal to the conduction current plus the time derivative of the electric displacement current through any surface bounded by the path.

2. The emf around a closed path is equal to the time derivative is equal to the time derivative of the magnetic displacement through any surface bounded by the path.

3. The total electric displacement through the surface enclosing a volume is equal to total charge within the volume

4. The net magnetic flux emerging through any closed surface is zero.Codes :

a b c d(A) 1 3 2 4(B) 4 3 2 1(C) 4 2 3 1(D) 1 2 3 4

MCQ 6.3.22 The equation of continuity defines the relation between(A) electric field and magnetic field

(B) electric field and charge density

(C) flux density and charge density

(D) current density and charge density

MCQ 6.3.23 What is the generalized Maxwell’s equation t

H J Dc 2

2#d = + for the free space ?

(A) 0H#d = (B) H Jc#d =

(C) t

H D22

#d = (D) H D#d =

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For View Only Shop Online at www.nodia.co.inMCQ 6.3.24 Magnetic field intensity is H 2y xa a a3 6x y z= + + A/m. What is the current density

/A mJ 2 ?(A) 2ay- (B) 7az-

(C) 3ax (D) 12ay

MCQ 6.3.25 A circular loop placed perpendicular to a uniform sinusoidal magnetic field of frequency 1w is revolved about an axis through its diameter at an angular velocity

2w rad/sec ( )<2 1w w as shown in the figure below. What are the frequencies for the e.m.f induced in the loop ?

(A) and1 2w w

(B) , and1 2 2 2w w w w+

(C) , and2 1 2 2w w w w-

(D) and1 2 1 2w w w w- +

MCQ 6.3.26 Which one of the following is not a Maxwell’s equation ?(A) jH Es we#d = +^ h

(B) QF E v B#= +^ h

(C) d dtdH l J S D S

c s s: : :

22= +# # #

(D) dB S 0S

: =#

MCQ 6.3.27 Consider the following three equations :

1. t

E Bd22

# =-

2. t

H J D22

#d = +

3. 0B:d =

Which of the above appear in Maxwell’s equations ?(A) 1, 2 and 3 (B) 1 and 2

(C) 2 and 3 (D) 1 and 3

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MCQ 6.3.28 A straight current carrying conductor and two conducting loops A and B are shown in the figure given below. What are the induced current in the two loops ?

(A) Anticlockwise in A and clockwise in B

(B) Clockwise in A and anticlockwise in B

(C) Clockwise both in A and B

(D) Anticlockwise both in A and B

MCQ 6.3.29 Which one of the following equations is not Maxwell’s equation for a static electromagnetic field in a linear homogeneous medium ?

(A) 0B:d = (B) 0Dd# = v

(C) d IB lc

0: m=# (D) A J20md =

MCQ 6.3.30 In free space, if 0vr = , the Poisson’s equation becomes(A) Maxwell’s divergence equation 0B:d =

(B) Laplacian equation 0V2d =

(C) Kirchhoff’s voltage equation 0VS =


MCQ 6.3.31 Match List I with List II and select the correct answer using the codes given below :

List I List II

a Continuity equation 1.t

H J D22

#d = +

b Ampere’s law 2.t

J D22=

c Displacement current 3.t

E B22

#d =-

d Faraday’s law 4.t

J v

22r

#d =-

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For View Only Shop Online at www.nodia.co.inCodes : a b c d(A) 4 3 2 1(B) 4 1 2 3(C) 2 3 4 1(D) 2 1 4 3

MCQ 6.3.32 Match List I (Type of field denoted by A) with List II (Behaviour) and select the correct answer using the codes given below :

List I List II

a A static electric field in a charge free region 1. 0A:d =

0A!#d

b A static electric field in a charged region 2. 0A: !d

0A#d =

c A steady magnetic field in a current carrying conductor

3. 0A: !d

0A!#d

d A time-varying electric field in a charged medium with time-varying magnetic field

4. 0A:d =

0A#d =Codes :

a b c d(A) 4 2 3 1(B) 4 2 1 3(C) 2 4 3 1(D) 2 4 1 3

MCQ 6.3.33 Which one of the following pairs is not correctly matched ?

(A) Gauss Theorem : d dvD s Ds v

: :d=# #

(B) Gauss’s Law : d dvD sv

: r=# #

(C) Coulomb’s Law : V dtd mf=-

(D) Stoke’s Theorem : ( )d dl sl s

: :x x#d=# #

MCQ 6.3.34 Maxwell equation ( / )tE B2 2#d =- is represented in integral form as

(A) dt

dE l B l: :22=-# # (B) d

tdE l B s

s: :

22=- ##

(C) dt

dE l B l:22

# =-# # (D) dt

dE l B ls

:22

# =- ##

IES EE 2004

IES EE 2003

IES EE 2003




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MCQ 6.3.35 The magnetic flux through each turn of a 100 turn coil is ( )t t43 - milli-Webers where t is in seconds. The induced e.m.f at 2t s= is(A) 1 V (B) 1 V-

(C) 0.4 V (D) 0.4 V-

MCQ 6.3.36 Two conducting coils 1 and 2 (identical except that 2 is split) are placed in a uniform magnetic field which decreases at a constant rate as in the figure. If the planes of the coils are perpendicular to the field lines, the following statements are made :

1. an e.m.f is induced in the split coil 2

2. e.m.fs are induced in both coils

3. equal Joule heating occurs in both coils

4. Joule heating does not occur in any coil

Which of the above statements is/are true ?(A) 1 and 4 (B) 2 and 4

(C) 3 only (D) 2 only

MCQ 6.3.37 For linear isotropic materials, both E and H have the time dependence ej tw and regions of interest are free of charge. The value of Hd# is given by(A) Es (B) j Ewe

(C) jE Es we+ (D) jE Es we-

MCQ 6.3.38 Which of the following equations is/are not Maxwell’s equations(s) ?

(A) t

J v:

22r

d =- (B) D v: rd =

(C) t

E B:

22d =- (D) d

tdH l E E s

s: :

22s e= +b l##

Select the correct answer using the codes given below :(A) 2 and 4 (B) 1 alone

(C) 1 and 3 (D) 1 and 4

MCQ 6.3.39 Assertion (A) : The relationship between Magnetic Vector potential A and the current density J in free space is ( )A# #d d J0m=

IES EE 2002

IES EE 2002

IES EE 2002

IES EE 2002

IES EE 2001



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For View Only Shop Online at www.nodia.co.inFor a magnetic field in free space due to a dc or slowly varying current is A J2

0md =-Reason (R) : For magnetic field due to dc or slowly varying current 0A:d = .(A) Both A and R are true and R is the correct explanation of A




MCQ 6.3.40 Given that t

H J D22

#d = +

Assertion (A) : In the equation, the additional term tD22 is necessary.

Reason (R) : The equation will be consistent with the principle of conservation of charge.(A) Both A and R are true and R is the correct explanation of A




MCQ 6.3.41 Consider coils , ,C C C1 2 3 and C4 (shown in the given figures) which are placed in the time-varying electric field ( )tE and electric field produced by the coils ,C C2 3l l and C 4l carrying time varying current ( )I t respectively :

The electric field will induce an emf in the coils(A) C1 and C2 (B) C2 and C3

IES EE 2001

IES EE 2001




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(C) C1 and C3 (D) C2 and C4

MCQ 6.3.42 A circular loop is rotating about the y -axis as a diameter in a magnetic field sinB tB a Wb/mx0

2w= . The induced emf in the loop is(A) due to transformer emf only

(B) due to motional emf only

(C) due to a combination of transformer and motional emf

(D) zero

MCQ 6.3.43 Match List I (Law/quantity) with List II (Mathematical expression) and select the correct answer :

List I List II

a. Gauss’s law 1. D: rd =

b. Ampere’s law 2.t

E B22

#d =-

c. Faraday’s law 3. E HP #=

d. Poynting vector 4. qF E v B#= +^ h

5.t

H J Dc 2

2#d = +

Codes : a b c d(A) 1 2 4 3(B) 3 5 2 1(C) 1 5 2 3(D) 3 2 4 1

***********

IES EE 2001

IES EE 2001



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SOLUTIONS 6.1


SOL 6.1.2 Option (B) is correct.The line integral of magnetic field intensity along a closed loop is equal to the current enclosed by it.

i.e. dH l:# Ienc=So, for the constant current, magnetic field intensity will be constant i.e. magnetostatic field is caused by steady currents.

SOL 6.1.3 Option (D) is correct.From Faraday’s law the electric field intensity in a time varying field is defined as

E#d tB22=- where B is magnetic flux density in the EM field.

and since the magnetic flux density is equal to the curl of magnetic vector potentiali.e. B A#d=So, putting it in equation (1), we get

E#d t

A22

#d=- ^ h

or E#d tA#d

22= -b l

Therefore, E tA

22=-

SOL 6.1.4 Option (B) is correct.Since total magnetic flux through a surface S is defined as

F dB SS

:= #From Maxwell’s equation it is known that curl of magnetic flux density is zero B:d 0=

dB SS

:# ( )dvB 0v

:d= =# (Stokes Theorem)

Thus, net outwards flux will be zero for a closed surface.

SOL 6.1.5 Option (A) is correct.From Faraday’s law, the relation between electric field and magnetic field is

E#d tB22=-




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Since the electric field inside a conducting sphere is zero.i.e. E 0=So the rate of change in magnetic flux density will be

tB22 ( )E#d=- 0=

Therefore ( , )r tB will be uniform inside the sphere and independent of time.

SOL 6.1.6 Option (B) is correct.From the integral form of Faraday’s law we have the relation between the electric field intensity and net magnetic flux through a closed loop as

dE l:# dtdF=-

Since electric field intensity is zero (E 0= ) inside the conducting loop. So, the rate of change in net magnetic flux through the closed loop is

dtdF 0=

i.e. F is constant and doesn’t vary with time.

SOL 6.1.7 Option (A) is correct.A superconductor material carries zero magnetic field and zero electric field inside it. i.e. B 0= and 0E =Now from Ampere-Maxwell equation we have the relation between the magnetic flux density and electric field intensity as

B#d t

J E0 0 0 2

2m m e= +

So, J 0= (B 0= , 0E = )Since the net current density inside the superconductor is zero so all the current must be confined at the surface of the wire.

SOL 6.1.8 Option (A) is correct.According to Lenz’s law the induced current I in a loop flows such as to produce a magnetic field that opposes the change in ( )tB .Now the configuration shown in option (A) and (B) for increasing magnetic flux Bi , the change in flux is in same direction to Bi as well as the current I flowing in the loop produces magnetic field in the same direction so it does not follow the Lenz’s law.For the configuration shown in option (D), as the flux Bd is decreasing with time so the change in flux is in opposite direction to Bd as well as the current I flowing in the loop produces the magnetic field in opposite direction so it also does not follow the Lenz’s law.For the configuration shown in option (C), the flux density Bd is decreasing with time so the change in flux is in opposite direction to Bd but the current I flowing in the loop produces magnetic field in the same direction to Bd (opposite to the direction of change in flux density). Therefore this is the correct configuration.



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Induced emf in a conducting loop is given by

Vemf dtdF=- where F is total magnetic flux passing through the loop.

Since, the magnetic field is non-uniform so the change in flux will be caused by it and the induced emf due to it is called transformer emf.Again the field is in ay direction and the loop is rotating about z -axis so flux through the loop will also vary due to the motion of the loop. This causes the emf which is called motion emf. Thus, total induced voltage in the rotating loop is caused by the combination of both the transformer and motion emf.

SOL 6.1.10 Option (D) is correct.As the conducting loop is falling freely So, the flux through loop will remain constant. Therefore, the voltage induced in the loop will be zero.

SOL 6.1.11 Option (B) is correct.The magnetic flux density passing through the loop is given as B 4z t ax3 2=Since the flux density is directed normal to the plane x 0= so the total magnetic flux passing through the square loop located in the plane x 0= is

F dB S:= # ( )z t dydz4zy

3 2

0

1

0

1=

==## t2= ( ( )d dydzS ax= )

Induced emf in a loop placed in magnetic field is defined as

Vemf dtdF=-

where F is the total magnetic flux passing through the loop. So the induced emf in the square loop is

Vemf ( )dtd t

t22

=- =- ( t2F = )

Therefore at time t sec2= the induced emf is Vemf 4 volt=-

SOL 6.1.12 Option (A) is correct.Magnetic flux density produced at a distance r from a long straight wire carrying current I is defined as

B I a20

prm= f

where af is the direction of flux density as determined by right hand rule. So the flux density produced by straight wire at a distance r from it is

B I a2 n0

prm= (an is unit vector normal to the loop)

Therefore the total magnet flux passing through the loop is

F dB S:= # I ad2d

d a0

prm r=

+

# (d adS anr= )

where dr is width of the strip of loop at a distance r from the straight wire. Thus,




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F I d20

2

3

pm

rr= b l# lnI2 2

30

pm= b l

( )( . )ln2

51 50

pm=

( )( ) ( . )ln2 10 5 1 57#= - 4.05 10 Wb7

#= -

SOL 6.1.13 Option (D) is correct.Electric field intensity experienced by the moving conductor ab in the presence of magnetic field B is given as E v B#= where v is the velocity of the conductor.So, electric field will be directed from b to a as determined by right hand rule for the cross vector. Therefore, the voltage difference between the two ends of the conductor is given as

Vab dE la

b:=- #

Thus, the positive terminal of voltage will be a and Vab will be positive.

SOL 6.1.14 Option (B) is correct.Given magnetic flux density through the square loop is B 0.1 /Wb mtaz 2=So, total magnetic flux passing through the loop is F dB S:= . 0.1t t0 1 1= =^ ^h h

The induced emf (voltage) in the loop is given as

Vemf 0. Voltdtd 2f=- =-

As determined by Lenz’s law the polarity of induced emf will be such that V V1 2+ Vemf=-Therefore, the voltage drop in the two resistances are respectively,

V1 ( ) . 33.3 mVV2 42

30 1

emf= + - = =b l

and V2 ( ) 66.7 mVV2 44

emf= + - =b l

SOL 6.1.15 Option (D) is correct.Consider a magnet bar being dropped inside a pipe as shown in figure.Suppose the current I in the magnet flows counter clockwise (viewed from above) as shown in figure. So near the ends of pipe, it’s field points upward. A ring of pipe below the magnet experiences an increasing upward flux as the magnet approaches and hence by Lenz’s law a current will be induced in it such as to produce downward flux. Thus, Iind must flow clockwise which is opposite to the current in the magnet. Since opposite currents repel each other so, the force exerted on the magnet due to the induced current is directed upward. Meanwhile a ring above the magnet experiences a decreasing upward flux; so it’s induced current parallel to I and it attracts magnet upward. And flux through the rings next to the magnet bar is constant. So no current is induced in them.



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Thus, for all we can say that the force exerted by the eddy current (induced current according to Lenz’s law) on the magnet is in upward direction which causes the delay to reach the bottom. Whereas in the cases of unmagnetized bar no induced current is formed. So it reaches in fraction of time.Thus, A and R both true and R is correct explanation of A.


Voltage, V1 N dtd

1F=-

where F is total magnetic flux passing through it.

Again V2 N dtd

2F=-

Since both the coil are in same magnetic field so, change in flux will be same for both the coil.Comparing the equations (1) and (2) we get

VV

2

1 NN

2

1=

V2 (12)V N 50003000

11

2N= = 7.2 volt=

SOL 6.1.17 Option (A) is correct.The magnetic flux density inside a solenoid of n turns per unit length carrying current I is defined as B nI0m=Let the length of solenoid be l and its cross sectional radius be r . So, the total magnetic flux through the solenoid is F ( )( )( )nI r nl0

2m p= (1)Since the total magnetic flux through a coil having inductance L and carrying current I is given as F LI=So comparing it with equation (1) we get,




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L n I l02 2m p=

and as for a given solenoid, radius r and length l is constant therefore L n2

\

SOL 6.1.18 Option (A) is correct.The magnetic flux density inside the solenoid is defined as B nI0m=where n " no. of turns per unit length I " current flowing in it.So the total magnetic flux through the solenoid is

F dB S:= # ( )( )nI a02m p=

where a " radius of solenoidInduced emf in a loop placed in a magnetic field is defined as

Vemf dtdF=-

where F is the total magnetic flux passing through the loop. Since the resistance R is looped over the solenoid so total flux through the loop will be equal to the total flux through the solenoid and therefore the induced emf in the loop of resistance will be

Vemf a ndtdI2

0p m=-

Since current I flowing in the solenoid is constant so, the induced emf is Vemf 0=and therefore the induced current in the loop will be zero.

SOL 6.1.19 Option (B) is correct.It will be similar to the current in a solenoid.So, the magnetic field will be in circumferential while the electric field is longitudinal.

SOL 6.1.20 Option (B) is correct.In Assertion (A) the magnetic flux through each turn of both coils are equal So, the net magnetic flux through the two coils are respectively 1F N1F=and 2F N2F=where F is the magnetic flux through a single loop of either coil and N1, N2 are the total no. of turns of the two coils respectively.Therefore the induced emf in the two coils are

Vemf 1 dtd 1F=- N dt

d1F=-

Vemf 2 dtd 2F=- N dt

d2F=-

Thus, the ratio of the induced emf in the two loops are

VV

emf

emf

1

2 NN

1

2=



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For View Only Shop Online at www.nodia.co.inNow, in Reason (R) : a primitive transformer is similar to the cylinder core carrying wound coils. It is the device in which by choosing the appropriate no. of turns, any desired secondary emf can be obtained.So, both the statements are correct but R is not the explanation of A.

SOL 6.1.21 Option (B) is correct.Electric flux density in the medium is given as D Ee= cosE t0e w= ( cosE E t0 w= )Therefore the displacement current density in the medium is

Jd sintD E t02

2 we w= =-

and the conduction current density in the medium is Jc cosE E t0s s w= =So, the ratio of amplitudes of conduction current density and displacement current density is

JJ

d

c we

s=

SOL 6.1.22 Option (D) is correct.The displacement current density in a medium is equal to the rate of change in electric flux density in the medium.

Jd tD22=

Since the displacement current density in the medium is given as Jd 20 . /cos A mt a1 5 10 y

8 2#= ^ h

So, the electric flux density in the medium is

D dt CJd= +# (C" constant)

20 .cos t dt Ca1 5 10 y8

#= +^ h#As there is no D.C. field present in the medium so, we get C 0= and thus,

D .

.sin ta

1 5 1020 1 5 10

y8

8

#

#= ^ h

1.33 10 .sin t a1 5 10 y7 8

##= -^ h

1 3.3 . /sin nC mt a5 1 5 10 y8 2

#= ^ h

SOL 6.1.23 Option (A) is correct.Given the volume charge density, vr 0=So, from Maxwell’s equation we have D:d vr= D:d 0= (1)Now, the electric flux density in a medium is defined as D Ee= (where e is the permittivity of the medium)So, putting it in equation (1) we get, ( )E: ed 0=or, ( ) ( )E E: :e ed d+ 0=




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and since dee 0. 0&4 .e (given)

Therefore, E:d 0.

SOL 6.1.24 Option (C) is correct.The ratio of magnitudes of displacement current to conduction current in any medium having permittivity e and conductivity s is given as

Conduction currentDisplacement current s

we=

where w is the angular frequency of the current in the medium.Given frequency, f 50 GHz=Permittivity, e .4 4 8 85 100

12# #e= = -

Conductivity, s 1.14 10 /s m8#=

So, w f2 2 50 10 100 109 9# # #p p p= = =

Therefore, the ratio of magnitudes of displacement current to the conduction current is

IIc

d .

.1 14 10

100 10 4 8 85 108

9 12

#

# # # #p=-

.9 75 10 8#= -

SOL 6.1.25 Option (D) is correct.Given the electric field intensity in time domain as

E sin cos

rt kr

aq w

=-

q^ h

So, the electric field intensity in phasor form is given as

Es sinr e ajkrq= q

-

and Es#d r kdr rE a1s

2= q f^ h sinjk r e ajkrq= - f-

^ h

Therefore, from Maxwell’s equation we get the magnetic field intensity as

Hs j rEs0

#dw=- sin

rkr e ajkr

0wq= f

-

SOL 6.1.26 Option (B) is correct.In phasor form the magnetic field intensity can be written as Hs 0.1 /cos A my e a15 jbx

zp= -^ h

Similar as determined in MCQ 42 using Maxwell’s equation we get the relation b15 2 2p +^ h 2

0 0w p e=Here w 6 109

#p=

So, b15 2 2p +^ h 3 106 10

8

9 2

#

#p= c m

b15 2 2p +^ h 400 2p= b2 175 2p= & b .6 /rad m23!=

***********



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SOLUTIONS 6.2

SOL 6.2.1 Option (D) is correct.Induced emf. in the conducting loop formed by rail, bar and the resistor is given by

Vemf dtdF=-

where F is total magnetic flux passing through the loop.Consider the bar be located at a distance x from the resistor at time t . So the total magnetic flux passing through the loop at time t is

F dB S:= # Blx= (area of the loop is S lx= )

Now the induced emf in a loop placed in magnetic field is defined as

Vemf dtdF=-

where F is the total magnetic flux passing through the loop. Therefore the induced emf in the square loop is

Vemf ( )dtd Blx=- Bl dt

dx=- ( BlxF = )

Since from the given figure, we have l 2 m= and B 0.1 /Wb m2=and /dx dt = velocity of bar 5 /m s=So, induced emf is Vemf ( . )( )( )0 1 2 5=- 1 volt=-According to Lenz’s law the induced current I in a loop flows such as to produce magnetic field that opposes the change in ( )tB . As the bar moves away from the resistor the change in magnetic field will be out of the page so the induced current will be in the same direction of I shown in figure.Thus, the current in the loop is

I RVemf=-

( )10

1=- - 0. A23= (R 10W= )

SOL 6.2.2 Option (B) is correct.Magnetic flux density produced at a distance r from a long straight wire carrying current I is defined as

B I a20

prm= f

where af is the direction of flux density as determined by right hand rule. So, the magnetic flux density produced by the straight conducting wire linking through the




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loop is normal to the surface of the loop.Now consider a strip of width dr of the square loop at distance r from the wire for which the total magnetic flux linking through the square loop is given as

F dB SS

:= #

( )I ad21a0

pm

r r=r

r+

# (area of the square loop is dS adr= )

lnIa a20

pm

rr= +

b l

The induced emf due to the change in flux (when pulled away) is given as

Vemf dtdF=- lnIa

dtd a

20

pm

rr=- +

b l; E

Therefore, Vemf Ia

a dtd

dtd

21 10

pm

rr

rr=- + -c m

Given dtdr 5 /velocity of loop m s= =

and since the loop is currently located at 3 m distance from the straight wire, so after . sec0 6 it will be at r 3 (0.6) v#= + ( velocity of the loopv" ) .3 0 6 5#= + 6 m=

So, Vemf ( )

( ) ( )230 2

81 5 6

1 50 # #p

m=- -: D ( 2 , 30m Aa I= = )

25 10 volt7#= - 2.5 voltm=

SOL 6.2.3 Option (B) is correct.Since total magnetic flux through the loop depends on the distance from the straight wire and the distance is constant. So the flux linking through the loop will be constant, if it is pulled parallel to the straight wire. Therefore the induced emf in the loop is

Vemf dtdF=- 0= (F is constant)

SOL 6.2.4 Option (D) is correct.Magnetic flux density produced at a distance r from a long straight wire carrying current I is defined as

B I a20

prm= f

where af is the direction of flux density as determined by right hand rule. Since the direction of magnetic flux density produced at the loop is normal to the surface of the loop So, total flux passing through the loop is given by

F dB SS

:= # I ad20

2

4

prm r=

r=c ^m h# (dS adr= )

Ia d20

2

4

pm

rr= # lnI

22 20

pm= ( )lnI 20

pm=

The current flowing in the loop is Iloop and induced e.m.f. is Vemf .



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For View Only Shop Online at www.nodia.co.inSo, Vemf I R dt

dloop

F= =-

( )dtdQ R ( )ln dt

dI20

pm=-

where Q is the total charge passing through a corner of square loop.

dtdQ ( )ln dt

dI4 20

pm=- ( 4R W= )

dQ ( )ln dI4 20

pm=-

Therefore the total charge passing through a corner of square loop is

Q (2)ln dI40

4

0

pm=- # ( )( )ln4 2 0 40

pm=- -

( )ln44 4 10 2

7# #

pp=

- 2.77 10 C7

#= - 277 nC=

SOL 6.2.5 Option (A) is correct.Since the radius of small circular loop is negligible in comparison to the radius of the large loop. So, the flux density through the small loop will be constant and equal to the flux on the axis of the loops.

So, B Iz RR a2 2 2 / z

03 2

2m=+^ h

where R " radius of large loop 5 m= z " distance between the loops 12 m=

B ( ) a2

212 5

5/ z

02 2 3 2

2#m

#=+^ ^h h6 @

13a25z3

0m=^ h

Therefore, the total flux passing through the small loop is

F dB S:= # r1325

30 2#

m p=^ h

wherer is radius of small circular loop.

13

25 4 10 103

73 2# #

#p p=

--

^^

hh .9 fWb65=

SOL 6.2.6 Option (C) is correct.Electric field in any medium is equal to the voltage drop per unit length.

i.e. E dV=

where V " potential difference between two points. d " distance between the two points.The voltage difference between any two points in the medium is V 2cosV ft0 p=So the conduction current density in the medium is given as Jc Es= ( "s conductivity of the medium)

Er= ( "r resistivity of the medium)

2cosdV

dV ft0

r rp= = ( )cosV V ft20 p=




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or, Jc dV0

r=

and displacement current density in the medium is given as

Jd tD

22=

tE

22e=

( )cost dV ft20

22e p= ; E ( )cosV V ft20 p=

( )sindV ft ft2 20e p p= -6 @

or, Jd 2df V0p e=

Therefore, the ratio of amplitudes of conduction current and displacement current in the medium is

II

d

c JJ

d

C=

( )/( )( )/( )d f VV d

2 0

0

p er= f2

1p er=

2 (1.6 10 ) (54 8.85 10 ) 0.77

18 12p# # # # # #

= -

.2 7=

SOL 6.2.7 Option (C) is correct.Total magnetic flux through the solenoid is given as F nI0m=where n is the no. of turns per unit length of solenoid and I is the current flowing in the solenoid.Since the solenoid carries current that is increasing linearly with timei.e. I t\

So the net magnetic flux through the solenoid will be F t\

or, F kt= where k is a constant.Therefore the emf induced in the loop consisting resistances RA, RB is

Vemf dtdF=-

Vemf k=-and the current through R1 and R2 will be

Iind R Rk

1 2=- +

Now according to Lenz’s law the induced current I in a loop flows such as to produce a magnetic field that opposes the change in ( )tB .i.e. the induced current in the loop will be opposite to the direction of current in solenoid (in anticlockwise direction).

So, VA I R R RkR

ind AA B

A= =- +

and VB I R R RkR

ind BA B

B=- = +b l

Thus, the ratio of voltmeter readings is

VVB

A RRB

A=-



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Induced emf in the conducting loop formed by rail, bar and the resistor is given by

Vemf dtdF=-

where F is total magnetic flux passing through the loop.The bar is located at a distance x from the resistor at time t . So the total magnetic flux passing through the loop at time t is

F dB S:= # Blx= where l is separation between the railsNow the induced emf in a loop placed in magnetic field is defined as

Vemf dtdF=-

where F is the total magnetic flux passing through the loop. Therefore the induced emf in the square loop is

Vemf ( )dtd Blx=- Bl dt

dx=- ( BlxF = )

Since from the given figure, we have l 5 m= B 2 T=and /dx dt " velocity of bar 4 /m s=So, induced emf is Vemf ( )( )( )2 5 4=- volt40=-Therefore the current in the bar loop will be

I RVemf= 10

40=- A5=-

SOL 6.2.9 Option (B) is correct.As obtained in the previous question the current flowing in the sliding bar is I 4 A=-Now we consider magnetic field acts in ax direction and current in the sliding bar is flowing in az+ direction as shown in the figure.

Therefore, the force exerted on the bar is

F Idl B#= # ( 4 ) (2 )dza az x0

5

#= -# za16 y 0

5=- 6 @ 40 Nay=-




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i.e. The force exerted on the sliding bar is in opposite direction to the motion of the sliding bar.

SOL 6.2.10 Option (A) is correct.Given the magnetic flux density through the square loop is B 7.5 (120 30 )cos t azcp= -So the total magnetic flux passing through the loop will be

F dB SS

:= # 7.5 (120 30 ) (1 1)( )cos t a az zcp #= - - -6 @

. ( )cos t7 5 120 30cp= -Now, the induced emf in the square loop is given by

Vemf dtdF=- . ( )sin t7 5 120 120 30# cp p= -

The polarity of induced emf (according to Lenz’s law) will be such that induced current in the loop will be in opposite direction to the current ( )I t shown in the figure. So we have

( )I t RVemf=-

. (120 30 )sin t5007 5 120# cp p=- - ( 250 250 )R 500W= + =

.7 (120 30 )sin t4 cp=- -

SOL 6.2.11 Option (D) is correct.As shown in figure the bar is sliding away from origin.Now when the bar is located at a distance dx from the voltmeter, then, the vector area of the loop formed by rail and the bar is dS (20 10 )( )dx az2

#= -

So, the total magnetic flux passing through the loop is

F dB SS

:= # (8 )(20 10 )x dxa az z

x 2 2

0#= -#

. .t t3

1 6 1 0 4 2 3

=+^ h8 B

Therefore, the induced e.m.f. in the loop is given as

Vemf . 3 . (1 1.2 )dt

d t t t31 6 0 4 3 2 2F

# #=- =- + +^ h

Vemf 1.6 . . ( . )( . )0 4 0 4 1 1 2 0 44 2 2#=- + +^ ^h h6 6@ @ ( . sect 0 4= )

0.35 volt=-Since the voltmeter is connected in same manner as the direction of induced emf (determined by Lenz’s law).So the voltmeter reading will be V 0.35 voltVemf= =-

SOL 6.2.12 Option (C) is correct.Since the position of bar is give as x .t t1 0 4 2= +^ h

So for the position 12 cmx = we have



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For View Only Shop Online at www.nodia.co.in .0 12 .t t1 0 4 2= +^ h

or, t . sec0 1193=As calculated in previous question, the induced emf in the loop at a particular time t is Vemf . . .t t t1 6 0 4 1 1 23 2 2=- + +^ ^h h6 @

So, at t . sec0 1193= ,

Vemf . ( . ) . . . .1 6 0 1193 0 4 0 1193 1 1 2 0 11933 2 2=- + +^ ^ ^h h h7 6A @

.0 02344=- 23.4 mV=-Since the voltmeter is connected in same manner as the direction of induced emf as determined by Lenz’s law. Therefore, the voltmeter reading at 12 cmx = will be V Vemf= 23.4 mvolt=-

SOL 6.2.13 Option (D) is correct.Consider the mutual inductance between the rectangular loop and straight wire be M . So applying KVL in the rectangular loop we get,

Mdtdi1 L dt

di Ri22= + ...(1)

Now from the shown figure (b), the current flowing in the straight wire is given as i1 ( ) ( )I u t I u t T1 1= - - (I1 is amplitude of the current)

or, dtdi1 ( ) ( )I t I t T1 1d d= - - (2)

So, at t 0= dtdi1 I1=

and MI1 L dtdi Ri2

2= + (from equation (1))

Solving it we get

i2 LM I e /R L t

1= -^ h for t T0 < <

Again in equation (2) at t T= we have

dtdi1 I1=-

and MI1- L dtdi Ri2

2= + (from equation (1))

Solving it we get

i2 LM I e / ( )R L t T

1=- - -^ h for t T>

Thus, the current in the rectangular loop is

i2 LM I e t T

LM I e t T

0 < <

>

/

( / )( )

R L t

R L t T

1

1

=-

-

- -

^ hZ

[

\

]]

]]

Plotting i2 versus t we get




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comSOL 6.2.14 Option (D) is correct.

Total magnetic flux passing through the loop formed by the resistance, bar and the rails is given as:

F dB SS

:= # B S:= . 0.5(1 )cos t ya a0 2 x x:w= -6 6@ @

0.1 . cos cost t1 0 5 1 w w= - -^ h6 @ (y 0.5 cos mt1 w= -^ h ) . cos cost t0 05 1w w= +^ h . cos cost t0 05 2w w= +^ h

So, the induced emf in the loop is

Vemf dtdF=-

and as determined by Lenz’s law, the induced current will be flowing in opposite direction to the current i . So the current i in the loop will be

i RVemf=- R dt

d1 F=- -b l

. sin cos sint t t50 05 2w w w w w= - -6 @

0. sin cost t23 1 2w w w=- +^ h

SOL 6.2.15 Option (C) is correct.Given the electric flux density in the medium is D 1.33 . /sin C mt x a3 10 0 2 y

8 2# m= -^ h

So, the electric field intensity in the medium is given as

E De= where e is the permittivity of the medium

or, E Dr 0e e=

.. .sin t x

a10 8 85 10

1 33 10 3 10 0 2y12

6 8

# #

# #=-

-

-^ h

( 10re = )

1.5 10 .sin t x a3 10 0 2 y4 8

##= -^ h

Now, from maxwell’s equation we have

E#d tB22=-

or, tB22 E#d=-

xE ay z2

2=-



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For View Only Shop Online at www.nodia.co.in ( 0.2) . .cos t x a1 5 10 3 10 0 2 y

4 8# ##=- - -^ ^h h

3 10 .cos t x a3 10 0 2 y3 8

##= -^ h

Integrating both sides, we get the magnetic flux density in the medium as

B 3 10 .cos t x a3 10 0 2 y3 8

##= -^ h#

.sin t x a3 103 10 3 10 0 2 y8

38

#

##= -^ h

10 .sin t x a3 10 0 2 y5 8

#= --^ h Tesla

Therefore the magnetic field intensity in the medium is

H B Br 0m m m= =

.sin t x

2 4 1010 3 10 0 2

7

5 8

# #

#

p=

--

-^ h

2rm =

Thus H . /sin A mt x a2 3 10 0 2 y8

#= -^ h

SOL 6.2.16 Option (D) is correct.Given the magnetic field intensity in the medium is H /cos A mt bx a10 z

10= -^ h

Now from the Maxwell’s equation, we have

H#d tD22=

or, tD22

xH az y2

2=- sinb t bx a10 y10=- -^ h

D sinb t bx dt C1010= - - +^ h# where C is a constant.

Since no D.C. field is present in the medium so, we get C 0= and therefore,

D /cos C mb t bx a10

10 y1010 2= -^ h

and the electric field intensity in the medium is given as

E De=

.cosb bx a

0 12 10 1010 t

y9 1010

# #= -- ^ h (e 0.12 /nF m= )

Again From the Maxwell’s equation

E#d tB22=-

or, tB22 . cosb bx a1 2 10 t

y10

#d=- -^ h: D

. sinb bx a1 2 10 tz

210=- -^ h

So, the magnetic flux density in the medium is

B . sinb bx dta1 2 10 tz

210=- -^ h#

( . )

cosb t bx a1 2 10

10 z10

210

#= -^ h (1)

We can also determine the value of magnetic flux density as : B Hm=




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(3 10 )cos t bx a10 z5 10

#= --^ h (2)

Comparing the results of equation (1) and (2) we get,

( . )

b1 2 1010

2

# 3 10 5

#= -

b2 .6 105 4#=

b /rad m344!=


The magnetic vector potential for a direct current flowing in a filament is given as

A RI dxa4 x

0

pm= #

Here current ( )I t flowing in the filament shown in figure is varying with time as ( )I t 8 At=So, the retarded vector potential at the point P will be given as

A /

RI t R c

dxa4 x0

pm

=-^ h#

where R is the distance of any point on the filamentary current from P as shown in the figure and c is the velocity of waves in free space. So, we have R x 42= + and 3 10 /m sc 8

#=

Therefore, A /

Rt R c

dxa48

.

.

xx

0

0 1

0 1

pm

=-

=-

^ h#

xt dx c dx4

84

1.

.

.

.0

20 1

0 1

0 1

0 1

pm=

+-

--< F##

8 10 lnt x x x43 108 10

.

.

..7 2

0 1

0 1

8

7

0 10 1

#

##= + + --

-

-

-^ h8 6B @

. .

. . .lnt8 100 1 4 01

0 1 4 01 0 53 107 15# #=

- ++ -- -

e o

.t8 10 0 53 108 15# #= -- -

or, A . /nWb mt a80 5 3 10 x7

#= - -^ h (1)

So, when A 0= t 6.6 10 6.6 secn9#= =-

and when t 0= A 5.3 10 /nWb m4#=- -

From equation (1) it is clear that A will be linearly increasing with respect to time. Therefore the plot of A versus t is



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Note : Time varying potential is usually called the retarded potential.

SOL 6.2.18 Option (A) is correct.Given Retarded scalar potential, V volty x ct= -^ h

and retarded vector potential, A /Wb my cx t ax= -a k

Now the magnetic flux density in the medium is given as B A#d=

yA ay z2

2=- Teslat cx az= -a k (1)

So, the magnetic field intensity in the medium is

H B0m

= ( 0m is the permittivity of the medium)

/A mt cx a1

z0m

= -a k (2)

and the electric field intensity in the medium is given as

E VtA

22d=- -

x ct y ya a ay x x=- - - +^ h ct x ay= -^ h (3)So, the electric flux density in the medium is D E0e= ( 0e is the permittivity of the medium) /C mct x ay0

2e= -^ h (4)Now we determine the condition for the field to satisfy all the four Maxwell’s equation.(i) D:d vr=or, vr ct x ay0: ed= -^ h6 @ (from equation (4)) 0=It means the field satisfies Maxwell’s equation if 0vr = .(ii) B:d 0=

Now, B:d t cx az:d= -a k9 C 0= (from equation (1))

So, it already, satisfies Maxwell’s equation

(iii) H#d t

J D22= +




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Now, H#d xH az y2

2=- c a a1y y

0 0

0

m me= = (from equation (2))

and from equation (4) we have

tD22 ca ay y0

0

0e me= = (Since in free space c 1

0 0m e= )

Putting the two results in Maxwell’s equation, we get the condition J 0=

(iv) E#d tB22=-

Now E#d xE a ay

z z22= =-

tB22 az=

So, it already satisfies Maxwell’s equation. Thus, by combining all the results we get the two required conditions as 0J = and 0vr = for the field to satisfy Maxwell’s equation.

SOL 6.2.19 Option (B) is correct.Given the electric field in time domain as E 5 sin cosy bx a10 6 10 z

9#p p= -^ ^h h

Comparing it with the general equation for electric field intensity given as E cosE t x a0 zw b= -^ h

We get, w 6 109#p=

Now in phasor form, the electric field intensity is Es 5 sin y e a10 jbx

zp= -^ h (1)

From Maxwell’s equation we get the magnetic field intensity as

Hs j E1s

0#dwm=- ^ h

jyE

xEa asz

xsz

y0 22

22

wm= -< F

cos sinj y e j b y ea a50 10 5 10jbxx y

jbx

0wm p p p= +- -^ ^h h6 @

Again from Maxwell’s equation we have the electric field intensity as

Es j H1s

0#dwe= ^ h j x

HyH a1 sy sx

z0 22

22

we= -< F

( )( ) ( ) ( )( ) ( )sin sinj b jb y e y e a1 5 10 50 10 10jbx jbxz2

0 0w m ep p p p= - +- -

6 @

10sinb ye a1 5 500 jbxz2

0 0

2 2

w m ep p= + -

6 @

Comparing this result with equation (1) we get

b1 5 50020 0

2 2

w m ep+^ h 5=

or, b 1002 2p+ 20 0w m e=

b 1002 2p+ 6 103 10

19 28 2# #

#p= ^

^h

h , c6 10 19

0 0#w p m e= =b l

b 1002 2p+ 400 2p=



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For View Only Shop Online at www.nodia.co.in b2 300 2p= b /rad m300! p=

SOL 6.2.20 Option (D) is correct.The force experienced by a test charge q in presence of both electric field E and magnetic field B in the region will be evaluated by using Lorentz force equation as F q vE B#= +^ h

So, putting the given three forces and their corresponding velocities in above equation we get the following relations q a ay z+^ h q E a Bx#= +^ h (1) qay q E a By#= +^ h (2) 2q a ay z+^ h q E a Bz#= +^ h (3)Subtracting equation (2) from (1) we get az a a Bx y #= -^ h (4)and subtracting equation (1) from (3) we get ay a a Bz x #= -^ h (5)Now we substitute B B B Ba a ax x y y z z= + + in eq (4) to get az B B B Ba a a ay z z y x z z x= - + -So, comparing the , andx y z components of the two sides we get B Bx y+ 1=and Bz 0=Again by substituting B B B Ba a ax x y y z z= + + in eq (5), we get ay B B B Ba a a ax y y x y z z y= - - +So, comparing the , andx y z components of the two sides we get B Bx z+ 1=and By 0=as calculated above B 0z = , therefore B 1x =Thus, the magnetic flux density in the region is B /Wb max 2= (B 1x = , B B 0y z= = )

SOL 6.2.21 Option (A) is correct.As calculated in previous question the magnetic flux density in the region is B /Wb max 2=So, putting it in Lorentz force equation we get F q E V B#= +^ h

or, q a ay z+^ h q E a ax x#= +^ h

Therefore, the electric field intensity in the medium is E /V ma ay z= +

SOL 6.2.22 Option (D) is correct.Given the magnetic flux density through the loop is B /x a4 z=-So the total magnetic flux passing through the loop is given as




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F dB S:= # x dxdya a2z z

y

y

x

x 22:= - -

++

b ^l h##

ln xx2 2 2= +

b ^l h ln xx4 2= +

b l

Therefore, the circulation of induced electric field in the loop is

dE lC

:# dtdF=- lndt

dxx4 2=- +

b l; E

xx dt

dxx

24 2=- +

+

bb

ll

xx

x dtdx

24 2

2=- + -b l

x x 2

8 2=+^

^h

h x x 4

8=+^ h

dtdx v a2 x= =b l

SOL 6.2.23 Option (D) is correct.As the magnetic flux density for 4<r is 0B = so, the total flux passing through the closed loop defined by 4 mr = is

F 0dB S:= =#So, the induced electric field circulation for the region 4 m<r is given as

dE lC

:# 0dtdF=- =

or, E 0= for 4 m<r

SOL 6.2.24 Option (B) is correct.As the magnetic field for the region 4 m<r and 5 m>r is zero so we get the distribution of magnetic flux density as shown in figure below.

At any distance r from origin in the region 4 5 m< <r , the circulation of induced electric field is given as

dE lC

:# dtdF=- dt

d dB S:=- b l#

sindtd t2 42 2w pr p=- -^ h8 B



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For View Only Shop Online at www.nodia.co.in cos t2 162w w pr p=- -^ h

or, E 2pr^ h cos t2 162w w pr p=- -^ h

E cos t

22 162

rr w w

=--^ h

So, the induced electric field intensity at 4.5 mr = is

E . ( . ) cos t4 52 4 5 162 w w=- -^ h

cos t1317w w=-

SOL 6.2.25 Option (B) is correct.For the region 5 m>r the magnetic flux density is 0 and so the total magnetic flux passing through the closed loop defined by 5 mr = is

F dB S0

5:= # d dB S B S

4

5

0

4: := + ##

sin t da S0 2 z4

5:w= + ^ h#

sin t2 5 42 2w p p= -^ ^ ^h h h8 B sin t18p w=So, the circulation of magnetic flux density for any loop in the region 5 m>r is

dE l:# dtdy=-

( )E 2pr sindtd t18p w=- ^ h

cos t18pw w=-So, the induced electric field intensity in the region 5 m>r is

E cos t a218

prpw w= -

f

cos ta18r w w=- f

SOL 6.2.26 Option (D) is correct.Let the test charge be q coulomb So the force presence of experienced by the test charge in the presence of magnetic field is F q v B#= ^ h ...(i)and the force experienced can be written in terms of the electric field intensity as F qE=Where E is field viewed by observer moving with test charge.Putting it in Eq. (i) qE q v B#= ^ h

E 2a az#wr= f^ ^h hwhere w is angular velocity and r is radius of circular loop. a2 2 2= r^ ^ ^h h h 8 /V ma= r

SOL 6.2.27 Option (A) is correct.Let the point change located at origin be Q and the current I is flowing out of the




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page through the closed triangular path as shown in the figure.

As the current I flows away from the point charge along the wire, the net charge at origin will change with increasing time and given as

dtdQ I=-

So the electric field intensity will also vary through the surface and for the varying field circulation of magnetic field intensity around the triangular loop is defined as

dH l:# I Id enc c enc= +6 6@ @

where Ic enc6 @ is the actual flow of charge called enclosed conduction current and Id enc6 @ is the current due to the varying field called enclosed displacement current

which is given as

Id enc6 @ dtd dE S

S0 :e= ^ h# dt

d dD SS

:= # (1)

From symmetry the total electric flux passing through the triangular surface is

dD SS

:# Q8=

So, Id enc6 @ dtd Q

dtdQ I

8 81

8= = =-b l (from equation (1))

Where as Ic enc6 @ I=So, the net circulation of the magnetic field intensity around the closed triangular loop is

dH lC

:# I Id enc c enc= +6 6@ @

I I8=- + 78 8= ^ h A14= (I 8 A= )

SOL 6.2.28 Option (C) is correct.The distribution of magnetic flux density and the resistance in the circuit are same as given in section A (Q. 31) so, as calculated in the question, the two voltage drops in the loop due to magnetic flux density 0.1tB az= are V1 33.3 mV=and V2 66.67 66.7mV mV= =Now V3 (voltmeter) which is directly connected to terminal cd is in parallel to



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For View Only Shop Online at www.nodia.co.inboth V2 and V1. It must be kept in mind that the loop formed by voltmeter V3 and resistance 2W also carries the magnetic flux density crossing through it. So, in this loop the induced emf will be produced which will be same as the field produced in loop abcd at the enclosed fluxes will be same.Therefore as calculated above induced emf in the loop of V3 is Vemf 100 mV=According to lenz’s law it’s polarity will be opposite to V3 and so Vemf- V V1 3= +or, V3 .100 33 3= - 6.7 mV2=

SOL 6.2.29 Option (C) is correct.The induced emf in a closed loop is defined as

Vemf dtdF=-

where F is the total magnetic flux passing through the square loopAt any time t , angle between B and dS is q since B is in ay direction so the total magnetic flux passing through the square loop is

F dB S:= # cosB S q= ^ ^h h

cos5 10 20 10 20 103 3 3# # # # q= - - -

^ ^h h

cos2 10 6# q= -

Therefore the induced emf in the loop is

Vemf dtdF=- cosdt

d2 10 6# q=- -

^ h sin dtd2 10 6

# q q= -

and as dtdq = angular velocity 2 / secrad=

So, Vemf sin2 10 26# q= -

^ ^h h 4 10 /sin V m6 q#= - 4 /sin V m2 mq=

SOL 6.2.30 Option (B) is correct.As calculated in previous question the induced emf in the closed square loop is Vemf 4 /sin V mmq=So the induced current in the loop is

I RVemf= where R is the resistance in the loop.

sin40 10

4 103

6

#

#q= -

- ( 40 mR W= )

0.1 sin mAq=

SOL 6.2.31 Option (A) is correct.The total magnetic flux through the square loop is given as

F dB S:= # sin cosB t S0 w q= ^ ^h h

So, the induced emf in the loop is

Vemf dtdF=- ( )( )sin cosdt

d B t S0 w q=- 6 @




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sin cosB Sdtd t t0 w w=- 6 @ ( tq w= )

2cosB S t0 w=-Thus, the maximum value of induced emf is Vemf B S0 w=

SOL 6.2.32 Option (B) is correct.As calculated in previous question the maximum induced voltage in the rotating loop is given as Vemf B S0 w=From the given data, we have B0 0.25 /Wb m2= S 64 64 10cm m2 4 2

#= = -

and w 60 2 377 / secradp#= = (In one revolution 2p radian is covered)So, the r.m.s. value of the induced voltage is

V. .emf r m s6 @ V B S

21

21

emf 0 w= = .2

1 0 25 64 10 3774# # #= -

^ h

.0 4265=Since the loop has 50 turns so net induced voltage will be 50 times the calculated value.i.e. V

. .emf r m s6 @ .50 0 4265#= ^ h 21.33 volt=

SOL 6.2.33 Option (A) is correct.e.m.f. induced in the loop due to the magnetic flux density is given as

Vemf t22F=- cos

tt10 120 2

22 p pr=- ^ ^h h

sin t10 10 120 10 1202 2# #p p p=- --

^ ^ ^h h h

sin t12 1202p p=As determined by Lenz’s law the polarity of induced e.m.f will be such that b is at positive terminal with respect to a .i.e. Vba sinV t12 120emf

2p p= =or Vab sin t12 1202p p=- 118.43 120sin Volttp=-

SOL 6.2.34 Option (C) is correct.As calculated in previous question, the voltage induced in the loop is Vab sin t12 1202p p=-Therefore, the current flowing in the loop is given as

I t^ h V250ab=- sin t

25012 1202p p=

.47 120sin t2 p=

***********



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SOLUTIONS 6.3

SOL 6.3.1 Option (A) is correct.Given, the magnetic flux density in air as B B

x yx a

x yy ay x0 2 2 2 2=

+-

+c m ...(1)

Now, we transform the expression in cylindrical system, substituting x cosr f= and y sinr f= ax cos sina arf f= - f

and ay sin cosa arf f= + f

So, we get B B a0= f

Therefore, the magnetic field intensity in air is given as

H aBB

0 0

0

m m= = f , which is constant

So, the current density of the field is J 0H#d= = (since H is constant)

SOL 6.3.2 Option (C) is correct.Maxwell equations for an EM wave is given as B:d 0=

E:d v

er=

Ed# tB22=-

Hd# tD J22= +

So, for static electric magnetic fields B:d 0= E:d /vr e=

Ed# 0= 0tB22 =b l

Hd# J= tD 022 =b l


H#d JtD22= + Maxwell Equations

dH SS

# :d̂ h## tdJ D S

S:

22= +` j## Integral form




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dH l:# t

dJ D SS

:22= +b l## Stokes Theorem

SOL 6.3.4 Option (A) is correct.From Maxwells equations we have

Hd# tJD

22= +

Thus, Hd# has unit of current density J (i.e., /A m2)

SOL 6.3.5 Option (D) is correct.This equation is based on Ampere’s law as from Ampere’s circuital law we have

dH ll

$# Ienclosed=

or, dH ll

$# dJ SS

:= #

Applying Stoke’s theorem we get

dH SS

$#d̂ h# dJ SS

:= #

H#d J=Then, it is modified using continuity equation as

H#d t

J D22= +

SOL 6.3.6 Option (C) is correct.When a moving circuit is put in a time varying magnetic field induced emf have two components. One due to time variation of magnetic flux density B and other due to the motion of circuit in the field.

SOL 6.3.7 Option (A) is correct.From maxwell equation we have

H#d t

J D22= +

The term tD22 defines displacement current.

SOL 6.3.8 Option (A) is correct.Emf induced in a loop carrying a time varying magnetic flux F is defined as

Vemf dtdF=-

9 dtd t3

1 3l=- b l

9 t2l=-at time, 3 st = , we have 9 3 2l=- ^ h

l /Wb s2 2=-

SOL 6.3.9 Option (B) is correct.According to Lenz’s law the induced emf (or induced current) in a loop flows such as to produce a magnetic field that opposed the change in B . The direction of the magnetic field produced by the current is determined by right hand rule.



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For View Only Shop Online at www.nodia.co.inNow, in figure (1), B directed upwarded increases with time where as the field produced by current I is downward so, it obey’s the Lenz’s law.In figure (2), B directed upward is decreasing with time whereas the field produced by current I is downwards (i.e. additive to the change in B) so, it doesn’t obey Lenz’s law.In figure (3), B directed upward is decreasing with time where as current I produces the field directed upwards (i.e. opposite to the change in B) So, it also obeys Lenz’s law.In figure (4), B directed upward is increasing with time whereas current I produces field directed upward (i.e. additive to the change in B) So, it doesn’t obey Lenz’s law.Thus, the configuration 1 and 3 are correct.

SOL 6.3.10 Option (A) is correct.Faraday’s law states that for time varying field,

E#d tB22=-

Since, the curl of gradient of a scalar function is always zeroi.e. V#d d̂ h 0=So, the expression for the field, E Vd=- must include some other terms is

E VtAd

22=- -

i.e. A is true but R is false.

SOL 6.3.11 Option (B) is correct.Faraday develops the concept of time varying electric field producing a magnetic field. The law he gave related to the theory is known as Faraday’s law.

SOL 6.3.12 Option (C) is correct.Given, the area of loop S 5 m2=Rate of change of flux density,

tB

22 2 / /Wb m S2=

So, the emf in the loop is

Vemf tdB S:

22=- # 5 2= -^ ^h h 10 V=-

SOL 6.3.13 Option (C) is correct.The modified Maxwell’s differential equation.

H#d t

J D22= +

This equation is derived from Ampere’s circuital law which is given as

dH l:# Ienc=

dH S:#d̂ h# dJ S= # H#d J=




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SOL 6.3.14 Option (B) is correct.Electric potential of an isolated sphere is defined as C a4 0pe= (free space)The Maxwell’s equation in phasor form is written as H#d j E Ewe s= + j E Jwe= + J Es=^ h

So A and R both are true individually but R is not the correct explanation of A.

SOL 6.3.15 Option (D) is correct.If a coil is placed in a time varying magnetic field then the e.m.f. will induce in coil. So here in both the coil e.m.f. will be induced.

SOL 6.3.16 Option (B) is correct.Both the statements are individually correct but R is not explanation of A.

SOL 6.3.17 Option ( ) is correct.

Ampere’s law H#d t

J D22= + 3a "^ h

Faraday’ law E#d tB22= 4b "^ h

Gauss law D:d vr= 1c "^ h

Current continuity J:d t2

2r=- 2d "^ h

SOL 6.3.18 Option (B) is correct.Since, the magnetic field perpendicular to the plane of the ring is decreasing with time so, according to Faraday’s law emf induced in both the ring is

Vemf tdB S:

22=- #

Therefore, emf will be induced in both the rings.

SOL 6.3.19 Option (D) is correct.The Basic idea of radiation is given by the two Maxwells equation

H#d tD22=

E#d tB22=-

SOL 6.3.20 Option (B) is correct.The correct maxwell’s equation are

H#d t

J D22= + D: rd =

E#d tB22=- 0B:d =

SOL 6.3.21 Option (B) is correct.In List I

a. dSB:# 0=The surface integral of magnetic flux density over the closed surface is zero or in other words, net outward magnetic flux through any closed surface is zero. 4a "^ h

b. dSD:# dvvvr= #



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Total outward electric flux through any closed surface is equal to the charge enclosed in the region. 3b "^ h

c. dlE:# tdSB

22=- #

i.e. The line integral of the electric field intensity around a closed path is equal to the surface integral of the time derivative of magnetic flux density 2c "^ h

d. dH S:# t

dD J a22= +b l#

i.e. The line integral of magnetic field intensity around a closed path is equal to the surface integral of sum of the current density and time derivative of electric flux density. 1d "^ h

SOL 6.3.22 Option (C) is correct.The continuity equation is given as J:d vr=-i.e. it relates current density J^ h and charge density vr .

SOL 6.3.23 Option (A) is correct.Given Maxwell’s equation is

H#d JtD

c 22= +

For free space, conductivity, 0s = and so, Jc 0Es= =Therefore, we have the generalized equation

H#d tD22=

SOL 6.3.24 Option (D) is correct.Given the magnetic field intensity, H 3 7 2y xa a ax y z= + +So from Ampere’s circuital law we have J H#d=

y x

a a a

3 7 2

x

x

y

y

z

z= 22

22

22

a a a0 2 0 0x y z= - - +^ ^ ^h h h 2ay=-

SOL 6.3.25 Option (D) is correct.The emf in the loop will be induced due to motion of the loop as well as the variation in magnetic field given as

Vemf td dSB v B l#2

2=- + ^ h# #So, the frequencies for the induced e.m.f. in the loop is 1w and 2w .

SOL 6.3.26 Option (B) is correct. F Q E v B#= +^ h is Lorentz force equation.




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SOL 6.3.27 Option (D) is correct.All of the given expressions are Maxwell’s equation.

SOL 6.3.28 Option (D) is correct.The direction of magnetic flux due to the current ‘i ’ in the conductor is determined by right hand rule. So, we get the flux through A is pointing into the paper while the flux through B is pointing out of the paper.According to Lenz’s law the induced e.m.f. opposes the flux that causes it. So again by using right hand rule we get the direction of induced e.m.f. is anticlockwise in A and clockwise in B .

SOL 6.3.29 Option (C) is correct. A2d J0m=-This is the wave equation for static electromagnetic field.i.e. It is not Maxwell’s equation.

SOL 6.3.30 Option (B) is correct.Poission’s equation for an electric field is given as

V2d v

er=-

where, V is the electric potential at the point and vr is the volume charge density in the region. So, for 0vr = we get, V2d 0=Which is Laplacian equation.


Continuity equation J#d tv

22r=- 4a "^ h


J D22= + 1b "^ h

Displacement current J tD22= 2c "^ h

Faraday’ law E#d tB22=- 3d "^ h

SOL 6.3.32 Option (B) is correct.A static electric field in a charge free region is defined as E:d 0= 4a "^ h

and E#d 0=A static electric field in a charged region have

E:d 0v !er= 2b "^ h

and E#d 0=A steady magnetic field in a current carrying conductor have B:d 0= 1c "^ h

B#d 0J0 !m=A time varying electric field in a charged medium with time varying magnetic field



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E#d tB 022 !=- 3d "^ h

E:d 0v !er=


V dtd mF=-

It is Faraday’s law that states that the change in flux through any loop induces e.m.f. in the loop.

SOL 6.3.34 Option (B) is correct.From stokes theorem, we have

dE S# :d̂ h# dE l:= # (1)

Given, the Maxwell’s equation E#d ( / )tB2 2=-Putting this expression in equation (1) we get,

dE l:# t

dB Ss

:22=- #

SOL 6.3.35 Option (B) is correct.Induced emf in a coil of N turns is defined as

Vemf N dtdF=-

where F is flux linking the coil. So, we get

Vemf dtd t t100 23=- -^ h

t100 3 22=- -^ h 100 1000 mV3 2 22=- - =-^_ h i (at 2 st = ) V2=-

SOL 6.3.36 Option (C) is correct.Since, the flux linking through both the coil is varying with time so, emf are induced in both the coils.Since, the loop 2 is split so, no current flows in it and so joule heating does not occur in coil 2 while the joule heating occurs in closed loop 1 as current flows in it.Therefore, only statement 2 is correct.

SOL 6.3.37 Option (A) is correct.The electric field intensity is E eE j t

0= w where E0 is independent of timeSo, from Maxwell’s equation we have

H#d t

J E22e= +

j eE E j t0s e w= + w

^ h jE Es we= +

SOL 6.3.38 Option (A) is correct.Equation (1) and (3) are not the Maxwell’s equation.




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SOL 6.3.39 Option (D) is correct.From the Maxwell’s equation for a static field (DC) we have B#d J0m= A#d #d̂ h J0m= A A2:dd d-^ h J0m=For static field (DC), A:d 0=therefore we have, A2d J0m=-So, both A and R are true and R is correct explanation of A.

SOL 6.3.40 Option (D) is correct.For a static field, Maxwells equation is defined as H#d J=and since divergence of the curl is zeroi.e. H:d #d̂ h 0= J:d 0=but in the time varying field, from continuity equation (conservation of charges)

J:d t

0v

22

!r=-

So, an additional term is included in the Maxwell’s equation.

i.e. H#d t

J D22= +

where tD22 is displacement current density which is a necessary term.

Therefore A and R both are true and R is correct explanation of A.

SOL 6.3.41 Option (D) is correct.For any loop to have an induced e.m.f., magnetic flux lines must link with the coil.Observing all the given figures we conclude that loop C1 and C2 carries the flux lines through it and so both the loop will have an induced e.m.f.

SOL 6.3.42 Option (A) is correct.Since, the circular loop is rotating about the y -axis as a diameter and the flux lines is directed in ax direction. So, due to rotation magnetic flux changes and as the flux density is function of time so, the magnetic flux also varies w.r.t time and therefore the induced e.m.f. in the loop is due to a combination of transformer and motional e.m.f. both.

SOL 6.3.43 Option (A) is correct.Gauss’s law D:d r= a 1"^ h


J Dc 2

2= + b 5"^ h

Faraday’s law E#d tB22=- 2c "^ h

Poynting vector P E H#= 3d "^ h

***********

CHAPTER 7ELECTRONAGNETICS WAVES

406 Electronagnetics Waves Chap 7



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EXERCISE 7.1

MCQ 7.1.1 What will be the direction of wave propagation in a non magnetic medium in which magnetic field intensity at any point is given by H /cos A mt ky a4 xw= -^ h

(A) az+ direction (B) az- direction

(C) ax+ direction (D) ay+ direction

MCQ 7.1.2 In a certain medium electric field intensity of a propagating wave is given by ,x tE^ h /sin V mE e t z a2 x

y0 w b= -a-^ h

The electric field phasor of the wave will be(A) /V mE e aj x

y0a b- +^ h

(B) /V mjE e aj xy0

a b- +^ h

(C) /V mjE e aj xy0- a b- +^ h

(D) /V mjE e aj xy0- a b-^ h

MCQ 7.1.3 In air, magnetic field intensity is given by H 10 /cos A mt ky a6 10 z7

#= -^ h . Wave number k for the EM wave will be(A) 1.8 /rad m (B) 2 /rad m

(C) 0.2 /rad m (D) 5 /rad m

MCQ 7.1.4 An electromagnetic wave is propagating in certain non magnetic material such that the magnetic field intensity at any point is given by H /cos A mt z a3 10 5 x

9= -^ h

The phase velocity of the wave in the medium will be(A) 1.5 10 /m s9

# (B) 5 10 /m s9#

-

(C) 0.5 10 /m s8# (D) 2 10 /m s8

#

MCQ 7.1.5 Magnetic field intensity in a certain non-magnetic medium is given by

H /cos A mH t y ax0 w b= -^ h

If the wavelength of the EM wave in the medium be 12.6 m then what will be the phase constant b in that medium ?(A) 0.25 /rad m (B) 0.5 /rad m

(C) 1.12 /rad m (D) 6.3 /rad m

Chap 7 Electronagnetics Waves 407


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For View Only Shop Online at www.nodia.co.inMCQ 7.1.6 In a nonmagnetic material electric field intensity is given by

E /cos V mt x a16 4 10 2 y8

#= -^ h

The relative permittivity of the medium will be(A) .1 5 (B) .2 25

(C) .0 44 (D) 225

MCQ 7.1.7 Electric field intensity in free space is E /cos V mt z a24 5 10 x8

# b= -^ h . The time period of the wave will be(A) 7.96 ns (B) 1.26 ns

(C) sec8 107# (D) 12.57 ns

MCQ 7.1.8 In air, a propagating wave has electric field intensity given by E 9 /cos V mt x a4 10 z

8# b= -^ h

The time taken by the wave to travel one-fourth of it’s total wave length is(A) 61.42 ns (B) 3.05 ns

(C) 7.23 ns (D) 3.93 ns

MCQ 7.1.9 What will be the intrinsic impedance of a lossless, nonmagnetic dielectric material having relative permittivity .2 25re = ?(A) . 2235 6 W (B) 167.56W

(C) 8.95 mW (D) 251.33W

MCQ 7.1.10 A radio wave is propagating at a frequency of 0.5 MHz in a medium ( 3 10 /S m7s #=, 1r r .m e= ).The wave length of the radio wave in that medium will be(A) 0.8 mm (B) 0.26 mm

(C) 0.4 mm (D) 0.13 mm

MCQ 7.1.11 The skin depth in a poor conductor is independent of(A) Permittivity (B) Permeability

(C) Frequency (D) None of these

MCQ 7.1.12 Assertion (A) : sin cosE z ctE ax0= ^ ^h h represents the electric field of a plane wave in free space.Reason (R) : A plane wave f propagating with velocity vp in az+ direction must satisfy the equation

tf v

zf

p2

22

2

2

22

22- 0=

(A) Both A and R are true and R is the correct explanation of A.







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MCQ 7.1.13 A propagating wave in free space has magnetic field intensity H 0. /cos A mt y a2 10 z

9 b= -^ h

What will be the electric field intensity of the wave at 1 cmy = at time, 0.1 nst = ?(A) 37.7 /V max- (B) 37.6 /V max-

(C) 19.8 /V max- (D) 37.6 /V max

MCQ 7.1.14 Phasor form of magnetic field intensity of a uniform plane wave in free space is given as Hs /A mj a ja e2 5 4 2y z

j x= + + b-^ ^h h

The maximum electric field of the plane wave equals to(A) 24.1 /V m (B) 14.22 /kV m

(C) 9.08 /kV m (D) 0

MCQ 7.1.15 Electric field intensity of linearly polarized plane wave in free space is given by E /cos V mt za a6 5 50y x w= - -^ ^h h

The phasor form of magnetic field intensity of the wave will be

(A) 5 6 /V mea ax yj z

050h- + -

^ h (B) 5 6 /V mea ax z

j z

0

50

h--

^ h

(C) 5 6 /V mea ax y

j z

0

50

h--

^ h (D) 5 6 /V mea ax y

j z

0

50

h- +-

^ h

MCQ 7.1.16 In a perfect conductor , 0resistivity .r^ h magnetic field of any EM wave(A) lags electric field by 90c (B) leads electric field by 45c

(C) lags electric field by 45c (D) will be in phase with electric field

Statement for Linked Question 17 - 18 :An electromagnetic wave travels in free space with the electric field component Es /V mea a10 5x z

j x z4 2= + - -^

^h

h

MCQ 7.1.17 What will be the phasor form of magnetic field intensity of the wave ?(A) 29.66 /mA me j x z4 2- - -^ h

(B) 5 /mA me5 j x z4 2- - -^ h

(C) 29.66 /mA me j x z4 2- -^ h

(D) 29.66 /A me j x z4 2- - -^ h

MCQ 7.1.18 What will be the time average power density of the electromagnetic wave ?(A) . . /W ma a665 9 331 6x z

2-^ h

(B) . . /W ma a148 9 74 15x z2-^ h

(C) . . /W ma a331 6 665 9x z2- +^ h

(D) 74.15 148.9 /W ma ax z2- +^ h



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For View Only Shop Online at www.nodia.co.inMCQ 7.1.19 A propagating wave has the phasor form of its electric field intensity defined as

Es 2 /V mea a a3 3 .x y z

j x y z0 1 3 3 2= - + - p- - + -^ `h j

The wave is linearly polarized along the direction of(A) 3 2a a a3x y z- + - (B) 2 a a a3 3x y z- + -

(C) 3 2a a a3x y z- + (D) 2 a a a3 3x y z- +

Statement for Linked Question 20 - 21 :In free space an electric field intensity vector is given by E 00 cos t z a2 xw b= -^ h

where w and b are constants.

MCQ 7.1.20 If 0 0

bm ew= . what will be the magnetic flux density vector B ?

3 10 /m s10 0

8

m e#= -

c m

(A) 3 10 cos t z ay10 w b# -^ h

(B) 3.33 10 cos t z ay7 w b# --^ h

(C) 3 10 cos t z ay8 w b# -^ h

(D) 3.33 10 cos t z ay6 w b# --^ h

MCQ 7.1.21 The poynting vector of the E -M field will be

(A) 10 cos t z az4

0

0 2

me w b-^ h (B) 10 cos t z az4

0

0 2

em w b-^ h

(C) 10 cos t z az40

2m w b-^ h (D) cos t z a10z

0

42

mw b-^ h

MCQ 7.1.22 The electric field associated with a sinusoidally time varying electromagnetic field is given by E /sin sin V mx z a15 2 10 3 y

8#p p p= -^ h

The time average stored energy density in the electric field is

(A) sin x254

02e p (B) sin x4

25 0 2e p

(C) sin x45 20e p (D) sin x4

25 20e p

MCQ 7.1.23 Electric field associated with a sinusoidally time varying electromagnetic field is given by E ( ) /sin sin V my t x a20 6 10 3 z

8#p p p= -^ h

What will be the time average stored energy density in the magnetic field ?

(A) sin x10 25 509

2

p p+-

^ h (B) sin x14410 25 50

92

p p+-

^ h

(C) sin x144 10 25 509

2#p p+^ h (D) sin x144 25 50 2p p+^ h




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MCQ 7.1.24 An electromagnetic wave is propagating from free space to a certain medium having relative permittivity 5re = . If wavelength of the wave in the medium be 20 cm then what would be it’s wavelength in free space ?(A) 6.67 cm

(B) 60 cm

(C) 180 cm

(D) 18 cm

MCQ 7.1.25 If some free charge is being imbedded in a piece of glass, then the charge will flow out to the surface nearly after(relative permittivity of glass, .254re = conductivity of glass, 10 /S m12s = - )(A) sec2

(B) sec4

(C) 35 sec

(D) sec20

MCQ 7.1.26 An electromagnetic wave propagating in free space is incident on the surface of a dielectric medium ( 0m , 4 0e ). If the magnitude of the electric field of incident wave is E0 then what will be the magnitude of the electric field of the reflected wave ?(A) /E2 30-

(B) /E 30-

(C) /E 20

(D) E0-

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EXERCISE 7.2

MCQ 7.2.1 Magnetic field intensity of a propagating wave in free space is given by H 0.3 /cos A mt y axw b= -^ h

If the total time period of the wave be T then the plot of H versus y at time, t T8=

will be

MCQ 7.2.2 A uniform plane wave is propagating with a velocity of 7.5 10 /m s7# in a lossless

medium having relative permeability .82rm = . The electric field phasor of the wave is given by Es 5 /V me a.j x

z0 3=

What will be the magnetic field intensity of the wave ?(A) 11.05 . . /cos mA mt x a9 54 10 0 3 y

6# +^ h

(B) 22.13 . . /cos mA mt x a9 54 10 0 3 y6

# +^ h

(C) 22.13 . . /cos mA mt z a9 54 10 0 3 y6

# +^ h

(D) 11.05 . . /cos mA mt x a2 25 10 0 3 y7

# +^ h




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MCQ 7.2.3 In a certain medium ( 4re = , 4rm = ). A plane wave is propagating such that the electric field intensity of the wave is E /sin V mE e t x a10/x

y02 8 b= --

^ h . The loss tangent of the medium will be(A) .1 94 (B) .0 27

(C) .0 35 (D) .0 52

Statement for Linked Question 4 - 5 :In a lossy medium ( 8re = , .0 5rm = , 0.01 /S ms = a plane wave is travelling in az+ direction that has the electric field intensity E 0.5 /cos t a10 3 x

9p p= +^ h at z 0= .

MCQ 7.2.4 What will be the distance travelled by the wave to have a phase shift of 10c ?(A) 20.95 mm (B) 477.3 mm

(C) 8.33 V (D) 3.65 m

MCQ 7.2.5 After traveling a distance z , the amplitude of the wave is reduced by %40 . So, the value of z equals to(A) 481.5 mm (B) 542 mm

(C) 1.06 m (D) 2.08 m

MCQ 7.2.6 A uniform plane wave is propagating at a velocity of 7 10 /m s7# in a perfect

dielectric such that the electric and magnetic fields of the wave are given by ,x tE^ h 00 /cos V mt x a3 5 10 y

6# p b= -^ h

,x tH^ h 1.9 /cos V mt x a5 10 z6

# p b= -^ h

The relative permittivity and relative permeability of the medium will be respectively(A) .1 70, .2 69 (B) .3 4, .5 37

(C) .1 70, .1 58 (D) .2 37, .2 69

MCQ 7.2.7 An electromagnetic wave is propagating in free space in ax- direction with a frequency w and phase angle zero. The EM wave is polarized in az+ direction. If the amplitude of electric field of the wave is E0 then the magnetic field of the wave will be

(A) cosE t c x ay0

0

h w w+a k (B) cosE t cx ay0 0h w w+^ h

(C) cosE t c x ay0

0

h w w- +a k (D) cosE t c x ay0

0

h w w-a k

MCQ 7.2.8 What will be the electric field of a plane wave polarized parallel to the x -z plane and propagating in free space in the direction from origin to the point , ,1 1 1^ h, that has the amplitude E0 and frequency w with zero phase angle ?

(A) cosE tcx y z a a

3 2x z

0 w w- + + -^ bh l: D

(B) cosE tcx y z a a

3 2x z

0 w w+ + + -^ bh l: D



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(C) cosE t c x y z a a2

x z0 w w- + + +

^ ch m9 C

(D) cosE t c x y z a a2

x z0 w w+ + + +

^ ch m9 C

MCQ 7.2.9 A plane wave is propagating with frequency 50 kHzf = in a medium ( 2 /S ms = , 80re = , 4rm = ). What will be the skin depth of the medium ?

(A) 1.7 m (B) 0.4 m

(C) 0.8 m (D) 1.3 m

MCQ 7.2.10 With a thickness t , silver coating is done for a microwave experiment to operate at a frequency of 10 GHz. For the successful experiment t should be(for silver, 1r r .m e= , 6.25 10 /S m7s #= )(A) greater than 0.64 mm (B) less than 0.64 mm

(C) exactly equal to 0.64 mm (D) none of these

MCQ 7.2.11 In a nonmagnetic material of conductivity 2 10 /S m7s #= , electric field of a propagating plane wave is given by E . 2 . /cos sin V mt y t ya a5 10 0 2 10 0 2x z

7 7= - + -^ ^h h

What will be the value of complex permittivity of the medium ?(A) /F mj2 36 0e-^ h (B) /F mj36 20e -^ h

(C) /F mj36 2-^ h (D) /F mj36 20e +^ h

MCQ 7.2.12 Assertion (A) : All the metals are opaque.Reason (R) : Skin depth of metals are in the range of nanometers.(A) Both A and R are true and R is the correct explanation of A.




MCQ 7.2.13 In the plane z 0= , electric field of a wave propagating in az+ direction in free space is E0 which is varying with time t as shown in the figure.




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If the magnetic field intensity of the wave at 1 sect m= be H1 then the plot of H1 versus z will be

MCQ 7.2.14 Three different dielectrics of permittivities 4 0e , 9 0e and 3 0e are defined in the space as shown in figure. If the leading edge of a uniform plane wave propagating in ax direction is incident on the plane mx 3=- then how much time it will take to strike the interface defined by the dielectric 2 and dielectric 3 ?

(A) 6 secn

(B) 0.02 secm

(C) 3 secn

(D) 0.06 secm

MCQ 7.2.15 An electromagnetic wave propagating in medium 1 ( 0m , 1e ) is incident on medium 2 ( 0m , 2e ) as shown in figure such that the electric field of reflected wave is /1 5 times of the electric field of incident wave. The value of /1 2e e equals to



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(A) /2 3 (B) /3 2

(C) /4 9 (D) /9 4

Statement for Linked Question 16 - 17 :An electromagnetic wave of 50 MHz frequency is incident on a dielectric medium such that it’s skin depth is 0.32 mm. (permittivity of dielectric .6 28 10 7

#= - )

MCQ 7.2.16 What will be the conductivity of the dielectric ?

(A) 0.32 10 /S m2# (B) 1.01 10 /S m5

#-

(C) 320 /S m (D) 0.99 10 /S m5#

MCQ 7.2.17 If an electromagnetic wave of 8 GHz frequency travels a distance of 0. 75 mm2 in the dielectric medium then it’s field intensity will be reduced by(A) 20 dB (B) 60 dB

(C) 0 dB (D) 30 dB

MCQ 7.2.18 Electric field of an electromagnetic wave propagating in a medium in ax+ direction is given by Es E j ea ay z

j x0= - b-^ h

The wave is(A) left hand circularly polarized

(B) Right hand circularly polarized

(C) elliptically polarized

(D) linearly polarized

MCQ 7.2.19 An electromagnetic wave has the electric field intensity in the phasor form given by Es j ea a2 z x

j y= - b-^ h

The EM wave is incident on a perfect conductor located at y 0= . What will be the polarization of the reflected wave ?(A) left hand circular (B) Right hand circular

(C) elliptical (D) linear




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MCQ 7.2.20 An electromagnetic wave propagating in free space is incident on a perfectly conducting slab placed at x 0$ . The electric field of the incident wave in the phasor form is given by Eis /V mea5 z

j y x6 8= - +^ h

The net electric field of the total wave (incident and reflected both) in free space after reflection will be(A) 10 /V meaz j y x6 8- -^ h (B) 10 /V meaz j y x6 8- - -^ h

(C) 20 8 /sin V mj e xaz j y6- - (D) 20 8 /sin V mj e xaz j y6-

MCQ 7.2.21 Electric field intensity of an EM wave propagating in free space is given by Eis 25 /V meax j z y6 8= - +^ h

If the wave is incident on a perfectly conducting plane at y 0= then the magnetic field intensity of the reflected wave will be

(A) /A mea a8 6y z j z y6 8

p p- + - -a

^k

h (B) /A mea a8 6y z j z y6 8

p p+ - -a

^k

h

(C) ea a8 6y z j z y6 8

p p+ -a

^k

h (D) /A mea a8 6y z j z y6 8

p p- + -a

^k

h

MCQ 7.2.22 An electromagnetic wave propagating in free space has magnetic field intensity H 0. /cos A mt y a4 xw b= -^ h

What will be the total power passing through a square plate of side 20 cm located in the plane x y 2+ = ?(A) 0.53 Watt (B) 1.88 Watt

(C) 18.8 mW (D) 53.31 mW

MCQ 7.2.23 An electromagnetic wave propagating in a lossless medium ( 41 0m m= , 1 0e e= , 01s = ) defined in the region 0y > is incident on a lossy medium ( 2 0m m= , 42 0e e=

, 0.1 /S m2s = ) defined in the region y 0# . The electric field intensity of the incident wave in lossless medium is given by Eis /V me a6 j y

z5= -

What will be the standing wave ratio ?(A) .1 22 (B) .0 8186

(C) .0 0997 (D) .10 025

MCQ 7.2.24 The complex electric field vector of a uniform plane wave propagating in free space is given by Es 2 /V mea a a3 3 .

x y zj x y z0 01 3 3 2= - - p- - + -

^ `h j

The unit vector in the direction of propagation of the wave will be

(A) 3 2a a a

163x y z- + -

(B) 3 2a a a

43x y z- + -

(C) 3 2a a a4 3x y z- - +^ h (D) 3 2a a a

43x y z-

+ -^ h



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For View Only Shop Online at www.nodia.co.inMCQ 7.2.25 Phasor form of electric field intensity of a uniform plane wave is given by

Es /V mea a23

2 .x y

j x y z0 04 2 3 3= - p- - - +d `n j

The wavelength along the direction of propagation is(A) 3.16 m (B) 0.08 m

(C) 12.5 m (D) 15.7 m

MCQ 7.2.26 In free space the complex magnetic field vector of a uniform plane wave is given by Hs /A mea a3 .

x zj x y z0 04 3 2 3=- + p- - -

^`

hj .

Frequency of the plane wave will be(A) 3.75 MHz (B) 2.4 10 Hz6

#

(C) 24 MHz (D) 2.4 MHz

Statement for Linked Question 27 - 28 :In free space complex electric field vector of a uniform plane wave is given by

Es /V mea a2 x zj x y z25

2 5 3= +p- - -

^ `h j

MCQ 7.2.27 The apparent wavelengths along the x , y and z axes are xl yl zl

(A) 16.7 m 25 m 28.87 m(B) 28.87 m 16.7 m 25 m(C) 16.7 m 28.87 m 25 m(D) 28.87 m 25 m 16.7 m

MCQ 7.2.28 The apparent phase velocities along the x , y and z axes are vPx vPy vPz

(A) 1.73 10 /m s10# 1.5 10 /m s10

# 1 10 /m s10#

(B) 6.93 10 /m s8# 6 10 /m s8

# 4 10 /m s8#

(C) 2.77 10 /m s7# 2.4 10 /m s7

# 1.6 10 /m s7#

(D) 1.2 10 /m s9# 1.2 10 /m s9

# 1.2 10 /m s9#

MCQ 7.2.29 Which of the following complex vector field represents the electric field of a uniform plane wave ?(A) 2j j ea a a3 .

x y zj y z0 6 3- - - p- +

_`

ij

(B) j ea a a2 3 .x y z

j x z0 05 3- - p- +_

^i

h

(C) j j j ea a a3 21 1 2

3 3 .x y z

j x y z0 02 3 3 2+ + + - p- + +b c

`l m

j= G

(D) j j j ea a a3 21 1 2

3 3 .x y z

j x y z0 02 3 3 2- - + - + p- + +b c

`l m

j= G




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MCQ 7.2.30 Which of the following pairs of vector Es and Hs field represents the complex electric and magnetic field vectors of a uniform plane wave ? Es Hs(A) 2 /V mj j ea a a3x y z

j x z3+ + p- +_

î

h 2 /A mj ea a a3x y zj x z3- - p- +

_^

ih

(B) /V mj j ea a3x zj x z3+ p- +

_^

ih /A mj j ea a3x z

j x z3- + p- +_

î

h

(C) /V mj j ea a3x zj x z3- p- +

_^

ih /A mj j ea a3 x z

j x z3- - p- +_

î

h

(D) 2 /V mj j ea a a3 .x y z

j x z0 1 3- - + p- +_

î

h 2 /A mj ea a a3 .x y z

j x z0 1 3- - p- +_

î

h

MCQ 7.2.31 The following fields exist in charge free regions 6 ( 10 )sin t yP a5 zw= +

( 2 )cos tQ a10r w r= - f

3 cot cosR a a12r f r f= +r f

( 6 )sin sinr t rS a6 q w= - q

The possible electromagnetic fields are(A) ,P Q (B) ,R S

(C) ,P R (D) ,Q S

MCQ 7.2.32 A uniform plane wave in region 1 is normally incident on the planner boundary separating regions 1 and 2. Both region are lossless and r r1 1

3e m= , r r2 23e m= . If the

20% of the energy in the incident wave is reflected at the boundary, the ratio /r r2 1e e is(A) 1.48

(B) 17.9

(C) 25.6

(D) 58.3

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EXERCISE 7.3

MCQ 7.3.1 A plane wave propagating in air with ( 6 5 ) /V meE a a a4 ( )x y z

j t x y3 4= + + w + - is incident on a perfectly conducting slab positioned at x 0# . The E field of the reflected wave is(A) ( 8 6 5 ) /V mea a a ( )

x y zj t x y3 4- - - w + +

(B) ( 8 6 5 ) /V mea a a ( )x y z

j t x y3 4- + - w + + -

(C) ( 8 6 5 ) /V mea a a ( )x y z

j t x y3 4- - - w - -

(D) ( 8 6 5 ) /V mea a a ( )x y z

j t x y3 4- + - w - -

MCQ 7.3.2 The electric field of a uniform plane electromagnetic wave in free space, along the positive x direction is given by 10( )j eE a ay z

j x25= + - . The frequency and polarization of the wave, respectively, are(A) 1.2 GHz and left circular (B) 4 Hz and left circular

(C) 1.2 GHz and right circular (D) 4 Hz and right circular

MCQ 7.3.3 Consider the following statements regarding the complex Poynting vector P for the power radiated by a point source in an infinite homogeneous and lossless medium. Re(P) denotes the real part of ,SP denotes a spherical surface whose centre is at the point source, and an denotes the unit surface normal on S . Which of the following statements is TRUE?(A) Re(P) remains constant at any radial distance from the source

(B) Re(P) increases with increasing radial distance from the source

(C) ( )Re dSaP ns

:^ h## remains constant at any radial distance from the source

(D) ( )Re dSaP ns

:^ h## decreases with increasing radial distance from the source

MCQ 7.3.4 The electric field component of a time harmonic plane EM wave traveling in a nonmagnetic lossless dielectric medium has an amplitude of 1 V/m. If the relative permittivity of the medium is 4, the magnitude of the time-average power density vector (in /W m2) is

(A) 301p (B) 60

1p

(C) 1201p (D) 240

1p

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MCQ 7.3.5 A plane wave having the electric field components cos yE a6 3 10i x8

# b= -^ h V/m and traveling in free space is incident normally on a lossless medium with

0m m= and 9 0e e= which occupies the region y 0$ . The reflected magnetic field component is given by

(A) 101 (3 10 ) /cos A mt y a8

xp # + (B) 20 (3 10 ) /cos A mt y a1x

8

p # +

(C) 20 (3 10 ) /cos A mt y a1x

8

p #- + (D) 10 (3 10 ) /cos A mt y a1x

8

p #- +

MCQ 7.3.6 A uniform plane wave in the free space is normally incident on an infinitely thick dielectric slab (dielectric constant 9re = ). The magnitude of the reflection coefficient is(A) 0 (B) 0.3

(C) 0.5 (D) 0.8

MCQ 7.3.7 A plane wave of wavelength l is traveling in a direction making an angle 30c with positive x -axis and 90c with positive y -axis. The E field of the plane wave can be represented as (E0 is constant)

(A) E eaE yj t x z

03

= w lp

lp- -c m (B) E eaE y

j t x z0

3= w l

plp- -c m

(C) E eaE yj t x z

03

= w lp

lp+ +c m (D) E eaE y

j t x z0

3= w l

plp- +c m

MCQ 7.3.8 The H field (in A/m) of a plane wave propagating in free space is given by

( )cos t z t za aH 5 52x y

0hw b w b p= - + - +a k.

The time average power flow density in Watts is

(A) 1000h (B) 100

0h

(C) 50 02h (D) 50

0h

MCQ 7.3.9 A right circularly polarized (RCP) plane wave is incident at an angle 60c to the normal, on an air-dielectric interface. If the reflected wave is linearly polarized, the relative dielectric constant r2e is

(A) 2 (B) 3

(C) 2 (D) 3

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For View Only Shop Online at www.nodia.co.inMCQ 7.3.10 The electric field of an electromagnetic wave propagation in the positive direction

is given by ( ) ( /2)sin sinE t z t za a2 x yw b w b p= - + - + . The wave is(A) Linearly polarized in the z-direction

(B) Elliptically polarized

(C) Left-hand circularly polarized

(D) Right-hand circularly polarized

MCQ 7.3.11 When a plane wave traveling in free-space is incident normally on a medium having .4 0re = then the fraction of power transmitted into the medium is given by

(A) 98 (B)

21

(C) 31 (D)

65

MCQ 7.3.12 A medium of relative permittivity 2r2e = forms an interface with free - space. A point source of electromagnetic energy is located in the medium at a depth of 1 meter from the interface. Due to the total internal reflection, the transmitted beam has a circular cross-section over the interface. The area of the beam cross-section at the interface is given by(A) 2p m2 (B) 2p m2

(C) 2p m2 (D) p m2

MCQ 7.3.13 A medium is divided into regions I and II about x 0= plane, as shown in the figure below.

An electromagnetic wave with electric field 4 5E a a a5x y z1 = + + is incident normally on the interface from region I . The electric file E2 in region II at the interface is(A) E E2 1= (B) 4 0.75 1.25a a ax y z+ -

(C) 3 3 5a a ax y z+ + (D) 3 3 5a a ax y z- + +

MCQ 7.3.14 The magnetic field intensity vector of a plane wave is given by ( , , , )x y z tH 10 (50000 0.004 30)sin t x ay= + +where ay , denotes the unit vector in y direction. The wave is propagating with a phase velocity.(A) 5 104

# m/s (B) 3 108#- m/s

(C) 1.25 107#- m/s (D) 3 108

# m/s

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MCQ 7.3.15 Refractive index of glass is 1.5. Find the wavelength of a beam of light with frequency of 10 Hz14 in glass. Assume velocity of light is 3 108

# m/s in vacuum(A) 3 mm (B) 3 mm

(C) 2 mm (D) 1 mm

MCQ 7.3.16 If ( )j eE a ax yjkz k t= + w- and ( / )( )k j eH a ay x

jkz j twm= + w- , the time-averaged Poynting vector is(A) null vector (B) ( / )k azwm

(C) (2 / )k azwm (D) ( /2 )k azwm

MCQ 7.3.17 A plane electromagnetic wave propagating in free space is incident normally on a large slab of loss-less, non-magnetic, dielectric material with > 0e e . Maxima and minima are observed when the electric field is measured in front of the slab. The maximum electric field is found to be 5 times the minimum field. The intrinsic impedance of the medium should be(A) 120p W (B) 60p W

(C) 600p W (D) 24p W

MCQ 7.3.18 The depth of penetration of electromagnetic wave in a medium having conductivity s at a frequency of 1 MHz is 25 cm. The depth of penetration at a frequency of 4 MHz will be(A) 6.25 dm (B) 12.50 cm

(C) 50.00 cm (D) 100.00 cm

MCQ 7.3.19 A uniform plane wave traveling in air is incident on the plane boundary between air and another dielectric medium with 5re = . The reflection coefficient for the normal incidence, is(A) zero (B) 0.5 180c

(B) 0.333 0c (D) 0.333 180c

MCQ 7.3.20 If the electric field intensity associated with a uniform plane electromagnetic wave traveling in a perfect dielectric medium is given by ( , )E z t 10 (2 10 0.1 )cos t z7p p= - V/m, then the velocity of the traveling wave is(A) 3.00 108

# m/sec (B) 2.00 108# m/sec

(C) 6.28 107# m/sec (D) 2.00 107

# m/sec

MCQ 7.3.21 A plane wave is characterized by (0.5 )e ea aE /x y

j j t jkz2= + p w - . This wave is(A) linearly polarized (B) circularly polarized

(C) elliptically polarized (D) unpolarized

MCQ 7.3.22 Distilled water at 25 Cc is characterized by 1.7 10 4s #= - mho/m and 78 oe e= at a frequency of 3 GHz. Its loss tangent tan d is ( 36

10 9

e = p-

F/m)

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For View Only Shop Online at www.nodia.co.in(A) 1.3 10 5

#- (B) 1.3 10 3

#-

(C) 1.3 10 /784#

- (D) 1.3 10 /7850e#

-

MCQ 7.3.23 If a plane electromagnetic wave satisfies the equation zE c

tEx x

2

22

2

2

22

22= , the wave

propagates in the(A) x -direction

(B) z -direction

(C) y -direction

(D) x -y plane at an angle of 45c between the x and z direction

MCQ 7.3.24 A material has conductivity of 10 2- mho/m and a relative permittivity of 4. The frequency at which the conduction current in the medium is equal to the displacement current is(A) 45 MHz (B) 90 MHz

(C) 450 MHz (D) 900 MHz

MCQ 7.3.25 A uniform plane electromagnetic wave incident on a plane surface of a dielectric material is reflected with a VSWR of 3. What is the percentage of incident power that is reflected ?(A) 10% (B) 25%

(C) 50% (D) 75%

MCQ 7.3.26 A uniform plane wave in air impinges at 45c angle on a lossless dielectric material with dielectric constant re . The transmitted wave propagates is a 30c direction with respect to the normal. The value of re is(A) 1.5 (B) .1 5

(C) 2 (D) 2

MCQ 7.3.27 Two coaxial cable 1 and 2 are filled with different dielectric constants r1e and r2e respectively. The ratio of the wavelength in the cables ( / )1 2l l is(A) /r r1 2e e (B) /r r2 1e e

(C) /r r1 2e e (D) /r r2 1e e

MCQ 7.3.28 Identify which one of the following will NOT satisfy the wave equation.(A) 50e ( )j t z3w - (B) [ (10 5 )]sin z tw +

(C) ( 5 )cos y t2 + (D) ( ) ( )sin cosx t

MCQ 7.3.29 A plane wave propagating through a medium [ 8, 2, 0]andvr re s= = = has its electric field given by 0.5 (10 ) /sin V me t zE X ( / )z 3 8 b= -- . The wave impedance, in ohms is(A) 377 (B) 198.5 180c

(C) 182.9 14c (D) 133.3

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MCQ 7.3.30 The intrinsic impedance of copper at high frequencies is(A) purely resistive

(B) purely inductive

(C) complex with a capacitive component

(D) complex with an inductive component

MCQ 7.3.31 The time average poynting vector, in /W m2, for a wave with /V meE a12 ( )j t zy= w b+

in free space is

(A) . a2 4zp- (B) . a2 4

zp

(C) . a4 8zp (D) . a4 8

zp-

MCQ 7.3.32 The wavelength of a wave with propagation constant (0.1 0.2 )mj 1p p+ - is

(A) .

m0 052

(B) 10 m

(C) 20 m (D) 30 m

MCQ 7.3.33 The depth of penetration of wave in a lossy dielectric increases with increasing(A) conductivity (B) permeability

(C) wavelength (D) permittivity

MCQ 7.3.34 The polarization of wave with electric field vector E eE a aj t zx y0= +w b+

^^

hh is

(A) linear (B) elliptical

(C) left hand circular (D) right hand circular

MCQ 7.3.35 The skin depth at 10 MHz for a conductor is 1 cm. The phase velocity of an electromagnetic wave in the conductor at 1,000 MHz is about(A) 6 10 / secm6

# (B) 6 10 / secm7#

(C) 3 10 / secm8# (D) 6 10 / secm8

#

MCQ 7.3.36 A uniform plane wave in air is normally incident on infinitely thick slab. If the refractive index of the glass slab is 1.5, then the percentage of incident power that is reflected from the air-glass interface is(A) 0% (B) 4%

(C) 20% (D) 100%

MCQ 7.3.37 Some unknown material has a conductivity of 10 /mho m6 and a permeability of 4 10 /H m7p#

- . The skin depth for the material at 1GHz is(A) 15.9 mm (B) 20.9 mm

(C) 25.9 mm (D) 30.9 mm

MCQ 7.3.38 The plane wave travelling in a medium of 1re = , 1rm = (free space) has an electric field intensity of 100 /V mp . Determine the total energy density of this field.

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For View Only Shop Online at www.nodia.co.in(A) 13.9 /nJ m3 (B) 27.8 /nJ m3

(C) 139 /nJ m3 (D) 278 /nJ m3

MCQ 7.3.39 For a plane wave propagating in an unbounded medium (say, free space), the minimum angle between electric field and magnetic field vectors is(A) 0c (B) 60c

(C) 90c (D) 180c

MCQ 7.3.40 For no reflection condition, a vertically polarized wave should be incident at the interface between two dielectrics having 4 and 71 2e e= = , with an incident angle of

(A) tan 491-

b l (B) tan 231-

b l

(C) tan 321-

b l (D) tan 941-

b l

MCQ 7.3.41 The electric field component of a wave in free space is given by E 10 (10 ) /cos V mt kZ ay7= +Following is a list of possible inferences :1. Wave propagates along ay2. Wavelength 188.5 ml =

3. Wave amplitude is 10 /V m

4. Wave number 0.33 /rad m=

5. Wave attenuates as it travels

Which of these inferences can be drawn from E ?(A) 1, 2, 3, 4 and 5 (B) 2 and 3 only


MCQ 7.3.42 A plane wave is generated under water ( 81 )and 00e e m m= = . The wave is parallel polarized. At the interface between water and air, the angle a for which there is no reflection is

(A) 83.88c (B) .83 66c

(C) .84 86c (D) .84 08c

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MCQ 7.3.43 An elliptically polarized wave travelling in the positive z -direction in air has andx y components Ex 3 ( ) /sin V mt zw b= - Ey ( 75 ) /sin V mt z3 cw b= - +

If the characteristic impedance of air is 360W, the average power per unit area conveyed by the wave is(A) 8 /W m2 (B) 4 /W m2

(C) 62.5 /mW m2 (D) 125 /mW m2

MCQ 7.3.44 The intrinsic impedance of copper at 3 GHz (with parameters : 4 10 / ;H m7m p#= -

10 /36 ;79e p= - and 5.8 10 /mho m7s #= ) will be(A) 0.02 ohme /j 4p (B) 0.02 ohme /j 2p

(C) 0.2 ohme /j 2p (D) 0.2 ohme /j 4p

MCQ 7.3.45 Consider the following statements regarding depth of penetration or skin depth in a conductor :1. It increases as frequency increases.

2. It is inversely proportional to square root of m and s .

3. It is inversely proportional to square root of f

4. It is directly proportional to square root of m and s .

Which of the above statements are correct ?(A) 1 and 2 only (B) 3 and 4 only

(C) 2 and 3 only (D) 1, 2, 3 and 4

MCQ 7.3.46 Consider the following statements :1. (Electric or magnetic) field must have two orthogonal linear components.

2. The two components must have the same magnitude.

3. The two components must have a time-phase difference of odd multiple of 90c.

Which of these are the necessary and sufficient conditions for a time-harmonic wave to be circularly polarized at a given point in space ?(A) 1 and 2 only (B) 2 and 3 only

(C) 1, 2 and 3 (D) 1 and 3 only

MCQ 7.3.47 Assertion (A) :The velocity of light in any medium is slower than that of vacuum.Reason (R) : The dielectric constant of the vacuum is unity and is lesser than that of any other medium.(A) Both A and R are individually true and R is the correct explanation of A


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For View Only Shop Online at www.nodia.co.in(C) A is true but R is false


MCQ 7.3.48 In which direction is the plane wave /sin V mt zE a35 10 2 y8= +^ h , (where ay is the

unit vector in y -direction), travelling ?(A) along y direction (B) along y direction

(C) along z direction (D) along z direction

MCQ 7.3.49 According to Poynting theorem, the vector product E H# is a measure of which one of the following?(A) Stored energy density of the electric field

(B) Stored energy density of the magnetic

(D) Power dissipated per unit volume

(E) Rate of energy flow per unit area

MCQ 7.3.50 If ( )j ea aE x yj z= + b- , then the wave is said to be which one of the following ?

(A) Right circularly polarized (B) Right elliptically polarized

(C) Left circularly polarized (D) Left elliptically polarized

MCQ 7.3.51

What must be angle q of a corner reflector, such that an incident wave is reflected in the same direction ?(A) 30c (B) 45c

(C) 60c (D) 90c

MCQ 7.3.52 Poynting vector is a measure of which one of the following ?(A) Maximum power flow through a surface surrounding the source

(B) Average power flow through the surface

(C) Instantaneous power flow through the surface

(D) Power dissipated by the surface

MCQ 7.3.53 The electric field component of a wave in free space is given by E (10 ) /sin V mt kz a25 y

7= +Which one of the following is the correct inference that can be drawn from this expression ?

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(A) The wave propagates along y -axis

(B) The wavelength is 188.5 m

(C) The wave number .k 0 33= rad/m

(D) The wave attenuates as it travels

MCQ 7.3.54 For an electromagnetic wave incident on a conducting medium, the depth of penetration(A) is directly proportional to the attenuation constant

(B) is inversely proportional to the attenuation constant

(C) has a logarithmic relationship with the attenuation constant

(D) is independent of the attenuation constant

MCQ 7.3.55 Which one of the following statements is correct ?A right circularly polarised wave is incident from air onto a polystyrene 2.7re =^ h

. The reflected wave is(A) right circularly polarised (B) left circularly polarised

(C) right elliptically polarised (D) left elliptically polarised

MCQ 7.3.56 The electric field of a wave propagating through a lossless medium ( ,81 )0 0m e is (6 10 )cos t bx aE 50 y

8p#= -What is the phase constant b of the wave ?(A) 2 /rad mp (B) 9 /rad mp

(C) 18 /rad mp (D) 81 /rad m

MCQ 7.3.57 If the phase velocity of a plane wave in a perfect dielectric is 0.4 times its value in free space, then what is the relative permittivity of the dielectric ?(A) 6.25 (B) 4.25

(C) 2.5 (D) 1.25

MCQ 7.3.58 In free space ( , ) 60( 2 ) /V mx t t xE ayw= - . What is the average power crossing a circular area of radius 4 m in the plane tancons tx = ?(A) 480 W (B) 340 W

(C) 120 W (D) 60 W

MCQ 7.3.59 What is the effect of the earth’s magnetic field in the reflected wave at frequencies in the vicinity of gyro-frequency ?(A) No attenuation in the reflected wave

(B) Decreased attenuation in the reflected wave

(C) Increased attenuation in the reflected wave

(D) Nominal attenuation in the reflected wave

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For View Only Shop Online at www.nodia.co.inMCQ 7.3.60 A plane electromagnetic wave travelling in a perfect dielectric medium of intrinsic

impedance 1h is incident normally on its boundary with another perfect dielectric medium of characteristic impedance 2h . The electric and magnetic field strengths of the incident wave are denoted by E1 and H1 respectively whereas Er and Hr denote these quantities for the reflected wave, and Et and Ht for the transmitted wave.Which of the following relations are correct ?1. E Hi i1h=

2. E Hr r1h=

3. E Ht t2h=

Select the correct answer using the codes given below(A) 1, 2 and 3 (B) 1 and 2

(C) 1 and 3 (D) 2 and 3

MCQ 7.3.61 A plane electromagnetic wave travelling in a perfect dielectric medium of dielectric constant 1e is incident on its boundary with another perfect dielectric medium of dielectric constant 2e . The incident ray makes an angle of 1q with the normal to the boundary surface. The ray transmitted into the other medium makes an angle of

2q with the normal.If 21 2e e= and 601 cq = , which one of the following is correct ?(A) 452 cq =

(B) 0.433sin21q = -

(C) 0.612sinq2 1= -

(D) There will be no transmitted wave

MCQ 7.3.62 Match List I (Nature of Polarization) with List II (Relationship Between X and Y Components) for a propagating wave having cross-section in the XY plane and propagating along z -direction and select the correct answer :

List-I List-II

a. Linear 1. X and Y components are in same phase

b. Left circular 2. X and Y components have arbitrary phase difference

c. Right circular 3. X component leads Y by 90c

d. Elliptical 4. X component lags behind Y by 90cCodes :

a b c d(A) 1 4 2 3(B) 4 1 2 3(C) 1 4 3 2(D) 4 1 3 2

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MCQ 7.3.63 Assertion (A) : For an EM wave normally incident on a conductor surface the magnetic field H undergoes a 180c phase reversal and the phase of electric field E remains same.Reason (R) : The direction of propagation of incident wave will reverse after striking a conductor surface.(A) Both A and R are true and R is the correct explanation of A




MCQ 7.3.64 Match List I with List II and select the correct answer :

List-I List-II

a. Propagation constant 1. /2wms

b. Radiation intensity 2. ( )Er2

22

h

c. Wave impedance 3. /E Ht t

4. E H#

Codes :

a b c(A) 1 2 3(B) 4 3 2(C) 1 3 2(D) 4 2 3

MCQ 7.3.65 If the E field of a plane polarized EM wave travelling in the z -direction is :E EE a ax x y y= + then its H field is :

(A) ZE

ZEa ay

xx

y

0 0- (B) Z

EZEa ay

xx

y

0 0+

(C) ZE

ZEa ax

yy

x

0 0- (D) Z

EZEa ax

yy

x

0 0- -

MCQ 7.3.66 Consider the following statements :For electromagnetic waves propagating in free space :1. electrical field is perpendicular to direction of propagation

2. electrical field is along the direction of propagation

3. magnetic field is perpendicular to direction of propagation

4. magnetic field is along the direction of propagation

Which of these statements are correct ?(A) 1 and 3 (B) 1 and 4

(C) 2 and 3 (D) 2 and 4

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For View Only Shop Online at www.nodia.co.inMCQ 7.3.67 In a uniform plane wave, the value of /E H is

(A) /m e (B) /e m

(C) 1 (D) me

MCQ 7.3.68 The phenomenon of microwave signals following the curvature of earth is known as(A) Faraday effect (B) ducting

(C) wave tilt (D) troposcatter

MCQ 7.3.69 Which one of the following statements is NOT correct for a plane wave with 0.5 (10 2 ) /cos A me t xH a. x

z0 1 6= --

(A) The wave frequency is 106 r.p.s

(B) The wavelength is 3.14 m

(C) The wave travels along + x -direction

(D) The wave is polarized in the z -direction

MCQ 7.3.70 Skin depth is the distance from the conductor surface where the field strength has fallen to(A) p of its strength at the surface (B) e of its strength at the surface

(C) /e1^ h of its strength at the surface (D) / e1 p^ h of its strength at the surface

MCQ 7.3.71 The vector magnetic potential of a particular wave traveling in free space is given by sinA t zA ax x w b= -^ h where Ax is a constant. The expression for the electric field will be(A) sinA t zax xb w b- -^ h (B) sinA t zay xb w b- -^ h

(C) cosA t zay xw w b- -^ h (D) cosA t zax xw w b- -^ h

MCQ 7.3.72 The depth of penetration of a wave in a lossy dielectric increases with increasing(A) conductivity (B) permeability

(C) wavelength (D) permittivity

MCQ 7.3.73 When a plane wave propagates in a dielectric medium(A) the average electric energy and the average magnetic energy densities are not

equal.

(B) the average electric energy and the average magnetic energy densities are equal

(C) the net average energy density is finite

(D) the average electric energy density is not dependent on the average magnetic energy density

MCQ 7.3.74 In free space H field is given as , cosz t t zH a61

yp w b=- +^ ^h h ,z tE^ h is

(A) 20cos t z axw b+^ h (B) 20cos t z azw b+^ h

(C) 20 sin t z ayw b+^ h (D) 20 sin t z axw b+^ h

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MCQ 7.3.75 If electric field intensity phasor of an EM wave in free space is /V meE a6 j yx

4= -

.The angular frequency w , in rad/s, is(A) 4 3 108

# # (B) y4 3 108# #

(C) t 3 108# # (D) 10 3 108

# #

MCQ 7.3.76 Assertion (A) : Electromagnetic waves propagate being guided by parallel plate perfect conductor surface.Reason (R) : Tangential component of electric field intensity and normal component of magnetic field intensity are zero on a perfect conductor surface.(A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is





MCQ 7.3.77 A uniform plane wave is propagating in a material for which 4 0e e= , 7 0m m= and 0s = . The skin depth for the material is

(A) zero (B) infinity

(C) 28 m (D) 14 m

MCQ 7.3.78 Consider the following statements :1. In conducting medium the field attenuates exponentially with increasing depth.

2. Conducting medium behaves like an open circuit to the electromagnetic field.

3. In lossless dielectric relaxation time is infinite.

4. In charge-free region, the Poisson’s equation becomes Laplace’s equation.

(A) 1, 2 and 3 only (B) 1, 3 and 4 only

(C) 2, 3 and 4 only (D) 1, 2, 3 and 4

MCQ 7.3.79 In free space ( , )Z tE 60 /cos V mt z axp w b= -^ h .The average power crossing a circular area of p square metres in the plane z =constant is(A) 16 /watt m2p (B) 15 /watt m2p

(C) 14 /watt m2p (D) 13 /watt m2p

MCQ 7.3.80 In free space ( , )Z tE 120 cos mt Z Vax 1p w b= - -

^ h

What is the average power in Wm 2- ?(A) 30 azp (B) 60 azp

(C) 90 azp (D) 120 azp

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For View Only Shop Online at www.nodia.co.inMCQ 7.3.81 The electric field of a uniform plane wave is given by :

E sin cos Vmt Z t Za a35 3 10 45 3 10x y8 8 1

# #p p p p= - + - -^ ^h h

What is the corresponding magnetic field H

(A) ( )sin cos Amt Z t Za a37710 3 10 377

10 3 10y x8 8 1

# #p p p p- + - - -^ ^h h

(B) 3 10 ( ) 3 10 ( )sin cos Amt Z t Za a37710

377108 8

y x1p p p p# #- - + - - -

^ ^h h

(C) 3 10 3 10 ( )sin cos Amt Z t Za a37710

377108 8

y x1p p p p# #- + - -

^ ^h h

(D) 3 10 ( ) 3 10 ( )sin sin Amt Z t Za a37710

377108 8

y x1p p p p# #- - + - - -

^ ^h h

MCQ 7.3.82 Consider the following statements in connection with electromagnetic waves :1. Conducting medium behaves like an open circuit to the electromagnetic field.

2. At radio and microwave frequencies the relaxation time is much less than the period

3. In loss-less dielectric the relaxation time is finite.

4. Intrinsic impedance of a perfect dielectric medium is a pure resistance.

Which is these statements is/are correct ?(A) 1 only (B) 1 and 2 only

(C) 2 and 3 only (D) 2, 3 and 4

MCQ 7.3.83 What causes electromagnetic wave polarization ?(A) Refraction

(B) Reflection

(C) Longitudinal nature of electromagnetic wave

(D) Transverse nature of electromagnetic wave

MCQ 7.3.84 Assertion (A) : The velocity of electromagnetic waves is same is same as velocity of light.Reason (R) : Electrons also travel with the same velocity as photons.(A) Both A and R are true and R is the correct explanation of A

(B) Both A and R are true but R is not the correct explanation of A



MCQ 7.3.85 Fields are said to be circularly polarized if their magnitudes are(A) Equal and they are in phase

(B) Equal and they differ in phase by 90! c

(C) Unequal and they differ in phase by 90! c

(D) Unequal and they are in phase

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MCQ 7.3.86 Which of the following is zero as applied to electromagnetic field ?(A) grad div A (B) div grad V

(C) div curl A (D) curl curl A

MCQ 7.3.87 What is the Poynting’s vector on the surface of a long straight conductor of radius b and conductivity s which carries current I in the z -direction ?

(A) bI i

2 r2 3

2

sp- (B)

bI i

2 r2 2

2

sp

(C) bI iz2

2

sp (D) b

I i2p f

MCQ 7.3.88 Consider the following statements regarding EM wave1. An EM wave incident on a perfect dielectric is partially transmitted and

partially reflected

2. An EM wave incident on a perfect conductor is fully reflected

3. When an EM wave is incident from a more dense medium to less dense medium at an angle equal to or exceeding the critical angle, the wave suffers total internal reflection

Which of the statements given above are correct ?(A) Only 1 and 2 (B) Only 2 and 3

(C) Only 1 and 3 (D) 1, 2 and 3

MCQ 7.3.89 A uniform plane wave has a wavelength of 2 cm in free space and 1 cm in a perfect dielectric. What is the relative permittivity of the dielectric ?(A) 2.0 (B) 0.5

(C) 4.0 (D) 0.25

MCQ 7.3.90 With the increase in frequency of an electromagnetic wave in free space, how do the velocity vc and characteristic impedance Zc change ?(A) vc increase and Zc decreases

(B) vc decreases and Zc increases

(C) Both vc and Zc increase

(D) Both vc and Zc remain unchanged

MCQ 7.3.91 The E field of a plane electromagnetic wave travelling in a non-magnetic non-conducting medium is given by 5 10 30cos t ZE a 9

x= +^ h. What is the dielectric constant of the medium ?(A) 30 (B) 10

(C) 9 (D) 81

MCQ 7.3.92 In the wave equation t t

E E E22

2

22

22me msd = + which term is responsible for

attenuation of the wave ?

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(A) E2d (B) tE2

2

22me

(C) tE

22ms (D) All of the above three

MCQ 7.3.93 Consider the following statements :1. Poisson’s equation finds application in vaccum tube and gaseous discharge

problems

2. Gauss’s law is useful for determining field and potential distribution about bodies having unsymmetrical geometry.

3. For the propagation of electro-magnetic waves, the time varying electric fields must support time varying magnetic fields.

4. The unit of Poynting’s vector is W/m2


(C) 2, 3 and 4 (D) 1, 2 and 4

MCQ 7.3.94 What is the phase velocity of plane wave in a good conductor ?

(A) fp ms (B) ( )fmsp s

(C) ( )f

msp (D) 2

( )f

msp

MCQ 7.3.95 The instantaneous electric field of a plane wave propagating in z -direction is ( )tE cos sinE t E t ea ax y

jkz1 2w w= - -

6 @

This wave is(A) Linearly polarised (B) Elliptically polarised

(C) Right hand circularly polarised (D) Left hand circularly polarised

MCQ 7.3.96 Assertion (A) : Skin depth is the depth by which electromagnetic wave has been increased to 37% of its original value.Reason (R) : The depth of penetration of wave in a lossy dielectric increases with increasing wavelength.(A) Both A and R are true and R is the correct explanation of A




MCQ 7.3.97 Which one of the following is the correct electromagnetic wave equation in terms of vector potential A ?

(A) t

A A J22

2

22d - =- (B)

tA A J2

2

2

22

em md - =-

(C) t

A A J22

2

22 md - =- (D)

tA A J2

2

2

22me md - =-

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MCQ 7.3.98 Which one of the following statements is correct ? The wavelength of a wave propagating in a wave guide is(A) smaller than the free space wavelength

(B) greater than the free space wavelength

(C) directly proportional to the group velocity

(D) inversely proportional to the phase velocity

MCQ 7.3.99 Which one of the following statements is correct ? For a lossless dielectric medium, the phase constant for a travelling wave, b is proportional to(A) re (B) re

(C) 1/ re (D) 1/ re

MCQ 7.3.100 In a lossless medium the intrinsic impedance 60h p= and 1rm = . What is the value of the dielectric constant re ?(A) 2 (B) 1

(C) 4 (D) 8

MCQ 7.3.101 An electromagnetic field is said to be conservative when(A) ( / )E E t2 2 2d 2 2me=

(B) ( / )H H t2 2 2d 2 2me=

(C) Curl on the field is zero

(D) Divergence of the field is zero

MCQ 7.3.102 Given that 0.5 . (10 2 ) ( )exp sinx t xH a0 1 A/mz6= - -6 @ , which one of the following

statements is not correct ?(A) Wave is linearly polarized along az(B) The velocity of the wave is 5 10 m/s5

#

(C) The complex propagation constant is (0.1 2)j+

(D) The wave is travelling along ax

MCQ 7.3.103 For a conducting medium with conductivity s , permeability m, and permittivity e, the skin depth for an electromagnetic signal at an angular frequency w is proportional to(A) s (B) 1/w

(C) 1/ s (D) 1/m

MCQ 7.3.104 The electric field of a uniform plane wave is given by 10 (10 )sinE t zw p= - 10 ( ) ( )cos t za a V/mx yw p+ -

The polarization of the wave is (A) Circular (B) Elliptical

(C) Linear (D) Undefined

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For View Only Shop Online at www.nodia.co.inMCQ 7.3.105 In free space ( , ) 0. 0 (4 10 )cosz t t z aH 2 A/mx

7 b#= - . The expression for ( , )z tE is

(A) ( , ) 37.7 (4 10 )cosz t t z aE y7 b#= -

(B) ( , ) 2.65 10 (4 10 )cosz t t z aE z7 b# #= -

(C) ( , ) 37.7 (4 10 )cosz t t z aE x7 b#= -

(D) ( , ) 37.7 (4 10 )cosz t t z aE y7 b#=- -

MCQ 7.3.106 A plane wave whose electric field is given by 100 ( 6 )cos t x aE zw p= - passes normally from a material ‘A’ having 4,re = 1rm = and 0s = to a material ‘B’ having 9,re = 4rm = and 0s = . Match items in List I with List II and select the correct answer :

List I List II

a Intrinsic impedance of medium ‘B’ 1. 6p

b Reflection coefficient 2. 80p

c Transmission coefficient 3. /1 7

d Phase shift constant of medium ‘A’ 4. /8 7Codes : a b c d(A) 4 1 2 3(B) 2 3 4 1(C) 4 3 2 1(D) 2 1 4 3

MCQ 7.3.107 In free space ( , ) 50 ( )cosz t t zE axw b= - V/m and ( , ) 5/12 ( )cosz t t zH a A/myp w b= - . The average power crossing a circular area of

radius 24 m in plane z = constant is

(A) 200 W (B) 250 W

(C) 300 W (D) 350 W

MCQ 7.3.108 Consider a plane electromagnetic wave incident normally on the surface of a good conductor. The wave has an electric field of amplitude 1 V/m and the skin depth for the conductor is 10 cm.Assertion (A) : The amplitude of electric field is (1/ )e (V/m)2 after the wave has travelled a distance of 20 cm in the conductor.Reason (R) : Skin depth is the distance in which the wave amplitude decays to (1/e) of its value at the surface.(A) Both A and R are true and R is the correct explanation of A




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MCQ 7.3.109 Three media are characterised by1. 8, 2, 0r re m s= = =

2. 1, 9, 0r re m s= = =

3. 4, 4, 0r re m s= = =

re is relative permittivity, rm is relative permeability and s is conductivity.The value of the intrinsic impedances of the media 1, 2 and 3 respectively are(A) 188 W, 377 W and 1131 W

(B) 377 W, 1131 W and 188 W

(C) 188 W, 1131 W, and 377 W

(D) 1131 W, 188 W, and 377 W

MCQ 7.3.110 A plane EM wave ( , )E Hi i travelling in a perfect dielectric medium of surge impedance ‘Z ’ strikes normally on an infinite perfect dielectric medium of surge impedance Z2 . If the refracted EM wave is ( , )E Hr r , the ratios of /E Ei r and /H Hi r are respectively(A) 3 and 3- (B) /3 2 and 1/3

(C) 3/4 and 3/2 (D) 3/4 and 2/3

MCQ 7.3.111 For a perfect conductor, the field strength at a distance equal to the skin depth is %X of the field strength at its surface. The value ‘ %X ’ is

(A) Zero (B) 50%

(C) 36% (D) 26%

***********

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SOLUTIONS 7.1

SOL 7.1.1 Option (B) is correct.Given magnetic field intensity in the non magnetic medium is H 3 /cos A mt kz axw= -^ h

The negative coefficient of z in t kzw -^ h shows that the wave is propagating in az+ direction.

SOL 7.1.2 Option (C) is correct.From the property of phasor, we know that the instantaneous electric field is the real part of eEs j tw" ,.i.e. ,x tE^ h Re eEs j t= w

" , (1)where Es is the phasor form of electric field.Given the electric field intensity in time domain, ,x tE^ h sinE e t x ax

y0 w b= -a-^ h

E e je e a2

xj t x j t x

y0= -aw b w b

-- - -^ ^h h

; E

.jE e e C Ca2x j t x

y0=- +a w b- -^ h .

where . .C C is complex conjugate of the 1st part.So, using the property of complex conjugates we get

,x tE^ h Re jE e e a2 20 ( )x j t x

y= - a w b- -& 0

Re jE e e e ax j x j ty0= - a b w- -

" ,Comparing it with equation (1), we get Es /V mjE e a2 j x

y0=- a b- +^ h

SOL 7.1.3 Option (C) is correct.Given the magnetic field intensity, H 10 /cos A mt ky a6 10 z

7#= -^ h

Comparing it with the general equation of magnetic field. H /cos A mH t ky az0 w= -^ h

We get, w 6 107#=

So, the wave no is,

k 0.2c 3 106 10

8

7

#

#w= = = (c is the velocity of wave in free space)




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SOL 7.1.4 Option (D) is correct.Given magnetic field intensity in the non magnetic medium is H 1.5 /cos A mt z a10 5 x

9= -^ h

Comparing it with the general equation of magnetic field intensity H /cos A mH t z ax0 w b= -^ h

We get, w 10 / secrad9=and b 5= .So, the phase velocity of the wave in the medium is given as

vP 10 /m s510 4

98

bw

#= = =

SOL 7.1.5 Option (A) is correct.Wavelength of an electromagnetic wave with phase constant b in a medium is defined as

l 2bp=

So, the phase constant of the wave in terms of wavelength can be given as

b .2

12 62

lp p= = 0.5 /rad m= ( 12.6 ml = )

SOL 7.1.6 Option (A) is correct.Given the electric field intensity in the nonmagnetic material as E 8 /cos V mt x a4 10 2 y

8#= -^ h

Comparing it with the general equation of electric field E /cos A mE t x ay0 w b= -^ h

We get, w 4 10 /rad s8#=

and b 2 /rad m=So, the phase velocity of the wave in the medium is given by

vp 10 /m s3 8

bw

#= =

Since the medium is non magnetic so, 0m m= and the relative permittivity of the medium is given as

re vcp

2= b l

2 103 10

8

8 2

#

#= c m .2 25=

SOL 7.1.7 Option (D) is correct.The general equation of electric field intensity of an EM wave propagating in az direction in a medium is given as E /cos A mE t x ay0 w b= -^ h Comparing it with the given expression of electric field intensity, we get w 5 10 /rad s8

#=So, the time period of the EM wave is

, T 25 10

28

#wp p= = 12.57 ns=



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The general equation of electric field intensity of an EM wave propagating in ax direction in a medium is given as E /cos A mE t x ay0 w b= -^ h Comparing it with the given expression of electric field intensity, we get w 10 /rad s4 8

#=So, the time period of the wave in air is given as

T 24 10

28

#wp p= =

. ns15 71=Since in one time period the wave travels its one wavelength (l) so, time taken by the wave to travel /4l distance is

t .93 ns4 4T= =

SOL 7.1.9 Option (D) is correct.Intrinsic impedance of any material is given as

h jj

s wewm= +

where m is permeability, s is conductivity and e is permittivity of the medium.Since the given material is lossless, nonmagnetic and dielectric so, we have s 0= (lossless) m 0m= (non magnetic)and e .2 25r 0 0e e e= = ^ h ( .2 25re = )Therefore the intrinsic impedance of the material is

h .j

j0 2 25 0

0

w ewm=

+ ^ h

. .1 5 1 53770h= = 3 .23 3W= 3770

0

0h em W= =b l

SOL 7.1.10 Option (B) is correct.Given,Frequency of the wave propagation, f 0.5 0.5 10MHz Hz6

#= =Conductivity of medium, s 3 10 /S m7

#=Relative permeability of medium, rm 1r .e=So, the angular frequency of the wave propagation is w f2p= .2 0 5 106

# #p= 106#p=

and we get

wes

.10 8 85 103 10

6 12

7

# # #

#p

= - .0 1 10 1>>13#=

Therefore, the phase constant of the propagating wave is given as

b 2wms= ( / 1>>s we )

210 4 10 3 106 7 7

# # # # #p p=-




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7695.29 /rad m=So, the wavelength of the radio wave in the medium is

l .8 mm2 2bp= =

SOL 7.1.11 Option (C) is correct.Attenuation constant for a plane wave with angular frequency w in a certain medium is given as

a 2 1 12w mewes= + -a k: D (1)

Since for a poor conductor, conductivity is very lowi.e. s << we

or, wes 1<<

So, in equation (1) using binomial expansion we get,

a 2 1 21 12w me

wes= + -a k: D ( / 1<<s we )

2 21w me

wes= 2

sem=

Therefore, the skin depth of the poor conductor is

d 1 2a s m

e= =

which is independent of frequency (w).

SOL 7.1.12 Option (B) is correct.Wave equation for a plane wave propagating in az+ direction is given as

tf v

zf

p2

22

2

2

22

22- 0= where vp is the velocity of wave propagation

Now from Assertion (A) the electric field is E sin cosE z ct ax0= ^ ^h h

It represents the electric field of a plane wave if it satisfies the wave equation

i.e. t

cz

E E2

22

2

2

22

22- 0= where c is velocity of wave in free space

From the given expression of field intensity we have

tE

22 sin sinc z ct0E=- ^ ^h h

or, tE2

2

22 sin cosc E z ct2

0=- ^ ^h h

and zE

22 cos cosE z ct0= ^ ^h h

or, zE2

2

22 sin cosE z ct0=- ^ ^h h

Thus, we get

, t

cz

E E2

22

2

2

22

22- 0=

Since, the electric field E satisfies the wave equation so it represents the field of a



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For View Only Shop Online at www.nodia.co.inplane wave.Therefore, A and R both are true and R is the correct explanation of A.

SOL 7.1.13 Option (A) is correct.Given the magnetic field intensity in free space is H 0.1 /cos A mt y a10 z

9 b= -^ h (1)The general equation of magnetic field intensity of the EM wave propagating in ay direction is given as H /cos A mH t y az0 w b= -^ h (2)

Comparing equations (1) and (2) we get,direction of wave propagation, ak ay=and angular frequency, w 10 / secrad9=So, the phase constant of the wave is

b c 3 1010

8

9

#

w= = (c is velocity of wave in free space)

3.33 /rad m=Now, electric field intensity in free space is defined as E a Hk0h #=-where 0h is intrinsic impedance in free space and ak is direction of wave propagation. So, E 377 0.1 cos t ya a10y z

9 b#=- -^ ^h h ( 3770h W= ) 37.7 .cos t y a10 3 33 x

9=- -^ h

Therefore, electric field intensity of the wave at 1 cmy = at 0.1 nst = is E 37.7 .cos a10 10 3 33 10 x

9 10 2=- -- -^ ^ ^ ^h h h h6 @

.6 /V ma46 x=-

SOL 7.1.14 Option (C) is correct.Given the magnetic field intensity of the plane wave in free space is Hs /A mj j ea a2 5 4 2y z

j x= + + b-^ ^h h

From the Maxwell’s equation, the maximum electric field intensity of the plane wave is given as E

max H

max0h=

where 0h is intrinsic impedance in air and Hmax

is the maximum magnetic field intensity of the plane wave.Now, the maximum magnetic field intensity of the plane wave is given as H

max H Hs s:= )

where Hs) is the complex conjugate of the magnetic field phasor.So, H

max 4 2 4 2j j j ja a a a2 5 2 5y z y z:= + + - -^ ^ ^ ^h h h h6 6@ @

j j j j2 5 2 5 4 4 2 2= + - -^ ^ ^ ^ ^ ^h h h h h h6 @

29 20#= 24.1 /A m=Therefore, the maximum electric field intensity of the plane wave is E

max H

max0h= .377 24 1#= 9.08 /kV m=




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SOL 7.1.15 Option (D) is correct.Given the instantaneous electric field in the free space is E 5 6 /cos V mt za a 50y x w= - -^ ^h h

So, the phasor form of electric field intensity is Es 5 6 /V mea ay x

j z50= - -^ h

The phasor form of magnetic field is given in the terms of electric field intensity as

Hs a E1k

0#h= ^ ^h h

where ak is the unit vector in the direction of wave propagation and 0h is the intrinsic impedance in free space.

So, Hs 5 6 /V mea a a1z y x

j z

0

50

h #= - -^ ^h h (ak az= )

5 6 /V mea a1x y

j z

0

50

h= - - -^ h

/V mea a1 15 36x yj z

0

50

h=- + -^ h

SOL 7.1.16 Option (C) is correct.For any electromagnetic wave propagating in a medium electric field leads magnetic field by an angle nq , where nq is the phase angle of intrinsic impedance given as

tan2 nq wes=

Now, for a perfect conductor

s 1 3.r=

i.e. tan2 nq 3.

2 nq 90c= nq 45c=So, electric field leads magnetic field by 45cor in other words magnetic field lags electric field by 45c.

SOL 7.1.17 Option (B) is correct.Given, the electric field intensity of the wave in phasor form Es 5 10 /V mea ax z

j x z4 2= + - -^

^h

h

So we get the direction of wave propagation as

ak 4 24 2 4 2a aa a a a

20x z

x z x z= -- = - 2a a

5x z= -

Therefore, the phasor form of magnetic field intensity of the plane wave is given as

Hs a E1k s

0h #= where 0h is intrinsic impedance in free space

2 5 10 /V mea a a a1201

5x z

x zj x z4 2

p #= - + - -c ^

^m h

h

29.66 /mA me j x z4 2=- - -^ h

SOL 7.1.18 Option (A) is correct.The time average power density of the EM wave is given as



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Pave E a2 k

0

2

h=

where E is the magnitude of the electric field intensity of the wave, ak is the unit vector in the direction of wave propagation and 0h is the intrinsic impedance in the free space. So, we get

Pave 20

4 2a a2 1205 10 x z

2 2

p= + -

^ ch m

18.9 4.15 /Watt ma a3x z2= -

SOL 7.1.19 Option (A) is correct.As the given electric field vector has the amplitude E0 2 a a a3 3x y z= - + -_ iSo in the same direction the wave will be polarized.

SOL 7.1.20 Option (B) is correct.From Maxwells’ equation we have

E#d tB22=-

Given E 100cos t z axw b= -^ h

or, E#d 100 sin t z ayb w b= -^ h

So, tB22=- E#d= 100 sin t z ayb w b= -^ h

Therefore the magnetic flux density vector is

B 100 sin t z a dtyb w b= -^ h#

cos t z a100yw

b w b= -^ h

cos t z a100y

0 0m ew b= -^ h

0 0

bm ew=d n

3 10 cos t z ay10 w b#= -^ h

SOL 7.1.21 Option (B) is correct.Poynting vector in an EM field is defined as P E H#=where E is electric field intensity and H is the magnetic field intensity in the region.Now, the electric field intensity in the region is given as E 100cos t z axw b= -^ h

and as calculated in previous question the magnetic field intensity in the region is B 3 10 cos t z ay10 w b#= -^ h

So, the poynting vector in the field is

P E B0m#= (H B

0m= )




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1003 10

coscos

t zt z

aa

x

y

0

10

#w b mw b#

= --

^^

hh

68

@B

cos t z a3 10z

0

122#

m w b= -^ h 10 cos t z az4

0

0 2

me w b= -^ h

SOL 7.1.22 Option (A) is correct.Time average stored energy density in electric field is defined as

we E E41 *

s s0 :e=

where Es is the electric field intensity in phasor form and E *s is its conjugate.

Therefore, the average stored energy density in the region is

we 5 5sin sinxe e xe ea a4/ /j j z

yj j z

y0 2 3 2 3

:e p p= p p p p- - +_ _i i

sin x425 0 2e p=

SOL 7.1.23 Option (A) is correct.Given the electric field E 10 /sin sin V my t x a6 10 3 z

8#p p p= -^ h

In phasor form, Es 10 sin ye e a/j j xz

2p= p b p- -

So, from Maxwell’s equation, the magnetic flux density in the phasor form is given as

Bs j E1sw #d= ^ h

where 6 108#w p= as determined from the given expression of E .

So, Bs sin cosy e e j y e ea a6 1010 3

6 1010j j x

yj j x

y8 2 38 2 3

# #pp p

pp p=- +

p p p p- - - -^ ^h h

Therefore, the time average energy density stored in the magnetic field will be

wm B B41 *

s s0

:m= ^ h where B *s is the conjugate of Bs

or, wm sin x14410 25 50

92

p p= +-

^ h

SOL 7.1.24 Option (A) is correct.For an EM wave propagating in two mediums, the wavelengths of the wave in two mediums are related as

1

2ll

1

2ee=

where 1l and 2l are the wavelengths of EM wave in two mediums with permittivity 1e and 2e respectively. So, the wavelength of plane wave in free space is given as

0

ll 1

re=

0l rl e=where l is the wavelength of the wave in the medium with relative permittivity re .So, 0l 20 9= 60 cm= (l 20 cm= , re 9= )



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Given,Conductivity of the glass, s 10 /S m12= -

and relative permittivity of the glass, re .2 25=So, the permittivity of glass is e r0e e= .2 25 0e=Therefore, the time taken by the charge to flow out to the surface is

t . se

. .10

2 25 8 85 1012

12# #= -

-^ ^h h

. sec19 9 20.=

SOL 7.1.26 Option (A) is correct.The reflection coefficient of the wave propagating from medium 1 to medium 2 is defined as

G 2 1

2 1

h hh h= +

-

where 1h and 2h are the intrinsic impedance of the two mediums respectively. So, the reflection coefficient for the wave propagating from free space to a dielectric medium is given as

G 0

0

h hh h= +-

where h is intrinsic impedance of the dielectric medium and 0h is intrinsic impedance in free space. Since the intrinsic impedance of the dielectric medium is given as

h 4 20

0 0

em

em h= = =

So, we have G /2/2

0 0

0 0

h hh h= +

-

//

1 2 11 2 1

31= +

- =-

Therefore, the magnitude of electric field of reflected wave is

Er E0G= E3

0=- (E0 is the magnitude of incident field)

***********




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SOLUTIONS 7.2

SOL 7.2.1 Option (D) is correct.Time period of wave propagating in a medium is given as :

T 2wp= where w is the angular frequency of the wave.

Given the magnetic field intensity in the free space is H 0.3 /cos A mt y axw b= -^ h

So, at /8t T= the magnetic field intensity is

H 0.3 cos T y a8 xw b= -b l 0.3 cos y a4 xp b= -a k (T 2 /p w= )

or, H 0. /cos y5 4b p= -^ h

Therefore we get the plot of H versus y as shown below

SOL 7.2.2 Option (D) is correct.Phase velocity of the medium, vp 7.5 10 /m s7

#=Relative permeability, rm .4 8=Conductivity s 0= (lossless medium)Since phase velocity of an EM wave in a medium is defined as

vp cr rm e

=

where c is the velocity of wave in air, rm is the relative permeability of the medium and re is the relative permittivity of the medium. So, we have

.7 5 107#

.4 83 10

r

8#

e=

^ h ( 3 10 /m sc 8

#= )

or, re .3 33=Now the intrinsic impedance of the medium is given as



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h jj

s wewm= +

r

r

0

0

e em m= ( 0s = )

377 .4. 452.43 33

8 W= = 37700

0h em W= =b l

Given the electric field intensity in the phasor form is Es 5 /V me a.j x

z0 3= (1)

and the general equation of electric field phasor of an EM wave propagating in ax direction is Es /V mE e aj x z0= b- (2)So, comparing the equations (1) and (2) we getdirection of wave propagation, ak ax=-and phase constant, b 0.3 /rad m=and from the Maxwell’s equation, the magnetic field phasor of the wave is given as

Hs a E1k sh #= ^ h

where h is the intrinsic impedance of the medium and ak is the unit vector in the direction of wave propagation.

So, Hs . 5ea a45 241 .

xj x

z0 3

#= -^ ^h h (ak ax=- )

. e a452 45 .j x

y0 3= 11.05 /mA me a.j x

y0 3=

and the angular frequency of the wave is given as w vpb= . .0 3 7 5 107

#= ^ ^h h .2 25 107#=

So, the magnetic field intensity of the EM wave in time domain is ,x tH^ h Re eHs j t= w

" , 11.05 .cos t x a0 3 yw= +^ h

11.05 . . /cos mA mt x a2 25 10 0 3 y7

#= +^ h

SOL 7.2.3 Option (D) is correct.Given the electric field intensity of the propagating wave, E /sin V mE e t x a10/x

y03 8 b= --

^ h (1)The general equation of electric field intensity of plane wave propagating in ax direction is given by E /sin V mE e t x a0

xyw b= -a-

^ h (2)Comparing equation (1) and (2) we get,

a /NP m31= and 10 / secrad8w =

So, the attenuation constant of a propagating wave is given as

a 12 12w mewes= + +a k: D

Let x0 1 2

wes= + a k

Therefore, a x2 1r r0 0

0w m e m e= -^ h

or, x 10 -^ h 2r r

20 0

2

w m e m ea=




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Now, we put /1 3a = , 4r rm e= = , 108w = so, we get

x 10 - / c

10 42 1 3

8 2 2

202

#

#=^ ^

^

h h

h /m sc 1 3 100

0 0

8#

m e= =c m

x 10 - /

210 4

1 3 3 108

8 2

#

# #= e o

x0 81 1= +

x0 89=

1 2

wes+ a k 8

9=

wes 64

81 1= -

Thus, loss tangent .0 52wes= =

SOL 7.2.4 Option (C) is correct.From the field intensity we get, w 109p=and it is given that, .0 5rm = , 0.01 /S ms = , 8re = .So, the phase constant,

b t2 1 12w mews= + +a k: D

. .10 2

8 0 51

10 80 01 19 0 09

0

2p

m ep e

= + +^ ^c

h hm= G

.20 95=Let the distance travelled by the wave be z to have a phase shift of 10c.

So, zb 10 rad18010c p= =

z .

. mm9 20 95

16 66#

p= =^ h

SOL 7.2.5 Option (A) is correct.The attenuation constant of a propagating wave in a medium is defined as

a 2 1 12w mewes= + -a k: D

Now, from the given data we have .0 5rm = , 0.01 /S ms = , 8re = .

So, a .

110 8

0.0110 28 0 5

190

29 0 0p

m ep e

= + -^ ^c

h hm= G

.0 9425=Initially the amplitude of the electric field .0 5= So, after travelling distance z amplitude of wave . e0 5 z= a- .Therefore, the distance travelled by the wave for which the amplitude of the wave



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For View Only Shop Online at www.nodia.co.inreduced by %40 is evaluated as

. e0 5 za- .0 5 10060

#= (amplitude reduces to %60 )

e . z0 9425-^ h .0 6=

or, z . .ln0 94251

0 61= b l 542 mm=

SOL 7.2.6 Option (A) is correct.Given the field intensities of the plane wave as ,x tE^ h 900 /cos V mt x a5 10 y

6# p b= -^ h

,x tH^ h 1.9 /cos V mt x a5 10 z6

# p b= -^ h

So, we get E 900= , .H 1 9= , 5 106#w p=

Now, the intrinsic impedance in the medium is

h . .HE

1 9900 473 7= = =

and phase constant of the wave in the medium is

b v 7 105 10

p7

6

#

#w p= = 0.224 m 1= -

Since, for a perfect dielectric 0s =

Therefore, h r

r0e

m h em= = (1)

and b v cpr r

w w m e= = (2)

Comparing the equation (1) and (2) we get,

rm c0wh

bh= c m . .

5 10 3770 224 473 7 3 10

6

8

#

#

p=

^ ^

^ ^ ^

h h

h h h> H

.5 37=Again from equation (1)

re r0 2

hh m= b l . .37 .4473 7

377 8 42

#= =b l

SOL 7.2.7 Option (B) is correct.General equation of electric field intensity of a plane wave propagating in free space in ax- direction having amplitude E0 and frequency w is given as : E cosE t x an0 w b= +^ h

where b is phase constant of the wave and an is the unit vector in the direction of polarization of wave and since the EM wave is polarized in az+ direction. So, an az=

and we get, E cosE t c x az0 w w= +a k (in free space cb w= )

Therefore, the magnetic field intensity of the wave is given as

H a E1k

0#h= ^ ^h h

where ak is the unit vector in the direction of wave propagation and 0h is the intrinsic impedance of the wave in the medium.




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So, H cosE t c xa a1x z

00h w w

#= - +^ ah k9 C (ak ax=- )

cosE t c x ay0

0

h w w= +a k

SOL 7.2.8 Option (B) is correct.General equation of electric field intensity of a plane wave propagating in free space is given as : E cosE t k r an0 :w= -^ h (1)where an is unit vector in direction of polarization, k is the wave number in the

direction of wave propagation with amplitude k cb w= = , and x y zr a a ax y z= + + is the position vector.Since, the wave is propagating in the direction from origin to point , ,1 1 1^ h.

So, k ca a a

1 1 1

0x y z

2 2 2w=

+ +

+ + -a

^k

h c

a a a3

x y zw= + +c m

and since the field is polarized parallel to x -z plane

So, an m n

m na ax z2 2

=++ where m and n are constants.

Now, the electric field of wave is always perpendicular to the direction of propagation of EM wave. So, we have k an: 0=

c m nm na a a a a

3x y z x z

2 2:

w + +++

c m= ;G E 0=

m n+ 0= m n=-Therefore, the unit vector in the direction of polarization of the wave is

an m m

m ma ax z

2 2=

+ -

+ -

^

^

h

h a a

2x z= - (m n=- )

Putting all the values in equation (1), we get the electric field of the wave as

E cosE t c x y za a a a a a a a

3 2x y z

x y zx z

0 :w w= - + + + + -c ^ bm h l= G

cosE tcx y z a a

3 3x z

0 w w= - + + -^ bh l: D

SOL 7.2.9 Option (C) is correct.Skin depth ( )d of any medium is defined as the reciprocal of attenuation constant ( )a of a plane wave in the medium

i.e. d 1a=

The attenuation constant of the plane wave in the medium is given as

a 2 1 12w mewes= + -a k: D

Now, wes

.f22

2 50 10 80 8 85 102

r 03 12

# # # # #p e e p= = -



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For View Only Shop Online at www.nodia.co.in .8991 8 1>>=

i.e. wes 1>>

So, a 2wms= ( /s we 1>> )

22 50 10 4 4 10 23 7

# # # # # #p p=-

.0 4p=

Therefore, d . 0.796 m10 4

1a p= = =

SOL 7.2.10 Option (B) is correct.For the microwave experiment the angular frequency is w f2p= 2 10 109

# #p= ( 10 GHzf = ) 2 1010

#p=

So, wes

..

2 10 1 8 85 106 25 10

10 12

7

# # # #

#p

= -

.1 12 10 1>>8#=

Therefore, the skin depth of the material is

d 1a= 2

wms= ( / 1>>s we )

.2 10 1 4 10 6 25 10

210 7 7

# # # # # #p p= -

6.36 10 m7#= - 0.636 mm=

Thus, for the successful experiment, width of coating must be greater than skin depthi.e. t .0 636> t 0.64 m> m

SOL 7.2.11 Option (A) is correct.Given, the electric field intensity of the plane wave E 3 . 2 . /cos sin V mt y t ya a10 0 2 10 0 2x z

7 7= - + -^ ^h h

Comparing it with the general equation of electric field of a plane wave, we getAngular frequency, w 107=Phase constant, b .0 2=So, the phase velocity of the propagating wave is

vp . 5 10 /m s0 2107

7

bw

#= = =

or, cre 5 107

#=

where c is velocity of wave in air and re is the relative permittivity of the medium.

So, re 5 103 10

7

8 2

#

#= c m 36=

Therefore, permittivity of the medium is e r 0e e= 36 0e=Now, the complex permittivity of the medium is given as




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ce je e= -l m

where el 36 0e e= =

and em 10

2 10 27

7#

ws= = =

Thus, ce /F mj24 50e= -^ h

SOL 7.2.12 Option (B) is correct.Conductivity of all the metals are in the range of mega siemens per meter and frequency of the visible waves are in the range of 10 Hz15 . So, we can assume Conductivity of a metal 10 /S m6.

Frequency of a visible wave 10 Hz15.

Now, the attenuation constant of a wave in a certain medium is given as :

a 2 1 12w mewes= + -a k: D

Since for a metal, >>s we

So, a 2w mewes= 2

wms=

Therefore, the skin depth of a metal is

d 1 210 4 10 10

215 7 6# # #a wms p

= = = -

24 1014

#p= 10 1 nm2

1 7 .p #= -

Thus, the skin depth is in the range of nanometers for a metal and that’s why the wave (visible wave) can’t penetrate inside the metal and the metals are opaque.i.e. (A) and (R) both are true and (R) is the correct explanation of (A).

SOL 7.2.13 Option (D) is correct.Since, the wave is propagating in free space so, the velocity of the wave is 3 10 /m s8

# and the amplitude of magnetic field intensity in z 0= plane is given as

H0 E

0

0

h=

Therefore, the plot of magnetic field intensity H0 versus time t in z 0= plane is as shown in the figure below :



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For View Only Shop Online at www.nodia.co.inSince, the wave is propagating in az+ direction so, an amplitude which exists in the plane z 0= at any time t must exist in the plane z 3 10 mt1 10 6 8

# # #= --^ h at sect 1 m= .

So, the amplitude of H0 will be equal to the H1 at sect 1 m= for the plane z 3 10 mt106 8

# #= -^ h

Thus, the plot of H1 versus z will be as shown in figure below

SOL 7.2.14 Option (D) is correct.Velocity of the wave in free space is

c 3 10 /m s0

0 8

em

#= =

So, the velocity of the wave in dielectric 1 is

vP1 c

4 20

0

em= =

The velocity of wave in dielectric 2 is

vP2 c

9 30

0

em= =

The velocity of wave in dielectric 3 is

vP3 c

3 30

0

em= =

Therefore, the time t taken by the wave to strike the interface at 5 mx = is t t t t1 2 3= + +

/ /c c3 10

62

33

28

#= + +

. . . 100 04 0 06 0 08 6#= + + -

^ h 0.06 secm=

SOL 7.2.15 Option (D) is correct.Intrinsic impedance of 1st medium is

1h 1

0

em=

and the intrinsic impedance of 2nd medium is

2h 2

0

em=

So, the reflection coefficient at the interface of the two medium is given as

G 2 1

2 1

h hh h= +

-




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G

2

0

1

0

2

0

1

0

em

em

em

em

=+

-

or, 51

1 1

1 1

2 1

2 1

e e

e e=+

- (given 5

1G = )

51

1

1

1

2

1

2

eeee

=+

-

5 15 1-+

2

2

1

2ee= (by rationalisation)

46

2

1ee=

2

1ee 4

9=

SOL 7.2.16 Option (D) is correct.GivenFrequency of the propagating wave, , f 50 50 10MHz Hz6

#= =Skin depth of the dielectric medium, d 0.32 0.32 10mm m3

#= = -

Permittivity of dielectric, m .6 28 10 7#= -

So, the conductivity of the dielectric medium is given as

s f1

2p md=

. ( ) ( . ) .3 14 50 10 6 28 10 0 32 101

6 7 3 2# # # # # #

= - -^ ^h h

0.99 10 /S m5#=

SOL 7.2.17 Option (A) is correct.Frequency of the wave, f 10GHz Hz8 8 9

#= =Distance travelled by the wave, z 0.175 0.175 10mm m3

#= = -

Permittivity of dielectric, m .6 28 10 7#= -

and as calculated in previous question the conductivity of the dielectric medium is s 0.99 10 /S m5

#=So, the attenuation constant of the wave in the dielectric medium is a fp ms= . ( ) ( . ) ( . )3 14 8 10 6 28 10 0 99 109 7 5

# # # # # #= -^ h

3.95 10 /NP m4#=

Therefore, the reducing factor of the field intensity in dB after travelling distance z is 20 log e z

10a- log e20 . .

103 95 10 0 175 104 3

= # # #- -^ ^h h 0 dB4=-

SOL 7.2.18 Option (A) is correct.Given the electric field intensity in phasor form Es E j ea ay z

j x0= - b-^ h

So, the instantaneous expression of electric field intensity will be,



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For View Only Shop Online at www.nodia.co.in E Re E j ea ay z

j t x0= - w b-^

^h

h$ .

Re cos sinE a ja t j t ey zj x

0 w w= - + b-^ ^ ^h h h6 @" ,

cos sinE a t a t ey zj x

0 w w= + b-^ ^^ h hh

Therefore, the magnitude of the field is E cos sinE t E t0

20

2w w= +^ ^h h

or, E E12

22+ E0=

which is a circular equation i.e. the wave is circularly polarized.Now, the instantaneous angle q that the field E makes with y -axis is given as

tanq cossin

E tE t

0

0

ww=

or, q tw=Therefore as the time increases, E rotates from y to z as shown in figure below :

and since the direction of wave propagation is in ax+ direction so, the rotation from y to z obeys the right hand rule. Thus, we conclude that the field is Right hand circularly polarized.

SOL 7.2.19 Option (B) is correct.Given the phasor form of electric field intensity, Es j ea a4 z x

j y= - b-^ h

So, the electric field intensity of the reflected wave will be Ers 4( )a ja ez x

j yG= - b6 @

where G is the reflection coefficient at the interface. Therefore, Ers j ea a4 z x

j y= - + b^ h (for perfect conductor 1G =- )

and the instantaneous expression of the electric field of reflected wave will be E Re cos sinj t j t ea a4 z x

j yw w= - + + b^ ^h h" ,

cos sint t ea a4 z xj yw w= - - b

^ ^^ h h h

Therefore, the magnitude of the reflected field is E cos sint t4 42 2w w= +^ ^h h

or, E E12

22+ 4=

which is a circular equation i.e. the wave is circularly polarized.Now, the instantaneous angle q that E makes with z -axis is given as

tanq cossin

tt

44

ww= -

-




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q tw=So, as time increases, electric field E rotates from z to x as shown in the figure below :

Since the direction of wave propagation is along ay- , so, the rotation from z to x follows left hand rule. Thus, we conclude that the EM wave is LHC (left hand circularly) polarized.

SOL 7.2.20 Option (C) is correct.Given the electric field intensity of incident wave, Eis 10 eaz j y x6 8= - +^ h

So, the direction of wave propagation is K 6 8a ay x= +Since the wave is incident on the perfect conductor so, the magnitude of the reflected wave is given as, Er0 Ei0=- 10az=- ( 1G =- for perfect conductor)The direction of wave propagation of reflected wave will be along 6 8a ay x-^ h as shown in figure below :

Therefore, the field intensity of the reflected wave is Ers 10 eaz j y x6 8=- - -^ h

Thus, the net electric field intensity of the total wave in free space after reflection will be Es E Eis rs= + 10 10e ea az

j y xz

j y x6 8 6 8= + -- + - -^ ^h h6 @

10 e e eaz j y j x x6 8 8= -- - -^ h /sin V mj e xa10 4z

j y4=- -



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Given, the electric field intensity of the incident wave, Eis 25 /V meax j z y6 8= - +^ h

So, the direction of the wave propagation is k 6 8a az y= +Since the wave is incident on a perfect conductor so, the magnitude of the electric field of the reflected wave is Er0 Ei0=- (reflection coefficient, 1G =- ) 25ax=-The reflected wave will propagate in 6 8a az y- direction as shown in figure below:

So, we get the electric field intensity of reflected wave as Ers 25 /V meax j z y6 8=- - -^ h

Since, the magnetic field intensity of a plane wave in terms of electric field intensity is defined as

H a E1k

0#h= ^ h

where ak is unit vector in the direction of wave propagation and 0h is the intrinsic impedance of free space. So, the magnetic field intensity of the reflected wave is given as

Hrs a E1k rs

0#h= ^ h

where, ak 6 86 8

. .k a aa a

a ak 0 6 0 8z y

z yz y= = -

- = -^ h

So, we get Hrs 0.6 0.8 25 ea a a1201

z y xj z y6 8

#p= - - - -^ ^ ^h hh6 @

15 20 ea a1201

y zj z y6 8

p= - - - -^ ^h h6 @

/A mea a4 3y z j z y2 3

p p=- + - -a

^k

h

SOL 7.2.22 Option (D) is correct.Given, the magnetic field intensity of the EM wave propagating in free space, H 0. /cos A mt y a2 xw b= -^ h

So, the time average power density of the EM wave is given as




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Pave H a21

x02h=

where 0h is the intrinsic impedance in free space and H is the magnitude of magnetic field intensity in free space.

So, Pave . a21 120 0 1 x

2p= ^ ^h h ( , .H120 0 10h p= = )

0.6 axp=Therefore, the total power passing through the square plate of side 20 cm is given as

Ptotal dSPave:= # SaPave n:=where S is the area of the square plate given as S . 0.04 m0 2 2 2= =^ h (Side of square 0.2 m= )and an is the unit vector normal to the plate given as

i.e. an a a

2x y= +

So, Ptotal 0.6 .a a a0 04

2x

x y:p= +

^ ch m= G

0.0 331 Watt9= 53.31 mW=

SOL 7.2.23 Option (D) is correct.Given, the electric field intensity of the incident wave, Eis 5 /V me aj y z

5= -

So, we get the phase constant of the wave as 1b 5=

or, c r r1 1w m e 5= ( vp

b w= )

c 4 1w^ ^h h 5=

w c25=

Now, the intrinsic impedance of the lossless medium is given as

1h 2 2 7541

1

0

00e

mem h= = = =

and the intrinsic impedance of lossy medium is 2h 2h q= h2

where, the magnitude of the intrinsic impedance is given as

2h /

./

c1 1

25 4

0 14

/ /

2

2 2 1 42 2

0

1 40 0

wes

m e

e

m e=+

=+a

b ^fk

l h p:

>D

H

.15 1860

/1 4p=

^ h



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For View Only Shop Online at www.nodia.co.in .95 48=and the phase angle of the intrinsic impedance is

tan2 2qh .3 772

2wes= =

or 2qh .37 57c=So, the reflection coefficient of the wave is given as

G . .. .

95 48 37 57 75495 48 37 57 754

2 1

2 1

cc

h hh h= +- =

+-

. .0 1886 171 08c=Therefore, the standing wave ratio is

S 11

GG

= -+

..

1 0 81861 0 8186= -+

1 .0252=

SOL 7.2.24 Option (A) is correct.The general expression for phasor form of electric field vector is Es E e j x y z

0x y z= b b b- + +^ h

Comparing the given field with this expression we get, x y zx y zb b b+ + . 3x y z0 01 3 2p= - + -^ h

So, the propagation vector is k . 3 3x y z a a a0 01 3x y z x y zd b b b p= + + = - + -^ ^h h

Therefore, the direction of the propagation of the wave is

ak kk=

3 2a a a9 3 4

3x y z=+ +

- + -

a a a41 5 2 2x y z= - + -^ h

SOL 7.2.25 Option (C) is correct.As calculated in previous question we have the propagation vector from the given data as k . 2 3a a a0 04 3x y zp= - - +^ h

and the direction of wave propagation is

ak kk=

0.04 2 3 30.04

a a aa a a2 3 3

x y z

x y z

pp

=- - +- - +^

^

h

h

( ) ( ) ( )

a a a

2 3 3

2 3 3x y z

2 2 2=

- + - +

- - +^ h

2 3a a a

43x y z= - - +

Therefore, the phase constant along the direction of propagation is b k ak:=

0.04 2 3 3 42 3 3

a a aa a a

x y zx y z

:p= - - + - - +_ ci m9 C

.0 16p=




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So, the wavelength along the direction of wave propagation is

l .5 m2 25bp= =

SOL 7.2.26 Option (C) is correctFrom the given expression of magnetic field vector we get, x y zx y zb b b+ + . x y z0 04 3 2 3p= - -^ h

So, the propagation vector of the plane wave is k x y zx y zd b b b= + +^ h

. 2 3a a a0 04 3 x y zp= - -^ h

and the direction of wave propagation is

ak kk=

0.04 3 2 30.04 3 2 3

a a aa a a

x y z

x y z

pp

=- -- -

^

^

h

h

a a a

43 2 3x y z=

- -^ h

Therefore, the phase constant along the direction of wave propagation is b 0.16k ak: p= =Since the wave is propagating in free space so it’s phase velocity will be vp 3 10 /m s8

#=

or, bw 3 108

#=

So, the frequency of the plane wave is

f .

2.4 10 Hz23 10 0 168

7#p

p#= =^ ^h h

( f2w p= )

MHz12=

SOL 7.2.27 Option (D) is correct.From the given expression of the field vector, we have the propagation vector,

k 2 3a a a25 3 x y zp= - -^ h

So the phase constants along x , y and z -axes are

xb 253 p= ; 25

2yb p=- ; 25

3zb p=-

Therefore, the apparent wave lengths along the three axes are

xl 28.87 m2

253

23

50xbp

pp= = = =

c m

yl 25 m2

252

2ybp

pp= = - =+

b l

zl 6.7 m2

253

2350 2

zbp

pp= = - =+ =

b l

SOL 7.2.28 Option (A) is correct.As determined in previous question, the propagation vector of the plane wave is



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For View Only Shop Online at www.nodia.co.in k 2 3a a a25 3 x y z

p= - -^ h

Therefore, the direction of wave propagation is

ak kk=

2 3a a a4

3 x y z= - -

So the phase constant along the direction of wave propagation is b 0.16k ak: p= =Therefore, the angular frequency of the propagating wave is w vpb= .3 10 0 168

# # p= ^ ^h h (In free space 3 10 /m svp 8#= )

1.51 10 / secrad8#=

So, for the determined values of apparent phase constants in previous question, the apparent phase velocities are given as

vpxxbw= . 6.93 10 /m s

253

1 51 1088#

p#= =

c m

vpyybw= . 6 10 /m s

252

1 51 1088#

p #=-

=

and vpzzbw= . 4 10 /m s

253

1 51 1088#

p #=-

=

SOL 7.2.29 Option (D) is correct.The necessary condition for the vector field eEE j

0= b- to represent the electric field intensity of a uniform plane wave is k E0: 0=where k is the propagation vector of the wave and E0 is the amplitude of the electric field intensity of the plane wave. Now, we check all the given options for this condition.(A) From given data we have k a a3 y z= + E0 2j ja a a3x y z=- - +So, k E0: j2 3 3 0!=- +(B) From given data we have E0 2ja a a3x y z= - - k a a3x z= +So, k E0: 1 3 0!= -(C) From given data we have

E0 j j ja a a3 21 1 2

3 3x y z= + + + -b cl m

k 3 2a a a3 x y z= + +

So, k E0: j j j3 23 3 2

3 3 2 3 0!= + + + -

(D) From given data we have




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E0 j j ja a a3 21 1 2

3 3x y z= - - + - +b cl m

k 3 2a a a3 x y z= + +

So, k E0: 3j j j3 2 3 23 3 2 3 0=- - + - + =

So the vector represents electric field vector of a uniform plane wave.

SOL 7.2.30 Option (D) is correct.For the field vectors Es and Hs defined as Es eE j

0= b-

and Hs eH j0= b-

The condition that it represents the field vectors of a uniform plane wave is 0E H0 0: = , 0kE0 : = and 0kH0 : =where k is the propagation vector of the plane wave.Now, we check the all given pairs for this conditionIn Option (D) E0 2j ja a a3x y z=- - + H0 2ja a a3x y z= - -and k a a3 x z= +So E H0 0: j j j4 3 0=- + - = kE0 : j j3 3 0=- + = kH0 : 3 3 0= - =Therefore, it represents the field vectors of a uniform plane wave.

SOL 7.2.31 Option (B) is correct.For a propagating electromagnetic wave, the field satisfies the following Maxweell’s equation. E:d 0=

E#d tB 022 !=-

Now, we check the condition for the given fields as below. P 60 sin t x a10 zw= +^ h

So, P:d 0=and P#d 600 0cos t x a10 y !w=- +^ h

i.e. P is a possible EM field.

again, Q cos t a10 2r w r= - f^ h

So, Q:d 0=

and Q#d 0cos t a1 10 2 z22 !r r w r= -^ h6 @

i.e. Q is a possible EM field

R 3 cot cosa a12r f r f= +r f

So, R:d cot sin1 3 02

22 !r r r f r

f= ^ h



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For View Only Shop Online at www.nodia.co.ini.e. R is not a possible EM field.

S sin sinr t r a1 6q w= - f^ h

So, S:d sin

sinsin

rt r

r1 6 02

2

2

2!

qw

f= -^

^h

h

i.e. S is not an EM field.Thus, the possible EM fields are P and Q .

SOL 7.2.32 Option ( ) is correct.Since, %20 of the energy in the incident wave is reflected at the boundary. So, we have,

2G 10020=

or, G . .0 2 0 447!= =Where G is the reflection coefficient at the medium interface. Therefore, we have

2 1

2 1

h hh h+- .0 447!=

r

r

r

r

r

r

r

r

02

20

1

1

02

20

1

1

h em h e

mh e

m h em

+

- 0.447!=

r

r

r

r

r

r

r

r

232

131

232

131

mm

mm

mm

mm

+

- .0 447!= ,r r r r1 1

32 2

3e m e m= =^ h

r r

r r

1 2

1 2

m mm m

+- .0 447!=

r

r

2

1

mm .

.1 0 4471 0 447"!=

r

r

2

1

mm .2 62= or 0.38

So, r

r

1

2

ee .0 056

r

r

1

2 3

mm= =b l or .17 9

***********




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SOLUTIONS 7.3

SOL 7.3.1 Option (C) is correct.Electric field of the propagating wave in free space is given as Ei (8 6 5 ) /V mea a a ( )

x y zj t x y3 4= + + w + -

So, it is clear that wave is propagating in the direction ( 3 4 )a ax y- + .Since, the wave is incident on a perfectly conducting slab at x 0= . So, the reflection coefficient will be equal to 1- .i.e. Er 0 ( 1)Ei0= - 8 6 5a a ax y z=- - -Again, the reflected wave will be as shown in figure below :

i.e. the reflected wave will be along the direction a a3 4x y+ . Thus, the electric field of the reflected wave will be Er ( 8 6 5 ) /V mea a a ( )

x y zj t x y3 4= - - - w - -

SOL 7.3.2 Option (B) is correct.Given, the electric field intensity of the EM wave as E 10( )j ea ay z

j x25= + -

So, we conclude that the wave is propagating in ax direction and the y and z-components of the field are same. Therefore, the wave is circularly polarized.Now, the angle formed by the electric field with the z -axis is given as q tw=So, with increase in time the tip of the field magnitude rotates from z to y -axis and as the wave is propagating in ax direction so, we conclude that the wave is left circular (i.e., left circular polarization).The phase constant of the field is given as

b cw=

25 cf2p= ( 25b = )



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f .c

225

2 3 1425 3 108

### #

p= = 1.2 GHz=

SOL 7.3.3 Option (C) is correct.Power radiated from any source is constant.

SOL 7.3.4 Option (C) is correct.Intrinsic impedance of EM wave

h 4 0

0

em

em= = 02

120 3p p= =

Time average power density of the EM wave is given as

Pave EH E21

21 2

h= = 2 601

1201

# p p= = ( 1 /V mE = )

SOL 7.3.5 Option (B) is correct.In the given problem

Reflection coefficient at the medium interface is given as

G 2 1

2 1

h hh h= +- 40 120

400 12021

p pp p= +- =-

As, given the electric field component of the incident wave is Ei 24 cos y a3 10 x

8# b= -^ h

So, we conclude that the incident wave is propagating along ay direction and the angular frequency of the wave is w 3 10 /rad s8

#=So, the phase constant of the wave is given as

b c 3 103 10 18

8

#

#w= = =

Therefore, the reflected wave will be propagating in ay- direction and its electric field component is given as Er (3 10 )cos yEi0 8G #= + ( 1 /rad mb = )where Ei0 is the maximum value of the field component of incident wave.i.e. Ei0 24ax=So, we have

Ei0 cos y a21 24 3 10 x

8#=- +^ h8 B

12 cos y a3 10 x8

#=- +^ h




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Therefore, the magnetic field component of the reflected wave is given as

Hr ( )a E1k rh #=

where 1h is the intrinsic impedance of medium 1, and ak is the unit vector in the direction of wave propagation. So, we get

Hr cos ya a1201 12 3 10y x

8# #p= - +^_ h i8 B

cos y a51 3 10 z

8#p= +^ h

SOL 7.3.6 Option (C) is correct.The intrinsic impedance of the wave is defined as

h em=

where m is permeability and e is permittivity of the medium.Now, the reflection coefficient at the medium interface is given as

G 2 1

2 1

h hh h=

+-

Substituting values for 1h and 2h we have

t o r

o

o

o

o r

o

o

o

=+-

e em

em

e em

em

11

1 91 9

r

r

ee=

+- =

+- ( 9re = )

.0 5=-or, G .0 5=

SOL 7.3.7 Option (B) is correct.Since, the wave is propagating in a direction making an angle 90c with positive y-axis. So, the y -component of propagation constant will be zero. As the direction of propagation makes an angle 30c with positive x -axis so, we have the propagation constant of the wave as g 30 30cos sinx y!c cb b=where b is the phase constant of the wave. So, we get

g x y223 2

21

!lp

lp= x y3

!lp

lp=

Now, in all the given options the direction of electric field of the wave is given along ay . So, considering that direction we get the field intensity of the wave as

E E ea ( )y

j t0= w g- E eax j t x y

02

= !w lp

lp-c m= G

SOL 7.3.8 Option (D) is correct.Since, the given field intensity have components in ax and ay direction so, the magnitude of the field intensity of the plane wave is

H 2 H Hx y2 2= + 5 3 5 10

0

2

0

2

0

2

h h h= + =c b bm l l

So, the time average power density of the EM wave is given as

Pave H2

02h

= 210 500

0

2

0

hh h= =b l watts



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The Brewster angle is given as

tan nq r

r

1

2

ee=

60tan c 1r2e=

or r2e 3=

SOL 7.3.10 Option (C) is correct.Given, the electric field intensity of the propagating wave E ( ) ( /2)sin sint z t za ax yw b w b p= - + - +So, we conclude that the wave is propagating along az direction and the field components along ax and ay are equal.i.e. Ex Ey=Therefore, the wave is circularly polarized. Now we will determine the field is either right circular or left circular. The angle between the electric field E and x -axis is given as

q tan sincos

tt1

ww= -

a k t2p w= -

So, with increase in time the tip of the field intensity moves from y to x -axis and as the wave is propagating in az direction therefore, the wave is left hand circularly polarized.

SOL 7.3.11 Option (B) is correct.The reflection coefficient at the medium interface is given as

G 2 1

2 1

o r

o

o

o

o r

o

o

o

h hh h= +- =

+-

e em

em

e em

em

11

1 41 4

31

r

r

ee=

++ =

+- =-

So, the transmitted power is Pt (1 )Pi2G= -

Pt 191 P P

98

i i= - =` j

or, PPi

t 98=


sinq 12

1re

= =

or q 45 4c p= =

The configuration is shown below. Here A is point source.




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Now AO 1= mFrom geometry BO 1= mThus, area r OB2p p p#= = = m2

SOL 7.3.13 Option (C) is correct.Given, the electric field of the EM wave in medium 1 as E1 4 3 5a a ax y z= + +As the medium interface lies in the plane x 0= so, the tangential and normal components of the electric field are E t1 3 5a ay z= +and E n1 4ax=Now, from the boundary condition we know that the tangential component of electric field is uniform. So, we get E t2 3 5E a at y z1= = +Again from the boundary condition the normal component of displacement vector are equal.i.e. D n2 D n1=or E n2 2e E n1 1e=or 4 Eo n2e 3 4ao ze=or E n2 3ax=Thus, the net electric field intensity in medium 2 is E2 E Et n2 2= + 3 3 5a a ax y z= + +

SOL 7.3.14 Option (C) is correct.From the expression of the magnetic field intensity of the EM wave, we haveAngular frequency, w ,50 000=Phase constant, b 0.004=So, the phase constant of the wave is given as

vp 4 105 10

3

4

#

#bw= = - 1.25 10 /m s7

#=

SOL 7.3.15 Option (C) is correct.Refractive index of glass 1.5ng r rm e= =Frequency f 10 Hz14= c 3 108#= m/secThe wavelength of the 10 Hz14 beam of light is

l fc

103 10

14

8#= = 3 10 6

#= -

So, wavelength of the light beam in glass is given as

gl .n 1 53 10

g

6#l= =

- 2 10 m6

#= -

SOL 7.3.16 Option (B) is correct.The time average poynting vector of the EM wave is defined as

Pave Re E H21 *

s s#= 6 @



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For View Only Shop Online at www.nodia.co.inwhere, Es is the phasor form of the electric field intensity and Hs) is the complex conjugate of the phsor form of magnetic field intensity. So, we have

E H *s s# ( ) ( )j e k j ea a a ax y

jkz j tx y

jkz j t

wm#= + - +w w- - +

( )( ) 0k j j kaz wm wm= - - =; E

Thus, Pave Re E H21 *

s s#= 6 @ 0=


We have VSWR 5EE

11

min

max

GG

= = =+-

or G 32=

As the wave is normally incident on the interface so, the reflection coefficient will be real (either positive or negative). Now, for a wave propagating from medium 1 to medium 2 having permittivities 1e and 2e respectively.(i) If >2 1e e , the reflection coefficient is negative

(ii) If <2 1e e then, the reflection coefficient is positive.

Since, the given EM wave is propagating from free space to the dielectric material with > 0e e , therefore

G 32=-

or, 2 1

2 1

h hh h+- 3

2=-

or, 120120

2

2

h ph p+- 3

2=-

So, 2h 24p=

SOL 7.3.18 Option (A) is correct.The skin depth ( )d of a material is related to the operating frequency ( )f as

d f

1\

Therefore, 1

2

dd f

f2

1=

252d 4

1=

or 2d 2541 25#= = cm

SOL 7.3.19 Option (D) is correct.The intrinsic impedance of a medium with permittivity e and permeability m is defined as

h em=

So, the reflection coefficient at the boundary interface of the two mediums is given as




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G 2 1

2 1

h hh h= +-

o r

o

o

o

o r

o

o

0

=+-

e em

em

e em

em

11

1 41 4

r

r

ee=

+- =

+- since 4re =

.33331 2 180c= - =

SOL 7.3.20 Option (A) is correct.We have ( , )E z t 10 (2 10 0.1 )cos t z7p p#= -So, we get w 2 10 t7p#= b .0 1p=Therefore, the phase velocity of the wave is given as

vp . 2 100 12 107

8#bw

pp

#= = = m/s

SOL 7.3.21 Option (C) is correct.We have E (0.5 )e ea a ( )

x yj j t kz2= + w -p

So, its components along x and y -axis are Ex 0.5e ( )j t kz= w -

and Ey e e ( )j j t kz2=p w -

i.e. Ex Ey!

Since, the components are not equal and have the phase difference of /2p so, we conclude that the EM wave is elliptically polarized.

SOL 7.3.22 Option (B) is correct.Loss tangent of a medium is defined as

tan d wes=

where s is the conductivity e is permittivity of the medium and w is operating angular frequency. So, we get

tan d .2 3 10 78

1 7 10o

9

4

# # #

#p e

=-

( f2w p= )

.3 10 39

1 7 10 9 109

4 9

# #

# # #=-

.3 102 5#= -


We have ZEx

2

2

22 c

tEx22

2

22=

As the field component Ex changes with z so, we conclude that the EM wave is propagating in z - direction.

SOL 7.3.24 Option (B) is correct.The required condition is Ic Id=i.e. the conduction current equals to the displacement current. So, we get Jc Jd=



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For View Only Shop Online at www.nodia.co.in Es j Ewe=or, s f2 o rp e e= ( 2 fw p= , r 0e e e= )

or, f 2 42

o r o r#p e es

pe es= =

49 10 2 109 2# # #=

-

45 10 906#= = MHz

SOL 7.3.25 Option (A) is correct.VSWR (voltage standing wave ratio) of the transmission line is defined as

S 11

GG

= -+

where G is the reflection coefficient of the transmission line. So, we get

3 11

GG

= -+

(VSWR 3= )

or G .0 5=Therefore, the ratio of the reflected power strength to the incident power is given as

PPi

r 0.252G= =

Thus, 25% of incident power is reflected.

SOL 7.3.26 Option (C) is correct.The fig is as shown below :

As per snell law

sinsin

i

t

qq 1

re=

or sinsin

4530cc 1

re=

2

121

1re

=

or re 2=

SOL 7.3.27 Option (A) is correct.Since, the phase constant is defined as

b 2lp w me= =

So, the wavelength in terms of permittivity of the medium can be given as

l 2w me

p=




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or, l 1\

eSo, we get

2

1

ll

2

1ee=

SOL 7.3.28 Option (C) is correct.A scalar wave equation must satisfy following relation

tE v

zE

p2

22

2

2

22

22- 0= ...(1)

where vp bw= (Phase velocity of the wave)

Basically w is the multiply factor of frequency, f and b is multiply factor of or orz x y .

So, we can conclude that expression given in option (C) does not satisfy equation (1) (i.e. the wave equation).

SOL 7.3.29 Option (D) is correct.In a lossless dielectric ( )0s = medium, impedance is given by

h em=

where m is permeability and e is permittivity of the medium. So, we get

h r

r

0

0

e em m= 120

r

rp em

#=

120 82

#p= 88.42 W=

SOL 7.3.30 Option (D) is correct.Intrinsic impedance of a medium is given as

h jj

s wewm= +

Since, copper is good conductor i.e. >>s we so, we get

h j 45cswm

swm= =

Thus, the impedance will be complex with an inductive component.

SOL 7.3.31 Option (B) is correct.Given, the electric field intensity of the EM wave as E 24 /V me a( )j t z

y= w b+

Now, the time average poynting vector for the EM wave is defined as

P E H21

s s#= )^ h

Ea2

sk

2

h= H Es

s

h=c m

where h is the intrinsic impedance of the medium and ak is the direction of wave propagation. Since, from the given expression of the field intensity we conclude that the wave is propagating along az- So, we have

P ( )

( ) .a a2 12024 2 4

z z

2

# p p= - =- (a ak z=- , 24 /V mE = )



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Given the propagation constant of the wave g 0.1 0.2j ja b p p= + = +So, we get b 0.2p=

or, 2lp 0.2p=

Therefore, wavelength of the propagating wave is

l . m0 22 5= =

SOL 7.3.33 Option (C) is correct.The depth of penetration or skin depth is defined as

d f1p ms

=

i.e. d f

1\

or, d \ l ( /c fl = )So, the depth of penetration (skin depth) increases with increase in wavelength.

SOL 7.3.34 Option (B) is correct.Given, the electric field intensity of the wave ( , )z tE E e ea a( ) ( )

oj t z

xj t z

y0e= +w b w b+ + ...(1)Generalizing ( )zE ( ) ( )E z E za ax y1 2= + ...(2)Comparing (1) and (2) we can see that ( ) ( )andE z E z1 2 are in space quadrature but in time phase so, their sum E will be linearly polarized along a line that makes an angle f with x -axis as shown below.

SOL 7.3.35 Option (B) is correct.Skin depth of the conducting medium at frequency, 10 MHzf1 = is given as

d f11p ms

=

or 10 2- 10 10

16p ms# # #

( 10 MHzf1 = )

or, ms 10 3

p=-

Now, phase velocity at another frequency ( f 2 1000 MHz= ) is

vp f4 2

msp=

Putting ms 10 /3 p= - in the above expression, we get

vp 10 / secm10

4 1000 10 33

66# # # # -p p

#= -

SOL 7.3.36 Option (C) is correct.Reflected power Pr of a plane wave in terms of incident power Pi is defined as Pr Pi2G= (1)where, G is the reflection coefficient at the medium interface given as




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G 2 1

2 1

h hh h= +- (2)

where 1h and 2h are the intrinsic impedance of the two mediums (air and glass) respectively. Since, the refractive index of the glass is .1 5i.e. n2 1.5c 2 2m e= = (3)where 2m 0m= (Permeability of glass) 2e r 0e e= (Permittivity of glass)So, putting these values in equation (3) we get re .1 5=

and 2h .1 52

2 0

r

0

em

eh h= = =

Therefore, from equation (2) we have

G .

.

1 5 0

1 5 00

0

hh

=+-

h

h

1 1.51 1.5

51= +

- =- (for free space 1 0h h= )

Thus, from equation (1) the reflected power is given as

Pr 51 Pi

2

#= b l

or, PPi

r %4=

SOL 7.3.37 Option (B) is correct.Skin depth of a material is defined as

d f1p ms

=

Putting the given values in the expression, we get

d .3 14 1 10 4 10 10

19 7 6

# # # # #p=

- 15.9 mm=

SOL 7.3.38 Option (C) is correct.The energy density in a medium having electric field intensity E is defined as

wE E21 2e= where e is permittivity of the medium.

So, due to the field 100 /V mE p= in free space, the energy density is

wE .21 8 85 10 10012 2

# p= -^ ^h h

1.39 10 /J m7 3#= - 1 9 /nJ m8 3=

SOL 7.3.39 Option (C) is correct.For a uniform plane wave propagating in free space, the fields E and H are every where normal to the direction of wave propagation ak and their direction are related as a ak E# aH=i.e. the angel between electric field aE^ h and magnetic field vector aH^ h is always 90c.

SOL 7.3.40 Option (A) is correct.The incidence angle of an EM wave for which there is no reflection is called



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For View Only Shop Online at www.nodia.co.inBrewster’s angle. For the vertically polarized wave (parallel polarized wave) the Brewster angle is defined as

tan ||Bq 1

2ee=

So, for the given dielectric medium we get

tan ||Bq 49=

or, ||Bq tan 231= -

b l

SOL 7.3.41 Option (A) is correct.Given, the electric field component of the EM wave propagating in free space, E 10 /cos V mt kz a10 y

7= +^ h

The general equation of electric field component of an EM wave propagating in az direction is given as E /cos V mE t kz ay0 w= +^ h

So, we conclude that the EM wave is propagating in az direction. w 10 /rad s7=or f2p 107=

f 2107

p=

So, l fc

103 10 27

8#

# p= = 188.5 m=

i.e. wavelength of the wave iswave amplitude, E0 10 /V m=

wave number, k 2602

301

lp

pp= = =

0. 33 /rad m2=The wave doesn’t attenuate as it travels. So, statement (2) and (3) are correct.

SOL 7.3.42 Option (A) is correct.The incidence angle of a plane wave for which there is no reflection is called Brewster’s angle. For the parallel polarized wave, Brewster’s angle is given as

tan ||Bq 1

2ee=

where 1e and 2e are the permittivity of two mediums respectively.So, for the given parallel polarized plane wave the incidence angle iq^ h for no reflection is given as

tan iq 81 0

0

ee=

or, iq tan 911= -

b l

Therefore, the angle a for no reflection is a 90 ic q= - .83 66c=




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SOL 7.3.43 Option (C) is correct.Given, the characteristic impedance of air, h 360W= Ex 3 /sin V mt zw b= -^ h

Ey 6 /sin V mt z 75cw b= - +^ h

So, the time average power per unit area is

Pave E

21 2

h= E E

21

360x y

2 2

#=+_ i

21

3603 62 2

#=+^ h

6.25 10 /W m2 2#= - 2.5 /mW m8 2=

SOL 7.3.44 Option (B) is correct.Operating frequency f 3 3 10GHz Hz9

#= =Medium parameters, m 4 10 /H m7p#= -

e /10 369 p= -

s 5.8 10 /S m7#=

So, we have intrinsic impedance defined as

h /

1/2 1 4

wesm e=

+ a k: D

10.

/

12 3 10 36

5 8 10

10 364 10

9

/

9

7 2 1 4

9

7

# # #

#

#

p p

pp

=+ -

-

-

f

^

p

h

> H

2.02 10 2 W#= -

The phase angle of intrinsic impedance is given as

qh tan21 1

wes= -

a k .tan2

1

2 3 10 3610

5 8 101

99

7

#

# # #

#

p p

= --f p

21

2 4#p p= =

So, h ej nh= q 0. 2e2 /j 4 W= p

SOL 7.3.45 Option (C) is correct.The Skin depth of a conductor is defined as

d f1p ms

=

So, statement 2 and 3 are correct while are incorrect.

SOL 7.3.46 Option (C) is correct.For circular polarization the two orthogonal field components must have the same magnitude and has a phase difference of 90c.So, all the three statements are necessary conditions.

SOL 7.3.47 Option (B) is correct.Velocity of light in any dielectric medium is defined as

v 1me

= 1r0 0m e e

= cre

=



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For View Only Shop Online at www.nodia.co.inwhere c is velocity of light in vacuum and re is dielectric constant of the medium.Since re 1>So, v c<Therefore, both A and R are true and R is correct explanation of A.

SOL 7.3.48 Option (D) is correct.Given, the electric field of a plane wave, E 50 /sin V mt z a10 2 y

8= +^ h

Comparing it with the general expression electric field of a plane wave travelling in az direction givenas E sinE t z ay0 w b= -^ h

We get the direction of propagation of the given plane wave is az- .

SOL 7.3.49 Option (D) is correct.The poynting vector is the instantaneous power flow per unit area in an EM wave and defined as P E H#=So, E H# is rate of energy flow (power flow) per unit area.

SOL 7.3.50 Option (C) is correct.Given the electric field, E j eaax y

j z= + b-^ h

So, it is clear that y -component of field leads the x -component by 90c and the wave propagates in z -direction. The components are same. So, the tip of electric field traverse in circular path in the clockwise direction and wave propagates in z-direction as shown in figure.

Therefore, it is negative circularly polarized wave or (left hand polarized wave).

SOL 7.3.51 Option (D) is correct.Consider the reflector is of angle 90cq = for which the incident and reflected wave is shown in figure.




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the x -axis i.e. reflected wave in same direction.

SOL 7.3.52 Option (C) is correct.Poynting vector represents the instantaneous power density vector associated with the EM field at a given point.i.e. P E H#=

SOL 7.3.53 Option (A) is correct.Given, the electric field intensity of the wave in free space, E 50 /sin V mt kz a10 y

7= +^ h

Comparing it with the general expression of electric field defined as E /sin V mE t z ay0 w b= -^ h

We get,(1) The wave propagates in az- direction along z -axis.

(2) The wavelength is given as

l 188.5 mfc

103 10 2

7

8# # p= = =

^ h

(3) Wave number, k . 0. 332188 52 2l

p p= = =

(4) The wave doesn’t attenuate as it travels.

SOL 7.3.54 Option (A) is correct.An electromagnetic wave incident on a conducting medium has the depth of penetration (skin depth) defined as

d 1a=

i.e. inversely proportion to attenuation constant.

SOL 7.3.55 Option (B) is correct.Since, after reflection the phase of both x and y components will be reversed so the reflected wave will be also right circularly polarised.

SOL 7.3.56 Option (C) is correct.Given,Electric field intensity of the wave E 10 cos t bx a6 10 y

8#p= -^ h



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For View Only Shop Online at www.nodia.co.inPermeability of medium, m 0m=Permittivity of medium, e 81 0e=From the expression of the electric field, we get the angular frequency as w 6 108

#p=The phase velocity of the wave is given as

vp 1me

= 81

10 0#m e

= 93 10

3108 8

#= = ( 3 10 /m sc 10 0

8

m e#= = )

So, the phase constant of the EM wave is

b /v 10 3

6 10p

8

8#w p= =

18 /rad mp=

SOL 7.3.57 Option (B) is correct.Given, the phase velocity of the plane wave in dielectric is 0.4 times its value in free spacei.e. vp 0.4c= (1)Since, the phase velocity of a medium having permittivity e and permeability m is defined as

vp 1me

=

So, putting it in equation (1) we get

1r0 0m e e

0.4c= ( ,r0 0m m m e e= = )

re . .0 41 6 25

2= =b l (c 1

0 0m e= )

SOL 7.3.58 Option (A) is correct.Given, the electric field in free space, ,x tE^ h 60 /cos V mt x a2 yw= -^ h

So, we get the magnitude of the electric field as E0 60=The time average power density in the electric field is given as

Pave 21

0

02

hE= 2

112060 2

# p= ^ h

Therefore, the average power through the circular area of radius 4 m is Pave rPave

2# p= ^ ^h h

21

12060

42

2# #p p= ^

^h

h 0 W12=

SOL 7.3.59 Option (C) is correct.The gyro frequency is the frequency whose period is equal to the period of revolution of an electron in its circular orbit under the influence of earth’s magnetic field. So, the radio wave at frequency near fg is attenuated by the earth’s magnetic field. (Since, there is a resonance phenomena and oscillating electron receive more and more energy from incident wave.)




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SOL 7.3.60 Option (C) is correct.The relation between electric and magnetic field of the reflected, transmitted and incident wave is given below. Ei Hi1h= Er Hr1h=- Et Ht2h=So, (1) and (3) are correct while (2) is incorrect.

SOL 7.3.61 Option (D) is correct.From snell’s law, sinn1 1q sinn2 2q= sin0 1 1m e q sin0 0 2m e q= sin2 600 2 cm e^ h sin0 2 2m e q=

sin 2q . 14 23 2 5 >#= =c m

which is not possible so there will be no transmitted wave.

SOL 7.3.62 Option (C) is correct.(1) Consider E1 is x -component and E2 is y -component so, when E1 and E2 will

be in same phase. The wave will be linearly polarized. 1a "^ h

(2) When E1 and E2 will have any arbitrary phase difference then it will be elliptically polarized. 2d "^ h

(3) When E1 leads E2 by 90c then tw increases counter clockwise and so the wave is right circularly polarized. 3c "^ h

(4) When E1 lags E2 by 90c then the tip of field vector E will traverse circularly in clockwise direction and left circularly polarized. 4b "^ h

SOL 7.3.63 Option (D) is correct.An incident wave normal to a perfect conductor is completely reflected in the reverse direction. The magnetic field intensity of reflected wave is same as the incident wave whereas the electric field intensity of reflected wave has the 180c phase difference in comparison to the incident field. ( 1G =- for conducting surface).

SOL 7.3.64 Option (B) is correct.(a) Propagation constant for a perfect conductor is

g ja b= +

where 2a b wms= = 1a "

(b) Radiation intensity of an antenna is defined as

,U q f^ h r Pave2=

rE r E2 2

22 2

2

h h= = c m 2b "

(c) Wave impedance of an EM wave is defined as



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For View Only Shop Online at www.nodia.co.in Eq Hh= f

h HE=

f

q 3c "

SOL 7.3.65 Option (B) is correct.Given,Electric field intensity, E E Ea ax x y y= +The direction of wave propagation, ak az=So, the magnetic field intensity of the EM wave is given as

H a Ek

h #=

where, h is the intrinsic impedance of the medium. Putting the expression for electric field in equation, we get

H E Ea a azx x y y#h= +^ h E Ea a1

x y y xh= -^ h

SOL 7.3.66 Option (B) is correct.An EM wave propagating in free space consists of electric and magnetic field intensity both perpendicular to direction of propagation.

SOL 7.3.67 Option (B) is correct.In a uniform plane wave the field intensities are related as E Hh=where h is intrinsic impedance given as

h jj

s wewm= +

Assume the medium is perfectly dielectric 0s =^ h. So, we get

h em=

or, HE e

m=

SOL 7.3.68 Option (A) is correct.The higher frequency (microwave) signal is continuously refracted on the ground as shown in figure.

This phenomenon is called ducting.

SOL 7.3.69 Option (D) is correct.Given, the magnetic field intensity of a plane wave, H 0.5 cose t x a10 2. x

z0 1 6= --

^ h (1)




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The general expression for magnetic field intensity of a plane wave travelling in positive x -direction is H cosH e t x ax

z0 w b= -a-^ h (2)

Comparing the equation (1) and (2) we get,Wave frequency, w 10 / secrad6=

Wavelength, l 3.14 m222

bp p= = =

and the wave travels in x+ -direction.Since, the magnetic field intensity points toward az direction and the wave propagates in ax+ direction. So, direction of electric field intensity will be aE a a a a ak H x z y##=- =- =^ hTherefore, the wave is polarized in ay direction (direction of electric field intensity).

SOL 7.3.70 Option (C) is correct.Skin depth d^ h is the distance through which the wave amplitude decreases to a factor e 1- or /e1 .

SOL 7.3.71 Option (D) is correct.From Maxwell’s equation, For a varying magnetic field B , the electric field intensity E is defined as

E#d tB22=-

Since, the magnetic flux density B in terms of magnetic vector potential is given as B A#d=So, from the two equations we have

E tA

22=- (For V 0d = )

Given, A sinA t zax x w b= -^ h

So, E sintA t zax x2

2 w b=- -^ h6 @ cosA t zax xw w b=- -^ h

SOL 7.3.72 Option (C) is correct.The depth of penetration of wave (skin depth) in a lossy dielectric (conductor) is given as

d f

1 1a p ms

= =

So, the skin depth increases when(1) permeability decreases

(2) conductivity decreases

(3) frequency decreases

Since, the wavelength of the wave is given as

l fvp= i.e. l f

1\

So, as l increases, f decreases and therefore, skin depth increases.



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For a good conductor, a fb p ms= =Since, the skin depth is defined as

d 1a=

or, d 1b= a b=^ h

Now, the phase constant of the wave is given as

b 2lp=

So, we have d 12b pl= = It is defined for a good conductor.

SOL 7.3.74 Option (B) is correct.Given, the magnetic field intensity of the wave propagating in free space,

,z tH^ h cos t z a61

yp w b=- +^ h

So, we conclude.direction of propagation, ak az=-direction of magnetic field, aH ay=So, the direction of electric field intensity is given as aE a aH k#= a ay z#= -^ h ax=-and the electric field amplitude is given as, E H0h=

cos t z120 61p p w b= - +^ ^bh hl

cos t z20 w b=- +^ h

So, the electric field vector of EM wave is ,z tE^ h 0 cos t z a1 xw b= +^ h

SOL 7.3.75 Option (B) is correct.Given, the electric field intensity of EM wave in phase form as Es 10e aj y x

4= -

So, we getthe phase constant, b 4 /rad m=Since, the wave is propagating in free space, therefore, the angular frequency w of the wave is given as w cb= 3 10 48

#= ^ ^h h 4 3 10 /rad s8# #=

SOL 7.3.76 Option (B) is correct.A and R both true and R is correct explanation of A.

SOL 7.3.77 Option (A) is correct.Skin depth of a material is defined as

d f1p ms

=




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Since, conductivity of the material is 0s = .So, we get d " infinity

SOL 7.3.78 Option (A) is correct.(1) In a conducting medium as the wave travels its amplitude is attenuated by the

factor e za- (i.e. attenuated exponentially).

(2) Conducting medium doesn’t behave as on open circuit to the EM field.

(3) In lossless dielectric 0s =^ h relaxation time is defined as

Tr " 3se=

(4) In charge free region 0vr =^ h. Poissions equation is generalised as

V2d v

er=-

V2d 0=which is Laplace equation. Therefore only statement 2 is incorrect.

SOL 7.3.79 Option (A) is correct.For a given electric field in free space the average power density is defined as

Pave E

21

0

2

h= 21

12060 2

pp

= ^ h 15 /Watt m2p=

SOL 7.3.80 Option (A) is correct.Given, E 120 cos t z axp w b= -^ h

Since, the wave is propagating in az directions so, the magnetic flux density of the propagating wave is

H 120 cos t za E a ak z x

0 0

#h h

p w b#= =-^ h6 @

cos t z ayw b= -^ h (a ak z= )

Therefore, the average power density of an EM wave is defined as

Pave Re E H21 *

#= " , 120 cos cost z t za a21

x y#p w b w b= - -^^ ^^h h h h8 B

60 azp=

SOL 7.3.81 Option (B) is correct.Given, the electric field intensity is E 10 10sin cost z t za a3 10 3 10x y

8 8# #p p p p= - + -^ ^h h

So, the magnetic field intensity is given as

H a Ek

0h#= (Direction of propagation is a ak z= )

sin cost z t za a5 3 10 37710 3 10y x

8 8# #p p p p= - + - -^ ^ ^h h h

SOL 7.3.82 Option (D) is correct.(1) For a perfect conducting medium the transmission coefficient is zero but a

medium having finite conductivity transmission coefficient has some finite value. So it doesn’t behave like an open circuit to the electromagnetic field.

(2) Relaxation time in a medium is defined as



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For View Only Shop Online at www.nodia.co.in Tr s

e=

Which in turn given the values in the range of sec10 20- . While the radio frequency wave has the time period ‘T ’ in the range of nsec to psec. (10 9- to 10 12- ) So the relation time at radio frequency/microwave frequency is much less than the period.(3) For a lossless dielectric 0s =^ h and so,

Tr " 3se=

(4) Intrinsic impedance of a perfect dielectric 0s =^ h is

h em= which is a pure resistance.

So, the statement (2), (3) and (4) are correct.

SOL 7.3.83 Option (D) is correct.The polarization of a uniform plane wave described the time varying behaviour of the electric field intensity vector so for polarization the field vector must be transverse to the propagation of wave.i.e. Transverse nature of electromagnetic wave causes polarization.

SOL 7.3.84 Option (B) is correct.In free space electrons and photon both have the same velocity 3 10 /m s8

# . So, the velocity of electromagnetic waves is same as velocity of light.So A and R both are true and R is the correct explanation of A.

SOL 7.3.85 Option (A) is correct.Fields are said to be circularly polarized if their components have same magnitudes but they differ in phase by 90! c.

SOL 7.3.86 Option (C) is correct.From Maxwell’s equation for an EM field, the divergence of the magnetic flux density is zero.i.e. B:d 0= A:d #d̂ h 0= div curl A 0=

SOL 7.3.87 Option (B) is correct.Electric field intensity due to the current element is defined as

E Js=

bI az2p s

=

The magnetic flux density due to the current element is given as

H bI a2p= f

So, the poynting vector of the field is P E H#=

bI a

2 2 3

2

p s=- r b

I i2 r2 3

2

sp=-




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SOL 7.3.88 Option (D) is correct.All the three statements are correct.

SOL 7.3.89 Option (C) is correct.Wavelength of a plane wave in any medium is defined as

l fvp=

where vp = phase velocity f = frequency of the wave

Since, vp cre

=

So, l 1r

\e

dielectric

air

ll

,

,

r air

r dielectric

ee=

12 1

re=

re 4=

SOL 7.3.90 Option (D) is correct.The velocity of an EM wave in free space is given as vc 3 10 /m sC 8

#= =and the characteristic impedance (intrinsic impedance) is given as

Zc 1200

0

em p= =

So both the terms are independent of frequency of the wave i.e. remain unchanged.

SOL 7.3.91 Option (D) is correct.Given, electric field intensity E 5 cos t z a10 30 x

9= +^ h

So, we conclude that, 109w = , and 30b =

and since b vpw=

b /c rew=

_ i (For non magnetic medium v c

pre

= )

re c 2

wb= b l

1030 3 10

9

8 2# #= c m 81=

SOL 7.3.92 Option (C) is correct.For attenuation of the wave the medium must have some finite conductivity s . In

the given wave equation the term tE

22ms involves s so this term is responsible for

the attenuation of the wave.

SOL 7.3.93 Option (A) is correct.The statement 1, 3 and 4 are correct while statement 2 is incorrect as Gauss’s law is applicable only for symmetrical geometry.



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In a Good conductor b fp ms=

So, phase velocity vp f2b

wmsp= =

SOL 7.3.95 Option (A) is correct.Given, the electric field intensity of the plane wave is tE^ h cos sinE t E t ea ax y

jkz1 2w w= - -

6 @

Since the components of the field are Ex E1=and Ey E2=i.e. Ex Ey! So, the wave is elliptically polarized.

SOL 7.3.96 Option (D) is correct.For a lossy dielectric, skin depth is defined as

d 2pl=

So, as the wavelength increases the depth of penetration of wave also increases.i.e. Reason (R) is correct.

The Skin depth is the depth by which electric field strength reduces to %e1 37= of

its original value.i.e. Assertion (A) is false.

SOL 7.3.97 Option (D) is correct.The electromagnetic equation in terms of vector potential A is given as

t

A A22

2

22med - Jm=-

SOL 7.3.98 Option (B) is correct.The wavelength of an EM wave propagating in a waveguide is defined as

l

ff1 c

2l=-

l

c m

where ll is the wavelength of the wave in unbounded medium(free space), fc is the cutoff frequency of the waveguide and f is the operating frequency.Now, for a propagating wave in the waveguide, the operating frequency is higher than the cutoff frequency.i.e. f /f f f 1<c c&

Putting it in equation (1) we get l < lli.e. Wavelength of a propagating wave in a wave guide is smaller than the free space wavelength.

SOL 7.3.99 Option (A) is correct.For a lossless dielectric medium s 0=




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and propagation constant, g j j ja b wm s we= + = +^ h

ja b+ jw me= b w me= i.e. b r\ e

SOL 7.3.100 Option (C) is correct.For a lossless medium 0s =^ h intrinsic impedance is defined as

h r

r

0

0

em

em

em= =

60p 120 1r

p e=

re 4=

SOL 7.3.101 Option (C) is correct.A field is said to be conservative if the curl of the field is zero.

SOL 7.3.102 Option (B) is correct.Given, the magnetic field intensity, H 0.5 /sin A me t x a10 2. x

z0 1 6= --

^ h

Comparing it with general expression of magnetic field intensity of wave propagating in ax direction given as H sinH e t x ax

z0 w b= -a-^ h

We get(i) the direction of wave propagation is ax(ii) .0 1a = , 2b =

So, propagation constant .j j0 1 2g a b= + = +

(iii) phase velocity, vp 5 10 /m s2106

5

bw

#= = =

(iv) a aH z= , a ak x=So, direction of polarization, aE a ak H#=-^ h a a ax z y#=-^ h i.e. wave is polarized along ay .

SOL 7.3.103 Option (C) is correct.Skin depth of any conducting medium is defined as

d f1p ms

=

So, at a given frequency f2w p=

1\d

m and 1

\ds

SOL 7.3.104 Option (B) is correct.Given, the electric field intensity of the plane wave, E 10 10 10sin cost z t za ax yw p w p= - + -^ ^h h

So, the field components are Ex sin t z10 10w p= -^ h



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For View Only Shop Online at www.nodia.co.in Ey cos t z10 w p= -^ h

and since, Ex Ey= So the polarization is circular.

SOL 7.3.105 Option (D) is correct.In free space electric field intensity is defined as E a Hk0h #=- ^ h

where ak is unit vector in the direction of propagation.Given, H 0.10 /cos A mt z a4 10 x

7# b= -^ h

So, the direction of propagation, a ak z=and we have, E 0.10cos t za a377 4 10z x

7# b#=- -^^ h h6 @ ( 3770h W= )

37.7 cos t z a4 10 y7

# b=- -^ h

SOL 7.3.106 Option (A) is correct.Given the electric field in medium A is E 100cos t x z6w p= -^ h

In medium A, 4re = , 1rm = , 0s =In medium B , 9re = , 4rm = , 0s =So,(a) intrinsic impedance of medium ‘B ’ is

Bh 94

32 120 80

0

0#e

mem p p= = = = 2a "^ h

(b) Intrinsic impedance of medium ‘A’ is

Ah 4 21 120 60

0

0#e

mem p p= = = =

So, reflection coefficient,

G B A

B A

h hh h= +

- 80 6080 60

51

p pp p= +- = 3b "^ h

(c) Transmission coefficient,

t 2B A

B

h hh= + 80 60

2 8078#

p pp= + = 4c "^ h

(d) Phase shift constant of medium A is given from the field equation as

b 12p= 1d "^ h

SOL 7.3.107 Option (A) is correct.Average power density in an EM wave is defined as

Pave Re E H21 *

s s#= ^ h 21 50 12

5# # p= .3 316=

So, the average power crossing a circular area of radius m24 is Pave rPave

2p= ^ h .3 316 242p= ^ ^_h h i 250 Watt=

SOL 7.3.108 Option (B) is correct.Electric field amplitude, E0 1 /V m=Skin depth, d 10 0.1cm m= =So, the attenuation constant of the wave in the conductor is




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a 1 10d= =

Now, the electric field intensity after travelling a distance z inside a conductor is E E e z

0= a-

where, E0 is the field intensity at the surface of the conductor. So, the distance travelled by the wave for which amplitude of electric field changes to (1/ )e (V/m)2 is given as

E eE

20=

E e z0

10- eE

20=

z10 2= z 20 cm=Alternatively, since the skin depth is the distance in which the wave amplitude decays to /e1^ h of its value at surface. So, for the amplitude to be /e1 2 of the field at its surface the wave penetrates a length of 2 20 cmd = .So A and R both are true and R is correct explanation of A.

SOL 7.3.109 Option (C) is correct.For any media having conductivity, 0s = . the intrinsic impedance is given as

h r

r0e

mem h= =

For media 1, 1h 2 1888 377 W= =^ h

For media 2, 2h 113119 377 W= =^ h

for media 3, 3h 37744 377 W= =^ h

SOL 7.3.110 Option (B) is correct.For an EM wave a medium incident on another medium, reflection coefficient is defined as

G EE

HH

i

r

i

r= =-

and G Z ZZ Z

22

31

2 1

2 1

h hh h= +

- = +- =

So, EEi

r HH

31

i

r=- =

EEr

i 3= and HH 3r

i =-

SOL 7.3.111 Option (B) is correct.For a perfect conductor conductivity , s 3=So, the skin depth of the perfect conductor is

d f1 0p ms

= =

***********

CHAPTER 8TRANSMISSION LINES

494 Transmission Lines Chap 8



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EXERCISE 8.1

MCQ 8.1.1 Assertion (A) : A sinusoidal voltage cosv V t2 10i 04

# p= ^ h is applied to the input terminal of a transmission line of length 20 cm such that the wave propagates with the velocity 3 10 /m sc 8

#= on the line. It’s output voltage will be in the same phase to the input voltage.

Reason (R) : Transmission line effects can be ignored if .l 0 01#l .

where l is the length of transmission line and l is the wavelength of the wave.(A) A and R both are true and R is correct explanation of A.

(B) A and R both are true but R is not correct explanation of A.



MCQ 8.1.2 A transmission line is formed of coaxial line with an inner conductor diameter of 1 cm and an outer conductor diameter of 4 cm. If the conductor has permeability

2c am m= and conductivity 11.6 10 /S mc7s #= then it’s resistance per unit length

for the operating frequency of 4 GHz will be(A) 4.95 /mW (B) 78.8 /mW

(C) 0.788 /mW (D) 0.495 /mW

MCQ 8.1.3 A transmission line formed of co-axial line with inner and outer diameters 1.5 cm and 3 cm respectively is filled with a dielectric of permeability 2 0m m= . It’s line parameter Ll will be equal to(A) 277 /nH m (B) 2.77 /nH m

(C) 872 /nH m (D) 8.7 /nH m

MCQ 8.1.4 A co-axial transmission line is filled with a dielectric having conductivity, 2 10 /S m3s #= - . If the inner and outer radius of the co-axial line are 1/4 cm and

1/2 cm respectively then the conductance per unit length of the transmission line will be(A) 9.1 /mS m (B) 1.45 /mS m

(C) 911 /S m (D) 145 /S m

MCQ 8.1.5 If Permittivity of the dielectric filled inside the coaxial transmission line having inner and outer diameter 2 cm and 5 cm respectively is 9 0e e= then the capacitance

Chap 8 Transmission Lines 495


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For View Only Shop Online at www.nodia.co.inper unit length of the line will be(A) 361 /pF m (B) 3.61 /nF m

(C) 5.74 /nF m (D) 57.4 /pF m

MCQ 8.1.6 A parallel plate transmission line consists of 1.2 cm wide conducting strips having conductivity, 1.16 10 /S m8s #= and permeability 0m m= is operating at 4 GHz frequency. What will be the line parameter Rl ?(A) 1.38 /mW (B) 0.69 /mW

(C) 0.97 /mW (D) 1.97 /mW

MCQ 8.1.7 A parallel plate transmission line is formed by copper strips of width . cmw 1 2= separated by a distance 0.3 cmd = . If the dielectric filled between the plates has permeability, 2 0m m= then what will be the inductance per unit length of the transmission line ?(A) 157 /nH m (B) 1.57 /H mm

(C) 0.78 /nH m (D) 78.1 /H mm

MCQ 8.1.8 The space between the strips of a parallel plate transmission line is filled of a dielectric of permittivity, .1 3re = and conductivity, 0.s . If the width of the strips is 9.6 cm and the separation between them is 0.6 cm then the line parameters Gl and Cl will be respectively(A) 0, 0.02 /nF m (B) 0.02 / , . /mS m nF m0 14

(C) 0, 0.18 /nF m (D) 1.8 / ,mS m 0

MCQ 8.1.9 Which one of the following statement is not correct for a transmission line ?(A) Attenuation constant of a lossless line is always zero.

(B) Characteristic impedance of both lossless and distortionless line is real

(C) Attenuation constant of a distortionless line is always zero.

(D) Both (A) and (C).

MCQ 8.1.10 Inductance and capacitance per unit length of a lossless transmission line are 250 /nH m and 0. /nF m2 respectively. The velocity of the wave propagation and characteristic impedance of the transmission line are respectively.(A) 2 10 /m s8

# , 100W (B) 3 10 /m s8# , 50W

(C) 2 10 /m s8# , 50W (D) 3 10 /m s8

# , 100W

MCQ 8.1.11 A 1GHz parallel plate transmission line consists of brass strips of conductivity 6.4 10 /S m7s #= separated by a dielectric of permittivity 6 0e e= . If the axial

component and transverse component of the electric field in the transmission line is Ez and Ey respectively then /E Ez y equals to(A) .2 16 10 4

#- (B) .4 167 10 5

#-

(C) .1 25 10 4#

- (D) .7 22 10 5#

-




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MCQ 8.1.12 A transmission line operating at a frequency 6 10 /rad s8# has the parameters

0.2 /k mR W=l , 4 /H mL m=l , 8 /S mG m=l , /pF mC 6=l . The propagation constant, g will be(A) . . mj0 5 1 2 1+ -

^ h (B) . . mj0 10 2 4 1+ -^ h

(C) . . mj1 2 0 5 1+ -^ h (D) . . mj2 4 0 10 1+ -

^ h

MCQ 8.1.13 The parameters of a transmission line are given as 10 /mR W=l , 0.1 /H mL m=l , 10 /pF mC =l , 40 /S mG m=l . If the transmission line is operating at a frequency,

1.2 10 /rad s9w #= then the characteristic impedance of the line will be(A) 50 2j W- (B) 4 100j W-

(C) 100 4j W- (D) 100 4j W+

MCQ 8.1.14 After travelling a distance of 20 m along a transmission line, the voltage wave remains %13 of it’s source amplitude. What is the attenuation constant of the transmission line ?(A) 0.13 /NP m (B) 0.10 /NP m

(C) 0.20 /NP m (D) 0.06 /NP m

MCQ 8.1.15 Amplitude of a voltage wave after travelling a certain distance down a transmission line is reduced by %87 . If the propagation constant of the transmission line is

. .j0 3 2 9+^ h then the phase shift in the voltage wave is(A) 61c (B) 561c

(C) 73c (D) 273c

MCQ 8.1.16 A parallel plate lossless transmission line consists of brass strips of width w and separated by a distance d . If both w and d are doubled then it’s characteristic impedance will(A) halved (B) doubled

(C) not change (D) none of these

MCQ 8.1.17 Phase velocity of voltage wave in a distortion less line having characteristic impedance, 0. kZ 20 W= and attenuation constant, 10 /mNP ma = is 0.5 10 /m svp 8

#= . The line parameters Rl and Ll will be respectively(A) 1 /mW , 0.5 /nH m (B) 10 /k mW , 2 /H mm

(C) 2 /mW , 1 /H mm (D) 1 /mW , 2 /H mm

MCQ 8.1.18 A distortionless line has parameters 4 /mR W=l and 4 10 /S mG 4#= -l . The

attenuation constant and characteristic impedance of the transmission line will be respectively(A) 25 /NP m, 0.01W (B) 100 /NP m, 4 10 2 W#

-

(C) 4 10 /NP m2#

- , 100W (D) 0.01 /NP m, 25W



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For View Only Shop Online at www.nodia.co.inMCQ 8.1.19 A transmission line operating at 5 GHz frequency has characteristic impedance

80Z0 W= and the phase constant 1.5 /rad mb = . The inductance per unit length of the transmission line will be(A) 3.81 /nH m (B) 38.1 /nH m

(C) 2.61 /nH m (D) 26.1 /nH m

MCQ 8.1.20 A 150W transmission line is connected to a 3 03 W resistance and to a 0 V5 DC source with zero internal resistance. The voltage reflection coefficients at the load end and at the source and of the transmission line are respectively(A) 1- , /1 3 (B) 1- , 1-

(C) /1 3, /1 3 (D) /1 3, 1-

MCQ 8.1.21 The voltage wave in a lossless transmission line has the maximum magnitude of 6 volt and minimum magnitude of 2.4 volt. The reflection coefficient of the transmission line is(A) .0 43 (B) .2 33

(C) .1 40 (D) .0 71

MCQ 8.1.22 An insulating material of permittivity 9 0e e= is used in a 25W lossless co-axial line . If the inner radius of the coaxial line is 0.6 mm then what will be it’s outer radius ?(A) 6.004 mm (B) 2.1 mm

(C) 3.002 mm (D) 4.2 mm

MCQ 8.1.23 A lossless transmission line of characteristic impedance 5Z 30 W= is connected to a load impedance Z j15 25L W= -^ h . What will be the standing wave ratio an the line ?(A) .0 57 (B) .3 65

(C) .0 27 (D) .1 22

MCQ 8.1.24 A purely resistance load ZL is connected to a 150W lossless transmission line. Such that it has a voltage standing wave ratio of 3. The possible value of ZL will be(A) 50W (B) 450W

(C) (A) and (B) both (D) none of these

MCQ 8.1.25 A voltage generator with 3 cos voltv t t10g9

#p=^ ^h h is applied to a 05 W lossless air spaced transmission line. If the line length is 10 cm and it is terminated in a load impedance Z j200 200L W= -^ h then the input impedance of the transmission line will be(A) .j50 50 8 W-^ h (B) . .j12 5 12 7 W-^ h

(C) . j25 4 25 W-^ h (D) .j25 25 4 W-^ h




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MCQ 8.1.26 The wavelength on a lossless transmission line terminated in a short circuit is l. What is the minimum possible length of the transmission line for which it appears as an open circuit at it’s input terminals ?(A) l

(B) /2l

(C) 4l

(D) /4l

MCQ 8.1.27 A lossless transmission line is operating at a frequency of 4 MHz. When the line is short circuited at it’s output end, the input impedance appears to be equivalent to an inductor with inductance of 32 nH but when the line is open circuited at it’s output end, the input impedance appears to be equivalent to a capacitor with capacitance of 20 pF. What is the characteristic impedance of the transmission line ?(A) 10W

(B) 1.6 kW

(C) 40W-

(D) 40W

MCQ 8.1.28 A /4l section of a 05 W lossless transmission line terminated in a 150W resistive load is preceded by another /4l section of a 200W lossless line as shown in figure. What is the input impedance, Zin ?

(A) 600W (B) 400W

(C) 267 W (D) 300W

MCQ 8.1.29 A transmission line of length l is short circuited at one end and open circuited at the other end. The voltage standing wave pattern in the transmission line will be



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MCQ 8.1.30 Phase velocity of a voltage wave in a transmission line of length l is vp . If the transmission line is open circuited at one end and short circuited at the other end then the natural frequency of the oscillation of the wave will be

(A) lnv2p ; , .......n 0 1 3= (B) l

n v4

2 1 p+^ h; , , .......n 0 1 3=

(C) ln v

42 1 p+^ h

; , , , ....n 1 2 3 3= (D) lnv2p ; , , .....n 1 2 3 3=

MCQ 8.1.31 At an operating frequency of 500 Hz, length of a transmission line is given by /l 4l= . For the same transmission line the length at 1 kHz will be given by

(A) l 8l= (B) l 4

l=

(C) l 2l= (D) none of these

MCQ 8.1.32 A lossless transmission line is terminated in a short circuit. The minimum possible length of the line for which it appears as a short circuit at its input terminals is(A) /2l

(B) /4l

(C) l

(D) 0

MCQ 8.1.33 A transmission line is operating at wavelength ‘l’. If the distance between successive voltage minima is 10 cm and distance between load and first voltage minimum is 7.5 cm then the distance between load and first voltage maxima is(A) /8l (B) /3 8l

(C) /5 8l (D) /4l

***********




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EXERCISE 8.2

MCQ 8.2.1 A z -polarized transverse electromagnetic wave (TEM) propagating along a parallel plate transmission line filled of perfect dielectric in ax+ direction. Let the electric and magnetic field of the wave be E and H respectively. Which of the following is correct relation for the fields.

(A) yE 0z

22 = (B)

zH

0y

22 =

(C) Both (A) and (B) (D) none of these

Statement for Linked Question 2 - 3 :A load impedance . . kZ j0 3 0 5L W= -^ h is being connected to a lossless transmission line of characteristic impedance 0.5 kZ0 W= operating at wavelength cm2l = .

MCQ 8.2.2 The distance of the first voltage maximum from the load will be(A) 0.44 cm

(B) 2.44 cm

(C) 1.56 cm

(D) 0.44 cm-

MCQ 8.2.3 The distance of the first current maximum from the load will be(A) 3.56 cm (B) 0.56 cm

(C) 1.44 cm (D) 2.56 cm

MCQ 8.2.4 Distance of the first voltage maximum and first current maximum from the load on a 50W lossless transmission line are respectively 4.5 cm and 1.5 cm. If the standing wave ratio on the transmission line is S 3= then the load impedance connected to the transmission line will be(A) j90 120 W-^ h (B) 10W

(C) j30 40 W-^ h (D) j40 30 W-^ h

MCQ 8.2.5 Total length of 50W lossless transmission line terminated in a load impedance Z j30 15L W= +^ h is /l 7 20l= as shown in figure. The total input impedance across the terminal AB will be



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(A) . .j38 3 64 8 W-^ h (B) . .j19 2 32 4 W-^ h

(C) . .j64 8 38 3 W-^ h (D) . .j32 4 19 2 W-^ h

MCQ 8.2.6 Assertion (A) : The input impedance of a quarter wavelength long lossless line terminated in a short-circuit is infinity.Reason (R) : The input impedance at the position where the magnitude of the voltage on a distortionless line is maximum is purely real.(A) A and R both are true and R is correct explanation of A.




Statement for Linked Question 7 - 8 :A voltage generator with 5 cosv t t2 4 10 30g

7#p= -^ ^h h and an internal impedance

30Zg W= is applied to a 30W lossless transmission line that has a relative permittivity .2 25re = and length, 6 ml = .

MCQ 8.2.7 If the line is terminated in a load impedance, Z j30 10L W= -^ h , then what will be the input impedance of the transmission line ?(A) . .j0 05 0 01 W-^ h (B) . .j50 62 23 48 W+^ h

(C) . .j92 06 21 80 W-^ h (D) . .j23 14 5 48 W+^ h

MCQ 8.2.8 The input voltage of the transmission line will be(A) . .cos t4 4 8 10 22 567

# cp +^ h

(B) 4.4 .cos Vt8 10 37 447# cp -^ h

(C) 4.4 .cos Vt8 10 22 567# cp -^ h

(D) 4.4 cos Vt8 10 307# cp -^ h

Statements for Linked Question 9 - 10 :Two equal load impedances of 150W are connected in parallel through a pair of transmission line, and the combination is connected to a feed transmission lien as shown in figure. All the lines are lossless and have characteristic impedance

100Z0 W= .




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MCQ 8.2.9 The effective load impedance of feedline (ZLl) equals to(A) . .j7 04 17 24 W-^ h

(B) . .j35 20 8 62 W+^ h

(C) . .j35 20 8 62 W-^ h

(D) . .j8 62 35 20 W+^ h

MCQ 8.2.10 The total input impedance of the feedline (line 3) will be(A) . .j2 15 1 13 W-^ h (B) . .j215 14 113 4 W-^ h

(C) . .j215 14 113 4 W+^ h (D) . .j107 57 56 7 W-^ h

MCQ 8.2.11 A 0.3 GHz voltage generator with 150 voltVsg = and an internal resistance 100Zg W= is connected to a 100W lossless transmission line of length .l 0 375l= .

If the line is terminated in a load impedance Z j100 100L W= -^ h then what will be the current flowing in the load ?(A) . .cos t0 67 3 10 108 48

# c-^ h

(B) . .cos t0 67 6 10 108 48# cp -^ h

(C) .cos t75 3 10 108 48# c-^ h

(D) . cos t0 67 6 10 1358# cp -^ h

MCQ 8.2.12 A voltage generator 150 VVsg = with an internal resistance 100Zg W= is connected to a load 150ZL W= through a .0 15l section of a 100W lossless transmission line. What is the average power delivered to the transmission line ?(A) Watt54

(B) Watt30

(C) 27 Watt

(D) 60 Watt

MCQ 8.2.13 A voltage generator 500 voltVsg = with an internal resistance 100Zg W= is applied to a configuration of lossless transmission lines as shown in figure. The power delivered to the loads ZL1 and ZL2 will be respectively



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(A) 612.23 Watt (B) 105.33 Watt

(C) 153.10 Watt (D) 306.11 Watt

MCQ 8.2.14 The input impedance of an infinitely long transmission line is equal to it’s characteristic impedance. The transmission line will be(A) slightly lossy

(B) lossless

(C) Distortion less

(D) (B) and (C) both

MCQ 8.2.15 An infinitely long lossy transmission line with characteristic impedance 200Z01 W= is feeded by a /2l section of 80W lossless transmission line as shown in figure. If a voltage generator 4 VVsg = with an internal resistance 100Zg W= is applied to the whole configuration then the average power transmitted to the infinite transmission line will be

(A) 2.2 mWatt (B) 22.2 mWatt

(C) 17.8 mWatt (D) 2.2 mWatt-

Common Data for Question 16 - 17 :A unit step voltage generator is applied to a 90W airspaced lossless transmission line at time, t 0= . At any time, t 0$ the voltage waveform at the sending end of the transmission line is shown in the figure below :




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MCQ 8.2.16 The length of the transmission line will be(A) 1200 m (B) 600 m

(C) 150 m (D) 300 m

MCQ 8.2.17 The unit step generator voltage connected to the line has an internal resistance 100Rg W= . What will be the load impedance connected to the transmission line ?

(A) 21.43W (B) 93.16W

(C) 42.86W (D) 233W

MCQ 8.2.18 At time t 0= unit step voltage generator Vg with an internal resistance Rg is applied to a 100W shorted transmission line filled with dielectric of permittivity

4 0e e= as shown in figure

The voltage waveform for any time t 0$ at the sending end is sown in figure below

Vg and Rg will be respectively equal to(A) 30 volt, 19.2W (B) 38.4 volt, 60W

(C) 60 volt, 38.4 W (D) 19.2 volt, 30W



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For View Only Shop Online at www.nodia.co.inStatement for Linked Question 19 - 20 :A 2.5 m section o an airspacd lossless transmission line is fed by a unit step voltage generator 30 voltVg = with internal resistance 200Rg W= . The transmission line is terminated in a resistive load 50ZL W= and characterized by 100Z0 W= .

MCQ 8.2.19 The bounce diagram of the transmission line will be

MCQ 8.2.20 The instantaneous voltage waveform v t^ h at the sending end of the transmission line will be




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MCQ 8.2.21 The SWR circle ‘L L1 2’ is shown on the smith chart for a lossless transmission line. If line is terminated in a load 50ZL W= then the possible value of the characteristic impedance of the line will be(A) 125W (B) 250W

(C) 20W (D) (A) and (C) Both

Common Data for Question 22 - 25 :A lossless transmission line characterized by 0Z 50 W= is terminated in a load Z j50 75L W= +^ h

MCQ 8.2.22 The reflection coefficient of the line will be(A) . e4 4 .j7 6c- (B) . e0 24 j76c

(C) . e4 4 j76c (D) . e0 24 j76c

MCQ 8.2.23 The input impedance at a distance of .0 35l from the load will be(A) . .j0 61 0 22 W-^ h (B) .j61 2 2 W+^ h

(C) .j61 2 2 W-^ h (D) . .j0 61 0 022 W+^ h

MCQ 8.2.24 The shortest length of the transmission line for which the input impedance appears to be purely resistive will be(A) .0 25l (B) .0 456l

(C) .0 106l (D) .0 544l

MCQ 8.2.25 The first voltage maximum will occur at a distance of(A) .0 106l from load (B) .0 144l from load

(C) .0 106l from Generator (D) .0 144l from generator

MCQ 8.2.26 A transmission line of characteristic impedance 50W is terminated by an inductor as shown in the figure.



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A positive wave with constant voltage 1 voltV0 = is incident on the load terminal at t 0= . At any time t the resulting negative wave voltage at the load terminal will be(A) Volte1 2 t25- -

^ h (B) Volte2 1t25 --^ h

(C) 2 Volte t25- (D) Volte 1t25 --^ h

MCQ 8.2.27 A transmission line has the characteristic impedance Z0 and the voltage standing wave ratio is S . The line impedance on the transmission line at voltage maximum and minimum are respectively.

(A) Z S0 , SZ0 (B) S

Z0 , Z S0

(C) Z S0 , Z S0 (D) SZ0 , SZ0

MCQ 8.2.28 Consider the three mediums of intrinsic impedances 1h , 2h and 3h respectively as shown in the figure. What will be the thickness ‘t ’ and intrinsic impedance ‘ 2h ’ of the medium 2 for which the reflected wave having wavelength ‘l’ is eliminated in medium 1 are thickness ‘t ’ intrinsic impedance 2h(A) /4l 1 3h h(B) /2l /1 3h h(C) /4l /1 3h h(D) /2l 1 3h h

Statement for Linked Question 29 - 30 :A quarter wave dielectric of thickness ‘t ’ and permittivity ‘e’ eliminates reflections of uniform plane waves of frequency .5 GHz2 incident normally from free space onto a dielectric of permittivity 16 0e . (Assume all media to have 0m m= )

MCQ 8.2.29 The permittivity of the dielectric coating equals to(A) /20e (B) /40e

(C) 4 0e (D) 2 0e

MCQ 8.2.30 What is the thickness ‘t ’ of the dielectric coating ?(A) 25 cm (B) 2.5 cm

(C) 1 cm (D) 10 cm




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MCQ 8.2.31 A transmission line has characteristics impedance 100W and standing wave ratio 3. The distance between the first voltage maximum and load is .0 125l. Load impedance of the transmission line is(A) j30 40 W+^ h (B) j60 80 W+^ h

(C) j30 40 W-^ h (D) j60 80 W-^ h

MCQ 8.2.32 A 100W lossless transmission line with it’s parameter 0.25 /H mL m=l and 100 /PF mC =l is terminated by it’s characteristic impedance. A 15 V voltage

source with internal resistance 50W is connected to the transmission line at t 0= . Plot of the voltage on the line at a distance 5 m from the source against time will be

MCQ 8.2.33 A lossless transmission line terminated by a load impedance Z ZL 0! is connected to a D.C. voltage source. The height of the first forward voltage pulse is V1

+. If the voltage reflection coefficients at the load and source are respectively LG and gG then the steady state voltage across the load is

(A) V 11

L

L1 G

G-++

; E (B) V 11

L

g L1 G

G G+

-+c m

(C) V 11

L

L1 G

G+-+

b l (D) V 11

g L

L1 G G

G-++

c m

MCQ 8.2.34 A 60W transmission line, terminated by a load of 180W is connected to a 100 V DC source at t 0= . The internal resistance of the source is 120W. The steady state voltage across the load will be



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For View Only Shop Online at www.nodia.co.in(A) 50 V (B) 100 V

(C) 120 V (D) 60 V

MCQ 8.2.35 At t 0= a 50 Volt D.C. source with an internal resistance 30W is connected to a transmission line of 15W characteristic impedance having a load of 45W. The steady state load current for the transmission line is(A) 0.67 A (B) 1.5 A

(C) 0.33 A (D) 1.3 A

Statement for Linked Question 36 - 37 :A transmission line of an unknown length terminated in a resistance is connected to a 5 V battery with zero internal resistance. The plot of input current to the line is shown in the figure below

MCQ 8.2.36 The characteristic impedance of the transmission line will be(A) 1.2 kW (B) 80W

(C) 8W (D) 12.5W

MCQ 8.2.37 The load resistance terminated to the transmission line will be(A) 263W (B) 80W

(C) 150W (D) 43W

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EXERCISE 8.3

MCQ 8.3.1 A coaxial-cable with an inner diameter of 2 mm and outer diameter of 2.4 mm is filled with a dielectric of relative permittivity 10.89. Given 4 10 / ,H m0

7m p#= -

/F m3610

0

9

e p=-

, the characteristic impedance of the cable is

(A) 330W (B) 100W

(C) 143.3W (D) 43.4 W

MCQ 8.3.2 A transmission line with a characteristic impedance of 100W is used to match a 50W section to a 200W section. If the matching is to be done both at 429 MHz and 1GHz, the length of the transmission line can be approximately(A) 82.5 cm (b) 1.05 m

(C) 1.58 cm (D) 1.75 m

MCQ 8.3.3 A transmission line of characteristic impedance 50W is terminated by a 50W load. When excited by a sinusoidal voltage source at 20 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be /4p radians. The phase velocity of the wave along the line is(A) 0.8 10 /m s8

# (B) 1.2 10 /m s8#

(C) 1.6 10 /m s8# (D) 3 10 /m s8

#

MCQ 8.3.4 A transmission line of characteristic impedance 50W is terminated in a load impedance ZL . The VSWR of the line is measured as 5 and the first of the voltage maxima in the line is observed at a distance of l/4 from the load. The value of ZL is(A) 10 W (B) 250 W

(C) (19.23 46.15)j W+ (D) (19.23 46.15)j W-

MCQ 8.3.5 If the scattering matrix [ ]S of a two port network is [ ]S . 0. 90

. 90

. 900 20 9

0 90 1

c

c

c

c= > H, then

the network is(A) lossless and reciprocal

(B) lossless but not reciprocal

(C) not lossless but reciprocal

(D) neither lossless nor reciprocal

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For View Only Shop Online at www.nodia.co.inMCQ 8.3.6 A transmission line has a characteristic impedance of 50W and a resistance of

0.1 /mW . If the line is distortion less, the attenuation constant(in Np/m) is(A) 500 (B) 5

(C) 0.014 (D) 0.002

MCQ 8.3.7 In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ration(VSWR) on the 60 W line is

(A) 1.00 (B) 1.64

(C) 2.50 (D) 3.00

MCQ 8.3.8 A transmission line terminates in two branches, each of length /2l , as shown. The branches are terminated by 50W loads. The lines are lossless and have the characteristic impedances shown. Determine the impedance Zi as seen by the source.

(A) 200W (B) 100W

(C) 50W (D) 25W

MCQ 8.3.9 One end of a loss-less transmission line having the characteristic impedance of 75W and length of 2 cm is short-circuited. At 5 GHz, the input impedance at the other end of transmission line is(A) 0 (B) Resistive

(C) Capacitive (D) Inductive

MCQ 8.3.10 A load of 50W is connected in shunt in a 2-wire transmission line of Z 500 W= as shown in the figure. The 2-port scattering parameter matrix (S-matrix) of the shunt element is

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(A) 21

21

21

21

--> H (B)

01

10= G

(C) 31

32

32

31

--> H (D) 4

1

43

43

41-

-> H

MCQ 8.3.11 The parallel branches of a 2-wire transmission line are terminated in 05 W and 200W resistors as shown in the figure. The characteristic impedance of the line is

50Z0 W= and each section has a length of 4l . The voltage reflection coefficient G

at the input is

(A) j 57- (B) 7

5-

(C) j75 (D) 7

5

MCQ 8.3.12 A transmission line is feeding 1 watt of power to a horn antenna having a gain of 10 dB. The antenna is matched to the transmission line. The total power radiated by the horn antenna into the free space is(A) 10 Watts (B) 1 Watts

(C) 0.1 Watts (D) 0.01 Watt

MCQ 8.3.13 Characteristic impedance of a transmission line is 50W. Input impedance of the open circuited line is 100 150Z joc W= + . When the transmission line is short circuited, then value of the input impedance will be(A) 50W (B) 100 150j W+

(C) 7.69 11.54j W+ (D) 7.69 11.54j W-

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For View Only Shop Online at www.nodia.co.inStatement of Linked Questions 14 - 15 :Voltage standing wave pattern in a lossless transmission line with characteristic impedance 50 and a resistive load is shown in the figure.

MCQ 8.3.14 The value of the load resistance is(A) 50 W (B) 200 W

(C) 12.5 W (D) 0

MCQ 8.3.15 The reflection coefficient is given by(A) 0.6- (B) 1-

(C) 0.6 (D) 0

MCQ 8.3.16 Many circles are drawn in a Smith Chart used for transmission line calculations. The circles shown in the figure represent

(A) Unit circles (B) Constant resistance circles

(C) Constant reactance circles (D) Constant reflection coefficient circles.

MCQ 8.3.17 Consider a 200 W, quarter - wave long (at 1 GHz) transmission line as shown in Fig. It is connected to a 20 V, 50 W source at one end and is left open circuited at the other end. The magnitude of the voltage at the open circuit end of the line is

(A) 10 V (B) 5 V

(C) 60 V (D) 60/7 V

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MCQ 8.3.18 Consider an impedance Z R jX= + marked with point P in an impedance Smith chart as shown in Fig. The movement from point P along a constant resistance circle in the clockwise direction by an angle 45c is equivalent to

(A) adding an inductance in series with Z

(B) adding a capacitance in series with Z

(C) adding an inductance in shunt across Z

(D) adding a capacitance in shunt across Z

MCQ 8.3.19 A lossless transmission line is terminated in a load which reflects a part of the incident power. The measured VSWR is 2. The percentage of the power that is reflected back is(A) 57.73 (B) 33.33

(C) 0.11 (D) 11.11

MCQ 8.3.20 A short - circuited stub is shunt connected to a transmission line as shown in fig. If 50Z0 W= , the admittance Y seen at the junction of the stub and the transmission

line is

(A) (0.01 0.02)j- mho (B) ( . . )j0 02 0 01- mho

(C) ( . . )j0 04 0 02- mho (D) ( . )j0 02 0+ mho

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For View Only Shop Online at www.nodia.co.inMCQ 8.3.21 The VSWR can have any value between

(A) 0 and 1 (B) 1- and 1+

(C) 0 and 3 (D) 1 and 3

MCQ 8.3.22 In an impedance Smith chart, a clockwise movement along a constant resistance circle gives rise to(A) a decrease in the value of reactance

(B) an increase in the value of reactance

(C) no change in the reactance value

(D) no change in the impedance

MCQ 8.3.23 A transmission line is distortionless if

(A) RL GC1= (B) RL GC=

(C) LG RC= (D) RG LC=

MCQ 8.3.24 The magnitudes of the open-circuit and short-circuit input impedances of a transmission line are 100W and 25W respectively. The characteristic impedance of the line is,(A) 25 W (B) 50 W

(C) 75 W (D) 100 W

MCQ 8.3.25 In a twin-wire transmission line in air, the adjacent voltage maxima are at 25 m and 12.4 m. The operating frequency is(A) MHz300 (B) GHz1

(C) GHz2 (D) . GHz6 28

MCQ 8.3.26 In air, a lossless transmission line of length cm50 with /H mL 10 m= , 40 pFC /m= is operated at MHz25 . Its electrical path length is(A) 0.5 meters (B) metersl

(C) /2 radiansp (D) 180 deg rees

MCQ 8.3.27 A transmission line of 50W characteristic impedance is terminated with a 100W resistance. The minimum impedance measured on the line is equal to(A) 0W (B) 25W

(C) 50W (D) 100W

MCQ 8.3.28 A very lossy, /4l long, 50W transmission line is open circuited at the load end. The input impedance measured at the other end of the line is approximately(A) 0 (B) 50W

(C) 3 (D) None of the above

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MCQ 8.3.29 A lossless transmission line having 50W characteristic impedance and length /4l is short circuited at one end and connected to an ideal voltage source of 1 V at the other end. The current drawn from the voltage source is(A) 0 (B) 0.02 A

(C) 3 (D) none of these

MCQ 8.3.30 The capacitance per unit length and the characteristic impedance of a lossless transmission line are C and Z0 respectively. The velocity of a travelling wave on the transmission line is

(A) Z C0 (B) Z C10

(C) CZ0 (D) Z

C0

MCQ 8.3.31 A /4l line, shorted at one end, presents impedance at the other end equal to(A) Z0 (B) Z2 0

(C) 3 (D) 0

where Z0 is characteristic impedance of the line.

MCQ 8.3.32 A 100W transmission line is first short-terminated and the minima locations are noted. When the short is replaced by a resistive load RL , the minima locations are not altered and the VSWR is measured to be 3. The value of RL is(A) 25W (B) 50W

(C) 225W (D) 250W

MCQ 8.3.33 If maximum and minimum voltage on a transmission line are 2 V and 5 Vrespectively, VSWR is(A) 0.5 (B) 2

(C) 1 (D) 8

MCQ 8.3.34 An ideal lossless transmission line of 60Z0 W= is connected to unknown ZL . If SWR 4= , find ZL .(A) 240W (B) 480W

(C) 120W (D) 100W

MCQ 8.3.35 Loading of a cable is done to1. Increase its inductance

2. Increase its leakage resistance

3. Decrease its leakage resistance

4. Achieve distortionless condition

(A) 1, 2, 3 and 4 (B) 1 and 3 only


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For View Only Shop Online at www.nodia.co.inMCQ 8.3.36 Given a range of frequencies, which of the following systems is best for transmission

line load matching ?(A) Single stub (B) Double stub

(C) Single stub with adjustable position (D) Quarter wave transformer

MCQ 8.3.37 A line of characteristic impedance 50W is terminated at one end by 50j W+ . The VSWR on the line is(A) 1 (B) 3

(C) 0 (D) j

MCQ 8.3.38 At UHF short-circuited lossless transmission lines can be used to provide appropriate values of impedance. Match List I with List II and select the correct answer using the code given below the lists :

List I List II

a. /l 4< l 1. Capacitive

b. /4 /2l< <l l 2. Inductive

c. /l 4l= 3. 0

d. /l 2l= 4. 3

Codes :

a b c d(A) 2 1 4 3(B) 3 1 4 2(C) 2 4 1 3(D) 3 4 1 2

MCQ 8.3.39 Consider the following statements regarding a transmission line :1. Its attenuation is constant and is independent of frequency

2. Its attenuation varies linearly with frequency

3. Its phase shift varies linearly with frequency

4. Its phase shift is constant and is independent of frequency

Which of the above statements are correct for distortion less line ?(A) 1, 2, 3, and 4 (B) 2 and 3 only


MCQ 8.3.40 The reflection coefficient on a 200 m long transmission line has a phase angle of 150c- . If the operating wavelength is 250 m, what will be the number of voltage

maxima on the line ?(A) 0 (B) 3

(C) 6 (D) 7

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MCQ 8.3.41 With regard to a transmission line, which of the following statements is correct ?(A) Any impedance repeats itself every /4l on the Smith chart.

(B) The SWR 2= circle and the magnitude of reflection coefficient .0 5= circle coincide on the Smith chart.

(C) At any point on a transmission line, the current reflection coefficient is the reciprocal of the voltage reflection coefficient.

(D) Matching eliminates the reflected wave between the source and the matching device location.

MCQ 8.3.42 It is required to match a 200W load to a 450W transmission line. To reduce the SWR along the line to 1, what must be the characteristic impedance of the quarter-wave transformer used for this purpose, if it is connected directly to the load ?(A) 90 kW (B) 300W

(C) 49 W (D) 2

3 W

MCQ 8.3.43 The load end of a quarter wave transformer gets disconnected thereby causing an open-circuited load. What will be the input impedance of the transformer ?(A) Zero (B) Infinite

(C) Finite and positive (D) Finite and negative

MCQ 8.3.44 A lossless transmission line of characteristic impedance Z0 and length /4l < l is terminated at the load end by an open circuit. What is its input impedance Zin?(A) tanZ jZ lin 0 b= (B) cotZ jZ lin 0 b=

(C) tanZ jZ lin 0 b=- (D) tanZ jZ lin 0 b=-

MCQ 8.3.45 Which one of the following statements for a short circuited loss free line is not correct ?(A) The line appears as a pure reactance when viewed from the sending end

(B) It can be either inductive or capacitive

(C) There are no reflections in the line

(D) Standing waves of voltage and current are set up along length of the lines

MCQ 8.3.46 Match List I (Load impedance) with List II (Value of Reflection Coefficient) and select the correct answer using the code given below the lists :

List-I List-II

a. Short Circuit 1. 0

b. Open Circuit 2. 1-

c. Line characteristics impedance 3. +1

d. 2# line characteristic impedance 4. +1/3

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For View Only Shop Online at www.nodia.co.inCodes :

a b c d(A) 2 1 3 4(B) 4 3 1 2(C) 2 3 1 4(D) 4 1 3 2

MCQ 8.3.47 When the reflection coefficient equals 1 0c what is the VSWR?(A) Zero (B) 1

(C) 3 (D) Infinite

MCQ 8.3.48 If the reflection coefficient is 1/5, what is the corresponding VSWR ?(A) 3/2 (B) 2/3

(C) 5/2 (D) 2/5

MCQ 8.3.49 Which one of the following is the characteristic impedance of lossless transmission line ?(A) /R G (B) /L G

(C) /L C (D) /R C

MCQ 8.3.50 Match List I (Quantity) with List II (Range of Values) and select the correct answer using the code given below the lists :

List-I List-II

a. Input Impedance 1. 1 1to- +

b Reflection coefficient 2. 1 to3

c. VSWR 3. 0 to3

Codes : a b c(A) 2 3 1(B) 3 2 1(C) 3 1 2(D) 2 1 3

MCQ 8.3.51 A quarter wave impedance transformer is terminated by a short circuit. What would its input impedance be equal to ?(A) The line characteristic impedance

(B) Zero

(C) Infinity

(D) Square root of the line characteristic impedance

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MCQ 8.3.52 Scattering parameters are more suited than impedance parameters to describe a waveguide junction because(A) the scattering parameters are frequency invariant whereas the impedance

parameters are not so

(B) scattering matrix is always unitary

(C) impedance parameters vary over unacceptably wide ranges

(D) scattering parameters are directly measurable but impedance parameters are not so

MCQ 8.3.53 In a transmission line the reflection coefficient at the load end is given by 0.3e j30c- . What is the reflection coefficient at a distance of 0.1 wavelength towards source ?(A) 0.3e j30c+ (B) . e0 3 j102c+

(C) . e0 3 j258c+ (D) . e0 3 66c+

MCQ 8.3.54 To couple a coaxial line to a parallel wire, it is best to use a :(A) Balun

(B) Slotted line

(C) Directional coupler

(D) Quarter wave transformer

MCQ 8.3.55 A plane wave having x -directed electric field propagating in free space along the z-direction is incident on an infinite electrically conducting (perfect conductor) sheet at z 0= plane. Which one of the following is correct ?(A) The sheet will absorb the wave

(B) There will be x -directed surface electric current on the sheet

(C) There will be y -directed surface electric current on the sheet

(D) There will be magnetic current in the sheet.

MCQ 8.3.56 For sea water with 5 /mho ms = and 80re = , what is the distance for which radio signal can be transmitted with 90% attenuation at 25 kHz ?(A) 0.322 m (B) 3.22 m

(C) 32.2 m (D) 322 m

MCQ 8.3.57 Consider the following statements regarding Smith charts :1. A normalized Smith chart applies to a line of any characteristic resistance and

serves as well for normalized admittance

2. A polar coordinate Smith chart contains circles of constant z and circles of constant z

3. In Smith chart, the distance towards the load is always measured in clockwise direction.

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For View Only Shop Online at www.nodia.co.inWhich of the statements given above are correct ?(A) 1, 2 and 3 (B) 2 and 3

(C) 1 and 3 (D) 1 and 2

MCQ 8.3.58 A j100 75 W-^ h load is connected to a co-axial cable of characteristic impedance 75 ohms at 12 GHz. In order to obtain the best matching, which one of the following will have to be connected ?(A) A short-circuited sub at load

(B) Inductance at load

(C) A capacitance at a specific distance at load

(D) A short-circuited stub at some specific distance from load

MCQ 8.3.59 In a line VSWR of a load is 6 dB. The reflection coefficient will be(A) 0.033 (B) 0.33

(C) 0.66 (D) 3.3

MCQ 8.3.60 Z 200L W= and it is desired that 50Zin W= . The quarter wave transformer should have a characteristic impedance of(A) 100 W (B) 40 W

(C) 10,000 W (D) 4 W

MCQ 8.3.61 Consider the following :For a lossless transmission line we can write :1. Z jZin 0=- for a shorted line with /8l l=

2. Z jin ! a= for a shorted line with /4l l=

3. Z Zin 0= for a matched line of any length

Select the correct answer using the codes given below :(A) 1 and 2 (B) 2 and 3

(C) 1 and 3 (D) 2 and 4

MCQ 8.3.62 The input impedance of a short circuited quarter wave long transmission line is(A) purely reactive

(B) purely resistive

(C) dependent on the characteristic impedance of the line

(D) none of the above

MCQ 8.3.63 A transmission line of output impedance 400W is to be matched to a load of 25W through a quarter wavelength line. The quarter wave line characteristic impedance must be(A) 40W (B) 100W

(C) 400W (D) 425W

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MCQ 8.3.64 The input impedance of /8l long short-circuited section of a lossless transmission line is(A) zero (B) inductive

(C) capacitive (D) infinite

MCQ 8.3.65 Match List I (Parameters) with List II (Values) for a transmission line with a series impedance /mZ R j Lw W= +l l and a shunt admittance /mho mY G j Cw= +l l , and select the correct answer :

List-I List-II

a. Characteristic impedance Z0 1. ZY

b. Propagation constant g 2. /Z Y

c. The sending-end input impedance Zin when the line is terminated in its characteristic impedance Z0

3. /Y Z

Codes : a b c(A) 3 1 1(B) 2 3 3(C) 2 1 2(D) 1 2 2

MCQ 8.3.66 Which of the following conditions will not guarantee a distortionless transmission line ?(A) 0R G= =

(B) RC GL=

(C) Very low frequency range ( , )R L G C>> >>w w

(D) Very high frequency range ( , )R L G C<< <<w w

MCQ 8.3.67 In an air line, adjacent maxima are found at 12.5 cm and 37.5 cm. The operating frequency is(A) 1.5 GHz

(B) 600 MHz

(C) 300 MHz

(D) 1.2 GHz

MCQ 8.3.68 Fig. I shows an open circuited transmission line. The switch is closed at time 0t = and after a time t the voltage distribution on the line reaches that shown in Fig. II. If c is the velocity in the line, then

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(A) /t l c< (B) /t l c=

(C) / /l c l c1 > (D) /t l c2<

MCQ 8.3.69 A 75W transmission line is first short-terminated and the minima locations are noted. When the short is replaced by a resistive load RL , the minima locations are not altered and the VSWR is measured to be 3. The value of RL is(A) 25W (B) 50W

(C) 225W (D) 250W

MCQ 8.3.70 For a lossy transmission line, the characteristic impedance does not depend on(A) the operating frequency of the line

(B) theconductivity of the conductors

(C) conductivity of the dielectric separating the conductors

(D) length of the line

MCQ 8.3.71 If the maximum and minimum voltages on a transmission line are 4 V and 2 V, respectively for a typical load, VSWR is(A) 1.0 (B) 0.5

(C) 2.0 (D) 8.0

MCQ 8.3.72 A transmission line is distortionless if(A) RG LC= (B) RC GL=

(C) CR

LG= (D) R G=

MCQ 8.3.73 If reflection coefficient for voltage be 0.6, the voltage standing wave ratio (VSWR) is(A) 0.66 (B) 4

(C) 1.5 (D) 2

MCQ 8.3.74 A signal of V10 is applied to a 80 ohm coaxial transmission line, terminated in a 100 ohm load. The voltage reflected coefficient is(A) 1/4 (B) 1/3

(C) 1/2 (D) 1

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MCQ 8.3.75 A transmission line of characteristic impedance of 50 ohm is terminated by a load impedance of (15 20)j- ohm. What is the normalized load impedance ?(A) . .j0 6 0 8- (B) . .j0 3 0 6-

(C) . .j0 3 0 4- (D) . .j0 3 0 4+

MCQ 8.3.76 Two lossless resistive transmission lines each of characteristic impedance Z are connected as shown in the circuit below. If the maximum voltage on the two lines is the same and the power transmitted by line A is W1, then what is the power transmitted by the line B ?

(A) W4 1 (B) W3 1

(C) W2 1 (D) W1 1

MCQ 8.3.77 A transmission line section shows an input impedance of 36W and 64 W respectively, when short circuited and open circuited. What is the characteristic impedance of the transmission line ?(A) 100W (B) 50W

(C) 45W (D) 48W

MCQ 8.3.78 Consider the following statements for transmission lines :1. When a transmission line is terminated by its characteristic impedance the line

will not have any reflected wave.

2. For a finite line terminated by its characteristic impedance the velocity and current at all points on the line are exactly same.

3. For a lossless half wave transmission line the input impedance is not equal to load impedance.

Which of the statements given above are correct ?(A) 1 and 2 (B) 2 and 3

(C) 1 and 3 (D) 1, 2 and 3

MCQ 8.3.79 What does the standing wave ratio (SWR) of unity imply ?(A) Transmission line is open circuited

(B) Transmission line is short circuited

(C) Transmission line’s characteristic impedance is equal to load impedance

(D) Transmission line’s characteristic impedance is not equal to load impedance

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For View Only Shop Online at www.nodia.co.inMCQ 8.3.80 h = half centre to centre spacing, r = conductor radius and e = permittivity of the

medium. Which one of the following is equal to the capacitance per unit length of a two-wire transmission line ?

(A) log r

hrh 1e 2

2pe

+ -b cl m) 3

(B) log r

hrh 1

2

e 2

2pe

+ -b cl m) 3

(C) log r

hrh 1

3

e 2

2pe

+ -b cl m) 3

(D) log r

hrh 1

4

e 2

2pe

+ -b cl m) 3

MCQ 8.3.81 For a line of characteristic impedance Z0 terminated in a load of ZR such that /Z Z 3R 0= , what is the reflection coefficient LG ?

(A) 1/3 (B) 2/3

(C) /1 3- (D) /1 2-

MCQ 8.3.82 A transmission line has , , ,R L G C distributed parameters per unit length of line. If g is the propagation constant of the line, which one of the following expressions represents the characteristics impedance of the line ?

(A) R j Lwg

+ (B) R j Lgw+

(C) G j Cgw+ (D) R j L

G j Cww

++

MCQ 8.3.83 What is the value of standing wave ratio (SWR) in free space for reflection for reflection coefficient 1/3G =-(A) 2/3 (B) 0.5

(C) 4.0 (D) 2.0

MCQ 8.3.84 What is the attenuation constant a for distortionless transmission line ?

(A) 0a = (B) R LCa =

(C) R CLa = (D) C

RLa =

MCQ 8.3.85 A 75 W distortionless transmission line has a capacitance of 10 10- f/m. What is the inductance per meter ?(A) 0.25 Hm (B) 500 Hm

(C) 5000 Hm (D) 50 Hm

MCQ 8.3.86 The open circuit and short circuit impedances of a line are 50 W each. What is the characteristic impedance of the line ?(A) 100 2W (B) 100W

(C) /100 2W (D) 50 W

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MCQ 8.3.87 A load impedance of (75 50)j- is connected to a transmission line of characteristic impedance Z 750 W= . The best method of matching comprises(A) A short circuit stub at load

(B) A short circuit stub at some specific distance from load

(C) An open stub at load

(D) Two short circuited stubs at specific distances from load

MCQ 8.3.88 When a lossless transmission line is terminated by a resistance equal to surge impedance, then what is value of the reflection coefficient ?(A) 1 (B) 1-

(C) 0 (D) 0.5

MCQ 8.3.89 A lossless transmission line of length 50 cm with 10L H/mm= , 40C pF/m= is operated at 30 MHz. What is its electric length ( )lb ?(A) 20l (B) 0.2l

(C) 108c (D) 40p

MCQ 8.3.90 Which one of the following is the correct expression for the propagation constant in a transmission line ?

(A) G j CR j L

ww

++

^

^

h

h (B)

G j CR j L

ww

--

^

^

h

h

(C) R j L G j Cw w- -^ ^h h (D) R j L G j Cw w+ +^ ^h h

MCQ 8.3.91 Assertion (A) : In a lossless transmission line the voltage and current distributions along the line are always constant.Reason (R) : The voltage and current distributions in an open line are such that at a distance /4l from the load end, the line looks like a series resonant circuit.(A) Both A and R are true and R is the correct explanation of A




MCQ 8.3.92 Consider the following statements :Characteristic impedance of a transmission line is given by

1. G j CR j L

ww

++ , (R, L, G and C are line constants)

2. Z Zoc sc , (Z Zandoc sc are the open and short circuit impedances of the line)

3. /V Il l, ( V Iandl l are the voltage and current of the wave travelling in the positive y direction)

Which of these are correct ?(A) 1,2 and 3 (B) 1 and 2

(C) 2 and 3 (D) 1 and 3

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For View Only Shop Online at www.nodia.co.inMCQ 8.3.93 A loss-less transmission line of characteristic impedance Z0 and /4l < l is terminated

at the load end by a short circuit. Its input impedance Zs is(A) tanZ jZ ls 0 b=- (B) cotZ jZ ls 0 b=

(C) tanZ jZ ls 0 b= (D) cotZ jZ ls 0 b=-

MCQ 8.3.94 A loss-less transmission line with characteristic impedance of 600 ohms is terminated in a purely resistive load of 900 ohms. The reflection coefficient is(A) 0.2 (B) 0.5

(C) 0.667 (D) 1.5

MCQ 8.3.95 A transmission line has R, L, G and C distributed parameters per unit length of the line, g is the propagation constant of the lines. Which expression gives the characteristic impedance of the line ?

(A) R j Lwg

+ (B) R j Lgw+

(C) G j Cgw+ (D) R j L

G j Cww

++

MCQ 8.3.96 The open circuit impedance of a certain length of a loss-less line is 100 W. The short circuit impedance of the same line is also 100 W. The characteristic impedance of the line is(A) 100 2 W (B) 05 W

(C) /100 2 W (D) 100 W

MCQ 8.3.97 In the relations S 1

1= G

G-

+; the values of S and G (where S stands for wave ratio

and G is reflection coefficient), respectively, vary as(A) 0 to 1 and 1- to 0 (B) 1 to 3 and 1- to 1+

(C) 1- to 1+ and 1 to 3 (D) 1- to 0 and 0 to 1

MCQ 8.3.98 Consider the following statements :The characteristic impedance of a transmission line can increase with the increase in1. resistance per unit length

2. conductance per unit length

3. capacitance per unit length

4. inductance per unit length

Which of these statements are correct ?

(A) 1 and 2 (B) 2 and 3

(C) 1 and 4 (D) 3 and 4

***********

IES EE 2003

IES EE 2003

IES EE 2002

IES EE 2001

IES EE 2001

IES EE 2001




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SOLUTIONS 8.1

SOL 8.1.1 Option (D) is correct.Giventhe input voltage, vi cosV t4 100

4# p= ^ h

and length of transmission line, l 20 20 10cm m2#= = -

So, the angular frequency of the applied voltage is w 4 104

# p=and the wavelength of the voltage wave is

l fv v2p p

wp= = 2f p

w=a k

Therefore, ll v2

20 10p

2#p

w=

-^ h

( )

( ) ( )2 3 10

4 10 20 108

4 2

# #

# # #

pp=

-

(in free space 3 10 /m svp 8#= )

.1 33 10 5#= -

Since, ll .0 01#

So the effect of transmission line on the voltage wave is negligible i.e. the output voltage will be in the same phase to the input voltage.Thus, A and R both are true and R is correct explanation of A.

SOL 8.1.2 Option (C) is correct.Given,Inner diameter of coaxial line, a2 1 cm= a& 0.5 10 m2

#= -

and outer diameter of coaxial line, b2 2 cm= b& 10 m2= -

Permeability of conductor, cm 2 0m=Conductivity of conductor, cs 11.6 10 /S m7

#=Operating frequency, f 4 GHz= 4 10 Hz9

#=So, the resistance per unit length of transmission line is given as :

Rl fa b2

1 1 1c

c

p sp m= +b l

.

( ) ( ).2

111 6 10

4 10 2 4 100 5 10

1101

7

9 7

2 2#

# # # # #

#pp p= +

-

- -b l

0.788 /mW=

SOL 8.1.3 Option (D) is correct.Inner diameter of coaxial line, a2 1.5 cm= &a 0.75 10 m2

#= -



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For View Only Shop Online at www.nodia.co.inOuter diameter of coaxial line, b2 3 cm= b& 1.5 10 m2

#= -

Permeability of the filled dielectric, m 2 0m=So, it’s inductance per unit length is given as

Ll ln ab

2pm= b l

( )..ln2

2 4 100 75 101 5 107

2

2# #

#

#p

p=-

-

-

c m

2.77 10 /H m7#= - 277 /nH m=

SOL 8.1.4 Option (D) is correct.Given,Inner radius of the coaxial line, a 1/ cm8= 1.25 10 m3

#= -

Outer radius of the coaxial line, b 1/ cm2= 5 10 m3#= -

Conductivity of dielectric, s 2 10 /S m3#= -

So, the conductance per unit length of the transmission line is given as

Gl /

.

( )ln lnb a

2

1 25 105 10

2 2 10

3

3

3

#

#

# #ps p= =-

-

-

^c

hm

.1 /mS m3=

SOL 8.1.5 Option (D) is correct.Given,Inner diameter of coaxial line, a2 1 cm= &a 0.5 10 m2

#= -

Outer diameter of coaxial line, b2 4 cm= &b 2 10 m2#= -

Permittivity of the dielectric, 9 0e e=So, the capacitance per unit length of the line is given as

Cl /

.

.ln lnb a

2

0 5 102 10

2 9 8 85 10

2

2

12

#

#

# # #pe p= =-

-

-

^c

hm

3.61 10 /F m10#= - /pF m323=

SOL 8.1.6 Option (C) is correct.Given,Width of strips, w 2.4 10 m2

#= -

Conductivity of strips, s 1.16 10 /S m8#=

Permeability of strips, m 0m=Operating frequency, f 4 GHz= 4 10 Hz9

#=So, the parameter Rl is given as

Rl wf2sp m=

. .2 4 102

1 16 104 10 4 10

2 8

9 7

# #

# # # #p p= -

-

0.9722 /mW=

SOL 8.1.7 Option (D) is correct.Strips width, w 4.8 4.8 10cm m2

#= = -

Separation between the plates, d 0.3 0.3 10cm m2#= = -

Permittivity of dielectric, m 2 0m=So, the inductance per unit length is given as




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Ll wdm=

..

4 8 102 4 10 0 3 10

2

7 2

#

# # # #p= -

- -

.57 10 /H m2 5#=

SOL 8.1.8 Option (C) is correct.The width of strips, w 9.6 9.6 10cm m2

#= = -

Separation between the strips, d 0.6 0.6 10cm m2#= = -

Relative permittivity of dielectric, re .1 3=Conductivity of dielectric, s 0.

So, the conductance per unit length of line is given as

Gl 0dws= = 0.s

and the capacitance per unit length of the line is given as

Cl . 1.3..

dw

dw 8 85 10

0 6 109 6 10

r012

2

2

##

#e e e # #= = = --

-

^ h

1.84 10 /F m10#= - 0. 8 /nF m2=

SOL 8.1.9 Option (C) is correct.Characteristic impedance of a transmission line is defined as

Z0 G j CR j L

ww=

++

l ll l

and the propagation constant of the transmission line is defined as g ja b= + R j C G j Cw w= + +l l l l^ ^h h

where, a is attenuation constant b is phase constant Rl is resistance per unit length of the line Gl is conductance per unit length of the line Ll is inductance per unit length of the line Cl is capacitance per unit length of the lineNow, for lossless line, Rl G 0= =lSo, the characteristic impedance of lossless transmission line is

Z0 CL=ll

and the propagation constant of lossless transmission line is g ja b= + j L Cw= l l

or a 0=Therefore, the attenuation constant of lossless line is always zero (real).i.e. statement (A) is correct.Again for distortionless line,

LRll CG=ll

So, the characteristic impedance of distortionless line is

Z0 CL

GR= =

ll

ll



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For View Only Shop Online at www.nodia.co.inand the propagation constant of the distortionless line is g ja b= + R G j L Cw= +l l l l

or, a R G= l l 0!

Therefore, the attenuation constant of distortion less line is not zero but it is real.Thus, (A) and (B) is correct statement while (C) is not a correct statement.

SOL 8.1.10 Option (C) is correct.Inductance per unit length, Ll 250 / 250 10 /nH m H m9

#= = -

Capacitance per unit length Cl 0.1 / 0.1 10 /nF m F m9#= = -

So, the velocity of wave propagation along the lossless transmission line is given as

vp .L C

1250 10 0 1 10

19 9

# #= =

- -l l ^ ^h h 10 /m s4 8

#=The characteristic impedance of the lossless transmission line is given as

Z0 .CL

0 1 10250 10

9

9

#

#= = -

-

ll (for lossless line, R G 0= =l l )

50W=

SOL 8.1.11 Option (A) is correct. Operating frequency, f 1 10GHz Hz9= = Conductivity,s 6.4 10 /S m7

#= Permittivity,e 6 0e= Axial component of electric field Ez= Transverse component of electric field Ey=So, the ratio of the two components for the transmission line is

EEy

z .6 4 10

2 10 67

90

#

# #swe p e= = ( f2w p= )

.2 109 3 4#= -

SOL 8.1.12 Option (B) is correct.Given the operating angular frequency of the transmission line is w 6 10 /rad s8

#=and the parameters of transmission line are Rl 0.2 / 200 /k m mW W= = Ll 4 / 4 10 /H m H m6m #= = -

Gl 8 / 8 10 /S m S m6m #= = -

Cl 4 / 4 10 /pF m F m12#= = -

So, the propagation constant of the transmission line is given as g R j L G j Cw w= + +l l l l^ ^h h

j j200 6 10 4 10 8 10 6 10 4 108 6 6 8 12# # # # #= + +- - -

^ ^ ^ ^ ^h h h h h6 6@ @

j j200 24 10 8 10 24 102 6 4# # #= + +- -

^ ^ ^h h h6 6@ @

. .j2 10 1 7= +^ h per meter

SOL 8.1.13 Option (C) is correct.Given the Operating angular frequency of the transmission line,




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w 1.2 10 /rad s9#=

and the parameters of transmission line are Rl 10 /mW= Ll 0.4 / 0.1 10 /H m H m6m #= = -

Cl 10 / 10 10 /pF m F m12#= = -

Gl 40 / 40 10 /S m S m6m #= = -

So, the characteristic impedance of the line is given as

Z0 G j CR j L

ww=

++

l ll l

.

. .j

j40 10 1 2 10 10 10

10 1 2 10 0 1 106 9 12

9 6

# # #

# #=+

+- -

-

^ ^

^ ^

h h

h h

j75 5W= -

SOL 8.1.14 Option (B) is correct.The amplitude of voltage wave after travelling a distance l along a transmission line is given as V1 V e l

0= a-

where V0 is the amplitude of the source voltage wave Now, in the given problem, after travelling 20 m distance along the transmission line the voltage wave remains %13 of it’s source amplitude. So, we get V1 %V e 13l

0= =a- of V0

e 20a- ^ h 0.13= ( 20 ml = ) a 0.10 /NP m=

SOL 8.1.15 Option (B) is correct.Given the propagation constant of the voltage wave g ja b= + . .j0 5 2 4= +So, we get the attenuation constant of the wave a 0.5= and phase constant of the wave along the transmission line is b 2.4=Since, the amplitude of voltage wave after travelling a distance l along a transmission line is given as V1 V e l

0= a-

where V0 is the amplitude of the source voltage wave. Since the amplitude of a voltage wave after travelling a certain distance down a transmission line is reduced by %87 so, for the given transmission line we have

V1 V e l0= a- V1 100

870= -b l

e la- 0.13=

l .ln10 131

a= b l 4.08 m=

Therefore, the shift in phase angle for the travelled distance is given as

f l 2360cb p= b l . .2 4 4 08 2

360cp= ^ ^ bh h l 561c=



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The width of strips w= Separation between strips d=So, the characteristic impedance of lossless transmission line is given as

Z0 wd

em=

When d and W is doubled, the characteristic impedance of the transmission line will be given as

Z0l Wd

22

em= Z0=

Therefore, the characteristic impedance will remain same.

SOL 8.1.17 Option (A) is correct.Attenuation constant, a 10 / 10 /mNP m NP m2= = -

Characteristic impedance, Z0 0.1 100kW W= =Phase velocity, vp 0.5 10 /m s8

#=Since the transmission line is distortion less so, the resistance per unit length of the transmission line is given as Rl Z0a= 1 /m10 1002 W= =-

^ ^h h

and the inductance per unit length of the lossless transmission line is given as

Ll .

/H mvZ

0 25 10100 4

p

08

#m= = =

SOL 8.1.18 Option (C) is correct.Given the parameters of distortionless transmission line are Rl 4 /mW=and Gl 4 10 /S m4

#= -

So, the attenuation constant of the distortion less transmission line is given as a R G= l l 4 4 10 4

# #= - 4 10 /NP m2#= -

and the characteristic impedance of the distortionless transmission line is given as

Z0 GR=ll

15

6 104 04

#W= =- distortionless line

SOL 8.1.19 Option (D) is correct.Operating frequency, f 5 5 10GHz Hz9

#= =Characteristic impedance, Z0 80W=Phase constant, b 1.5 /rad m=So, the inductance per unit length of the transmission line is given as

Ll Z0

wb= .

2 5 101 5 80

9# #

#p

= ( f2w p= )

.8 /nH m4 8=

SOL 8.1.20 Option (A) is correct.Load impedance, ZL 300W=Characteristic impedance, Z0 150W=




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So, the reflection coefficient at the load terminal is given as

LG Z ZZ ZL

L

0

0= +- 300 150

300 15031= +

- =

and the reflection coefficient at generator end is given as

gG Z ZZ Zg

g

0

0= +-

where Zg is internal impedance of the generator. Since, it is given that the internal resistance of the generator is zero (i.e., Zg 0= ) so, we get

gG 0 1500 150 1= +- =-

SOL 8.1.21 Option (D) is correct.The maximum magnitude of voltage wave, Vmax 6 volt=The minimum magnitude of voltage wave, Vmin 2.4 volt=So, the standing wave ratio on the transmission line is given as

S . .VV

2 46 2 5

min

max= = =

Therefore, the reflection coefficient of the transmission line is evaluated as

G .. .S

S11

2 5 12 5 1 0 43= +

- = +- =

SOL 8.1.22 Option (B) is correct.Characteristic impedance, Z0 25W=Inner radius of the coaxial line, a 0.6 0.6 10mm 3

#= = -

Permittivity of insulated material, e 9 0e= & 9re =Now, the characteristic impedance of a lossless coaxial line is given as

Z0 ln ab60

re= b l

where b is the outer radius of the coaxial line. So, we get

25 .

ln b9

600 6 10 3

#= -b l

or, b . e0 6 10 /3 25 9 60#= -

^ h 0.0021 .1m mm3= =

SOL 8.1.23 Option (B) is correct.Load impedance, ZL j15 25 W= -^ h

Characteristic impedance Z0 25W=So, the reflection coefficient of the transmission line is given as

G Z ZZ ZL

L

0

0= +-

jj

15 25 2515 25 25

=- +- -

^

^

h

h

. e0 57 .j79 8= c-

Therefore, the standing wave ratio of the transmission line is determined as

S .. .1

11 0 571 0 57 3 65G

G= -

+= -

+ =

SOL 8.1.24 Option (C) is correct.Characteristic impedance, Z0 50W=Voltage standing wave ratio, S 3=



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For View Only Shop Online at www.nodia.co.inSince, the load connected to the lossless transmission line is purely resistive so, phase angle of the reflection coefficient of the line will be qG 0= or pNow, the magnitude of the reflection coefficient is given as

G .SS

11

3 13 1 0 5= +

- = +- =

So, reflection coefficient of the transmission line is G ejG= qG

. e0 5 j0= or . e0 5 jp

.0 5= or .0 5-For G 0.5= the load impedance of the transmission line is given as

ZL 150 .. 450Z 1

11 0 51 0 5

0 WGG= -

+ = -+ =; ;E E

and for G .0 5=- the load impedance of the transmission line is given as

ZL 150 ..Z 1

11 0 51 0 5 750 WG

G= -+ = +

- =; ;E E

Therefore, the possible values of load impedance connected to the transmission line are ZL 50W= or 450W

SOL 8.1.25 Option (A) is correct.Load impedance, ZL j200 200 W= -^ h

Characteristic impedance, Z0 100W=Length of transmission line, l 10 10 10 0.1cm m2

#= = =-

Generator voltage, v tg ^ h 3 cos voltt109#p= ^ h

So, we get the angular frequency w 109

#p=and the phase constant of the wave on the transmission line is

b v 3 1010

310

p8

9

#

#w p p= = = (in air 3 10 /m svp 8#= )

or lb .310 0 1 3#p p= =

Therefore, the input impedance of the lossless transmission line is given as

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m

/

/tan

tanj jj j

100100 200 200 3200 200 100 3

pp

=+ -- +^ ^

^f

h h

hp

.j35 35 4 W= -^ h

SOL 8.1.26 Option (A) is correct.Load impedance, ZL 0= (Short circuit)Input impedance, Zin 3= (Open circuit)and, wave length l=Now, the input impedance of lossless transmission line is defined as




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Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m

where, l is the length of the transmission line and b is the phase constant of the voltage wave along the transmission line. So, we get

3 tantan

Z j lZ jZ l

000

0 0

bb

=++

^

^

h

h

or, 3 tanjZ l0 b=Since, for a practical transmission line, Z0 3! so, we have tan lb 3=or, lb /2p= (for minimum length)Therefore, the minimum required length of the transmission line is

l 21

#p

b= 2 2#p

pl= 2b l

p=b l

/2l=

SOL 8.1.27 Option (A) is correct.Given,Operating frequency, f 2 MHz= 2 10 Hz6

#=So, the angular frequency of voltage wave is w 2 fp= 4 10 / secrad6p#=When the line is short circuited, input impedance is Zinsc j Lw= (equivalent to 32 nH inductance) j 4 10 32 106 9

# #p= -^ ^h h 0.4j W=

When the line is open circuited, input impedance is

Zinoc j C1w= (equivalent to 20 pF capacitance)

j 4 10 20 10

16 12

# #p= -

^ ^h h 3979.9j W=-

Therefore, the characteristic impedance of the transmission line is given as Z0 Z Zinoc in

sc= . .j j0 4 3978 9= -^ ^h h

02 W=

SOL 8.1.28 Option (D) is correct.Given, the length of the transmission lines 1 and 2 l1 /l 42 l= =So, the input impedance for line 1 is given as :

Zin1 ZZ

150100

L

012 2

= = ^ h 3

200 W=

From the shown arrangement of the transmission line it is clear that the effective load for line 2 will be equal to the input impedance of line 1.

i.e. ZLl Z 3200

in1 W= =

Therefore, the input impedance for the whole combination is



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Zin /ZZ

200 3200

L

022 2

= =l ^

^

h

h 003 W=

SOL 8.1.29 Option (C) is correct.Since the transmission line has one short circuited and one open circuited end so at the short circuit end voltage must be zero while at open circuit end voltage must be maximum. So the voltage standing wave pattern will be half sinusoids with zeros at short circuited end and maxima at the open circuited end.

SOL 8.1.30 Option (C) is correct.The natural frequency of oscillation of a wave in a transmission line of length l which is open circuited at one end and short circuited at other end is given as

fn ln v

42 1 p=

+^ h, , , , ....n 1 2 3 3=

where vp is phase velocity of the wave.

SOL 8.1.31 Option (C) is correct.The dimension of the transmission line will remain same at all frequencies i.e. l will be constant but as it is defined in terms of wavelength which changes with the frequency so , the expression for length will vary in terms of wavelength l. The wavelength of a wave is defined in terms of frequency f as

l fc=

where, c is the velocity of wave in free space so, at 500 Hzf = we have

l c500=

Therefore, the length of transmission line is

l 4l= c

2000= (1)

Now, the wavelength at frequency, 1 1000kHz Hzf = = is given as

l c1000= (2)

Since, the length of the transmission line will be same as determined in equation (1). So, we get

l c2000=

/c21000

= ^ h 2

l= (from eq. (2))




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SOL 8.1.32 Option (D) is correct.Given, the transmission line is terminated in short circuit i.e., ZL 0=and line should be short circuited at its input terminal i.e. Zin 0= .The input impedance of a lossless transmission line is defined as

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m

So, 0 tanZ ZjZ l

00

00

0 b= ++

c m ,Z Z0 0L in= =^ h

tanj lb 0= lb 0= , p , 2p ,..........Since, length of transmission line cant be zero i.e., l 0! so, we get

l /

l l2 2& &b

pp lp l= = =

^ hSOL 8.1.33 Option (D) is correct.

Given, the distance between successive maxima and minima is 10 cm.i.e. /2l 10 cm= l 20 cm=Now, the distance between first minima and load is lmin 7.5 cm=

lmin 4> l

So, the distance between first maxima and load will be

lmax l 4minl= -

7.5 .4 20

7 54

l l l#= - = - 2

l=

***********



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SOLUTIONS 8.2

SOL 8.2.1 Option (C) is correct.Since, the TEM wave is z -polarized i.e. the electric field of the wave is directed along az+ .i.e. aE az=and the direction of wave propagation is along axi.e. ak ax=So, the direction of magnetic field intensity will be aH a ak E#= a ax z#= ay=-As E is in az+ direction and H is in ay- direction so, we can consider the two vectors as E E az z= (1)and H H ay y=- (2)Now, from the maxwell’s equation in phasor form we have H#d j Ewe= (for perfect dielectric 0s = )

H

a a a

0 0

x

x

y

y

y

z

z

-22

22

22 j E az zwe= using equation (1) and (2)

zH

xHa ay

xyz2

222- j E az zwe=

It gives the result as

zHy

22

0=

Again from Maxwell’s equation we have E#d j Hwm=-

E

a a a

0 0

x

x

y

y

z

z

z

22

22

22 j H ay ywm=+ using equation (1) and (2)

yE

xEa az

xzy2

222- j H ay ywm=

So, it gives the result as

yEz22 0=

Thus, Both (A) and (B) are correct.




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SOL 8.2.2 Option (C) is correct.The voltage maximum exists at the point where the incident and the reflected voltage wave both are in same phase and the distance of voltage maximum from the load is given as

lmax n

4 2pq l l= +G (1)

where qG is phase angle of reflection coefficient, l is the wavelength of the voltage wave and , , , ....n 0 1 2=Now, the reflection coefficient of a transmission line is given as

G Z ZZ ZL

L

0

0= +-

. . .

. . .jj

0 3 0 5 0 50 3 0 5 0 5

=- +- -

^

^

h

h

. .. . .jj e0 8 0 5

0 2 0 5 0 57 .79 8= -- - = c-

i.e. qG .79 8c=-So, from equation (1) for n 0= we have

lmax .

4 479 8 4 10

180

2# #

#c

cpq l

pp= = -G

-

0.44 10 m2#=- -

which is negative (i.e. the point doesnt exist). Therefore, the 1st maximum voltage will exist for n 1= and the distance of the 1st maximum from the load is

i.e. lmax 4 2pq l l= +G (n 1= )

.0 44 10 2 102 2# #=- +- - 1.56 10 m2

#= -

.56 cm2=

SOL 8.2.3 Option (B) is correct.In a lossless transmission line, the current maximum lies at the same point where the voltage minima lies and similarly, the current minima lies at the same point where the voltage maxima lies as shown in the figure below :

Now, it is clear from the figure that the distance between two adjacent maxima and minima is /4l



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For View Only Shop Online at www.nodia.co.ini.e. l lmax min- 4

l=

Since the maximum voltage wave lies at a distance lmax 1.56 cm=So, the distance of 1st voltage minimum (the distance of 1st current maxima) from the load will be

lmin l 4maxl= - 1.56 0. cm4

4 24= - =

Thus, the distance of 1st current maximum from the load is 0.56 cm.

SOL 8.2.4 Option (C) is correct.Given,The position of first voltage maximum, lmax 4.5 cm=Position of first current maximum(voltage minima), lmin 1.5 cm=Standing wave ratio, S 3=Characteristic impedance, Z0 50W=Since, the distance between a maximum and an adjacent minimum is /4l as discussed in previous question.i.e. l lmax min- /4l= 4.5 1.5- /4l=

So, l 12 cm=Again the distance of first voltage maximum from the load is given as

lmax n

4 2pq l l= +G

4.5 412

0pq

= +G ^ h (For n 0= )

qG 23p=

Now, the magnitude of reflection coefficient is given as

G .SS

11

3 13 1

42 0 5= +

- = +- = =

So, the reflection coefficient of the transmission line is G qG= G 0.5 3 /2 . e0 5< /j3 2p= = p .j0 5=-Therefore, the load impedance of the transmission line is given as

ZL ..Z jj

11 50 1 0 5

1 0 50 G

G= -+ = +

-: ;D E

j30 40 W= -^ h

SOL 8.2.5 Option (A) is correct.Characteristic impedance, Z0 50W=Load impedance, ZL j30 15 W= +^ h

Length of transmission line, l 7 /20l=

Since, the transmission line is lossless so, the attenuation constant is zero




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i.e. a 0=or, g j ja b b= + =Therefore, the input impedance of the lossless transmission line is given as

Zin tanhtanhZ Z Z l

Z Z lL

L0

0

0

gg= +

+c m tan

tanZ Z jZ lZ jZ l

L

L0

0

0

bb= +

+< F ( jg b= )

50( )

tan

tan

j j

j j

50 30 15 2207

30 15 50 2207

lp llp l

=+ +

+ +J

L

KKKK ^ b

bN

P

OOOOh l

l 2b l

p=b l

50( )

tan

tan

j j

j j

50 30 15 107

30 15 50 107

p

p

=+ +

+ +J

L

KKKK ^ b

bN

P

OOOOh l

l

. .j18 4 19 2 W= -^ h

SOL 8.2.6 Option (B) is correct.In the assertion (A) given,Length of the transmission line, l /4l=Load impedance, ZL 0=

So, we get lb 24 2l

p l p= =b bl l ( /2b p l= )

The input impedance of the lossless transmission line is given as

Zin tantanZ Z jZ l

Z jZ l0

L

L0

0 bb= +

+c m

tan

Z ZjZ

j20

0

03

p= =f p

Now, we consider the reason part,Distance of the maxima from load is given as lmax /2n2q p b= +G^ h

where, qG is the phase angle of reflection coefficient b is the phase constant of the voltage waveand n , , , ....0 1 2=Therefore, the input impedance at the point of maxima is given as

Zin Zee

11

j l

j l

0 2

2

max

max

GG=

-+

b

b

-

-

c m Ze e

e e

1

10 j j n

j j n

2

2

G

G=

-

+q q p

q q p

- +

- +

G G

G G

f^

^

ph

h

(G ejG= qG )

Z 11

0 GG

= -+

f p

So, Zin is real if Z0 is real and since, Z0 is always real for a distortionless line. Thus, Zin will be purely real at the position of voltage maxima in a distortionless line.i.e. A and R both are true but R is not the explanation of A.

SOL 8.2.7 Option (A) is correct.Length of transmission line, l 6 m=



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For View Only Shop Online at www.nodia.co.inCharacteristic impedance, Z0 30W=Relative permittivity, re .2 25=Load impedance, ZL j30 10 W= -^ h

So, we get the angular frequency, w 8 107

#p=and the phase constant of the volatge wave along the transmission line is

b /v c

8 10p r

7#w

ep= = 1v c

prme e

= =c m

.3 10

8 10 2 258

7

#

# #p= .3 10

8 10 1 58

7

#

# #p=

52p=

or, lb 6 2.4 rad52p p#= =

Therefore, the input impedance of the lossless transmission line is given as

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m .

.tan

tanj jj j

3030 30 10 2 430 10 30 2 4

pp

=+ -- +^ ^

^f

h h

hp

. .j12 14 5 48 W= +^ h

SOL 8.2.8 Option (C) is correct.Given the generator voltage to the transmission line, ( )V tg cos t10 8 10 307

# cp= -^ h

So, in phasor form the generator voltage is Vsg e10 j30= c-

and as determined in previous question, the input impedance of transmission line is Zin . .j23 14 5 48 W= +^ h

So, for determining the input voltage, we draw the equivalent circuit for the transmission line as shown in figure below :

Using voltage division, we get the input voltage to the transmission line as

, Vin V Z ZZ

gin g

in#= +c m

or, V ,s in Z ZZVsgin g

in= +c m (in phasor form)

. .. .e j

j10 30 23 14 5 4823 14 5 48j30= + +

+-c m

.e e15 0 44 .j35 9 44= c-^ ^h h




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. e4 4 .j22 56= c-

Thus, the instantaneous input voltage of the transmission line is v tin ^ h Re eV ,s in

j t= w6 @

4.4 .cos voltt8 10 22 567# cp= -^ h

SOL 8.2.9 Option (C) is correct.Given,Load impedances to the line 1 and 2, ZL1 150ZL2 W= =Length of the transmission lines 1 and 2, l1 /5l2 l= =Now, we consider the input impedance of line 1 and line 2 be Zin1 and Zin2 respectively. Since, the transmission line are identical so, the input impedances of the transmission lines 1 and 2 will be equal and given as

Zin1 tantanZ Z Z jZ l

Z jZ lin

L

L2 0

0 1

1 0 1

bb= = +

+c m (lossless transmission line)

tan

tan

j

j100

100 150 25

150 100 25

lp llp l

=+

+J

L

KKKK b

bN

P

OOOOl

l ( /2b p l= )

2

tan

tan

j

j100

100 150 52

150 100 5p

p

=+

+J

L

KKKK b

bN

P

OOOOl

l

. .j70 4 17 24 W= -^ h

Therefore, the effective load impedance of the feedline will be equal to the equivalent input impedance of the parallel combination of the line 1 and 2.i.e. ZLl ||Z Zin in1 2=

. .Z j

2 270 4 17 24in1= =

-^ h Z Zin in1 2=

. .j35 20 8 62 W= -^ h

SOL 8.2.10 Option (B) is correct.Given the length of the feed line, l .0 3l=and as calculated in above question, the effective load impedance of the feedline is ZLl . .j35 20 8 62 W= -^ h

So, lb . .2 0 3 0 6lp l l= =b ^l h

Therefore, input impedance of the feedline (lossless transmission line) is given as

Zin tantanZ

Z jZ lZ jZ l

L

L0

0

0

bb= +

+l

lc m

. . .

. . 0.tan

tanj jj j

100100 35 20 8 62 0 635 20 8 62 100 6

pp

=+ -

- +^ ^

^f

h h

hp

. .j215 14 113 4 W= -^ h




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For View Only Shop Online at www.nodia.co.inOperating frequency f 0.3 0.3 10GHz Hz9

#= =Load impedance, ZL j100 100 W= -^ h

Characteristic impedance Z0 100W=Generator voltage in phasor form, Vsg 150 volt=Internal resistance of generator Zg 100W=Length of the transmission line, l .0 375l=So, the input impedance of the lossless transmission line is given as

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m

.

.

tan

tan

j j

j j100

100 100 100 2 0 375

100 100 100 2 0 375

lp l

lp l

=+ -

- +J

L

KKKK ^ b

bN

P

OOOOh l

l

j200 100 W= +^ h

Now, for determining the load current, we draw the equivalent circuit for the transmission line as shown in the figure below :

Therefore, the voltage across the input terminal of the transmission line is given as

V ,s in Z ZZVsgg in

in= +c m

.jj e150 100 200 100

200 100 106 1 .j8 13= + ++ = c

c m

Since, at any point, on the transmission line voltage is given as zVs ^ h V e ej z j z

0 G= +b b+ -^ h

where V0+ is the voltage due to incident wave, G is the reflection coefficient of the

transmission line at load terminal and z is the distance of the point from load as shown in figure. So, for z l=- V ,s in V e ej l j l

0 G= +b b+ -^ h (1)

Now, the reflection coefficient of the transmission line at load terminal is




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G Z ZZ ZL

L

0

0= +- j

j100 100 100100 100 100= - +

- - . e0 45 .j63 43= c-

Putting the value of G and V ,s in in equation (1), we get

. e106 1 .j8 13c .V e e e0 45. .j j j0

2 0 375 6343 2 0 375= + clp l l

p l+ - -_ b ^ b ^ il h l h

V0+

..

e e ee

0 45106 1

.

.

j j j

j

135 63 43 135

8 13

=+c c c

c

- -

e75 j135= c-

The current at any point on the transmission line is given as

zIs ^ h ZV e ej z j z

0

0 G= -b b+

-^ h

So, the current flowing in the load (at 0z = ) is

IsL ZV 1

0

0 G= -+

^ h .e e10075 1 0 45 .

jj

13563 43= -

cc

--

^ h

. e0 67 .j108 4= c-

Therefore, the instantaneous current at the load terminal will be i tL ^ h Re eIsL j t= w

" , . . .cos t0 67 2 0 3 10 108 49# # cp= -^ h

0. .cos t75 3 10 108 48# cp= -^ h

SOL 8.2.12 Option (C) is correct.Generator voltage in phasor form, Vsg 150 V=Internal impedance of generator, Zg 100W=Load impedance ZL 150W=Length of transmission line, l 0.15l=Characteristic impedance Z0 100W=So, the input impedance of the lossless transmission line is given as

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m

.tan

tan

j

j100

100 150 2 0 15

150 100 2 15

lp l

lp l

=+

+J

L

KKKK b

bN

P

OOOOl

l

tantan

jj100

100 50 54150 100 54

cc= +

+c m 80.5 32.j 7 W= -^ h

Now, for determining the power delivered, we draw the equivalent circuit for the transmission line as shown in figure below :

Using voltage division, we get the input voltage as



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V ,s in V Z ZZ

gin g

in= +c m . .. .jj150 82 5 32 7 100

82 5 32 7= - +-

c m . e71 8 .j11 46= c-

So, the current at the input current is

I ,s in ZV ,

in

s in= . ..je

82 5 32 771 8 .j11 46

= -c-

. e0 81 .j10 16= c

Therefore, the average input power delivered to the transmission line is given as

Pin Re V I21

, ,s in s in= )6 @ . .Re e e2

1 71 8 0 81. .j j11 46 10 16= c c- -^ ^h h6 @

27 Watt=

SOL 8.2.13 Option (C) is correct.Since, the lengths of line 1 and line 2 are l1 /l 22 l= =So, the input impedance of the line 1 is given as

Zin1 tantanZ Z jZ l

Z jZ lL

L0

0 1

1 0

bb= +

+c m

tan

tanZZ jZ

Z jZ

22

22

L

L

0

0 1

1 0

lp llp l

=+

+J

L

KKKK b

bN

P

OOOOl

l ( /2b p l= )

Z ZZ Z0

0LL0

0

11= +

+ =b l

50W=Similarly, the input impedance of line 2 is given as Zin2 ZL2= 150W=The effective load for line 3 will be equal to the equivalent impedance of the parallel combination of input impedances of line 1 and line 2.i.e. ZLl ||Z Zin in1 2=

2150= 75W=

So, the input impedance for line 3 is given as Zin ZL= l 75W= (length of line 3, /l 2l= )Therefore, the input voltage of line 3 is

V ,s in Z ZZVsgin g

in= +c m 500 75 10075= +b l

214.28 volt=and so the current at the input terminal of line 3 is

I ,s in 2.86 AZV ,

in

s in= =

Thus, the average power delivered to the lossless transmission line 3 is given as Pin )Re V I, ,s in s in= 6 @

. .21 214 28 2 86# #= ^ ^h h 306.11 Watt=

Since, the transmission line is lossless so, the power delivered to each load will be same and given as




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P1 P2= P2in= 306.11 153.1 Watt2

1#= =

SOL 8.2.14 Option (D) is correct.Given, transmission line is of infinite length i.e. l 3= .and input impedance of the transmission line is equal to its characteristic impedancei.e. Zin Z0=Since, the input impedance of a transmission line is defined as

Zin tanhtanhZ Z Z l

Z Z lL

L0

0

0

gg= +

+c m

So, Z0 tanhtanhZ Z Z l

Z Z lL

L0

0

0

gg= +

+c m

Solving the equation, we get tanh lg 1=

e ee el l

l l

+-

g g

g g

-

- 1=

e lg- 0=Since, l 3= . So for satisfying the above condition propagation constant g must have a real part.i.e. real part of g 0!

or, a 0! ( jg a b= + )As the attenuation constant of the voltage wave along the transmission line is not equal to zero therefore, it is a lossy transmission line.

SOL 8.2.15 Option (C) is correct.As discussed in previous question the input impedance of infinitely long lossy transmission line is equal to it’s characteristic impedance. So, the input impedance to line 1 will be Zin1 200Z01 W= =From the shown arrangement of the transmission line it is clear that the effective load impedance for line 2 will be equal to the input impedance of line 1.i.e. ZL2 200Zin1 W= =Since the length of the line 2 is /2l so, the input impedance of line 2 will be equal to its load i.e. Zin2 ZL2= 200W= ( /l 2l= )Therefore, the reflection coefficient at the load terminal of line 2 is given as

G Z ZZ Z

200 100200 100

L

L

2 02

2 02= +- = +

- 31=

Now, the input voltage of line 2 is determined by using voltage division rule as

V ,s in Z ZZV ,s gin g

in

2

2= +c m

4 200 100200= +b l volt3

8=

Again, the voltage at any point on line 2 is given as



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For View Only Shop Online at www.nodia.co.in zVs ^ h V e ej z j z

0 G= +b b+ -^ h (lossless line)

where V0+ is voltage of incident wave b is phase constant of the voltage wave and

z is distance from load. So, for z /2l=-

zVs ^ h V e ej j0

22

22G= +l

p llp l+ - - -_ b b il l

38 V e ej j

0 G= +p p+ -^ h ( zV V ,s s in=^ h at z /2l=- )

V0+ 3

81 3

11

#=- -b l

31G =b l

2 volt=-Therefore, the incident average power to the line 2 is given as

Pavi 20 mWattZV2 2 100

40

02

2

#= = =

+

So, the reflected average power at the input terminal of line 1 (load terminal of line 2) is

Pavr 20 2.2 mWattP 31

avi2

2G #= = =b l

Thus, we get the transmitted power to the line 1 as Pavt 20 2.2 17.8 mWattP Pav

iavr= - = - =

SOL 8.2.16 Option (B) is correct.Consider the length of the transmission line is l as shown in figure below

The generator voltage is applied to the transmission line at time t 0= for which the voltage at the sending end is v 0^ h 10 volt= (at t 0= )After time 4 st mD = the voltage v t^ h at the sending end changes to 6 V. This change in the voltage will be caused only if the reflected voltage wave from the load comes to the sending end. So, the time duration for the change in voltage at sending end can be given as tT =(time taken by incident wave to reach the load) + (time taken by reflected wave to reach sending end from the load)

or, tT vlvl

p p= + v

l2p

= (1)

where l is the length of the transmission line (distance between load and sending terminal) and vp is phase velocity of the wave along the transmission line. Since,




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the line is air spaced so, vp 3 10 /m sc 8

#= =Putting it in equation (1) we get

4 sm l3 10

28

#= ( 4 st mD = )

Thus, length of the transmission line is

l 0 m23 10 4 10 24

8 6# # #= =

-

SOL 8.2.17 Option (C) is correct.Let the load impedance connected to the transmission line is ZL so the equivalent circuit for the transmission line will be as shown in figure below :

Since, the internal resistance of the generator is equal to the characteristic impedance of the linei.e. Rg Z 1000 W= =So, the reflection coefficient due to source resistance will be zero and therefore, the change in voltage at sending will be caused only due to the reflection coefficient at load terminal given as ( )v tD V0G= +

where, V0+ is amplitude of the incident voltage wave and G is the reflection coefficient

at the load terminal. Since, the change in voltage at 4 st m= is ( )v tD 6 10 4= - =-So, we get 4- 10G= ( 10 VV0 =+ )

G .104 0 4=- =-

Z ZZ ZL

L

0

0

+-

b l .0 4=-

ZZ

100100

L

L

+-

b l .0 4=- ( 100Z0 W= )

10Z 0L- . Z0 4 40L=- - ZL .8629 W=

SOL 8.2.18 Option (B) is correct.Observing the waveform we conclude that at the sending end voltage changes at t t1= . The changed voltage at the sending is given as v t1^ h V V VL g L0 0 0G G G= + ++ + + (1)



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For View Only Shop Online at www.nodia.co.inwhere V0

+ is voltage at sending end at t 0= , LG and gG are the reflection coefficients at the load terminal and the source terminal respectively. So, we get

LG Z ZZ ZL

L

0

0= +- 1=- (Z 0L = )

and gG R ZR Zg

g

0

0= +-

(Z Rg g= )

Putting these values in equation (1), we get v t1^ h V V Vg1 1 1G= - -+ + +

v t1^ h Vg 1G=- + (2)From the shown wave form of the voltage at sending end, we have v t1^ h 6 volt= V0

+ 24 volt=Putting these values in equation (2), we get 6 24gG=- ^ h

or, gG 4=-

R ZR Zg

g

0

0

+-

4=-

Rg 60W= 100Z0 W=^ h

At t 0= as the voltage just applied to transmission line, the input impedance is independent of ZL and equals to Z0 (i.e. Z Zin 0= at t 0= ). Therefore, using voltage division the voltage at the sending end is given as

V0+ V R Z

Zg

g 0

0= +c m

24 V 60 100100

g= +b l (V0+ 24 volt= )

Vg 10024 160#= 38.4 volt=

SOL 8.2.19 Option (D) is correct.Length of the transmission line, l 1.5 m=Internal resistance of generator, Rg 200W=Characteristic impedance, Z0 100W=Generator voltage, Vg 30 volt=Load impedance, ZL 50W=So, the reflection coefficient at the load terminal is

LG Z ZZ Z

50 10050 100

31

L

L

0

0= +- = +

- =-

and the reflection coefficient at the source terminal is

gG R ZR Z

200 100200 100

31

g

g

0

0= +- = +

- =

Again as discussed in previous question at time, t 0= as the voltage is just applied to the transmission line, the input impedance is independent of ZL and equals to Z0 (i.e. Z Zin 0= at t 0= ). Therefore, using voltage division the input voltage at the sending end is given as




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V0+ V R Z

Zg

g 0

0= +c m 30 200 100100 10 volt#= + =b l

Now, the time taken by the wave to travel from source terminal to the load terminal (or load terminal to source terminal) is given as

T cl=

where, l is length of transmission line and c is the velocity of the voltage wave along the transmission line. So, we get

T . ns3 10

1 5 58#

= =

Therefore, for the interval 0 5 nst <# , the incident wave will be travelling from source to load and will have the voltage V1

+ 10 volt=For the interval 5 10ns nst <# an additional reflected wave will be travelling from load to source and will have the voltage

V1- 3.33 voltV 3

10L 1G= =- =-+

For 10 15ns nst# # the wave reflected by source resistance travelling from source to load will be added to that has the voltage

V2+ . 1.11 voltV 3

3 33g 1G= =- =--

For 15 20ns nst# # again the wave reflected by load travelling from load to source will be added that has the voltage

V2- . 0.37 voltV 3

1 11L 2G= = =+

This will be continuous and the bounce diagram obtained between source at z 0=^ h and load (at 1.5 mz = ) will be as shown in figure below :

SOL 8.2.20 Option (C) is correct.From the bounce diagram that obtained between source terminal ( 0z = ) and load terminal ( 1. mz 5= ) in previous question, we can determine the voltage v t^ h at any



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For View Only Shop Online at www.nodia.co.ininstant by just summing all the voltage waves existing at any time t .Since, for interval 0 10 nst# # only a single voltage wave with 10 voltV1 =+ exists at sending end so, the voltage at the sending end (z 0= ) for the interval is v t^ h 10 voltV1= =+ for 0 10 nst <#

again for the interval 10 20ns nst <# , three voltage waves with 10 voltV1 =+ , 3.33 voltV1 =-- and 1.11 voltV2 =-+ exists at the sending end so, the voltage at

the sending end for the interval is v t^ h V V V1 1 2= + ++ - + . .10 3 33 1 11= - - 5.6 volt= for 10 20ns nst <#

Thus, the obtained voltage wave form is plotted in the figure below





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As shown in the smith chart, SWR circle meets the rG axis (real part of reflection coefficient) at L1 and L2 respectively. So, We have the two possible values of normalised impedance (real values of zL). zL1 .2 5= at L1

zL2 .0 4= at L2

Since, the normalised impedance is defined as

zL Characteristic impedanceLoad impedance=

So, we have zL1 .ZZ 2 5L

01= =

or, Z01 . . 20Z2 5 2 5

50L W= = =

Similarly, zL2 .ZZ 0 4L

02= =

or, Z02 . . 125Z0 4 0 4

50L W= = =

Therefore, the two possible values of the characteristic impedance of the lossless transmission line are 20W and 125W.

SOL 8.2.22 Option (B) is correct.We can determine the reflection coefficient of the transmission line using smith



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For View Only Shop Online at www.nodia.co.inchart as explained below :(1) First we determine the normalized load impedance of the transmission line as

zL .ZZ j j100

100 50 1 0 5L

0= = + = +

(2) Comparing the normalized impedance to its general form

zL r jx= + where r is the normalized resistance (real component) and x is the normalized

reactance (imaginary component). we get

r 1= and .x 0 5=(3) Now, we determine the intersection point of r 1= circle and .x 0 5= circle on

the smith charge and denote it by point P as shown in the smith chart. It gives the position of normalized load impedance.

(4) We join the point P and the centre O to form the line OP

(5) Extend the line OP to meet the r 0= circle at Q . The magnitude of the reflection coefficient of the transmission line is given as

G 9.42.1 .cm

cmOQOP 0 22= = =

(6) Angle of the reflection coefficient in degrees is read out from the scale at point Q as

qG 76.0c=(7) Thus, we get the reflection coefficient of the transmission line as

G 0.22ej76qG= = cG




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Alternate Method :Reflection coefficient of the transmission line is defined as

G Z ZZ ZL

L

0

0= +- j

j100 50 100100 50 100= + +

+ - . . e0 24 76 0 24 j76c= = c

which is same as calculated from smith chart.

SOL 8.2.23 Option (C) is correct.As shown in the smith chart in previous question normalized load impedance is located at point P . So, for determining the input impedance at a distance of .0 35l from the load we follow the steps as explained below :(1) First we draw a SWR circle (circle centered at origin with radius OP )

(2) For finding input impedance at a distance of .0 35l from load we move a distance of .0 35l on WTG scale (wave length toward generator) along the SWR circle.

(3) Since, the line OP corresponds to the reading of .0 144l on WTG scale so, after moving a distance of .0 35l on WTG scale we reach at . .0 144 0 35l l+ .0 494l= on WTG scale. The reading corresponds to the point A on the SWR circle.

(4) Taking the values of r and x -circle at point A we find out normalized input impedance as



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For View Only Shop Online at www.nodia.co.in zin r jx= + . .j0 61 0 022= + -^ h . .j0 61 0 022= -(5) Therefore, the input impedance at a distance of .0 35l from load is given as

Zin z Zin 0= . .j100 0 61 0 022= -^ h .j61 2 2 W= -^ h

Alternate Method :We can conclude the input impedance at .l 0 35l= directly by using formula

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m lossless transmission line

.

0.

tan

tan

j j

j j100

100 100 50 2 0 35

100 50 100 2 35

lp l

lp l

=+ +

+ +J

L

KKKK ^ b

bN

P

OOOOh l

l

.j87 2 2 W= -^ h

as calculated above using with chart.

SOL 8.2.24 Option (C) is correct.For determining the shortest length of the transmission line for which the input impedance appears to be purely resistive, we follow the steps as explained below :(1) First we determine the WTG reading of the point denoting the normalized

load impedance on the smith chart. From the above question, we have the reading of point P as .0 144l on WTG circle.

(2) Since, the resistive load lies on the real axis of reflection coefficient ( rG -axis). So, we move along the SWR circle to reach the rG -axis and denote the points as A and B .

(4) Since, point B is nearer to the point P so, it will give the shortest length of the transmission line for which the input impedance appears to be purely resistive.

(5) Now, we have the reading of point B on WTG scale as .0 25l. So, the shortest length for the input impedance to be purely resistive is given as the difference between the readings at point B and P . i.e.,

l . .0 25 0 144l l= - .0 106l=

SOL 8.2.25 Option (A) correct.The voltage maximum occurs at the point where the SWR circle intersects the positive rG axis on smith chart. The SWR circle of the load impedance intersects the positive rG axis at point B as shown in the Smith chart. So, the point B gives the position of first voltage maxima.As calculated in previous question the distance between point B and point A on the WTG scale is .0 106l. Therefore, the 1st voltage maximum occurs at a distance of .0 106l from load.

SOL 8.2.26 Option (B) is correct.At any time t , the currents of positive and negative waves are respectively I+ and I- and the voltages of positive and negative waves are respectively V+ and V- as shown in the figure.




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I+ ZV

Z1

0 0= =

+ 1 VoltV =+

^ h

and I- ZV

0=-

-

Now, the voltage and current across an inductor are related as

v Ldtdi=

V V++ - dtd I I2= ++ -^ h

V1 + - dtd V2 50

1= - -

: D ,V Z1 500= =+^ h

V1 + - dtdV

251=-

-

dt25- V

dV1

=+ -

-

Taking integration both sides we get ln V1 + -

^ h t C25 1=- + where C1 is a constant V1 + -

^ h Ae t25= - (1)Since, the voltage V+

^ h wave is incident at t 0= so, at t 0= + the current through inductor is zero and therefore, from the property of an inductor at t 0= + the current through inductor will be also zero.i.e. I I 0at t++ -

= +^ h 0=

ZV

ZV

0at t0 0-

+ -

= +; E 0=

Z ZV1

0at t0 0

0--

= +: D 0= 1 VoltV =+

^ h

So at t 0= +, V0- 1 volt=

Putting it in equation (1), we get 1 1+^ h A= A 2=Thus, the voltage of the reflected wave is V- Volte2 1t25= --

^ h

SOL 8.2.27 Option (D) is correct.The voltage of positive wave in transmission line is V0

+. So, at the voltage maxima, magnitude of the voltage is given as V maxs V 10 G= ++

6 @



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For View Only Shop Online at www.nodia.co.inand at the point of voltage maxima the current will be minimum and given as

I mins ZV

10

0G= -

+

6 @

So, the line impedance at the point of voltage maxima will be

Zmax IV

Z 11

min

max

s

s0 G

G= = -+

b l

Z S0= S 11

GG= -

+b l

Now, at the voltage minimum the voltage magnitude is V mins V 10 G= -+

6 @

and at the point of voltage minimum current will be maximum and given as,

I maxs ZV

10

0G= +

+

6 @

and the line impedance at the point will be

Zmin IV

Z 11

max

min

s

s0 G

G= = +-

b l SZ0= S 1

1GG= -

+b l

SOL 8.2.28 Option (A) is correctTo determine the required quantity, we note that for a particular line of characteristic impedance Z0, the product of the line impedances at two positions (two values of d ) separated by an odd multiple of /4l is given by

Z d Z d n2 1 4l+ -^ h6 :@ D" ', 1

/Z

dd

Zd n

d n

11

1 2 1 4

1 2 1 40 0 l

l

GG

G

G=

-+

- + -

+ + -

^

^f

^^

^b

h

hp

h h

h l

* *4 4

Z dd

d e

d e11

1

1j n

j n

02

2 2 14

2 2 14

GG

G

G= -

+

-

+b l

b l

- -

- -

^

^

^

^

^

^

h

h

h

h

h

h

> >H H

Zdd

d ed e

11

11

j n

j n

02

2 1

2 1

GG

GG

=-+

-+

p

p

- -

- -

^

^

^

^^

^

h

h

h

hh

h

> >H H

Zdd

dd

11

11

02

GG

GG

=-+

+-

^

^

^

^

h

h

h

h> >H H

Z02=

As the intrinsic impedance of medium 1 is 1h and that of medium 3 is 3h so, for required match, thickness t is /4l and the intrinsic impedance ( 2h ) of the medium 2 is given as 1 3h h 2

2h= or 2h 1 3h h=

SOL 8.2.29 Option (C) is correct.As determined in previous question, for a wave travelling through the three mediums of intrinsic impedances 1h , 2h and 3h , the condition for matching dielectric (the intrinsic impedance of medium 2 that eliminates the reflected wave in medium 1) is 2h 1 3h h=Since, all the media have 0m m= so, for the dielectrics 0s =^ h the above equation can be rewritten as




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2

0

em 160

0

0

0

em

em= b bl l h e

m=b l

where 2e is the permittivity of the medium 2. e 4 0e=

SOL 8.2.30 Option (B) is correct.The thickness ‘t ’ of the dielectric coating for the perfect matching (the condition for eliminating reflection) is given as

t 4l= (quarter wave)

where l is the wavelength of plane wave. The wavelength in terms of frequency is

l fvp=

where vp is the phase velocity of the wave in the propagation medium which is given as

vp .14

12

3 10 1 5 100 0

88#

#me m e

= = = =

So, at frequency f 1.5 GHz= the thickness of the dielectric coating is given as

So, t fv4p=

.. 0.25 2.5m cm

4 1 5 101 5 10

9

8

#

#= = =^ h

SOL 8.2.31 Option (B) is correct.Distance between load and first voltage maxima, lmax .0 125l=Characteristics impedance, Z0 100W=Standing wave ratio, S 3=Position of voltage maxima (lmax) in terms of reflection coefficient qG G is

lmax n

4 2pq l l= +G where n 0= , 1, 2,.......

So, for 1st voltage maxima we have n 0= and so, we get the position of first voltage maxima as

lmax 4pq l= G

.0 125l 4pq l= G & qG 2

p=

The magnitude of reflection coefficient is defined in terms of SWR as

G SS

11

3 13 1

21= +

- = +- =

So, the reflection coefficient of the transmission line is

G e j21

2/j 2qG= = =p

G

Therefore, the load impedance of the transmission line is given as

ZL Z 11

0 GG= -

+b l j

j100

1 2

1 2=-

+J

L

KKK

N

P

OOO j60 80 W= +^ h




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For View Only Shop Online at www.nodia.co.inGiven, the transmission line is terminated by its characteristic impedance i.e., ZL Z0=So, there will be no reflected wave and therefore, the height of the voltage pulse will be given as

V1+ Z Z

Z Vg

g

0

0= + (Zg " internal resistance of generator)

100 50100 15#= + 10 Volt=

As the wave travels in the Z+ direction along transmission line at velocity

vp .L C

10 25 10 100 10

16 12

# # #= =

- -l l ^ ^h h 2 10 /m s8

#=So, the voltage pulse will reach at 5 ml = at time,

t0 25 ns2 10

58

#= =

So, at 5 ml = for 0 25 nst< < , V 0=and for 25 nst$ V 10 VoltV1= =+

Therefore the plot of voltage against time at a distance 5 m from the source is as shown in graph below.

SOL 8.2.33 Option (A) is correct.As the first forward voltage pulse is V1

+ so, the first reflected pulse voltage is V1

- VL 1G= +

The 2nd forward pulse voltage is given as V2

+ V Vg g L1 1G G G= =- +

The 2nd reflected pulse voltage is given as V2

- V VL g L22

1G G G= =+ +

So, summing up all the pulses at load end for steady state (t" 3) we get the load voltage as VL ...V V V V1 1 2 2= + + + ++ - + -

....V 1 L g L g L12G G G G G= + + + ++

6 @

.... ....V 1 1g L g L L g L12 2G G G G G G G= + + + + + ++ ^ ^h h7 A

V 11

1g L g L

L1 G G G G

G= - + -

+c cm m< F V 1

1g L

L1 G G

G= -++

c m




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SOL 8.2.34 Option (A) is correct.Characteristic impedance, Z0 60W=Load impedance, ZL 180W=Voltage generator, Vg 100 V=Internal resistance, Zg 120W=So, the first forward voltage pulse will be

V1+ 100 VoltZ Z

Z V 60 12060

3100

gg

0

0= + = + =c bm l

The reflection coefficient at load terminal is given as

LG Z ZZ Z

180 60180 60

21

L

L

0

0= +- = +

- =

The reflection coefficient at source terminal is given as

gG Z ZZ Z

120 60120 60

31

g

g

0

0= +- = +

- =

Therefore, the voltage across the load at steady state is given by the expression as determined in previous question

i.e. VL V 11

g L

L1 G G

G= -++

c m /

3100

1 31

21

1 1 2=-

+

b bf

l lp

7 Volt3100

23

56 5# #= =

SOL 8.2.35 Option (D) is correct.Voltage generator, Vg 50 Volt=Internal impedance, Zg 30W=Characteristic impedance, Z0 15W=Load impedance, ZL 45W=So, first forward voltage pulse is

V1+ Z Z

Z Vg

g0

0= +c m 15 3015 50= +b l 3

50=

Now, the reflection coefficient at source terminal is

gG Z ZZ Z

30 1530 15

31

g

g

0

0= +- = +

- =

and the reflection coefficient at load terminal is

LG Z ZZ Z

45 1545 15

21

L

L

0

0= +- = +

- =

So, at steady state t 3=^ h voltage across load is given as

VL V 11

g L

L1 G G

G= -++

c m 350

1 21

31

1 21

=-

-J

L

KKK b b

N

P

OOOl l

350

56

23

# #= 30 Volt=

Therefore, the current through load at steady state is given as

IL AZV

4530

32

L

L= = =



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Since, the internal resistance of the battery is zero so, the 1st forward voltage pulse is V1

+ 6 VoltVg= =and from the plot we get the first forward pulse current as I1+ 75 mA=Therefore, the characteristic impedance of the transmission line is given as

Z0 80IV

75 106

1

13

#W= = =+

+

-

SOL 8.2.37 Option (D) is correct.Reflection coefficient at source and load end are given as

gG Z ZZ Z

1g

g

0

0= +- =-

and LG Z ZZ ZL

L

0

0= +-

Now, from the plot of input current (current at generator end) we get, V1

+ 75 mA= (1)and V V V1 1 2- ++ - + 5 mA=- (2)where, V1

+ is the first forward voltage pulse, V1- is the first reflected voltage pulse

and V2+ is the second forward voltage pulse. So, putting the values of these voltages

in terms of reflection coefficients we get V V VL L g1 1 1G G G- ++ + + 5 mA=- V 1 L L1 G G- -+

^ h 5 mA=- 1gG =-^ h

1 2 LG- 755=- 75 mAV1 =+

^ h

or, LG 158=

For determining load resistance of the line the reflection coefficient is written in the terms of impedances as

Z ZZ ZL

L

0

0

+- 15

8=

ZZ

8080

L

L

+- 15

8= ( 80Z0 W= as calculated in previous question)

Z 15 8L -^ h 80 8 15 80# #= +Thus, ZL 262.85W=

***********




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SOLUTIONS 8.3

SOL 8.3.1 Option (B) is correct.Characteristic impedance of a coaxial cable is defined as

Z0 ln ab

em= b l

where, b " outer cross sectional diameter a " inner cross sectional diameter

So, Z0 ln ab

r0

0

e em= b l

.

.ln10 10 89

4 10 361

2 49

7

#

# #p p= -

-

b l

05 W=

SOL 8.3.2 Option (C) is correct.Since, Z0 Z Z1 2= 100 50 340#=As this is quarter wave matching so, the length of the transmission line would be odd multiple of /4l .

Now, l ( )m2 1 4l= +

For 429 ,MHzf1 = l1 0.174 mfc

4 429 10 43 10

16

8

# # #

#= = =

For 1GHzf2 = , l2 0.075 mfc

4 1 10 43 10

29

8

# # #

#= = =

Now, only the length of the line given in option (C) is the odd multiple of both andl l1 2 as :

( )m2 1+ .l

1 58 91

= =

( )m2 1+ .l

1 58 342

-=

Therefore, the length of the line can be approximately 1.58 cm.

SOL 8.3.3 Option (C) is correct.Length on the transmission line, d 2 mm=Operating frequency, f 10= GHzPhase difference, q /4p=Since the phase difference between the two points on the line is defined as



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For View Only Shop Online at www.nodia.co.in q d2

lp=

where l is operating wavelength and d is the distance between the two points. So, we get

4p d2

lp=

or l 8 8 2 16mmd #= = = mmTherefore, the phase velocity of the wave is given as vp 10 10 16 10f 9 3l # # #= = - .6 10 / secm2 8

#=

SOL 8.3.4 Option (D) is correct.Since, voltage maxima is observed at a distance of /4l from the load and we know that the separation between one maxima and minima equals to /4l so voltage minima will be observed at the load.Now, the input impedance at the point of voltage minima on the line is defined as

Zminin6 @ S

Z0=

where, Z0 is characteristic impedance and S is the standing wave ratio on the line. Therefore, the load impedance of the transmission line (equal to the input impedance at load) is given as

ZL Zminin= 6 @ S

Z0= 550 10W= = ( 50Z0 W= , S 5= )

SOL 8.3.5 Option (C) is correct.For a lossless network, S S11

221

2+ 1=Since, from the given scattering matrix we have 0.2S 011 c= , .S 0 9 9012 c= .S 0 9 9021 c= , .S 0 1 9022 c=So, we get ( . ) ( . )0 2 0 92 2+ 1!

Therefore, the two port is not lossless.Now, for a reciprocal network, S12 S21=As for the given scattering matrix we have S12 .S 0 9 9021 c= =Therefore , the two port is reciprocal.

SOL 8.3.6 Option (A) is correct.For a distortion less transmission line characteristics impedance

Z0 GR= (1)

Attenuation constant for distortionless line is a RG= (2)

So, using equation (1) and (2) we get

a ZR

0= . .50

0 1 0 002= =




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SOL 8.3.7 Option (B) is correct.For a lossless transmission line, the input impedance is defined as

Zin tantanZ Z jZ l

Z jZ lo

o L

L o

bb= +

+< F

Now, for the quarter wave ( /4l ) line we haveLoad impedance, ZL 30W=Characteristic impedance, Zo 30W=

Length of the line, l 4l=

So, tan lb tan 24 3l

p l= =b l 2b lp=b l

Therefore, the input impedance of the quarter wave line is

Zin1

tan

tanZ

lZ jZ

lZ jZ

oo

L

Lo

b

b=+

+R

T

SSSS

V

X

WWWW

60ZZL

02

W= =

Now, for /8l transmission line we haveLoad impedance, ZL 0W= (short circuit)Characteristic impedance, Zo 30W=

Length of the line, l 8l=

So, we get

tan lb 1tan 28l

p l= =b l

Therefore, the input impedance of the /8l transmission line is given as Zin2 30tanjZ l jo b= =The equivalent circuit is shown below :

The effective load impedance of the 60W transmission line is ZL j60 30= +So, the reflection coefficient at the load terminal is

G Z ZZ ZL o

L o= +- j

j60 3 6060 3 60

171= + +

+ - =

Therefore, the voltage standing wave ratio of the line is given as

S 11

GG

= -+

.1 171 17 1 64=-+ =



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The transmission line are as shown below. Length of all line is 4l

The input impedance of a quarter wave ( /4l ) lossless transmission line is defined as

Zin ZZL

02

=

where, Z0 is the characteristic impedance of the line and ZL is the load impedance of the line. So, for line 1 we have the input impedance as

Zi1 200ZZ

50100

L1

012 2

W= = =

Similarly, for line 2, the input impedance is

Zi2 200ZZ

50100

L2

022 2

W= = =

So, the effective load impedance of line 3 is given as ZL3 || 200 || 200 100Z Zi i1 2 W W W= = =Therefore, the input impedance of line 3 is

Zi 5ZZ

10060 4

L3

02 2

W= = =

SOL 8.3.9 Option (A) is correct.The input impedance of the lossless transmission line is defined as

Zin tantan

ZZ jZ lZ jZ l

oo L

L o

bb

=++

^

^f

h

hp

Since, the given transmission line of characteristic impedance 75Z0 W= is short circuited (Z 0L = ) at its one end. Therefore, the input impedance of the line is Zin tanjZ lo b= ^ h

Now, the operating wavelength of the line is

l 0.1fc

3 103 10

9

8

#

#= = = m or 10 cm ( 3 GHzf = )

So, lb 1l2102

5lp p p

#= = = ( 1 cml = )

Therefore, Zin tanjZ 5op=

Since, ( /5)tanZo p is positive so, Zin is inductive.




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SOL 8.3.10 Option (C) is correct.The 2-port scattering parameter matrix is

S SSSS

11

21

12

22= > H

S11 ||||

Z Z ZZ Z ZL o

L o

0

0=+-

^

^

h

h

|| 50||

50 5050 50 50

31=

+-

=-^

^

h

h

S12 S21= ||

||Z Z ZZ Z2

L o o

L o=

+^

^

h

h

||||

50 50 502 50 50

32=

+=

^

^

h

h

S22 ||||

Z Z ZZ Z ZL o o

L o o=+-

^

^

h

h

||||

50 50 5050 50 50

32=

+-

=-^

^

h

h

SOL 8.3.11 Option (A) is correct.The input impedance of a quarter wave ( /l 4l= )lossless transmission line is defined as

Zin ZZL

o2

=

where, Z0 is characteristic impedance and ZL is the load impedance of the line. So, we have the input impedance of line 1 as

Zin1 25ZZ

10050

L

o

1

12 2

= = =

Similarly, the input impedance of line 2 is

Zin2 12.5ZZ

20050

L

o

2

22 2

= = =

The effective load impedance of the line 3 is given as ZL ||Z Zin in1 2=

25 || 12.5=325=

So, the input impedance of the 50W transmission line is

ZS /( )

30025 350 2

= =

Therefore, the reflection coefficient at the input terminal is given as

G Z ZZ Z

300 50300 50

72

S

S

0

0= +- = +

- =

SOL 8.3.12 Option (D) is correct.We have 10 logGp 10= dBor Gp 10=The power gain of the antenna is defined as

Gp PPin

rad=

where Prad is the radiated power of the antenna and Pin is the input power feed to the antenna. So, putting all the values we get

10 1WPrad=

or Prad 10= Watts



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The characteristic impedance of a transmission line is defined as Z0

2 Z Zoc sc=where Zoc and Zsc are input impedance of the open circuited and is short circuited line. So, we get

Zsc ZZoc

02

= j j100 15050 50

2 350#= + = +

( )j13

50 2 3= - .69 .54j8 15= -

SOL 8.3.14 Option (C) is correct.From the diagram, VSWR is given as

S 4VV

14

min

max= = =

Since, voltage minima is located at the load terminal so, the load impedance of the transmission line is given as

ZL 12.5Z SZ

450

minino W= = = =6 @ ( 50 ,Z S 40 W= = )

SOL 8.3.15 Option (D) is correct.The reflection coefficient at the load terminal is given as

G .. 0.6Z Z

Z Z125 5012 5 50

L O

L O= +- = +

- =-

SOL 8.3.16 Option (C) is correct.The given circles represent constant reactance circle.

SOL 8.3.17 Option (C) is correct.The ratio of the load impedance to the input impedance of the transmission line is given as

VVin

L ZZin

0=

or VL ZZ Vin

in0= 60 V50

10 300#= =

SOL 8.3.18 Option (D) is correct.Suppose at point P impedance is Z ( 1)r j= + -If we move in constant resistance circle from point P in clockwise direction by an angle 45c, the reactance magnitude increase. Let us consider a point Q at 45c from point P in clockwise direction. It’s impedance is Z1 0.5r j= -or Z1 0.5Z j= +Thus movement on constant r - circle by an 45c in CW direction is the addition of inductance in series with Z .

SOL 8.3.19 Option (A) is correct.The VSWR of a transmission line is defined as

S 11

GG

=+-




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where G is the reflection coefficient of the transmission line. So, we get

2 11

GG

=+-

(S 2= )

or G 31=

Thus, the ratio of the reflected and incident wave is given as

PPi

r 912G= =

or Pr P9i=

i.e. 11.11% of incident power is reflected.

SOL 8.3.20 Option (D) is correct.The input impedance of a lossless transmission line is defined as

Zin tantan

ZZ jZ lZ jZ l

oo L

L o

bb

=++

6

6

@

@

Now, for /2l transmission line we have l /2l=and ZL1 100W=So, the input impedance of the /2l transmission line is

Zin1 100tantan

ZZ jZZ jZ

Zoo L

L oL

1

11p

pW=

++

= =6

6

@

@ 2b l

p=b l

For /8l transmission line, we have l /8l=and ZL2 0= (short circuit)So, the input impedance of /8l line is

Zin2 0[ ]

50tan

ZZjZ

jZ j0

oo

oo

4 W=+

+ = =p

6 @ 2b l

p=b l

Thus, the net admittance at the junction of the stub is given as

Y Z Z1 1in in1 2

= + 0.01 0. 2j j1002

501 4= + = -



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VSWR (voltage standing wave ratio) of a transmission line is defined as

S 11

GG= -

+

where G is the reflection coefficient of the transmission line that varies from 0 to 1. Therefore, S varies from 1 to 3.

SOL 8.3.22 Option (B) is correct.Reactance increases, if we move along clockwise direction in the constant resistance circle.

SOL 8.3.23 Option (C) is correct.A transmission line is distortion less if LG RC=

SOL 8.3.24 Option (B) is correct. Zo Z Z 100 25OC SC #= = 10 5 50W#= =

SOL 8.3.25 Option (B) is correct.We know that distance between two adjacent voltage maxima is equal to /2l , where l is wavelength. So, we get

2l . .27 5 12 5= -

or, l 2 15 30 cm#= =Therefore, the operating frequency of the transmission line is

f 1GHzc30

3 1010#

l= = = ( 3 10 /cm sc 10#= )

SOL 8.3.26 Option (C) is correct.Electrical path length lb=

where b , 50 cml2lp= =

Now, the operating wavelength l of the transmission line is given as

l fu= f LC

1 1#=

LC1

a u =

25 10

110 10 40 10

16 6 12# # # ##=

- - 2 m

25 105 10

6

7

#

#= =

So, the electric path length is

lb 50 1022 2p

# #= - radian2p=

SOL 8.3.27 Option (B) is correct.The input impedance at the voltage minima on the transmission line is defined as

Zminin6 @ S

Z0=

where S is standing wave ratio along the transmission line. Since, the reflection coefficient LG of the transmission line is given as

LG Z ZZ Z

100 50100 50

15050

31

L

L

0

0= +- = +

- = =




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So, the standing wave ratio of the line is

S 11

L

L

GG

= -+

21 3

11 3

1=

-

+=

Therefore, the minimum input impedance measured on the line is equal to

Z minin 25250 W= =

SOL 8.3.28 Option (D) is correct.For a lossy transmission line the input impedance is given as

Zin tanhtanhZ Z jZ l

Z jZ lL

L0

0

0

gg= +

+< F

Load impedance, ZL 3= (open circuited at load end)Length of line, l /4l=

So, Zin limtanh

tanh

Z

ZZ j l

ZjZ l1

Z

L

L0

0

0

L g

g

=+

+

"3

R

T

SSSS

V

X

WWWW

tanhj lZ0

g= 0= tanh 4 " 3gp

a k

SOL 8.3.29 Option (D) is correct.Input impedance of a lossless transmission line is given by

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+< F

where Z0 Charateristic impedance of line"

ZL Load impedance"

l Length of transmission line"

and b 2 /p l=

So, we have lb 24l

p l= 2p=

ZL 0= (Short circuited at load)and Z0 50W=Therefore, the input impedance of the transmission line is

Zin 50 //

tantanjj

50 0 20 50 2

pp

= ++

= G 3=

i.e. infinite input impedance and thus, the current drawn from the voltage source will be zero.

SOL 8.3.30 Option (B) is correct.For lossless transmission line, the phase velocity is defined as

vp LC1

bw= = ...(1)

Characteristics impedance for a lossless transmission line is given as

Z0 CL= ...(2)



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For View Only Shop Online at www.nodia.co.inSo, from equation (1) and (2) we get

vp ( )C Z C Z C1 1

0 0= =

SOL 8.3.31 Option (C) is correct.Input impedance of a /4l^ h transmission line is defined as

Zin ZZL

02

=

where Z0 is characteristic impedance of the line and ZL is load impedance of the line. Since, the /4l line is shorted at one end (i.e. Z 0L = ) So, we get,

Zin limZZ

Z L0

02

L

3= ="

SOL 8.3.32 Option (D) is correct.Voltage minima of a short circuited transmission line is located at it’s load. As the location of minima is same for the load RL (i.e. the minima located at RL) so, the first voltage maxima will be located at /4l distance from load.

Now, lmax 4pq l= G ...(1)

where lmax is the distance of point of maxima from the load, Tq is phase angle of reflection coefficient and l is operating wavelength of line. So, putting the value of lmax is equation (1), we get

4l 4p

q l= G

Tq p=Now, the standing wave ratio of the line is given as

S 11

GG

=-+

or, 3 11

L

L

GG

=-+

S 3=] g

LG /1 2=

i.e. LG 21

21

L q pG= = =-G

The reflection coefficient at the load terminal is given as

LG Z ZZ ZL

L

0

0= +-

21- R

R7575

L

L= +- Z RL L=^ h, 75Z0 W=^ h

R 75L- - R2 150L= - R3 L 75= & RL 25W=

SOL 8.3.33 Option (B) is correct.The VSWR (voltage standing wave ratio) in terms of maxima and minima voltage is defined as

S VV

min

max= 24 2= =




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SOL 8.3.34 Option (D) is correct.Characteristic impedance, Z0 60W=SWR S 4=

So, we have 11

L

L

GG

-+ S 4= =

LG .53 0 6= =

The reflection coefficient at load is defined as

LG Z ZZ ZL

L

0

0= +-

So, .0 6 ZZ

6060

L

L= +-

ZL ..

0 41 6 60#= 012 W=

SOL 8.3.35 Option (A) is correct.Loading of a cable is done to increase the inductance as well as to achieve the distortionless condition.i.e. statement (1) and (4) are correct.

SOL 8.3.36 Option (C) is correct.Single stub with adjustable position is the best method for transmission line load matching for a given frequency range.

SOL 8.3.37 Option (B) is correct.The reflection coefficient at load terminal is defined as

LG Z ZZ ZL

L

0

0= +- j

j50 5050 50= +

+ -

j=Therefore, the standing wave, ratio is

VSWR 11

1 11 1

L

L3

GG

=-+

= -+ =

SOL 8.3.38 Option (D) is correct.Given, the load impedance is short circuiti.e. ZL 0=So, input impedance for lossless line is given as

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m tanjZ l0 b=

Now, for /l 4< l l& b 2< p

So, tan lb is positive and therefore, Zin is inductive 2a "

For l4 2< <l l l2 < <&p b p

tan lb is ve- and therefore, Zin is capacitive 1b "

For l 4l= l 2& b p=



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For View Only Shop Online at www.nodia.co.intan l 3b = and therefore Zin 3= 4c "

For l 2l= l& b p=

tan l 0b = and therefore, Z 0in = 3d "

SOL 8.3.39 Option (C) is correct.For distortionless transmission line, a RG= , LCb w=and for lossless transmission line, a 0= , LCb w=So, for both the type of transmission line attenuation is constant and is independent of frequency. Where as the phase shift b varies linearly with frequency w .i.e. statement 1 and 3 are correct.

SOL 8.3.40 Option (A) is correct.Given,Length of transmission line, l 500 m=Phase angle, qG 150c=-Operating wavelength, l 150 m=Consider the reflected voltage wave for the lossless transmission line terminated in resistive load as shown in figure.

Since, the reflection coefficient has a phase angle 150c- So, the wave lags by 150c angle.

The voltage wave has the successive maxima at each /2l distance,

So, the total no. of maxima tan maxDis ce between two imaTotal length=

/150 2500=

^ h 6 3

2=




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i.e. 6 maxima and remaining phase angle 32 360 240# c c= =

From the wave pattern shown above we conclude that the remaining phase 240c^ h will include one more maxima and therefore the total no. of maxima is 7.

SOL 8.3.41 Option (A) is correct.Reflection coefficient at load terminal is defined as

LG Z ZZ ZL

L

0

0= +-

For a matched transmission line we have ZL Z0=So, LG 0=i.e. matching eliminated the reflected wave between the source and the matching device location.

SOL 8.3.42 Consider the quarter wave transformer connected to load has the characteristic impedance Z 0l as shown in the figure.

So, we have the input impedance,

Zin tan

tanZ

Z jZ

Z jZ

24

24

L

L

0

0

0

lp llp l

=+

+l

lJ

L

KKKK b b

b bN

P

OOOOl l

l l ZZL

02

=l^ h

this will be the load to 450W transmission line

i.e. Z Ll ZZ Z

200L

02

02

= =l l^ ^h h

and for matching Z0 Z L= l

450 Z200

02

=l^ h

Z 0l 300450 200 W= =^ ^h h

SOL 8.3.43 Option (D) is correct.Given ZL 3= (open circuit)

and l 4l= (quarter wave)

Zin tantanZ Z jZ l

Z jZ l0 L

L0

0

bb= +

+c m cotjZ l0 b=- ZL" 3^ h



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For View Only Shop Online at www.nodia.co.in cotjZ 2

40 lp l=- b bl l

SOL 8.3.44 Option (A) is correct.Length of transmission line l /4< l (open circuit)Load impedance, ZL 3=So, the input impedance of the transmission line is given as

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m (lossless line)

tanZ j l1

0 b= c m ZL 3=^ h

cotjZ l0 b=-

SOL 8.3.45 Option (C) is correct.Given,Load impedance, ZL 0= (short circuit)Line parameters, R G= 0= (loss free line)Attenuation constant, a 0= (loss free line)So, the input impedance of the line is given as Zin tanjZ l0 b=i.e. pure reactanceStatement (A) is correct.Since tan lb can be either positive or negative So Zin can be either capacitive or inductive.Statement (B) is correct.The reflection coefficient at load is

LG Z ZZ Z 1 0L

L

0

0 != +- =-

So, the reflection exists.Statement (C) is incorrect.and since the standing waves of voltage and current are set up along length of the lines so, statement (D) is also correct.

SOL 8.3.46 Option (C) is correct.(a) short circuit Z 0L =^ h

So LG Z ZZ Z 1L

L

0

0= +- =- 2a "^ h

(b) Open circuit ZL 3=^ h

So, LG Z ZZ Z 1L

L

0

0= +- = 3b "^ h

(c) Line characteristic impedance Z ZL 0=^ h

LG Z ZZ Z 0

0 0

0 0= +- = 1c "^ h

(d) 2# line characteristic impedance Z Z2L 0=^ h

LG Z ZZ Z

22

31

0 0

0 0= +- = 4d "^ h




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SOL 8.3.47 Option (A) is correct.Given, reflection coefficient, LG 1 0c=

So, VSWR 11

1 11 1 3G

G=

-+

= -+ =


Given, reflection coefficient, 51G =

So, VSWR 11

46

23

GG

=-+

= =

SOL 8.3.49 Option (C) is correct.Characteristic impedance of transmission line is defined as

Z0 G j CR j L

ww= +

+

So, for lossless transmission line (R G 0= = )

Z0 CL=

SOL 8.3.50 Option (C) is correct.Input impedance has the range from 0 to 3. 3a "^ h

VSWR has the range from 1 to 3 2c "^ h

Reflection coefficient G^ h ranges from 1- to 1+ . 1b "^ h

SOL 8.3.51 Option (C) is correct.Input impedance of a quarter wave transformer is defined as

Zin ZZL

02

=

where Z0 is the characteristic impedance of the line and ZL is the load impedance. Since the quarter wave transformer is terminated by a short circuit (ZL 0= ) so, we get the input impedance of the transformer as Zin 3=

SOL 8.3.52 Option (D) is correct.The scattering parameters linearly relate the reflected wave to incident wave and it is frequency invariant so the scattering parameters are more suited than impedance parameters.

SOL 8.3.53 Option (C) is correct.Given, the reflection coefficient as LG . e0 3 j30= c-

At any point on the transmission line the reflection coefficient is defined as zG^ h eL z2G= g-

where z is the distance of point from load. z .0 1l= (Given)So, zG^ h e .

L2 0 1G= g l- ^ h . e e0 3 .j j30 2 0 1= c b l- -

^ ^ ^h hh (Assume 0a = ) . e e0 3 .j j30 0 4= c p- -

^ h



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For View Only Shop Online at www.nodia.co.in 0. e6 j152= c-

SOL 8.3.54 Option (D) is correct.Balun is used to couple a coaxial line to a parallel wire.

SOL 8.3.55 Option (B) is correct.The reflection coefficient of the conducting sheet is 1G =- where as the transmission coefficient is 0G = . So, there will be x -directed surface current on the sheet.

SOL 8.3.56 Option (B) is correct.Given,Operating frequency, f 25 kHz=Conductivity, s 5 /mho m=Relative permittivity, re 80=The attenuation constant for the medium is defined as

a 2wms= >>s we^ h

22 25 10 4 10 53 7

# # # #p p=

-^ ^ ^h h h

0m m=^ h

.0 7025=The attenuated voltage at any point is given as V V e l

0= a- (1)where V0 is source voltage and l is the distance travelled by waveSince, the radio signal is to be transmitted with 90% attenuation so, the voltage of the signal after 90% attenuation is V %V 900= - of V0

. V0 1 0=Comparing it with equation (1) we get .0 1^ h e l= a-

or, l ..

3.27ln

m0 70250 1

=- =^ h

SOL 8.3.57 Option (A) is correct.In Smith chart, the distance towards the load is always measured in anticlockwise direction. So, statement 3 is incorrect while statement 1 and 2 are correct.

SOL 8.3.58 Option (A) is correct.Given,Characteristic impedance, Z0 75W=Load impedance, ZL j100 75 W= -^ h

The condition for matching is Z Ll Z0=where Z Ll is the equivalent load impedance of the transmission line after connecting an additional circuit. So, the best matching will be obtained by a short circuited stub at some specific distance from load.




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SOL 8.3.59 Option (B) is correct.Given, the voltage standing wave ratio in decibels is VSWR indecibels 6 dB=or, 20 log S10 6= S 10 2/6 20= =^ h

So, the reflection coefficient at the load terminal is given as

G 0.33SS

11

2 12 1= +

- = +- =

SOL 8.3.60 Option (D) is correct.Input impedance of a quarter wave transformer (lossless transmission line) is defined as

Zin ZZL

02

=where Z0 is the characteristic impedance of the line and ZL is the load impedance of the line. So, we get Z0 100Z Z 50 200in L W= = =^ ^h h

SOL 8.3.61 Option (A) is correct.(1) Given,Length of line, l /8l=Load impedance, ZL 0=

So, lb 28 4l

p l p= =b bl l

Therefore the input impedance of the lossless transmission line is

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m

tanZ ZjZ 4

00

0p

= f p

jZ0= (i.e., incorrect statement)(2) Given,Length of line, l /4l=Load impedance, ZL 0=

So, lb 24 2l

p l p= =b bl l

Therefore the input impedance of the lossless transmission line is

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m

tan

Z ZjZ

j20

0

03

p= =f p (i.e., correct statement)

(3)Given,Length of line, l /2l=Load impedance, ZL 3=

So, lb 22l

p l p= =b bl l

Therefore, the input impedance of the lossless transmission line is



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Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m

tanZ j j10 3p= =-b l (i.e., incorrect statement)

(4) Matched line have the load impedance equal to its characteristic impedance

i.e. ZL Z0=So, for the matched line the input impedance is

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m Z0= (i.e., correct statement)

SOL 8.3.62 Option (C) is correct.Given,Length of line, l /8l=Load impedance, ZL 0=

So, lb 28 4l

p l p= =b bl l

Therefore, the input impedance of the transmission line is

Zin tanhtanhZ Z Z l

Z Z lL

L0

0

0

gg= +

+c m tanhZ l0 g=

If the line is distortion less (i.e. a 0= ) then, the input impedance of the line is Zin tanjZ l jZ0 0b= =So, it will depend on characteristic impedance as the line is resistive or reactive.

SOL 8.3.63 Option (B) is correct.Since, a transmission line of output impedance 400W is to be matched to a load of 25W through a quarter wavelength line. So, for the quarter wave line we haveInput impedance, Zin 400W= (same as the o/p impedance of the matched line)Load impedance, ZL 25W=Length of line, l /4l=The characteristic impedance of quarter wave transmission line is Z0 that connected between the load and the transmission line of output impedance 400W as shown in figure.

So,, the input impedance at AB is given as

Zin tan

tanZZ jZ

Z jZ

2

2

L

L0

0

0

p

p=

+

+f p Z

ZL

02

= ^ h




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Therefore, Z0 Z Zin L= 00800 50 2# W= =

SOL 8.3.64 Option (B) is correct.Given, Length of transmission line, l /8l=Load impedance, ZL 0= (Short circuited line)

So, we get lb 28 4l

p l p= =b bl l

Therefore, the input impedance of lossless transmission line is given as

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m Z Z

jZ jZ00

00= =c m which is inductive

So, the input impedance of /8l long short-circuited section of a lossless transmission line is inductive.

SOL 8.3.65 Option (C) is correct.In list I(a) Characteristic impedance of a transmission line is defined as

Z0 G j CR j L

YZ

ww= +

+ = (a 2" )

(b) Propagation constant of the line is given as g R j L G j C ZYw w= + + =^ ^h h (b 1" )(c) Sending end input impedance is

Zin tantanZ Z Z l

Z Z lL

L0

0

0

gg= +

+c m

Given, ZL Z0= (terminated in characteristic impedance, Z0)So, we get the input impedance as

Zin Z YZ

0= = (c 2" )

SOL 8.3.66 Option (C) is correct.For a distortionless transmission line, the attenuation constant (a) must be independent of frequency (w) and the phase constant (b ) should be linear function of w .(a) R G 0= =For this condition propagation constant is given as g j R j L G j Ca b w w= + = + +^ ^h h

i.e. a 0= and b LCw=As the attenuation constant is independent of frequency and the phase constant is linear function of w so, it is a distortionless transmission line.(b) RC GL=

LR C

G=

This is the general condition for distortionless line for which a RG= and LCb w=(c) R L>> w , G C>> w



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For View Only Shop Online at www.nodia.co.in g j RGa b= + =i.e. a RG= , and b 0=Since, b is not function of w so, it is not the distortionless line.(d) R L<< w , G C<< w g ja b= + j L j Cw w= ^ ^h h

i.e. a 0= and b LCw=

SOL 8.3.67 Option (B) is correct.Distance between adjacent maxima of an EM wave propagating along a transmission line is /2l . So, we get /2l . .37 5 12 5= -^ h

/2l 25 cm=i.e. l 50 cm=

Therefore, the operating frequency of the line is

f c50 103 10

2

8

#

#l= = -

00 MHz3=

SOL 8.3.68 Option (C) is correct.Forward voltage wave along the transmission line is given as

V0+ Z Z

Z E E20 0

0= + =

As the transmission line is open circuited at its load terminal ZL 3=^ h so, the reflection coefficient at the load terminal is

LG Z ZZ Z 1L

L

0

0= +- =

Therefore, the voltage travelling in reverse direction is

V0- V E

2L 0G= =+

The time taken by the wave to travel the distance between source and load terminal is given as

t1 cl=

where l is the length of transmission line and c is velocity of propagating wave. Now, from the plot we observe that at z 0= , voltage of the line is /2E where as at z l= , voltage is E therefore, it is clear that the voltage wave has been reflected




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from the load but not reached yet to the generator.

i.e. cl t c

l2< <


SOL 8.3.70 Option (A) is correct.The characteristic impedance for a lossy transmission line does not depend on the length of the line.

SOL 8.3.71 Option (D ) is correct.

SOL 8.3.72 Option (B) is correct.A distortionless transmission line has it’s parameters related as

LR C

G=

or RC GL=

SOL 8.3.73 Option (B) is correct.Given the reflection coefficient of the line is G 0.6=So, the voltage standing wave ratio is defined as

SWR 11

GG

=-+

..

1 0 61 0 6 4= -+ =

SOL 8.3.74 Option (B) is correct.Characteristic impedance, Z0 50W=Load impedance, ZL 100W=Forward voltage V + 10 V=So, the reflection coefficient of the line is given as

G Z ZZ Z

100 50100 50

L

L

0

0= +- = +

- 31=

SOL 8.3.75 Option (C) is correct.Characteristic impedance, Z0 50W=Load impedance, ZL 15 20j W= -So, the normalized load impedance is given as

zL ZZL

0= j50

155020= - 0. 0.j4 3= -

SOL 8.3.76 Option (D) is correct.Since both the transmission lines are identical except that the loads connected to them are Z2 and /Z 2 respectively. Let the maximum voltage across the loads be Vm So, the power transmitted to the loads are

PA ZV2m2

=

and PB /ZV

2m2

=

Given, PA W1=



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For View Only Shop Online at www.nodia.co.inSo, Vm2 Z W2 1= ^ h

and PB / /ZV

ZZW

2 22m

21= =

^ h W2 1=

SOL 8.3.77 Option (A) is correct.Given, the short circuited and open circuited input impedance as Z . .s c 36W= , 64Z . .o c W=So, the characteristic impedance of the transmission line is defined as Z0 Z Z. . . .o c s c= 36 64#= 48W=

SOL 8.3.78 Option (D) is correct.(1) ZL Z0= (line terminated by its characteristic impedance)So, reflection coefficient

G Z ZZ Z 0L

L

0

0= +- =

i.e. no any reflected wave.(2) ZL Z0= G 0=and so, there will be no reflected wave and the wave will have only forward voltage and current wave which will be equal at all the points on the line.(3) For a lossless half wave transmission line

Zin ZL=So, statement 3 is incorrect while statements 1 and 2 are correct.

SOL 8.3.79 Option (C) is correct.Since, the standing wave ratio of the wave is 1.i.e. SWR 1=So, expressing it in terms of reflection coefficient, we get

11

GG

-+

1=

G 0=

Z ZZ ZL

L

0

0

+- 0=

ZL Z0=i.e. characteristic impedance is equal to load impedance.





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Given, the two wire transmission line has

Half center to center spacing h d2= =

Conductor radius r=So, the capacitance per unit length of the line is defined as

C log r

drd

2 2 1e

2pe=

+ -b bel l o= G

log r

hrh 1e

2pe=

+ -b l= G


Reflection coefficient, G Z ZZ ZR

R

0

0= +-

Z Z

Z Z

3

30

00

0=

+

- 5=-

SOL 8.3.82 Option (B) is correct.Propagation constant, g R j L G j Cw w= + +^ ^h h

The characteristic impedance of the transmission line is given as

Z0 G j CR j L

ww= +

+

Z0 R j L

gw= +

SOL 8.3.83 Option (A) is correct.Given the reflection coefficient,

G 31=-

So, the standing wave ratio

S 11

GG

=-+

24 2= =

SOL 8.3.84 Option (B) is correct.For distortionless transmission line

GR C

L=

and so, the attenuation constant,

a RG= R LRC= b l R L

C=

SOL 8.3.85 Option (D) is correct.Capacitance per unit length, C 10 /F m10= -

Characteristic impedance, Z0 50W=Now, for distortionless line the characteristic impedance is given as



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Z0 CL=

50 L10 10= -

So, the inductance per unit length is L 0.25 /H m50 102 10 m#= =-

^ ^h h

SOL 8.3.86 Option (B) is correct.The characteristic impedance Z0 in terms of open circuit impedance Zoc and short circuit impedance Zsc is defined as Z0 Z Zoc sc= 100 100= ^ ^h h (Given Z Z 100oc sc= = ) 05 W=

SOL 8.3.87 Option (D) is correct.Given,Load impedance, ZL j75 50 W= -^ h

Characteristic impedance, Z0 75W=Since, for matching the load impedance is equal to the characteristic impedance (i.e., ZL Z0= ) so, we have to produce an additional impedance of j50+ at load to match it with transmission line. Therefore, for matching the transmission line a short circuit stub is connected at some specific distance from load.

SOL 8.3.88 Option (C) is correct.Given, The load impedance = Surge impedancei.e. ZL Z0=So, reflection coefficient of the line is given as

G Z ZZ ZL

L

0

0= +-

0=

SOL 8.3.89 Option (C) is correct.Given,Length of transmission line, l 50 0.5cm m= =Operating frequency, f 30 30 10MHz Hz6

#= =Line parameters, L 10 / 10 10 /H m H m6m #= = -

and C 40 / 40 10 /pF m F m12#= = -

So, the phase constant of the wave along the transmission line is b LCw= 2 30 10 10 10 40 106 6 12

# # # #p= - -^ ^h h

56p=

Therefore, lb . .56 0 5 0 6 108# cp p= = =




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SOL 8.3.90 Option (A) is correct.Propagation constant in a transmission line is defined as. g L j R R j Cw w= + +^ ^h h

SOL 8.3.91 Option (A) is correct.For a series resonant circuit the required conditions are(1) The angular frequency is

w LC1=

(2) The total equivalent impedance is pure resistivei.e. Z R=Now, the input impedance at a distance /4l from the load is defined as

Zin ZZL

02

=

And since the transmission line is open ZL 3=^ h

So, Zin 0= which is purely resistivei.e. R is correct statement.In a lossless line voltage or current along the line are not constant.i.e. A is not a correct statement.

SOL 8.3.92 Option (D) is correct.The characteristic impedance of a transmission line can be defined as below.

Z0 G j CR j L

ww= +

+

Z0 Z Zoc sc=

Z0 IV= +

+

So, all the three statements are correct.

SOL 8.3.93 Option (C) is correct.Given, load impedance of the transmission line is ZL 0= (Short circuit)So, the input impedance of the lossless line is given as

Zin tantanZ Z jZ l

Z jZ lL

L0

0

0

bb= +

+c m

tanZ ZjZ l

00

0 b= c m

tanjZ l2 0 b=

SOL 8.3.94 Option (D) is correct.Given,Characteristic impedance, Z0 600W=Load impedance, ZL 900W=So, the reflection coefficient of the transmission line is given as

LG Z ZZ ZL

L

0

0= +-

CHAPTER 9WAVE GUIDES

590 Wave Guides Chap 9



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EXERCISE 9.1

MCQ 9.1.1 An electromagnetic wave propagating in an airfilled 10 cm8# waveguide has it’s electric field in phasor form given as Ezs 5 /sin sin V mx y e20 25 j zp p= b-

^ ^h h

What is the mode of propagation of the EM wave ?(A) TM21 (B) TM12

(C) TE21 (D) TE12

MCQ 9.1.2 Assertion (A) : In a waveguide operating below cutoff frequency there is no net average power flow down the waveguide.Reason (R) : Propagation of energy requires a propagating mode.(A) A and R both are true and R is correct explanation of A.




MCQ 9.1.3 An airfilled rectangular waveguide is operating in TM mode at a frequency twice the cutoff frequency. What will be the intrinsic wave impedance ?(A) 892 W (B) 327 W

(C) 211W (D) 377 W

MCQ 9.1.4 An EM wave is propagating in TM21 mode in an air filled 10 4 cm# waveguide at a frequency of .5 GHz2 . What will be the phase constant of the EM wave ?(A) 1.865 /rad m (B) 1.207 /rad m

(C) 186.5 /rad m (D) 120.7 /rad m

MCQ 9.1.5 The electric field component of an electromagnetic wave propagating in a rectangular waveguide is given in phasor form as Ezs /sin sin V mE x y e50 40 rz

0 p p= -^ ^h h

The ratio of field components /E Exs ys will be equal to(A) . cot tanx y1 25 50 40p p^ ^h h

(B) . cot tanx y0 8 50 40p p^ ^h h

(C) . tan cotx y1 25 40 50p p^ ^h h

(D) . tan cotx y0 8 40 50p p^ ^h h

Chap 9 Wave Guides 591


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For View Only Shop Online at www.nodia.co.inMCQ 9.1.6 An EM wave is propagating in TEM mode in a parallel plate waveguide filled of a

dielectric ( .254re = , 1rm = ). If the waveguide operating at 20 GHz then the phase constant and the group velocity of the wave will be respectively(A) 4.5 10 /rad s8

# , 139.6 /m s

(B) 139.6 /rad s, 4.5 10 /m s8#

(C) 2 10 /rad m8# , 314.2 /m s

(D) 314.2 /rad m, 2 10 /m s8#

MCQ 9.1.7 An a b# rectangular waveguide is operating in four different modes as TM11, TM12, TE10 and TE20. If a b2= then the ascending order of the operating modes for their cut-off frequencies will be(A) TE TE TM TM< < <10 20 11 12

(B) TE TM TE TM< < <10 11 20 12

(C) TM TM TE TE< < <12 11 20 10

(D) TE TE TM TM< < <10 20 12 11

MCQ 9.1.8 An airfilled cm4 2# rectangular waveguide is operating at TE10 mode at frequency of 3.75 GHz. What will be the group velocity of the propagating wave in the waveguide ?(A) 1.8 10 /m s9

# (B) 2.4 10 /m s9#

(C) 2.4 10 /m s8# (D) 1.8 10 /m s8

#

MCQ 9.1.9 A rectangular waveguide with the dimensions 2.5 cma = , 5 cmb = is operating at a frequency 15 GHzf = . If the wave guide is filled with a lossless dielectric with 1rm = , 2re = then the wave impedance of propagating TE20 mode in the waveguide will be(A) 377 W (B) 323W

(C) 457 W (D) 470W

MCQ 9.1.10 Cutoff wavelength of a parallel plate waveguide for TM2 mode is 3 mm. If the guide is operated at a wavelength 0.1 cml = then the no. of possible modes that can propagate in the waveguide is(A) 4 (B) 5

(C) 8 (D) 9

MCQ 9.1.11 A lossless parallel plate waveguide is operating in TM3 mode at frequencies as low as 15 GHz. What will be the dielectric constant of the medium between plates if the plate separation is 20 mm ?(A) 1.73 (B) 3

(C) 6 (D) 9




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MCQ 9.1.12 A parallel plate wave guide has the plate separation 20 mmb = is made with glass .2 1re =^ h between it’s plates. If the guide is operating at a frequency GHzf 20=

then which of the following modes will propagate ?(A) TM1 (B) TM3

(C) TE2 (D) all the three

MCQ 9.1.13 The cutoff frequency of TM1 mode in an air filled parallel plate wave guide is 2.5 GHz. If the guide is operating at wavelength 3 cml = then what will be the group velocity of TE3 mode ?(A) 9.9 10 /m s7

# (B) 2 10 /m s8#

(C) 3.97 10 /m s8# (D) 1.5 10 /m s8

#

MCQ 9.1.14 A symmetric slab waveguide has a slab thickness md 15 m= with refractive indices n 31 = , .n 2 52 = as shown in figure. The phase velocity of the TE1 mode at cutoff will be

(A) 2.5 10 /m s8#

(B) 1.2 10 /m s8#

(C) 7.5 10 /m s8#

(D) 2.07 10 /m s8#

MCQ 9.1.15 In a symmetrical slab waveguide, the phase velocity of TE1 mode at cutoff is vp1. So, the phase velocity of TM2 mode at cutoff will be(A) vp1 (B) 2vp1

(C) v2p1 (D) v2 p1

Common Data for Question 16 - 17 :A strip line transmission line has the ground plane separation, 0.632 cmb = and filled of a material with .84re = .

MCQ 9.1.16 If characteristic impedance of the transmission line is 55 then what will be the width of conducting strip ?(A) 0.47 cm (B) 0.30 cm

(C) 0.15 cm (D) 0.62 cm



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For View Only Shop Online at www.nodia.co.inMCQ 9.1.17 If the transmission line is operating at a frequency 3 GHzf = then what will be it’s

guide wavelength ?(A) 29.7 cm (B) 3.37 cm

(C) 33.7 cm (D) 6.74 cm

MCQ 9.1.18 Consider the following statements1. TEM mode can not exist within a hollow waveguide.

2. Any of the TM mode can’t be the dominant mode of propagation in rectangular waveguide.

The correct statement is

(A) only 1 (B) only 2

(C) 1 and 2 both (D) None of these

MCQ 9.1.19 The lowest order TM mode that can exist in a cavity resonator is(A) TM111 (B) TM110

(C) TM011 (D) TM101

MCQ 9.1.20 If the dimensions of cavity resonator are equal (i.e., a b= ) then the lowest order TE mode will be(A) TE011 (B) TE100

(C) TE101 (D) (A) and (C) both

MCQ 9.1.21 An air filled cavity resonator has the dimensions a b c> > . Which of the following modes are arranged in ascending order with respect to their resonant frequencies ?(A) TM110, TE011, TE101

(B) TE011, TM110, TE101

(C) TM100, TM101, TM111

(D) TM110, TE101, TE011

MCQ 9.1.22 An airfilled, lossless cavity resonator has dimensions 0 cma 4= , 25 cmb = and 20 cmc = . What is the resonant frequency of TE101 mode ?

(A) 781MHz (B) 901.4 MHz

(C) 450.7 MHz (D) 960.4 MHz

MCQ 9.1.23 An airfilled cubic cavity resonator a b c= =^ h has dominant resonant frequency of 1 GHz5 . The dimension of the cavity resonator is(A) 1.41 cm (B) 70 cm

(C) 2.5 cm (D) 3.8 cm

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EXERCISE 9.2

MCQ 9.2.1 An electromagnetic wave is propagating in a parallel plate waveguide operating at TM1 mode. The magnetic field lines in the yz -plane will be (Assume the positive x -axis directs into the paper)

MCQ 9.2.2 An EM wave is propagating at a frequency ‘ f ’ in an air filled rectangular waveguide having the cutoff frequency ‘ fc ’. Consider the phase velocity of the EM wave in the waveguide is vp . The plot of ( / )c vp versus /f fc^ h will be. (c is the velocity of wave in air)



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MCQ 9.2.3 An electromagnetic wave propagating at a frequency ‘ f ’ in free space has the wavelength ‘l’. At the same frequency it’s wavelength in an airfilled waveguide is gl . If the cutoff frequency of the waveguide is fc then the plot of /gl l^ h versus

/f fc^ h will be

Statement for Linked Question 4 - 5 :A parallel plate waveguide is separated by a dielectric medium of thickness b with the constitutive parameters e and m

MCQ 9.2.4 If the cut-off frequencies of the waveguide for the modes TE1, TE2 and TE3 are respectively c1w , c2w , c3w then which of the following represents the correct relation between the cutoff frequencies ?(A) c c c1 2 3w w w= = (B) < <c c c1 2 3w w w

(C) > >c c c1 2 3w w w (D) c c c1 3 2w w w=




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MCQ 9.2.5 The f -b curve (graph of frequency f versus phase constant b ) of the waveguide for the modes TM2, TM3 and TM4 will be

MCQ 9.2.6 A parallel plate waveguide operating at a frequency of 5 GHz is formed of two perfectly conducting infinite plates spaced 8 cm apart in air. The maximum time average power that can be propagated per unit width of the guide for TM1 mode without any voltage breakdown will be (Dielectric strength of air = 3 10 /V m6

# )(A) 828 /MW m (B) 414 /MW m

(C) 104 /MW m (D) 207 /MW m

MCQ 9.2.7 An air filled parallel plate wave guide has the separation of 12 cm between it’s plates. The guide is operating at a frequency of 2.5 GHz. What is the maximum average power per unit width of the guide that can be propagated without a voltage breakdown for TEM mode ?(A) 358 /MW m (B) 143.2 /MW m

(C) 716 /MW m (D) 1.432 /GW m

MCQ 9.2.8 A parallel plate waveguide filled of a dielectric .8 4re =^ h is constructed for operation in TEM mode only over the frequency range 0 1.5 GHzf< < . The maximum allowable separation between the plates will be(A) 6.90 cm (B) 29 cm

(C) 3.45 cm (D) 1.20 cm



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For View Only Shop Online at www.nodia.co.inStatement for Linked Question 9 - 10 :A parallel plate waveguide having plate separation .1 mmb 18= is partially filled with two lossless dielectric with permittivity 2r1e = and 1.05r2e = .

MCQ 9.2.9 The frequency ‘ f0’ at which the TM1 mode propagates through the guide without suffering any reflective loss is(A) 12.8 GHz (B) 16.2 GHz

(C) 9.28 GHz (D) 8.44 GHz

MCQ 9.2.10 How many TM modes that can propagate in the guide at the frequency f0 ?(A) one (B) two

(C) three (D) four

MCQ 9.2.11 An a b# airfilled rectangular waveguide is operating at a frequency, 5 GHzf = . What will be it’s dimensions if the design frequency is %10 larger than the cutoff frequency of dominant mode while being 15 % lower than the cutoff frequency for the next higher order mode ?(A) 3.3 cma = , 2.7 cmb =

(B) 1.1 cma = , 0.9 cmb =

(C) 0.37 cma = , 0.3 cmb =

(D) 0.5 cma = , 0.4 cmb =

MCQ 9.2.12 A symmetric dielectric slab waveguide with it’s permittivities .2 2r1e = and .2 1r2e = is operating at wavelength, 2.6 mml = . If the slab thickness is 20 md m= then how many modes can propagate in the slab ?(A) 8 (B) 25

(C) 15 (D) 24

MCQ 9.2.13 A rectangular waveguide operating in TE10 mode has the phase constant b . If the average power density of the guide in this mode is Pav then what will be the relation between Pav and b ?(A) Pav \ b (B) Pav 2

\ b

(C) P 1av \ b (D) Pav is independent of b

MCQ 9.2.14 A symmetric dielectric slab waveguide with it’s refractive indices n1 and n2 is operating at wavelength, 3.1 mml = . If the slab thickness is 10 mm and .n 3 32 = then what will be the maximum value of n1 for which it supports only a single pair of TE and TM mode ?(A) 3.304 (B) 3.20

(C) 2.40 (D) 3.42




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Common Data for Question 15 - 16 :An asymmetric slab waveguide has the different mediums above and below the slab as shown in figure. The regions above and below the slab have refractive indices n2 and n3 respectively while the slab has refractive index n1

MCQ 9.2.15 If .n 2 81 = , .n 1 72 = , 2.1n3 = , then the minimum possible wave angle for the wave propagation will be(A) .48 6c (B) .37 4c

(C) .41 4c (D) .54 1c

MCQ 9.2.16 If the refractive indices of the mediums are related as n n n> >1 2 3 the maximum phase velocity of a guided mode will be(A) n c3 (B) /c n1

(C) /c n3 (D) /c n2

(c is velocity of wave in free space)

MCQ 9.2.17 An air filled waveguide is of square cross-section of 4.5 cm on each side. The waveguide propagates energy in the TE22 mode at 6 GHz. The wavelength of the TE22 mode wave in the guide is(A) 4.2 cm (B) 2.72 cm

(C) 5.3 cm (D) 1.18 cm

MCQ 9.2.18 An attenuator is formed by using a section of waveguide of length ‘l ’ operating below cutoff frequency. The operating frequency is 6 GHz and the dimension of the guide is 4.572 cma = as shown in figure. What will be the required length ‘l ’ to achieve attenuation of 100 dB between /I P and /O P guides ?



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For View Only Shop Online at www.nodia.co.in(A) 10.3 cm (B) 20.7 cm

(C) 5.2 cm (D) 12.7 cm

MCQ 9.2.19 A rectangular waveguide with the dimension, 1.07 cma = is operating in TE10 mode at a frequency, 10 GHzf = . If the waveguide is filled with a dielectric material having .8 8re = and .tan 0 002d = then the attenuation constant due to dielectric loss will be(A) 12.23 /dB m (B) 0.705 /dB m

(C) 6.12 /dB m (D) 3.03 /dB m

MCQ 9.2.20 The first four propagating modes of a circular waveguide are respectively(A) TM01, TE21, TE01, TM11

(B) TM11, TM21, TM02, TM12

(C) TE11, TE21, TM11, TE01

(D) TE11, TM01, TE21, TE01

Statement for Linked Question 21 - 22 :A microstrip line has the substrate thickness 0. 16 cmd 6= with .2 2re =

MCQ 9.2.21 If the characterisitc impedance of the guide is 100W then what will be the width of microstrip.(A) 2.83 cm (B) 0.28 cm

(C) 0.36 cm (D) 0.14 cm

MCQ 9.2.22 If the transmission line is operating at a frequency, 8 GHzf = then the effective permittivity ee and guide wavelength gl will be ee gl(A) 1.76 2.83 cm(B) 2.83 1.76 cm(C) 0.158 18.87 cm(D) 18.87 0.158 cm

Common Data for Question 23 - 24 :A rectangular cavity resonator with dimensions 2.5 cma = , 2 cmb = and 5 cmc =is filled with a lossless material ( 0m m= , 3 0e e= ).

MCQ 9.2.23 The resonant frequency of the cavity resonator for TE101 mode will be(A) 6.7 GHz (B) .34 GHz6

(C) 7.74 GHz (D) 3.87 GHz




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MCQ 9.2.24 If the resonator is made of copper then the quality factor for TE101 mode is (Conductivity of copper 5.8 10 /S mr 7

#= )(A) 7733 (B) 14358

(C) 6625 (D) 11075

MCQ 9.2.25 An air filled circular waveguide has it’s inner radius 1 cm. The cutoff frequency for TE11 mode will be .p 1 84111 =l^ h(A) 0.55 GHz (B) 9.3 GHz4

(C) 8.79 GHz (D) 4.71GHz

MCQ 9.2.26 A cylindrical cavity shown in the figure below is operating at a wavelength of 2 cm in the dominant mode.

The radius a of the cylindrical cavity will be .p 2 40501 =^ h

(A) 0.77 cm (B) 0.38 cm

(C) 1.53 cm (D) 0.19 cm

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EXERCISE 9.3

MCQ 9.3.1 The magnetic field among the propagation direction inside a rectangular waveguide with the cross-section shown in the figure is Hz ( . ) ( . ) ( . )cos cos cosx y t z3 2 094 10 2 618 10 6 283 102 2 10

# # # b= -

The phase velocity vp of the wave inside the waveguide satisfies(A) v c>p (B) v cp =

(C) v c0 < <p (D) v 0p =

MCQ 9.3.2 The modes in a rectangular waveguide are denoted by TE or TMmn mn where m and n are the eigen numbers along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is TRUE?(A) The TM10 mode of the waveguide does not exist

(B) The TE10 mode of the waveguide does not exist

(C) The TM10 and the TE10 modes both exist and have the same cut-off frequencies

(D) The TM10 and the TM01 modes both exist and have the same cut-off frequencies

MCQ 9.3.3 The electric and magnetic fields for a TEM wave of frequency 14 GHz in a homogeneous medium of relative permittivity re and relative permeability 1rm = are given by /V mE e aE ( )

pj t y

z280= w p- and 3 /A me aH ( )j t y

x280= w p- . Assuming the

speed of light in free space to be 3 10 /m s8# , the intrinsic impedance of free

space to be 120p , the relative permittivity re of the medium and the electric field amplitude Ep are(A) 3, 120Er pe p= =

(B) 3, 360Er pe p= =

(C) 9, 360Er pe p= =

(D) 9, 120Er pe p= =

GATE 2012

GATE 2011

GATE 2011




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MCQ 9.3.4 Which of the following statements is true regarding the fundamental mode of the metallic waveguides shown ?

(A) Only P has no cutoff-frequency

(B) Only Q has no cutoff-frequency

(C) Only R has no cutoff-frequency

(D) All three have cutoff-frequencies

MCQ 9.3.5 A rectangular waveguide of internal dimensions (a 4= cm and b 3= cm) is to be operated in TE11 mode. The minimum operating frequency is(A) 6.25 GHz (B) 6.0 GHz

(C) 5.0 GHz (D) 3.75 GHz

MCQ 9.3.6 The E field in a rectangular waveguide of inner dimension a b# is given by

E ( )sin sinh a H a

x t z a2y2 0

wm p p w b= -a bk l

Where H0 is a constant, and a and b are the dimensions along the x -axis and the y -axis respectively. The mode of propagation in the waveguide is(A) TE20

(B) TM11

(C) TM20

(D) TE10

MCQ 9.3.7 An air-filled rectangular waveguide has inner dimensions of 3 cm # 2 cm. The wave impedance of the TE20 mode of propagation in the waveguide at a frequency of 30 GHz is (free space impedance 3770h W= )(A) 308 W (B) 355 W

(C) 400 W (D) 461 W

MCQ 9.3.8 A rectangular wave guide having TE10 mode as dominant mode is having a cut off frequency 18 GHz for the mode TE30. The inner broad - wall dimension of the rectangular wave guide is

(A) 35 cm (B) 5 cm

(C) 25 cm (D) 10 cm

MCQ 9.3.9 Which one of the following does represent the electric field lines for the mode in the cross-section of a hollow rectangular metallic waveguide ?

GATE 2009

GATE 2008

GATE 2007

GATE 2007

GATE 2006

GATE 2005



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MCQ 9.3.10 The phase velocity of an electromagnetic wave propagating in a hollow metallic rectangular waveguide in the TE10 mode is(A) equal to its group velocity

(B) less than the velocity of light in free space

(C) equal to the velocity of light in free space

(D) greater than the velocity of light in free space

MCQ 9.3.11 In a microwave test bench, why is the microwave signal amplitude modulated at 1 kHz(A) To increase the sensitivity of measurement

(B) To transmit the signal to a far-off place

(C) To study amplitude modulations

(D) Because crystal detector fails at microwave frequencies

MCQ 9.3.12 A rectangular metal wave guide filled with a dielectric material of relative permittivity 4re = has the inside dimensions 3.0 cm # 1.2 cm. The cut-off frequency for the

dominant mode is(A) 2.5 GHz (B) 5.0 GHz

(C) 10.0 GHz (D) 12.5 GHz

MCQ 9.3.13 The phase velocity for the TE10 -mode in an air-filled rectangular waveguide is (c is the velocity of plane waves in free space)(A) less than c

(B) equal to c

(C) greater than c

(D) none of these

MCQ 9.3.14 The phase velocity of wave propagating in a hollow metal waveguide is(A) grater than the velocity of light in free space

(B) less than the velocity of light in free space

(C) equal to the velocity of light free space

(D) equal to the velocity of light in free

GATE 2004

GATE 2004

GATE 2003

GATE 2002

GATE 2001




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MCQ 9.3.15 The dominant mode in a rectangular waveguide is TE10, because this mode has(A) the highest cut-off wavelength

(B) no cut-off

(C) no magnetic field component

(D) no attenuation

MCQ 9.3.16 A TEM wave is incident normally upon a perfect conductor. The E and H field at the boundary will be respectively,(A) minimum and minimum

(B) maximum and maximum

(C) minimum and maximum

(D) maximum and minimum

MCQ 9.3.17 A rectangular waveguide has dimensions 1 cm # 0.5 cm. Its cut-off frequency is(A) 5 GHz (B) 10 GHz

(C) 15 GHz (D) 12 GHz

MCQ 9.3.18 Assuming perfect conductors of a transmission line, pure TEM propagation is NOT possible in(A) coaxial cable

(B) air-filled cylindrical waveguide

(C) parallel twin-wire line in air

(D) semi-infinite parallel plate wave guide

MCQ 9.3.19 Indicate which one of the following will NOT exist in a rectangular resonant cavity.(A) TE110 (B) TE011

(C) TM110 (D) TM111

MCQ 9.3.20 The ratio of the transverse electric field to the transverse magnetic field is called as(A) wave guide impedance (B) wave guide wavelength

(C) phase velocity (D) Poynting vector

MCQ 9.3.21 Consider a rectangular waveguide of internal dimensions 8 4cm cm# . Assuming an H10 mode of propagation, the critical wavelength would be(A) 8 cm (B) 16 cm

(C) 4 cm (D) 32 cm

MCQ 9.3.22 am

bn2 2 2g p p w me= + -a ak k represents the propagation constant in a rectangular

waveguide for(A) TE waves only (B) TM waves only

(C) TEM waves (D) TE and TM waves

GATE 2001

GATE 2000

GATE 2000

GATE 1999

GATE 1999

IES EC 2012

IES EC 2012

IES EC 2012



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For View Only Shop Online at www.nodia.co.inMCQ 9.3.23 With the symbols having their standard meaning, cut-off frequency (frequency

below which wave propagation will not (occur) for a rectangular waveguide is

(A) am

bn1

mep p+a ak k (B) a

mbn

21 2 2

p mep p+a ak k

(C) am

bn

21

p mep p+a ak k (D) a

mbn1 2 2

mep p+a ak k

MCQ 9.3.24 A standard air filled waveguide WR-187 has inside wall dimensions of 4.755 cma = and 2.215 cmb = . At 12 GHz, it will support(A) TE10 mode only (B) TE10 and TE20 modes only

(C) ,TE TE and TE10 20 01 modes only (D) , ,TE TE TE and TE10 20 01 11 modes

MCQ 9.3.25 Consider the following statements relating to the cavity resonator :1. The cavity resonator does not posses as many modes as the corresponding

waveguides does.

2. The resonant frequencies of cavities are very closely spaced.

3. The resonant frequency of a cavity resonator can be changed by altering its dimensions.

Which of the above statements is/are correct ?(A) 2 and 3 only (B) 2 only

(C) 3 only (D) 1, 2 and 3

MCQ 9.3.26 The correct statement is(A) Microstrip lines can support pure TEM mode of propagation but shielded

coaxial lines cannot

(B) Microstrip lines cannot support pure TEM mode of propagation but shielded coaxial lines can

(C) Both microstrip lines and shielded coaxial lines can support pure TEM mode of propagation

(D) Neither microstrip lines nor shielded coaxial lines can support pure TEM mode of propagation.

MCQ 9.3.27 An air-filled rectangular waveguide has dimensions of 6 cma = and 4 cmb = . The signal frequency is 3 GHz. Match List I with List II and select the correct answer using the code given below the lists :

List I List II

a. TE10 1. 2.5 GHz

b. TE01 2. 3.75 GHz

c. TE11 3. 4.506 GHz

d. TM11 4. 4.506 GHz

IES EC 2012

IES EC 2010

IES EC 2010

IES EC 2010

IES EC 2010




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Codes :

a b c d(A) 1 2 3 4(B) 4 2 3 1(C) 1 3 2 4(D) 4 3 2 1

MCQ 9.3.28 Assertion (A) : TEM (Transverse Electromagnetic) waves cannot propagate within a hollow waveguide of any shape.Reason (R) : For a TEM wave to exist within the waveguide, lines of H field must be closed loops which requires an axial component of E which is not present in a TEM wave.(A) Both A and R are individually true and R is the correct explanation of A




MCQ 9.3.29 For plane wave propagating in free space or two conductor transmission line, what must be the relationship between the phase velocity vr , the group velocity vg and speed of light c ?(A) v c v> >p g

(B) v c v< <p g

(C) v c vp g= =

(D) v v c< <p g

MCQ 9.3.30 Consider the following statements :In a microstrip line

1. Wavelength e

0l el= , where ee is the effective dielectric constant and 0l is the

free space wavelength.

2. Electromagnetic fields exist partly in the air above the dielectric substrate and partly within the substrate itself.

3. The effective dielectric constant is greater than the dielectric constant of the air.

4. Conductor losses increase with decreasing characteristic impedance.

Which of the above statements is/are correct ?(A) 1, 2 and 3

(B) 1 and 2

(C) 2, 3 and 4

(D) 4 only

IES EC 2010

IES EC 2009

IES EC 2009



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For View Only Shop Online at www.nodia.co.inMCQ 9.3.31 Match List I with List II and select the correct answer using the codes given below

the lists.

List I (Type of transmission structure) List II (Modes of propagation)

a. Strip line 1. Quassi TEM

b. Hollow rectangular waveguide 2. Pure TEM

c. Microguide 3. /TE TEM

d. Corrugated waveguide 4. Hybrid

Codes a b c d(A) 2 1 3 4(B) 4 1 3 2(C) 2 3 1 4(D) 4 3 1 2

MCQ 9.3.32 A standard waveguide WR90 has inside wall dimensions of 2.286 cma = and 1.016 cmb = . What is the cut-off waveguide for TE01 mode ?

(A) 4.572 cm (B) 2.286 cm

(C) 2.032 cm (D) 1.857 cm

MCQ 9.3.33 When a particular mode is exited in a waveguide, there appears an extra electric component, in the direction of propagation. In what mode is the wave propagating ?(A) Transverse electric

(B) Transverse magnetic

(C) Transverse electromagnetic

(D) Longitudinal

MCQ 9.3.34 Consider the following statements :For a square waveguide of cross-section 3 3m m# it has been found1. at 6 GHz dominant mode will propagate.

2. at 4 GHz all the mode are evanescent.

3. at 11GHz only dominant modes and no higher order mode will propagate.

4. at 7 GHz degenerate modes will propagate.

Which of the above statements are correct ?(A) 1 and 2

(B) 1, 2 and 4

(C) 2 and 3

(D) 2, 3 and 4

IES EC 2009

IES EC 2009

IES EC 2009

IES EC 2009




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MCQ 9.3.35 Match List with List II and select the correct answer using the codes given below the lists.

List I (Mode) List II (Characteristic)

a. Evanescent mode 1. Rectangular waveguide does not support

b. Dominant mode 2. No wave propagation

c. TM10 and TM01 3. Lowest cut-off frequency

Codes a b c(A) 1 2 3(B) 1 3 2(C) 2 3 1(D) 2 1 3

MCQ 9.3.36 Assertion (A) : A z -directed rectangular waveguide with cross-sectional dimensions 3 1cm cm# will support propagation at 4 GHz.

Reason (R) : k m n3 1

2z2 2 2p p

lp+ + =a a bk k l, where l is the wavelength.

(A) Both A and R are individually true and R is the correct explanation of A.




MCQ 9.3.37 Which one of the following is the correct statement ?A rectangular coaxial line can support(A) only TEM mode of propagation

(B) both TEM and TE modes of propagation

(C) either TE or TM mode of propagation

(D) TEM, TE or TM mode of propagation

MCQ 9.3.38

A rectangular waveguide (A) is gradually deformed first into a circular wave guide (B) and lack again into a rectangular waveguide (C) which is oriented through 90c

IES EC 2009

IES EC 2009

IES EC 2006

IES EC 2006



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For View Only Shop Online at www.nodia.co.inwith respect to (A) If the input mode is TE10, which mode is excited in the output waveguide (C) ?(A) TE10 (B) TE01

(C) TE11 (D) TM11

MCQ 9.3.39 The dominant mode in a circular waveguide is a :(A) TEM mode (B) TM01 mode

(C) TE21 mode (D) TE11 mode

MCQ 9.3.40 The cut-off frequency of the dominant mode of a rectangular wave guide having aspect ratio more than 2 is 10 GHz. The inner broad wall dimension is given by :(A) 3 cm (B) 2 cm

(C) 1.5 cm (D) 2.5 cm

MCQ 9.3.41 In a waveguide, the evanescent modes are said to occur if :(A) The propagation constant is real

(B) The propagation constant is imaginary

(C) Only the TEM waves propagate

(D) The signal has a constant frequency

MCQ 9.3.42 Assertion (A) : A microstrip line cannot support pure TEM mode of propagation.Reason (R) : A microstrip line suffers from various forms of losses.(A) Both A and R are true and R is the correct explanation of A




MCQ 9.3.43 Consider the following statements relating to the cavity resonators :1. For over-coupling the cavity terminals are at voltage maximum in the input

line at resonance

2. For over-coupling the cavity terminals are at the voltage minimum in the input line at resonance

3. For under-coupling the normalized impedance at the voltage maximum is the standing wave ratio

4. For over-coupling the input terminal impedance is equal to the reciprocal of the standing wave ratio

Which of the statements given above are correct ?(A) 1 and 2 (B) 3 and 4

(C) 1 and 3 (D) 2 and 4

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MCQ 9.3.44 Consider the following statements relating to the microstrip lines :1. Modes on microstrip lines are purely TEM

2. Microstrip line is also called open strip line

3. Radiation loss in microstrip line can be reduced by using thin high dielectric materials

4. Conformal transformation technique is quite suitable for solving microstrip problems

Which of the statements given above are correct ?(A) 1, 2 and 3

(B) 2, 3 and 4

(C) 1, 3 and 4

(D) 1, 2 and 4

MCQ 9.3.45 Match List I (Dominant Mode of Propagation) with List II (Type of transmission Structure) and select the correct answer :

List-I List-II

a. Coaxial line 1. TE

b. Rectangular waveguide 2. Quasi TEM

c. Microstrip line 3. Hybrid

d. Coplanar waveguide 4. TEMCodes : a b c d(A) 1 4 2 3(B) 4 1 3 2(C) 1 4 3 2(D) 4 1 2 3

MCQ 9.3.46 For TE or TM modes of propagation in bounded media, the phase velocity(A) is independent of frequency

(B) is a linear function of frequency

(C) is a non-linear function of frequency

(D) can be frequency-dependent or frequency-independent depending on the source

MCQ 9.3.47 A waveguide operated below cut-off frequency can be used as(A) A phase shifter

(B) An attenuator

(C) An isolator


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For View Only Shop Online at www.nodia.co.inMCQ 9.3.48 Assertion (A) : The quality factor Q of a waveguide is closely related to its

attenuation factor a.Reason (R) : Normally attenuation factors obtainable in waveguides are much higher than those obtainable in transmission lines.(A) Both A and R are true and R is the correct explanation of A



(D) A is false but R is

MCQ 9.3.49 Assertion (A) : The greater the ‘Q ’, the smaller the bandwidth of a resonant circuit.Reason (R) : At high frequencies the Q of a coil falls due to skin effect.(A) Both A and R are true R is the correct explanation of A

(B) Both A and R are true R is NOT the correct explanation of A



MCQ 9.3.50 For a wave propagation in an air filled rectangular waveguide.(A) guided wavelength is never less than free space wavelength

(B) wave impedance is never less than the free space impedance

(C) TEM mode is possible if the dimensions of the waveguide are properly chosen

(D) Propagation constant is always a real quantity

MCQ 9.3.51 When a particular mode is excited in a wave-guide, there appears an extra electric component in the direction of propagation. The resulting mode is(A) transverse-electric

(B) transverse-magnetic

(C) longitudinal

(D) transverse-electromagnetic

MCQ 9.3.52 For a hollow waveguide, the axial current must necessarily be(A) a combination of conduction and displacement currents

(B) time-varying conduction current and displacement current

(C) time-varying conduction current and displacement current

(D) displacement current only

MCQ 9.3.53 As a result of reflections from a plane conducting wall, electromagnetic waves acquire an apparent velocity greater than the velocity of light in space. This is called(A) velocity propagation (B) normal velocity

(C) group velocity (D) phase velocity

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MCQ 9.3.54 Assertion (A) : A thin sheet of conducting material can act as a low-pass filter for electromagnetic waves.Reason (R) : The penetration depth is inversely proportional to the square root of the frequency.(A) Both Assertion (A) and Reason (R) are individually true and Reason (R) is





MCQ 9.3.55 Consider the following statements in connection with cylindrical waveguides :1. At low frequency the propagation constant is real and wave does not propagate.

2. At intermediate frequency the propagation constant is zero and wave cut off.

3. At high frequency the propagation constant is imaginary and wave propagates.

4. At transition condition the cut-off frequency is inversely proportional to the eigen values of the Bessel function for the respective TEnr mode.

Which of these statements is/are correct ?(A) 1, 2 and 3 (B) 2 only

(C) 2 and 3 only (D) 2, 3 and 4

MCQ 9.3.56 How is the attenuation factor in parallel plate guides represented ?(A) Power lost/power transmitteda =

(B) 2 Power lost/power transmitteda #=

(C) /(2 )Power lost per unit length power transmitteda #=

(D) )Power lost/(power lost power transmitteda = +

MCQ 9.3.57 Which one of the following statements is correct? A wave guide can be considered to be analogous to a(A) low pass filter

(B) high pass filter

(C) band pass filter

(D) band stop filter

***********

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SOLUTIONS 9.1

SOL 9.1.1 Option (A) is correct.Given, electric field intensity of the propagating wave is Ezs 5 /sin sin V mx y e20 25 j zp p= b-

^ ^h h (1)So, we conclude that the wave is propagating in az direction. Since, the wave has it’s component of electric field in the direction of propagation so, the waveguide is operating in TMmn (Transverse magnetic) mode.Now for determining the value of m and n , we compare the phasor form of electric field to its general equation given as.

Ezs sin sinE am x

nn y eo

j zp p= b-a ak k (2)

where a and b are the dimensions of waveguide and since, the waveguide has the dimension 10 4 cm# so, we get a 10 cm= and 4 cmb =Now, comparing equation (1) and (2) we get

am xp x20p= & m 2=

bn yp y25p= & n 1=

Thus, the mode of propagation of wave is TM21.

SOL 9.1.2 Option (A) is correct.A wave mode propagates in a waveguide only if it’s frequency is greater than cutoff frequency. If there is no any propagating mode inside the waveguide then energy in the propagating mode is zero. So, average power flow down the waveguide below cutoff frequency is zero.i.e. Both the statements are correct and R is correct explanation of A.

SOL 9.1.3 Option (C) is correct.The intrinsic impedance of an airfilled waveguide for TM mode is defined as

TMmnh ff1 cmn

0

2

h= - c m

Since, the operating frequency is twice the cutoff frequencyi.e. f 2f ,c mn=So, we get the intrinsic wave impedance as

TMmnh 377 1 21 2

= - b l 32.6.49W= 327. W




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SOL 9.1.4 Option (D) is correct.The dimensions of wave guide, a 10 0.1cm m= =and, b 4 cm=The mode of propagation, m 2= , n 1=Operating frequency, f 7.5 7.5 10GHz Hz9

#= = ,Unbounded phase velocity, vp 3 10 /m sc 8

#= = (air filled)So, the cut-off frequency of the waveguide is given as

fc c

am

bn

22 2= +a ak k 2

3 10 20 258

2 2#= +^ ^h h

4.8 10 Hz9#=

Therefore, the phase constant of the wave inside the waveguide is defined as

b v ff1

p

c2w= - c m c

fff f2 c

2

2 2p= -

. .3 10

2 10 7 5 4 889 2 2

##

p= -^ ^h h

20.7 /rad m2=

SOL 9.1.5 Option (A) is correct.Since, the electric field component exists in the direction of propagation so it will be operating in TM (Transverse magnetic) mode. So, for the TM mode the electric field components in phasor form are given as

Exs hrxEzs

2 22=-

and Eys hryEzs

2 22=-

Since, the given electric field component is Ezs /sin sin V mE x y e50 400

rzp p= -^ ^h h

So, Exs cos sinhr E x y e50 50 40 rz2 0p p p=- -^ ^ ^h h h

and Eys sin coshr E x y e40 50 40o

rz2 p p p=- -^ ^ ^h h h

Therefore, the ratio of the components is

EEys

xs cot tanx y4050 50 40pp p p= ^ ^h h

. cot tanx y1 25 50 40p p= ^ ^h h

SOL 9.1.6 Option (D) is correct.Relative permittivity of dielectric re 2.25=Relative permeability of the dielectric, rm 1=Operating frequency, f 10 10GHz Hz10= =Since, the waveguide is operating in TEM mode so, the phase constant is given as

b w me= cf2

r rp m e= .

3 102 10 1 58

10

#

##

p= ^ h 314.2 /rad m=

The group velocity of the wave in TEM mode will be equal to its phase velocity in the unbounded dielectric medium



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For View Only Shop Online at www.nodia.co.ini.e. vg v c1

pr rme m e

= = = 2 10 /m s8#=

SOL 9.1.7 Option (A) is correct.In an a b# rectangular waveguide, cutoff frequency for TE mn^ h or TM mn^ h mode is defined as

fc mn^ h am

bn

21 2 2

me= +a ak k

Now, for TM11 mode

fc1 a b21 1 12 2

me= +b bl l

f a a21 1 22 2

m= +b bl l a b2=

a

52

1me

= c m

Similarly, for TM12 mode

fc2 a b21 1 22 2

me= +b bl l

a17

21me

= c m

For TE10 mode fc3 a21 1 0

2

me= +b l

a21me

=

For TE20 mode fc4 a21 2 0

2

me= +b l

a2

21me

= c m

So comparing cutoff frequencies of all the modes we get the modes in ascending order of cutoff frequencies as TE TE TM TM< < <10 20 11 12

SOL 9.1.8 Option (D) is correct.Dimensions of wave guide a 5 5 10cm m2

#= = -

and b 3 3 10cm m2#= = -

Operating frequency, f 3.75 3.75 10GHz Hz9#= =

Since operating mode of the waveguide is TE10 (i.e., m 1= and n 0= ) so, the cutoff frequency of the airfilled waveguide is given as

fc c

am

bn

22 2= +a ak k

3 102 5 10

1 0 3 108

2

29

##

#= + =-b l

The group velocity of the EM wave in the waveguide is given as

vg c ff1 c

2

= - c m .

3 10 13 75 103 108

9

9 2

##

#= - c m

1.8 10 /m s8#=

SOL 9.1.9 Option (C) is correct.Dimensions of waveguide, a 2.5 2.5 10cm m2

#= = -

and b 5 5 10cm m2#= = -

Operating frequency, f 15 15 10GHz Hz9#= =

Conductivity of medium, s 0= (lossless dielectric)




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Relative permittivity, re 2=Relative permeability, rm 1=The operating mode of the waveguide is TE20 mode (i.e., m 2= and n 0= )So, the cutoff frequency of the waveguide in the TE20 mode is given as

fc am

bn

21 2 2

me= +a ak k

.23 10

2 5 102

r r

8

2

2#

#m e= -b l

.8 5 109#=

The wave impedance for the TE20 mode is given as

TE20h

ff1 c

2

h=- c m

ff1

1r

r

c0 2

h em=

- cf

mp\

.

377 21

1 158 5

12

=- b

flp

323W=

SOL 9.1.10 Option (D) is correct.Given, the cutoff frequency for TM2 mode is c 2

l^ h 2 2 10mm m3#= = -

Since, the cutoff wavelength for TM or TEn n mode for a parallel plate waveguide is defined as

c nl^ h n

b2re= (1)

where b is the separation between parallel plates of the waveguide and re is relative permittivity of the medium. So, putting the known values in the expression, we get

2 10 3#

- b22

re= (n 2= )

b 2 10r

3#

e=

-

Now, for any n mode to propagate the operating wavelength must be less than or equal to the cutoff frequency.i.e. l c n

# l^ h

So, from equation (1) for the propagation of wavelength 0.1 cml = we have the relation as

.0 1 10 2#

- bb2

r# e

.0 1 10 2#

- n2 2 10

rr

3

###

ee

-

n .0 1 104 10

2

3

#

## -

-

n 4#

Therefore, the possible modes that can propagate in the waveguide areTEM, TE1, TE2, TE3, TE4, TM1, TM2, TM3 and TM4

Thus, there are nine possible modes that can propagate in the waveguide.



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Plate separation, b 10 10mm m2= = -

Minimum operating frequency, fmin 15 15 10GHz Hzfc 9#= = =

Since, for TMn mode of parallel plate waveguide, cutoff frequency is defined as

fc n^ h bn

2 me=

So, for TM 3^ h mode n 3=^ h we have the cutoff frequency as

fc 3^ h b2

3r0 0m e e

=

15 109#

2 103 3 10

r2

8

#

# #

e= -

^ h f fminc 3 =^ h

re 3=or re 9=

SOL 9.1.12 Option (D) is correct.Plate separation, b 20 20 10mm m3

#= = -

Relative permittivity of medium, re 2.1=Operating frequency, f 16 GHz=For propagation of wave the operating frequency must be greater than the cutoff frequency of TE n^ h or TM n^ h mode of parallel plate waveguidei.e. f f> c n^ h

f bn

2>

r0 0m e e n fb2< r0 0m e e

n .3 10

2 16 10 20 10 2 1< 8

9 3

#

# # # ##

-

n .3 09<So, the maximum allowed mode is n 3=Since, all the modes given in the option are in the range, therefore, all the three modes will propagate.

SOL 9.1.13 Option (C) is correct.Cutoff frequency of TM 1^ h mode, fc 1^ h 2.5 2.5 10GHz Hz9

#= =Operating wavelength, l 3 3 10cm m2

#= = -

The cutoff frequency of TE 3^ h mode of the parallel plate waveguide is given as fc 3^ h f3 1c= ^ h 3 2.5 10 Hz9

# #= 7.5 10 Hz9#=

Since, the operating frequency of the waveguide is defined as

f cl=

where l is the operating wavelength. So, the operating frequency of the parallel plate waveguide is

f 10 Hz3 103 10

2

810

#

#= =-




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Therefore, the group velocity of TE3 mode is given as

vg 3^ h ff1 1 c

0 0

32

m e= -

^e

ho

.3 10 110

7 5 10810

9 2

##= - c m 2 10 /m s8

#=

SOL 9.1.14 Option (C) is correct.At cutoff the mode propagates in the slab at the critical angle which means that the phase velocity will be equal to that of a plane wave in upper or lower media of refractive index n2. So, the phase velocity at cutoff will be

vp . 1.2 10 /m snc

2 53 10

2

88#

#= = =

SOL 9.1.15 Option (A) is correct.The phase velocity at cutoff is independent of the mode and equal to the phase velocity of a plane wave in unbounded media. Since, in the given problem the phase velocity of TM2 mode is to be determined for same waveguide so, the phase velocity of TM2 mode will be equal to that of TM1 mode.i.e., vp2 vp1=

SOL 9.1.16 Option (C) is correct.Relative permittivity of material, re .8 8=Separation between strip line, b 0.632 cm=Characteristic impedance, Z0 35=So, Zr 0e . ( ) .8 8 35 103 8= =Since, Zr 0e 120<Therefore, the width to separation ratio of strip line transmission line is given as

bw .

Z30 0 441r 0ep= -

.w

0 632 .

.8 8 3530 0 441p= -

^ ^h h w .0 295=

SOL 9.1.17 Option (C) is correct.Guide wavelength of a stripline is defined as,

gl fcre

=where, c is velocity of wave in free space, f is the operating frequency and re is the relative permittivity of the medium. So, we get

gl .8 8 3 103 10

9

8

#

#=^ ^h h

3.37 cm=

SOL 9.1.18 Option (B) is correct.Statement 1Suppose on the contrary the TEM mode existed. In this case the magnetic field must lie solely in the transverse xy -plane. The magnetic field lines must form closed paths in this transverse plane, since 0H:d = . From Ampere’s law, the integral



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For View Only Shop Online at www.nodia.co.inof this transverse magnetic field around these closed paths must yield the axial conduction or displacement current. But E 0z = for the TEM mode so, no axial displacement current can exist. Also, since there is no center conductor so, no axial conduction current can exist. Therefore statement 1 is correct.Statement 2The dominant mode is the mode that has lowest cutoff frequency. Now, f ,c mn is clearly minimized when either m or n is zero. Since, TM01 or TM10 mode doesn’t exist so, TM mode can’t be the dominant mode of propagation in rectangular waveguide.It is also correct statement.

SOL 9.1.19 Option (C) is correct.For a TMmnp mode, neither m nor n can be zero otherwise all field components vanish, however p can be zero. So, the lowest order TM mode is TM110.

SOL 9.1.20 Option (D) is correct.Since, TEmnp mode of cavity resonator can have either m 0= or n 0= (but not both at a time) where as p can’t be zero for TE mode so, the lowest order of TE mode is TE011 if a b< TE101 if a b>As the dimensions of the cavity resonator are equal (a b= ) so, both the TE101 and TE011 are lowest order mode.

SOL 9.1.21 Option (D) is correct.Given, the dimensions of cavity resonator are related as a b c> > (1)The condition for propagating TE and TM modes in a cavity resonator are as follows :(1) for TMmnp mode, neither m nor n can be zero however p can be zero.

(2) For TEmnp mode p can’t be zero but either m or n can be zero (but not both at a time)

The resonant frequency of TMmnp or TEmnp mode in a cavity resonator is defined as

fmnp am

bn

cp

21 /2 2 2 1 2

me= + +a a ak k k: D

So, comparing the resonant frequency for the different values of , andm n p using the relation defined in equation (1), we get the lowest order mode will be TM110 and the ascending order can be written as below :TM110; TE101; TE011; TE TM111 111=

SOL 9.1.22 Option (C) is correct.In an airfilled cavity resonator, resonant frequency is defined as

fmmp am

bn

cp

21 /

0 0

2 2 2 1 2

m e= + +a a ak k k: D




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So, for TE101 mode (m p 1= = , n 0= ) the resonant frequency is

f101 23 10

30 101

20 101 /8

2

2

2

2 1 2#

# #= +- -b bl l; E

9.01388 10 Hz8#= 901.4 MHz=

SOL 9.1.23 Option (A) is correct.The resonant frequency of a cavity resonator is defined as

fmnp am

bn

cp

21 /

0 0

2 2 2 1 2


Since, a b c= = so, the dominant modes are TE101 or TE011 or TM110. Therefore, taking any of ,m n or p equal to zero, we get the resonant frequency as

fmnp am

bn

cp

21 /

0 0

2 2 2 1 2


fmnp a23 10 2 /8

2

1 2#= ; E a b c= =^ h

15 109# a2

3 10 28#

#=

a 1.41 10 1.41m cm2#= =-

i.e. a b c= = 1.41 cm=

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SOLUTIONS 9.2

SOL 9.2.1 Option (B) is correct.Since, the waveguide is operating at TMn mode so, the phasor form of magnetic field of the EM wave will be given as

Hxs coshj H b

nxy e j z0we= b-

a k

Since, the waveguide is operating at TM1 mode (i.e. n 1= )

So, Hxs coshj H b

y e j z0we p= b-

a k

Therefore, the instantaneous magnetic field intensity of the wave is given as

Hx Re coshj H b

y e ej z j t0

we p= b w-a k' 1

cos sinh H by t z0

we p w b=- -a ^k h

at t 0= Hx cos sinh by zwe p b= a ^k h (1)

As the EM wave is propagating in y -z plane so, in TM mode the y and z -components of the magnetic field intensity will be zero.i.e. H Hy z= 0=Thus, the field will have the component only in z -direction for which we sketch the field lines in y -z plane. From equation (1), we conclude that the field intensity Hzdepends on the values cosines and sines of the two variables defined as

cos byp

a k 0 0.5

0.5 1

ve

ve

y

y

< << <

=+-)

sin zb ve

ve

z

z

0

2

< << <b p

p b p=+-*

Using these values we get the sketch of the field lines in the yz -plane as shown in the figure below where x -axis directs into the paper.




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The phase velocity of the EM wave in the guide is defined as

vp bw=

where w is the operating angular frequency and b is the phase constant inside the airfilled waveguide given as

b /c f f1 c2w= - ^ h waveguide

So, we get

vp /c f f1 c

2ww=- ^ h

& vc

ff

p

c2 2

+b cl m 1=

The above equation is the equation of a circle. So, the graph between ( / )c vp and /f fc^ h will be as plotted below :

SOL 9.2.3 Option (A) is correct.Wavelength for a propagating wave inside the waveguide is defined as

gl 2bp=

where b is the phase constant of the wave in the waveguide given as

b ff1 c

2

w me= - c m

where fc is the cutoff frequency of the waveguide and f is the operating frequency



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For View Only Shop Online at www.nodia.co.inof the waveguide. So, we get

gl

ff1

2c

2

w me

p=- c m

& gl

ff

2

1

1c0 0

2w m ep=

- c m

(for airfilled guide 0m m= , 0e e= )

& gl fc

ff1

1c

2=

-b

cfl

mp

& gl

ff1

1c

2l=

- cf

mp

& g

ll

/

/f ff f

1c

c2

2

=-^

^

h

h

Thus, the plot between /f fc^ h and /gl l^ h is as sketched below :

SOL 9.2.4 Option (C) is correct.The propagation constant g^ h in the parallel plate waveguide is defined as

2 2g w me+ bn 2p= a k (1)

Since, for lossless medium propagation constant is given as g jb= (attenuation constant, 0a = )Putting it in equation (1), we get

2 2b w me- + bn 2p= a k

At the cutoff frequency, cw w= phase constant is zero (i.e., b 0= ). So, we get

c2w me b

n 2p= a k

cw bnmep=

So, for TE1 mode c1w b mep=

So TE2 mode c2w b

2mep=




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For TE3 mode c3w b

3mep=

Comparing the three expressions we get, < <c c c1 2 3w w w

SOL 9.2.5 Option (A) is correct.As calculated in previous question, the expression for the operating frequency of the wave in the waveguide is given as

2 2b w pe- + bn 2p= a k

& 2w me bn2 2b p= + a k

& f 42 2p me^ h bn2 2b p= + a k

So, for TM2 mode (n 2= ) f 2 b41 42

2 2

p meb p

= + a k: D

For TM3 mode (n 3= ) f 2 b41 92

2 2

p meb p

= + a k: D

and for TM4 mode (n 4= ) f 2 b41 162

2 2

p meb p

= + a k: D

Thus, for the above obtained expressions for the frequencies at different modes, we sketch the f -b curve as shown below :

SOL 9.2.6 Option (C) is correct.Operating frequency, f 5 5 10GHz Hz9

#= =Separation between the plates b 6 6 10cm m2

#= = -

So, the cut off frequency for TM1 mode is given as

fc b2

10 0m e

= 2 6 10

3 102

8

# #

#= - 2.5 10 Hz9#=

For a parallel plate waveguide, phasor form of components of electric field and magnetic field intensity of a propagation wave are given as

Eys cosE bn y e z

0p= g-

a k

and Hxs cos

ff

Ebn y e

1 c

z

0

20

h

p=--

g-

c

a

m

k



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So, the average power is given as

Pave Re dSE H21

s s#= )" ,# ( )E H w dy2

1ys xs

b

0= -^ ^h h" ,#

cosw

ff

Eby dy2

1 c

b

0

202

0

2

h

p=- c

a

m

k#

cos

w

ff

E by

dy21

2

1 2

c

b

0

202

0h

p

=-

+

c

b

m

l

# wb

ff

E4

1 c0

202

h=

- c m

The maximum power propagation will be due to the maximum electric field in the medium (the dielectric strength of the medium). So, we have the maximum average power as

P maxave^ h wb

ff4

1

3 10

c0

2

6 2#

h=

- c

^

m

h ( E max0^ h 3 10 /V m6

#= )

Putting all the values, we get the average power per unit width as

wP maxave^ h

ff4

6 10

120 1

3 10

c

2

2

6 2#

##

p=

-

-

c

^

m

h

.4 135 108#= 414 /MW m=

SOL 9.2.7 Option (D) is correct.Separation between waveguide plates, b 12 0.12cm m= =Operating frequency, f 2.5 2.5 10GHz Hz9

#= =For the TEM mode, phasor form of electric and magnetic field components are given as Eys E e0 z= g-

Hxs E e z

0

0

h=- g-

So, the average power propagated in the waveguide is given as

, Pave Re dSE H21

s s#= )& 0#

Re E H ds21

ys xs= -^ ^h h" ,# E E wdy21b

00 0

0

h=- -^ ch m#

E wb2 0

02

h=

The maximum electric field, without any voltage breakdown is defined as the dielectric strength of the medium as given and as the dielectric strength of air is E max0^ h 3 10 /V m6

#=So, the maximum average power propagated in the waveguide is

P maxave^ h .w2 1203 10

0 126 2

#

#

p=

^

^^

h

hh




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Therefore, the maximum time average power propagated per unit width in the waveguide is

wP maxave^ h

.1 432 109#= 1.432 /GW m=

SOL 9.2.8 Option (B) is correct.Maximum operating frequency fmax 1.5 10 Hz9

#=Relative permittivity of medium, re .8 4=The cutoff frequency in TEM mode is f 0c = and the cutoff frequency in TE n^ h or TM n^ h mode is given as

fc n^ h bn

2 me=

So, for TE1 or TM1 mode n 1=^ h we get

fc 1^ h b2

1r0 0m e e

=

Since, the guide is to be operated only in TEM mode. So, the operating frequency must be less than fc 1^ h while it must be greater than 0 (cutoff frequency in TEM mode).i.e. 0 f f< < c 1^ h

or, f b2

1<r0 0m e e

b f2

1<r0 0m e e

As the frequency inside the waveguide ranges in 0 1.5 GHzf< < , therefore, the maximum allowable separation between the plates is

bmax fc

2 max re=

. .2 1 5 10 8 43 10

9

8

# # #

#= ( 1.5 GHzfmax = )

0.345 m= 3.45 cm=

SOL 9.2.9 Option (A) is correct.The Brewster’s angle for parallel polarized wave is given as

tan ||Bq 1

2ee=

||Bq .tan 21 051= -

b l .35 9c=

The cutoff frequency for TM1 mode in 1st medium (permittivity = r1e ) is given as

fc 1^ h b2

1r0 0 1m e e

= .2 14 1 10 23 10

3

8

# #

#= -

7.52 10 Hz9#=

So, the frequency for which there is no any reflective loss is given as

f0 cosfc 1

q= ^ h

where q is ray angle that has the value, 90 ||Bcq q= - . So, we get

f0 .

..

cos sin907 52 10

35 97 52 10

B11

9 9# #c cq

=-

=^ h

12.8 GHz=



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For View Only Shop Online at www.nodia.co.inNOTE :Brewster’s angle is the incident angle of a plane wave at the interface of two mediums for which there is no any reflection in the medium.

SOL 9.2.10 Option (A) is correct.As we have determined in the previous question, the value of f0 is f0 12.8 GHz=and the cutoff frequency for TM1 mode is fc 1^ h 7.52 GHz=So, the cutoff frequency for TM2 mode will be fc 2^ h f2 c 1= ^ h 15.04 GHz=Since, the operating frequency f0 is below the cutoff frequency for TM2 mode so, TM2 mode or the higher modes can’t propagate at the frequency f0. Therefore, only one mode TM1^ h can propagate at the frequency f0 through the waveguide.

SOL 9.2.11 Option (A) is correct.Consider the dominant mode of the waveguide is TE10. Since, the cut-off frequency for TEmn mode is defined as

fc f a

mbn

21 2 2

m= +a ak k

So, the cutoff frequency for the TE10 mode is

fc 10^ h ca a

c2

12#= = (for airfilled waveguide 1/c me= )

Now, the next higher order mode of the waveguide will be TE01 so, it’s cutoff frequency is given as

fc 01^ h bc2=

For the given condition design frequency will be f . .f f1 1 0 9c c10 01= =^ ^h h

Since, the operating frequency of the waveguide is f 5 5 10GHz Hz9

#= =

So, we get fc 10^ h .1 15 109#=

ac2 .1 1

5 109#=

a .

3.3 cm2 5 103 10 1 1

9

8

# #

# #= =^

^

h

h

and fc 01^ h .0 95 109#=

bc2 .0 9

5 109#=

b .

2.7 cm2 5 103 10 0 9

9

8

# #

# #= =^

^

h

h

SOL 9.2.12 Option (C) is correct.For a propagating mode TMn or TEn the cutoff wavelength of the symmetric




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dielectric slab is defined as

cl nd

12 r r1 2e e= -

-

where r1e and r2e are the permittivities of dielectrics and d is the slab thickness. So, we get

cl . .n 1

2 20 10 2 2 2 16# #= -

-- .

n 11 26 10 5

#= --

Since the operating wave length must be lower than or equal to the cutoff wavelengthi.e. l c# lTherefore, for the propagation of wavelength 2.6 mml = in the dielectric slab waveguide, we have the condition as

.2 6 10 6#

- .n 1

1 26 10 5## -

-

n 1- ..2 6 101 26 10

6

5

#

## -

-

n 1- .4 85#

n .5 85#

So, the possible values of n for which the wavelength 2.6 mml = can propagate in the waveguide are n 1= , 2, 3, 4, 5. Thus, we get the possible modes as follows :TE1, TE2, TE3, TE4, TE5

TM1, TM2, TM3, TM4, TM5

and as TEM doesn’t exist in the dielectric slab waveguide so, total 10 modes can propagate for the operating wavelength.

SOL 9.2.13 Option (A) is correct.For a rectangular waveguide operating in TE10 mode the phasor form of electric field is given as Eys sinE kx e j z0= b-

^ h

Hxs sinE kx e j z0wmb=- b-

^ h

Hzs cosj K E kx e j z0wm= b-^ h

Since, the wave is propagating in TE mode so, no any other field component exists in the waveguide.Now, the average power in an EM wave is defined as

Pav Re E H21

s s#= )" ,

Since, Hzs has a factor j . So it would lead to an imaginary part of the total power when cross product with Ey is taken. Therefore, the real power in the case is found through the cross product with complex conjugate of Hxs as below :

Pav Re E H21

ys xs#= )" , sinE kx a2

1z0

2 2

wmb= ^ h

Thus, Pav \ b



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Cutoff wavelength for symmetric slab waveguide is defined as,

c nl^ h nd

12 r r1 2e e= -

- (1)

where d is the thickness of slab n is the propagating mode, r1e and r2e are the relative permittivites of the mediums.Now, the refractive indices of the two mediums can be given as n1 r1e=and n2 r2e=So, the equation can be rewritten as

c nl^ h nd n n

12 1

222

= --

Since, the waveguide supports only a single pair of TE and TM modes. i.e. it supports n 1= mode and denies all the higher modes. Therefore, the operating wavelength l must be with in the range.i.e. c 1l^ h c 2$ $l l^ h (2)where ( )c 1l and ( )c 2l are the wavelengths for mode n 1= and n 2= respectively. Putting n 1= in equation (1) we get c 1l^ h 3=Therefore, the condition obtained in equation (2) reduces to l c 2$ l^ h

.3 1 10 6#

- .n

2 12 10 10 3 36

12 2

# #= ---^ h

.n 3 312 2- ^ h .

2 10 103 1 10

6

6

# #

## -

-

n1 .3 304#

Thus, the maximum value of n1 is 3.304.

SOL 9.2.15 Option (A) is correct.The wave angle must be equal to or greater than the critical angle of total reflection at both interfaces. So, the minimum wave angle in the slab is determined for the greater of the two critical angles determined at two interfaces.Since, n3 n> 2

It means the critical angle will be greater for n3 media and given as

c3q .sin nn 48 61

1

3 c= =-a k

Therefore, the minimum possible wave angle will be .48 6c.

SOL 9.2.16 Option (D) is correct.Phase velocity of a guided mode is defined as

vp bw=

So, maximum phase velocity for the guided mode is




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v maxp minbw= (1)

where minb is the minimum phase constant given as minb sinn k min1 0 q= (2)where minq is the minimum possible wave angle, k0 is the wave number in free space and n1 is the refractive index of propagating media (slab).Now, from the given relation for refractive index, we have n2 n> 3

So, as described in previous question the minimum wave angle will be determined by larger critical angle (i.e. at the interface of n1 and n2) which is given as

sin minq sin nn

,c 121

2q= =

Putting it in equation (2), we get

minb n k nn

1 01

2= n k2 0=

Again putting the value of minb in equation (1), we get

v maxp n k2 0

w= nc

2= (velocity of wave in air, c k0

w= )

SOL 9.2.17 Given, the cross-section dimension of the waveguide is a 4.5 cmb= =The cut off frequency of the rectangular waveguide is defined as

( )fc mn am

bn

21 /

0 0

2 2 1 2

m e= +a ak k: D

So, the cutoff frequency for TE22 mode of waveguide of square cross section is

fc22 a b21 2 2 /

0 0

2 2 1 2

m e= +b bl l; E

.3 10

0 04528

# #= 2 GHz=

The phase constant of the wave inside the waveguide is given as

b ff1

/c

0 0

2 1 2

w m e= - c m= G 3 10

2 6 10 1 62 /

8

9 2 1 2

#

# #p= - b l; E

1.1847 10 m2 1#= -

Therefore, the wavelength of the TE22 mode wave is

l 2bp= . 5.303 10 5.3 cm118 47

2 2p#= = =-

SOL 9.2.18 Option (C) is correct.Given, the operating frequency of the waveguide is f 6 6 10GHz Hz9

#= =So, the wave number in the waveguide of dimension ‘a ’ is given as

k cf2p=

3 102 6 10 408

9

#

# #p p= =

Now, the attenuation constant of section of waveguide (attenuator) with dimension /a 2 is given as

a /a

k2

2 2p= -b l .0 045752 40

2 2p p= -b ^l h 0.04572 ma =^ h



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For View Only Shop Online at www.nodia.co.in 55.63 /Np m=Since, the total required attenuation is 100 dB along the attenuator so, we have 100 dB- loge20 l= a-

where l is length of the attenuator. (length travelled by wave in the small section of waveguide). Therefore, solving the equation we get, 10 5- e l= a-

l .. 0.2067 20.67 cm55 63

11 5= = =

SOL 9.2.19 Option (B) is correct.Dimension of waveguide, a 1.07 0.0107cm m= =Operating frequency, f 10 10 10 10GHz Hz Hz9 10

#= = =Permittivity of dielectric, re .8 8=and tan d .0 002=The phase constant of the EM wave inside the waveguide is defined as b /k a2 2p= - ^ h where k is the wave number in the unbounded medium given as k kr 0e= (k0 is wave number in free space)

. cf8 8 2p= ^ h k c

f20

p=b l

.8 83 10

2 108

10

#

#p= ^ h (c 3 10 /m s8#= )

621.3 m 1= -

So, the phase constant of the wave along the waveguide is

b . .621 3 0 01072 2p= -^ ah k

547.5 m 1= -

Therefore, the attenuation constant due to dielectric loss is given as

da tank2

2

bd=

.. .

2 547 5621 3 0 0022

=^

^ ^

h

h h

0.705 /Np m= 6.12 /dB m=

SOL 9.2.20 Option (D) is correct.In a circular waveguide, cutoff frequency for TEmn mode is given as

fcmn aP

2mn

p pe= l

and the cutoff frequency for TMmn mode of the waveguide is given as

fcmn aP

2mn

p me=

where a is the cross sectional radius of waveguide, P mnl and Pmn are the roots of the Bessel’s equation. Their values are related as listed below in increasing order P P P P< < <11 01 21 01l l l P P P P< < <11 31 21 41= l l and so on.So, for the corresponding values of P mnl and Pmn , we get the increasing order of the modes with respect to their cutoff frequencies as shown below on the frequency axis :




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Thus, the first four propagating modes are respectively TE11, TM01, TE21, TE01 or TM11

SOL 9.2.21 Option (C) is correct.GivenThickness of substrate, d 0.316 cm=Relative permittivity of substrate, re .2 2=Characteristic impedance of line, Z0 100W=The width to thickness ratio ( / )w d is defined as

dw

ee

28A

A

2=-

for 2dw <

where . .A Z60 2

111 0 23 0 11r

r

r

r

0 eee

e= + + +- +c m

Now, we assume 2<dw . So, we get

A ... . .

.60100

22 2 1

2 2 12 2 1 0 23 2 2

0 11= + + +- +b bl l .2 21=

and therefore, the width to thickness ratio is

dw .

ee

28 0 896 2<.

.

2 2 21

2 21

=-

=#

As the obtained value of ( / )w d is less than 2 so, our assumption was correct and we have

dw .0 896=

or, w . d0 896 #= ^ h . .0 896 0 316#= ^ ^h h 0.283 cm=

SOL 9.2.22 Option (A) is correct.As calculated in previous question the width to thickness ratio is

dw .0 896=

So, the effective value of permittivity is given as

ee 21

21

11

12r r

wd

e e= + + -+

. .

.2

2 2 12

2 2 1

1 0 89612

1=+

+ -

+

^ h

.1 758=Therefore the guided wavelength of the EM wave is



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For View Only Shop Online at www.nodia.co.in gl

fcee

=

where f is operating frequency and c is velocity of wave in free space. So, we get

gl .1 758 8 10

3 109

8

#

#=^ ^h h

8 GHzf =

2.83 cm=

SOL 9.2.23 Option (D) is correct.The resonant frequency for TEmnp mode is defined as

fr am

bn

cp

21 /2 2 2 1 2

me= + +a a ak k k: D

So for TE101 mode the resonant frequency of the cavity resonator is

fr 3 . .2

3 100 025

10 051 /8 2 2 1 2

#= +b bl l; E ( 0m m= , 3 0e e= )

3.87 10 Hz9#=

3.87 GHz=

SOL 9.2.24 Option (A) is correct.The quality factor of TE101 mode is defined as

QTE101 b a c ac a ca c abc

2 3 3 2 2

2 2

d=

+ + ++

^ ^

^

h h

h

6 @

where d is skin depth given as

d f1r c0p m s

=

where fr " resonant frequency for the defined mode. 0m 4 10 7

#p= -

cs = Conductivity of copperSo, we get the skin depth as

d . .3 87 10 4 10 5 8 10

19 7 7

# # #p p=

-^ ^ ^h h h

.1 06 10 6#= -

Therefore, the quality factor of the resonator is

QTE101 . . . .

. .

1 06 10 2 2 2 5 5 2 5 5 2 5 5

2 5 5 2 5 2 5 106 3 3 2 2

2 2 2

# #

#=

+ + +

+-

-

^ ^ ^ ^ ^ ^ ^

^ ^ ^ ^ ^

h h h h h h h

h h h h h

8

8

B

B

" ", ,

.7732 7= 7733.

SOL 9.2.25 Option (B) is correct.Given, the inner radius of the guide is a 1 0.01cm m= =The cutoff frequency for TEmn mode of a circular waveguide is defined as

f ,c mn ap

2mn

p me= l

where p mnl is the mth root of Bessel’s function J 0n =l^ h.Now, from the given data we have




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P 11l .1 841=So, the cutoff frequency of TE11 mode in the circular waveguide is

fc11 .

2 101 841

20 0p m e

= -^ h

.2 10

3 10 1 8412

8

#

# #p

= -

8.79 10 Hz9#= 8.79 GHz=

SOL 9.2.26 Option (A) is correct.The resonant frequency of TMmnl mode in cylindrical cavity is defined as

f ,c mnl ap

dl

21 mn

2

p mep= - +a bk l

where a is radius of cylindrical cavity, d is height of the cylindrical cavity and pmn is the root of Bessel’s equation.Since, the dominant mode in cylindrical cavity is TM010 so, the cutoff frequency for dominant mode is

fc010 ap

ap c

2 201

0 0

01

p m e p= =

Therefore, the cutoff wavelength for dominant mode is given as

,c 010l fcc010

=

2 10 2#

- pa2

01

p=

a .

22 405 2 10 2

#p=

-^ ^h h

( 2 cm,c 010l = )

.7 65 10 3#= -

0.765 cm=

***********



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SOLUTIONS 9.3

SOL 9.3.1 Option (D) is correct.Given, the magnetic field component along the z -direction as Hz 3 (2.094 10 ) (2.618 10 ) (6.283 10 )cos cos cosx y t z2 2 10 b# # #= -So, xb .2 094 102

#= yb .2 618 102

#= w 6.283 10 /rad s10

#=For the wave propagation inside the rectangular waveguide,

b ( )c x y2

22 2w b b= - +

Substituting the values, we get

b . ( . . )3 10

6 283 10 2 094 2 618 108

10 22 2 4

#

##= - +c m

j261-

Since, b is imaginary so, mode of operation is non-propagatingi.e. vp 0=

SOL 9.3.2 Option (A) is correct.TM11 is the lowest order mode of all the TMmn modes.

SOL 9.3.3 Option (D) is correct.From the given expressions of E and H , we can write, b 280p=

or 2lp 280p= & 140

1l =

So, the wave impedance is given as

h E

HE

3120p

rep= = = (1)

Since, the operating frequency of the wave is f 14 GHz=So, the operating wavelength of the wave can also be given as

l fc

14 103 10

1403

r r r9

8

#

#

e e e= = =

or 1401

1403

re=

or, re 9=




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From equation (1) we have

E3p 120E

9120

p&p p= =

SOL 9.3.4 Option (A) is correct.Rectangular and cylindrical waveguide doesn’t support TEM modes and have cut off frequency.Coaxial cable support TEM wave and doesn’t have cut off frequency.

SOL 9.3.5 Option (A) is correct.Cut-off Frequency for TEmn mode of a rectangular waveguide is defined as

fc c

am

bn

22 2= +a ak k

So, for TE11 mode ( ,m n1 1= = ) the cutoff frequency is

fc 6.2523 10

41

31

10 2 2#= =+b bl l GHz ( 3 10 /cm sc 8#= )

SOL 9.3.6 Option (A) is correct.Given, the electric field intensity of the wave inside rectangular waveguide as

E ( )sin sinh a H a

x t z a2y2 0

wm p p w b= -a bk l

This is TE mode and we know that

Ey sin cosam x

bm y

\p p

a ak k

So, comparing it with the given expression we get m 2= and n 0= . Therefore, the propagating mode is TE20.

SOL 9.3.7 Option (B) is correct.The cut-off frequency for the TEmn mode of the waveguide is defined as

fc c

am

bn

22 2= +a ak k

So, the cutoff frequency of the TE20 ( 2m = , n 0= )mode is

fc 3 10

. 10cam

2 2 0 0328

##= = =a k GHz ( 3 cma = )

Therefore, the wave impedance of the TE20 mode is given as

'h

ff1

3 1010

377c

20

10

10 2

#

h=-

=1 -c cm m

400W= ( 30 GHzf = )

SOL 9.3.8 Option (B) is correct.The cut-off frequency of TEmn mode is defined as

fc c

am

bm

22 2= +a ak k

So, the cutoff frequency of TE30 (m 3= , n 0= ) mode is

fc cam

2= a k

or 18 109# a2

3 10 38#= ( 18 GHzfc = )

or a 401= m

25= cm



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SOL 9.3.10 Option (D) is correct.For any propagating mode inside a rectangular waveguide the velocities are related as vp c v> > g

i.e. the phase velocity of the wave inside the waveguide is greater than the velocity of light in the free space.

SOL 9.3.11 Option (D) is correct.In a microwave test bench, the microwave signal is modulated at 1 kHz because crystal detector fails at microwave frequencies.

SOL 9.3.12 Option (A) is correct.The cutoff frequency of TEmn mode in a rectangular waveguide is defined as

fc am

bn

21 2 2

me= +a ak k

Since in the given rectangular waveguide a b> so, the dominant mode is TE10 and the cutoff frequency for the dominant mode is given as

fc c

am

bn

2 r

2 2

e= +a ak k 1 c

0 0m e=c m

. b43 10

0 031 08 2 2#= +b bl l

.13 . GHz0 2

10 2 58

#= =

SOL 9.3.13 Option (B) is correct.Phase velocity of an EM wave inside an air-filled rectangular waveguide

vp

ff

cc

2=

1 - c m

where c is velocity of EM wave in free space fc is the cutoff frequency of the propagating mode and f is the operating frequency. Since, for a wave propagation the operating frequency must be greater than the cutoff frequency.i.e. f f> c

Therefore, the phase velocity of the wave will be always greater than the velocity of wave in free space.i.e. vp c>

SOL 9.3.14 Option (A) is correct.In a hollow metal wave guide v c v> >p g

where vp " Phase velocity c " Velocity of light in free space. vg " Group velocitySo, the phase velocity of a wave propagating in a hollow metal waveguide is greater than the velocity of light in free space.




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SOL 9.3.15 Option (A) is correct.In a wave guide dominant gives lowest cut-off frequency and hence the highest cut-off wavelength.

SOL 9.3.16 Option (B) is correct.As the impedance of perfect conductor is zero, electric field is minimum and magnetic field is maximum at the boundary.

SOL 9.3.17 Option (B) is correct.Cutoff frequency for TEmn mode in a rectangular waveguide is defined as

fc v

am

bn

2p 2 2= +a ak k

Since, for the given rectangular waveguide a b> so, the dominant mode is TE10 and the cutoff frequency of the dominant mode of rectangular waveguide is

fc 15 10av2 2 10

3 10p2

89

#

##= = =- (For air 3 10vp 8

#= )

15= GHz

SOL 9.3.18 Option (D) is correct.In TE mode E 0z = , at all points within the wave guide. It implies that electric field vector is always perpendicular to the waveguide axis. This is not possible in semi-infine parallel plate wave guide.

SOL 9.3.19 Option (A) is correct.In a rectangular resonant cavity TEmnp mode must have its p 1= . So, the mode TE110 doesn’t exist in the rectangular resonant cavity.

SOL 9.3.20 Option (A) is correct.The transverse electric field and transverse magnetic field inside a waveguide are related as E Hh= where h is intrinsic impedance

or h HE

=

i.e. the ratio of transverse electric field to the transverse magnetic field is called waveguide impedance.

SOL 9.3.21 Option (C) is correct.Cutoff wavelength for Hmn mode of a rectangular waveguide is defined as

cl

am

bn

22 2

=+a ak k

where a and b are the dimensions of waveguide.So, for the H10 mode (m 1= , n 0= ), the cutoff wavelength is

cl

81 0

22

=+b l

( 8 cma = )

16 cm=



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A rectangular waveguide supports TE and TM waves where as it doesn’t support TEM waves.The propagation constant for TE or TM waves inside a rectangular waveguide is defined as

g am

bn2 2 2p p w me= + -a ak k

SOL 9.3.23 Option (C) is correct.Cut-off frequency for TEmn or TMmn mode inside a rectangular waveguide is defined as

fc am

bn

21 2 2

p mep p= +a ak k

Where a and b are the dimensions of rectangular waveguide.

SOL 9.3.24 Option (D) is correct.Cut-off frequency for TE10 mode is

fc10 am

bn

21

0 0

2 2

m e= +a ak k

.2

3 104 755 10

18

2#

##

= -b l ( 1m = , n 0= )

3.16 GHz=Cut-off frequency for TE01 mode is

fc01 .23 10

2 215 1018

2#

##

= -b bl l

6.77 GHz=cut-off frequency for TE11 mode is

fc11 . .23 10

4 755 101

2 215 1018

2

2

2

2##

# #= +- -b bl l

7.47 GHz=and the cut-off frequency for TE20 mode is

fc20 .6.3 GHz2

3 104 755 10

28

2#

##= =-

^ h

Since the operating frequency 12 GHzf = so, we have, fc10, fc01, fc11, f f>c20 . Therefore, all the modes will propagate.Note : For avoiding so many calculation we should directly calculate the higher frequency modes first for higher operating frequency. As in this case if we calculates fc11 first then by getting f f> c11 it is clear that TE01, TE10 and TE11 all the three modes are propagating and by observing option we directly can say option (D) is correct.

SOL 9.3.25 Option (D) is correct.Consider a rectangular waveguide has dimensions a b= and the corresponding resonator has the dimensions a b d= = . now take and operating point that has frequency f just greater than the cut-off frequency for m n 1= = . So we have the




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propagating modes in waveguide. TE01, TE10, TE11 and TM11.Where as the propagating modes in resonator are TE011, TE101, TM110

Therefore the cavity resonator does not possess as many modes as corresponding waveguides.As the resonating frequency of a TEmnp or TMmnp mode is defined as

frmnp am

bn

dp

21 2 2 2

mep p p= + +a a ak k k

So for the different modes (different values of m , n andp) the resonant frequency are very closely spaced and also the resonant frequencies of cavity can be changed by altering its dimensions.

SOL 9.3.26 Option (C) is correct.Microstrip lines cannot support pure TEM mode but shielded coaxial lines can support pure TEM mode.

SOL 9.3.27 Option (A) is correct.Given,Operating frequency, f 3 3 10GHz Hz9

#= =Dimensions of waveguide a 6 cm=and b 4 cm=The cut-off frequency for /TE TMmn mn mode is defined as

fcmn am

bn

21

0 0

2 2

m e= +a ak k

So, for TE10, fc10 23 10

61 0

10 2#= +b l

2.5 GHz= a 1"^ h

for TE01, fc01 23 10 0 4

110 2#= + b l

3.75 GHz= 2b "^ h

For orTE TM11 11, fc11 23 10

61

4110 2 2#= +b bl l

4.506 GHz= 3, 4c d" "^ h

SOL 9.3.28 Option (A) is correct.A TEM wave doesn’t have an electric component in its direction of propagation consequently there is no longitudinal displacement current. The total absence of a longitudinal current inside a waveguide leads to the conclusion that there can be no closed loops of magnetic field lines in any transverse plane. Therefore, TEM waves cannot exist in a hollow waveguide of any shape.i.e. Both A and R are true and R is correct explanation of A.

SOL 9.3.29 Option (A) is correct.Phase velocity of a wave propagating in a waveguide is defined as



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ff

c

1 c2

=- c m

The group velocity of the wave propagating in waveguide is defined as

vg c ff1 c

2

= - c m

where c is the velocity of wave in free space, fc is the cutoff frequency and f is the operating frequency. As the operating frequency f is always grater than cutoff frequency fc . So, comparing the above two expressions we get vp c v> > g

SOL 9.3.30 Option (B) is correct.In a microstrip line operating wavelength is defined as l e0l e=where, 0l is free space wave length and ee is the effective dielectric constant. So, Statement 1 is correct.The electromagnetic fields exist partly in air above the dielectric substrate and partly within the substrate.Statement 2 is correct.The effective dielectric constant of microstrip line is ee and given as 1 < <e re ei.e. greater than dielectric constant of air (1). Statement 3 is correct.Conductor losses, increase with decreasing characteristic impedance in microstrip line. Statement 4 is correct.

SOL 9.3.31 Option (B) is correct.Stripline carries two conductors and a homogenous dielectric. So, it supports a TEM mode (Pure TEM). 2a "

Hollow rectangular waveguide can propagate TEM and TE modes but not TEMmode. 3b "

Microstripline has some of its field lines in the dielectric region and some fraction in the air region. So it cannot support a pure TEM wave instead the fields are quasi-TEM. 1c "

SOL 9.3.32 Option (B) is correct.Given, the dimension of waveguide is 2.286 cma = , 1.016 cmb = .The cut off wavelength of the guide, for TEmn mode is defined as

cl

am

bn

22 2

=+a ak k

So, for TE01 Mode the cut off wavelength of the guide is

a b0 1

22 2

=+b bl l

b2= 2.032 cm=




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SOL 9.3.33 Option (C) is correct.Since, the electric component is existed in the direction of propagation. So the electric field is not transverse to the propagating wave and therefore the mode is transverse magnetic (TM mode).

SOL 9.3.34 Option (C) is correct.Given, dimension of waveguide 3 cma b= = and so the dominant mode is either TE01 or TE10 mode.So, the cutoff frequency for dominant mode is given as

fc am

bn

21

0 0

2 2

m e= +a ak k

23 10

3 1018

2#

##

= - (for TE01 or TE10 mode)

5 GHz=So, at 6 GHz dominant mode will propagate.Statement 1 is correct.At 4 GHz no modes will propagate so the modes are evanescent at 4 GHz.Statement 2 is correct.At 11GHz along with the dominant mode TE11 mode f 5 2c =^ h will also propagate.Statement 3 is incorrect.Degenerate modes are the different modes that have the same cut off frequency and at 7 GHz frequency TE01 and TE10 propagates that has the same cut off frequency i.e. Degenerate modes propagate at 7 GHz.Statement 4 is correct.

SOL 9.3.35 Option (C) is correct.Evanescent mode " No wave propagation dominant mode is the mode that has lowest cutoff frequency.Rectangular waveguide does not support TM01 and TM10 mode.A 2" , B 3" , C 1"

SOL 9.3.36 Option (D) is correct.Assertion (A) : Given the dimension of waveguide, 3 cma = , 1 cmb =So, the dominant mode TE10^ h has the cutoff frequency.

fc 23 10

3 1018

2#

##= -b l

5 GHz= f 4 <GHz fc=So, at 4 GHz there is no propagating mode. i.e Assertion (A) is false.Reason (R) : The wave equation for the rectangular waveguide is defined as

k am

bn

z2 2 2p p- -a ak k 2

lp= b l

for a 3= , b 1= we have

k m n3 1z

2 2 2p p- -a ak k 2 2

lp= b l So, Reason (R) is also false.



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A rectangular coaxial line can support all the three modes (TE, TM or TEM).

SOL 9.3.38 Option (C) is correct.Consider the rectangular waveguide (A) has the dimension a b# after deforming into waveguide (C) the dimension is changed to b a# and so the input mode TE10 is charged to TE01. (Since the frequency of mode must remain same for both the waveguide dimensions).

SOL 9.3.39 Option (D) is correct.The dominant mode in a circular waveguide is TE11.

SOL 9.3.40 Option (B) is correct.Consider the dimension of inner broad wall of waveguide is a (i.e. a b> ). So, the dominant mode will be TE10.Since, the cutoff frequency of the TEmn mode is defined as

fc am

bn

21 /

0 0

2 2 1 2

m e= +a ak k: D

So, for dominant mode TE10^ h we have

10 109# a b2

1 1 0 /

0 0

2 2 1 2

m e= +b bl l; E ( 10 GHzfc = )

10 109# a2

3 10 18#

#=

a 1.5 cm2 103 10

10

8

#

#= =

SOL 9.3.41 Option (A) is correct.Propagation constant in a waveguide is defined as

g 2 f ff1cc

2

p me= - c m

Since, for the evanescent mode of waveguide the operating frequency is less than the cutoff frequency.i.e. f f< c

or ffc 1<

So, for this condition the propagation constant g is purely real.

SOL 9.3.42 Option (C) is correct.Microstrip lines consist no ground plate and so the electric field lines remain partially in air and partially in the lower dielectric substrate. This makes the mode of propagation quasi TEM (not pure TEM)Due to the open structure and presence of discontinuity in microstrip line, it radiates electromagnetic energy and therefore radiation losses take place.

SOL 9.3.43 Option (B) is correct.Statements 1 and 3 are correct.





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Modes on microstrip lines are quasi TEM (not purely TEM). So the 1st statement is incorrect while rest of the statements are correct.

SOL 9.3.45 Option (D) is correct.Coaxial line " The dominant mode of propagation is TEM. 4a "^ h

Rectangular waveguide " propagating mode is TE or TM. 1b "^ h

Microstrip line " The mode of propagation is Quasi TEM. 2c "^ h

Coplanar waveguide " The propagation mode is hybrid of propagation TE TMmn mn+^ h 3d "^ h

SOL 9.3.46 Option (B) is correct.The phase velocity of TE or TM mode is defined as

vp

ff

c

1 c2

=- c m

where c " Velocity of wave in free space fc " cutoff frequency f " operating frequencySo, vp is a nonlinear function of frequency.

SOL 9.3.47 Option (C) is correct.A waveguide operated below cut off frequency can be used as an attenuator.

SOL 9.3.48 Option (C) is correct.Quality factor of a waveguide is defined as

Q v2 gaw= i.e. Q closely related to a.

Also the attenuation factors obtained in waveguides are much higher than that in transmission lines.So, both statements are true but R is not the correct explanation of A.

SOL 9.3.49 Option (C) is correct.Quality factor (Q ) of a resonator is defined as

Q ( )Re

Bandwidthsonant freqeuncy fr=

or, Bandwidth Qf

Q1r

\=

Therefore, the greater the ‘Q ’, the smaller the bandwidth of resonatorQ is also defined for a resonator as

Q 2ab=

where b is phase constant and a is attenuation constant of a resonator given as

a 2 \wms w=

So Q 1\ w

So, at higher frequency the Q of coil falls due to skin effect.



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Guided wavelength of a propagating wave in rectangular waveguide is

gl 1

c

2

ll

l=- b l

where l is free space wavelength and cl is cutoff frequency.Since, for propagation the operating wavelength must be less than cut off frequencyi.e. l c# lSo, we get gl $ lSo, for a wave propagation in an air filled rectangular waveguide, guided wavelength is never less than free space wavelength.

SOL 9.3.51 Option (A) is correct.Transverse magnetic mode (TM mode) consists of magnetic field intensity perpendicular to the direction of propagation where as the electric field intensity may be in the direction of propagation.

SOL 9.3.52 Option (D) is correct.Since the conduction current requires conductor along the axis and a hollow waveguide doesn’t have a conductor along its axis. So, the axial current is due to displacement current only.

SOL 9.3.53 Option (D) is correct.Electromagnetic waves propagating in a medium (bounded that has the velocity greater than the velocity in free space (velocity of light in space) is given as

vp

ff

C

1 c2

=- c m

or vp C> The velocity vp is called phase velocity of the wave.

SOL 9.3.54 Option (A) is correct.A and R both true and R is correct explanation of A.

SOL 9.3.55 Option (A) is correct.Statement 1, 2 and 3 are correct.

SOL 9.3.56 Option (B) is correct.Attenuation factor in a parallel plate waveguide is defined as

a Power transmitted

Power lost per unit lengthPP2 2l

0 #= =

^

^

h

h

SOL 9.3.57 Option (C) is correct.Since the waveguide has a cutoff frequency fc below which no wave propagates while above fc all the waves propagates so it can be considered as high pass filter.

***********

CHAPTER 10ANTENNA AND RADIATING SYSTEMS

648 Antenna and Radiating Systems Chap 10



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EXERCISE 10.1

MCQ 10.1.1 A Hertzian dipole of length /25l is located at the origin. If a point P is located at a distance r from the origin then for what value of ‘r ’ the point will be in radiation zone.

(A) r 52l= (B) r 5

l=

(C) Both (A) and (B) (D) none of these

MCQ 10.1.2 A quarter wave monopole antenna is operating at a frequency, 25 MHzf = . The length of antenna will be(A) 48 m (B) 3 m

(C) 6 m (D) 12 m

MCQ 10.1.3 A half wave dipole antenna is located at origin as shown in figure below. The antenna is fed by a current ( ) 83.3 cos mAi t tw= . What will be the electric field strength at point P

(A) 25 /mV m (B) 50 /mV m

(C) 50 /V mm (D) 2.5 /V mm

MCQ 10.1.4 The transmitting antenna of a radio navigation system is a vertical metal mast 25 m in height inducted from the earth. A source current is supplied to it’s base such that the current amplitude in antenna decreases linearly toward zero at the top of the mast. The effective length of antenna will be(A) 50 m (B) 20 m

(C) 12.5 m (D) 25 m

Chap 10 Antenna and Radiating Systems 649


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For View Only Shop Online at www.nodia.co.inMCQ 10.1.5 A vertical antenna of length 8.5 m is operating at a frequency, 2 MHzf = . The

radiation resistance of the antenna is(A) 1.97 W (B) 0.51W

(C) 39.4 W (D) 26.3W

MCQ 10.1.6 The current in a short circuit element of length .l 0 03l= is given by

I z^ h 0for

for

I z l

I l z l2 4

4 2

<

<

0

0

#

#

=

Z

[

\

]]

]]

What will be the radiation resistance of the element ?(A) 0.71W (B) 0.6W

(C) 0.05W (D) 0.4 W

MCQ 10.1.7 A dipole antenna radiating at 200 MHz is fed from a 60W transmission line matched to the source. What will be the length of the dipole that matches the line impedance at the signal frequency ?(A) 0.83 m (B) 0.41 m

(C) 0.49 m (D) 0.24 m

MCQ 10.1.8 A certain antenna is used to radiate a 0.2 GHz signal to a satellite in space. Given the radiation resistance of the antenna is 31.6W. The antenna is(A) half wave dipole (B) quarter wave dipole

(C) one-fifth wave dipole (D) none of these

MCQ 10.1.9 A time harmonic uniform current cosI t2 1007

#p^ h flows in a small circular loop antenna of radius 30 cm. Radiation resistance of the antenna is(A) 92.3 mW (B) 325.05W

(C) 10.83W (D) 3.076 mW

Statement for Linked Question 10 - 11 :An antenna is a center fed rod having cross sectional radius 4 cm and conductivity

2.9 10 /S m7s #= . The length of the antenna is 30 m.

MCQ 10.1.10 If a 0. 5 MHz2 current flows in the antenna then the loss resistance of the antenna is(A) 1.93W (B) 1.97 W

(C) 0.022W (D) 0.031W

MCQ 10.1.11 The radiation efficiency of the antenna is(A) . %95 4 (B) . %96 8

(C) . %98 6 (D) . %93 5




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MCQ 10.1.12 A 200 MHz uniform current flows in a small circular loop of radius 20 cm. If the loop is made of copper wire of radius 5 mm then it’s loss resistance will be (conductivity of copper, 5.8 10 /S m7s #= )(A) 0.104 W (B) 6.52 10 5 W#

-

(C) 9.57 W (D) 1.53W

MCQ 10.1.13 A quarter wave monopole antenna is connected to a transmission line of characteristic impedance 7Z 50 W= . The standing wave ratio will be (Input impedance of quarter wave monopole is . .Z j36 5 21 25in W= +^ h )(A) 1.3874 (B) 1.265

(C) 2.265 (D) 2.583

MCQ 10.1.14 Radiated power of a vertical antenna is 0.2 kW. What will be the maximum electric field intensity at a distance of 10 km from the antenna ?(A) 3.8 /mV m (B) 1.9 /mV m

(C) 19 /mV m (D) 3.6 /mV m

MCQ 10.1.15 A quarter wave monopole antenna is fed by a current ( ) 41.7 cos mAi t tw= . The average power radiated by antenna is(A) 254 mW (B) 127 mW

(C) 63.5 mW (D) 31.7 mW

MCQ 10.1.16 A dipole antenna in free space has a linear current distribution. If the length of the dipole is .0 01l then the value of current I0 required to radiate a total power 250 mW is(A) 5.03 A (B) 2.53 A

(C) 7.56 A (D) 50.3 A

MCQ 10.1.17 A monopole antenna in free space has the length of the antenna .0 02l. The antenna is extending vertically over a perfectly conducting plane and has a linear current distribution. What value of I0 is required to radiate a total power of 2 W ?(A) 11.4 A (B) 7.1 A

(C) 14.2 A (D) 3.6 A

Statement for Linked Question 18 - 19 :A Hertzian dipole is operating at a frequency, 0.2 GHzf = .

MCQ 10.1.18 What will be the maximum effective area of the dipole ?(A) 0.54 m2 (B) 1.07 m2

(C) 0.18 m2 (D) 0.27 m2



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For View Only Shop Online at www.nodia.co.inMCQ 10.1.19 If the antenna receives 1.5 Wm of power then what is the power density of the

incident wave ?(A) 8.33 /W m2m (B) 5.56 /W m2m

(C) 1.40 /W m2m (D) 2.793 /W m2m

MCQ 10.1.20 Directivity of quarter wave monopole is(A) 1.64 (B) 1.22

(C) 3.28 (D) 0.609

MCQ 10.1.21 An antenna has a uniform radiation intensity in all directions. The directivity of the antenna is(A) 5 (B) 0.25

(C) 0.2 (D) 4

MCQ 10.1.22 The input power of a certain antenna with an efficiency of 90 % is 0.8 Watt. If the antenna has maximum radiation intensity of 1 /W Sr then it’s directivity will be(A) 5.26 (B) .16 53

(C) .0 76 (D) .9 55

MCQ 10.1.23 An antenna has maximum radiation intensity of 1.5 /W Sr. If the directivity of the antenna is .D 20 94= then radiated power of antenna will be(A) 1.11 W (B) 0.30 W

(C) 0.26 W (D) 0.90 W

MCQ 10.1.24 Normalized radiation intensity of an antenna is given by

,U q f^ h 0 /2, 0 2

0

sin

otherwise

# # # #q q p f p= )

The directivity of antenna will be(A) 2.55 (B) 8.0

(C) 0.81 (D) 1.27

MCQ 10.1.25 An antenna has the uniform field pattern given by

U q^ h /

/

4 0 3

0 3

< << <q p

p q p= *

where U q^ h is independent of f. The directivity of the antenna is(A) /1 4 (B) 4

(C) 16 (D) 1

MCQ 10.1.26 Three element array that has the current ratios : :1 2 1 as shown in figure




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The resultant group pattern of this array will be same as the two element antenna array with(A) 0a = , /d 4l= (B) 180ca = , /d 2l=

(C) 0a = , /d 2l= (D) 180ca = , d 2l=

MCQ 10.1.27 When the two three-element arrays with current ratio : :1 2 1 are displaced by /2l then it forms(A) Four element array with current ratio 1 : 3 : 3 : 1

(B) Three element array with current ratio 2 : 4 : 2

(C) Four element array with current ratio 3 : 1 : 1 : 3

(D) Three element array with current ratio 1 : 3 : 1

***********



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EXERCISE 10.2

MCQ 10.2.1 A Hertzian dipole of length /100l is located at the origin and fed with a current of ( ) 10sin Ai t t2 8= . A point P is located at a distance r from the dipole as shown

in figure. What will be the magnetic field at P ?

(A) . sin t1 15 10 908 c+^ h (B) . cos t1 15 10 908 c+^ h

(C) . sin t1 15 10 908 c-^ h (D) . sin t2 30 10 908 c+^ h

MCQ 10.2.2 Directivity of Hertizian monopole antenna is(A) 5 (B) 3

(C) 1/2 (D) 9

MCQ 10.2.3 Directive gain of Hertzian dipole antenna is(A) . sin1 5 2q (B) sin3 2q

(C) sin3

2q (D) sin32 2q

MCQ 10.2.4 Two Hertzian dipole antennas are placed at a separation of /d 2l= on z -axis to form an antenna array as shown in figure below :




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If the 1st antenna carries a current 0I Is1 0 c= and the 2nd antenna carries a current 180I Is2 0 c= then the resultant field pattern of the antenna array will be

MCQ 10.2.5 An antenna array is formed by two Hertzian dipoles placed at a separation of /4l as shown in figure. The current fed to the two antennas are I s1 and I s2 respectively.

If I s2 is lagging I s1 by an angle /2p then the resultant field pattern of antenna array will be



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MCQ 10.2.6 The group pattern function of a linear binomial array of N -elements as shown in figure is

(A) cos cosd2

N 1b q a+ -

b l; E (B) cos cosd2

Nb q a+b l; E

(C) cos cosd2

N 1b q a+ +

b l; E (D) cos cosd N 1b q a+ -^ h7 A

Statement for Linked Question 7 - 8 :Maximum electric field strength radiated by an antenna is 6 /mV m measured at 40 km from the antenna.

MCQ 10.2.7 If the antenna radiates a total power of 100 kW then the directivity of antenna is(A) 2.02 dB-

(B) 9.6 dB

(C) 0.0096 dB

(D) 20.18 dB-

MCQ 10.2.8 If the efficiency of the radiation is %95 then it’s maximum power gain is(A) .9 12 10 3

#- (B) .9 4 10 3

#-

(C) .0 11 10 3#

- (D) .9 6 10 3#

-




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Statement for Linked Question 9 - 10 :A radar with an antenna of 2.8 m in radius transmits 30 kW at a frequency 3 GHz. The effective area of the antenna is %70 of it’s actual area.

MCQ 10.2.9 If the minimum detectable power is 0.13 mW for a target of cross section 1.25 m2 then the maximum range of the radar is(A) 584.3 m (B) 1270 m

(C) 292.1 m (D) 977.8 m

MCQ 10.2.10 The average signal power density at half of the range of radar will be(A) 350.25 /W m2 (B) 69.80 /W m2

(C) 80.69 /W m2 (D) 250.35 /W m2

Statement for Linked Question 11 - 13 :A transmitting antenna is being fed by a current source of amplitude 50 AI0 = and frequency 180 kHzf = . The effective length of antenna is 20 m.

MCQ 10.2.11 What will be the maximum field intensity at a distance 80 km from the antenna ?(A) 3.39 /mV m (B) 1.41 /mV m

(C) 2.83 /mV m (D) 0.71 /mV m

MCQ 10.2.12 The time average radiated power of the antenna is(A) 0.43 kW (B) 0.29 kW

(C) 2.33 kW (D) 1.14 kW

MCQ 10.2.13 What will be the radiation resistance of the antenna ?(A) 0.23W (B) 2.91W

(C) 0.34 W (D) 1.7 W

Statement for Linked Question 14 - 15 :A metallic wire of cross sectional radius 6 mm is wound to form a small circular loop of radius 2 m with 10 turns. Conductivity of metallic wire is 2.9 10 /S m7s #= .

MCQ 10.2.14 If a 0.5 MHz uniform current flows in the loop then it’s radiation resistance will be(A) 2.37 10 6 W#

- (B) 1.42 10 3 W#-

(C) 2.37 10 4 W#- (D) 4.53 10 4 W#

-

MCQ 10.2.15 Radiation efficiency of the antenna will be(A) . %18 36 (B) . %0 101

(C) . %10 89 (D) . %0 055



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For View Only Shop Online at www.nodia.co.inMCQ 10.2.16 The polar radiation pattern of a /8l thin dipole antenna is

Statement for Linked Question 17 - 18 :Two short antennas at the origin in free space carry identical currents 4 cos Atw . one in the ax direction and other in the az direction.

MCQ 10.2.17 If both the antennas are of length 0.1 m and wavelength is 2 ml p= then the electric field Es at the distant points 0,0,1000P^ h and , ,Q 1000 0 0^ h will be at point P at point Q(A) 1. /V mj e a2 10 j

z2 1000

#- - -^ h . /V mj e a1 2 10 j

x2 1000

#- - -^ h

(B) . /V mj e a1 2 10 jx

2 1000#

- -^ h . /V mj e a1 2 10 j

z2 1000

#- -

^ h

(C) . /V mj e a1 2 10 jx

2 1000#- - -

^ h . /V mj e a1 2 10 jz

2 1000#- - -

^ h

(D) . /V mj e a1 2 10 jz

2 1000#

- -^ h . /V mj e a1 2 10 j

z2 1000

#- -

^ h

MCQ 10.2.18 E at point , ,0 1000 0^ h at t 0= will be(A) 9.92 /mV ma ax z+^ h (B) 9.92 /mV ma ax z- +^ h

(C) 1.2 /mV ma ax z+^ h (D) 12 /mV ma ax z- +^ h




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Statement for Linked Question 19 - 20In a free space short circuit vertical current element is located at the origin in free space. The radiation field due to the element at any point is given as

E sq /sin V me10 j r10

p q= p-

MCQ 10.2.19 E sq at point P (r 100= , /2q p= , /6f p= ) is(A) 0.2 /V mej1000p (B) 0.2 /V me j1000p-

(C) 0.1 /V me j1000p- (D) 0.1 /V mej1000p

MCQ 10.2.20 If the vertical element is shifted to a point . , ,0 1 2 2p p

a k then, E sq at point , ,P 100 2 6p p

a k changes to

(A) 0.1 /V me j1000p- (B) 0.1 /V me e .j j1000 0 5p p-

(C) 0.1 /V me .j0 5p- (D) 0.1 /V me e .j j1000 0 5p p- -

MCQ 10.2.21 A short circuit current element of length .l 0 06l= carries the current distributed as

I z^ h I ll z2

0=-

< F for l z l2 2# #-

The radiation resistance of the antenna will be(A) 0.71W (B) 2.84 W

(C) 2.13W (D) 0.18W

MCQ 10.2.22 An antenna is made of straight copper wire of length 1 cm carrying current of frequency 0.3 GHz. If the wire has a cylindrical cross section of radius 1 mm then

the ratio of the radiation resistance to the ohmic resistance of wire will be RRl

rad .

(A) 11 (B) 6

(C) 17 (D) 5

MCQ 10.2.23 A 2 cm long Hertzian dipole antenna radiates 2 W of power at a frequency of 0.6 GHz. The rms current in the antenna is(A) 1.78 A (B) 3.56 A

(C) 1.26 A (D) 0.89 A

***********



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EXERCISE 10.3

MCQ 10.3.1 The radiation pattern of an antenna in spherical co-ordinates is given by ( )U q ; /cos 0 24 # #q q p=The directivity of the antenna is(A) 10 dB (B) 12.6 dB

(C) 11.5 dB (D) 18 dB

MCQ 10.3.2 For a Hertz dipole antenna, the half power beam width (HPBW) in the E -plane is(A) 360c (B) 180c

(C) 90c (D) 45c

MCQ 10.3.3 At 20 GHz, the gain of a parabolic dish antenna of 1 meter and 70% efficiency is(A) 15 dB (B) 25 dB

(C) 35 dB (D) 45 dB

MCQ 10.3.4 A /2l dipole is kept horizontally at a height of 20l above a perfectly conducting

infinite ground plane. The radiation pattern in the lane of the dipole (E plane) looks approximately as

MCQ 10.3.5 A mast antenna consisting of a 50 meter long vertical conductor operates over a perfectly conducting ground plane. It is base-fed at a frequency of 600 kHz. The radiation resistance of the antenna in Ohms is

(A) 5

2 2p (B) 5

2p

(C) 5

4 2p (D) 20 2p

GATE 2012

GATE 2008

GATE 2008

GATE 2007

GATE 2006




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MCQ 10.3.6 Two identical and parallel dipole antennas are kept apart by a distance of /4l in the H - plane. They are fed with equal currents but the right most antenna has a phase shift of 90c+ . The radiation pattern is given as.

MCQ 10.3.7 Consider a lossless antenna with a directive gain of 6 dB+ . If 1 mW of power is fed to it the total power radiated by the antenna will be(A) 4 mW (B) 1 mW

(C) 7 mW (D) 1/4 mW

MCQ 10.3.8 Two identical antennas are placed in the /2q p= plane as shown in Fig. The elements have equal amplitude excitation with 180c polarity difference, operating at wavelength l. The correct value of the magnitude of the far-zone resultant electric field strength normalized with that of a single element, both computed for

0f = , is

(A) 2 cos s2lp

b l (B) 2 sin s2lp

b l

(C) 2 cos slp

a k (D) 2 sin slp

a k

MCQ 10.3.9 A person with receiver is 5 km away from the transmitter. What is the distance that this person must move further to detect a 3-dB decrease in signal strength(A) 942 m (B) 2070 m

(C) 4978 m (D) 5320 m

MCQ 10.3.10 A medium wave radio transmitter operating at a wavelength of 492 m has a tower antenna of height 124. What is the radiation resistance of the antenna?(A) 25 W (B) 36.5 W

(C) 50 W (D) 73 W

GATE 2005

GATE 2004

GATE 2003

GATE 2002

GATE 2001



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For View Only Shop Online at www.nodia.co.inMCQ 10.3.11 In uniform linear array, four isotropic radiating elements are spaced /4l apart. The

progressive phase shift between the elements required for forming the main beam at 60c off the end - fire is :(A) p- (B) 2- p radians

(C) 4- p radians (D) 8- p radians

MCQ 10.3.12 If the diameter of a /2l dipole antenna is increased from /100l to /50l , then its(A) bandwidth increases (B) bandwidth decrease

(C) gain increases (D) gain decreases

MCQ 10.3.13 For an 8 feet (2.4m) parabolic dish antenna operating at 4 GHz, the minimum distance required for far field measurement is closest to(A) 7.5 cm (B) 15 cm

(C) 15 m (D) 150 m

MCQ 10.3.14 An electric field on a place is described by its potential V 20( )r r1 2= +- - where r is the distance from the source. The field is due to(A) a monopole (B) a dipole

(C) both a monopole and a dipole (D) a quadruple

MCQ 10.3.15 A transmitting antenna radiates W251 isotropically. A receiving antenna located m100 away from the transmitting antenna, has an effective aperture of cm500 2.

The total received by the antenna is(A) W10 m (B) W1 m

(C) W20 m (D) W100 m

MCQ 10.3.16 The vector H in the far field of an antenna satisfies(A) 0 0andH H$ #d d= =

(B) 0 0andH H$ ! !#d d

(C) 0 0andH H$ !#d d=

(D) 0 0andH H$ ! #d d =

MCQ 10.3.17 The radiation resistance of a circular loop of one turn is .0 01W. The radiation resistance of five turns of such a loop will be(A) .0 002W (B) .0 01W

(C) .0 05W (D) .0 25W

MCQ 10.3.18 An antenna in free space receives 2 Wm of power when the incident electric field is 20 /mV m rms. The effective aperture of the antenna is(A) 0.005 m2 (B) 0.05 m2

(C) 1.885 m2 (D) 3.77 m2

GATE 2001

GATE 2000

GATE 2000

GATE 1999

GATE 1999

GATE 1998

GATE 1998

GATE 1998




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MCQ 10.3.19 The maximum usable frequency of an ionospheric layer at 60c incidence and with 8 MHz critical frequency is

(A) 16 MHz (B) MHz3

16

(C) 8 MHz (D) 6.93 MHz

MCQ 10.3.20 The far field of an antenna varies with distance r as

(A) r1

(B) r12

(C) r13 (D)

r1

MCQ 10.3.21 The critical frequency of an ionospheric layer is 10 MHz. What is the maximum launching angle from the horizon for which 20 MHz wave will be reflected by the layer ?(A) 0c (B) 30c

(C) 45c (D) 90c

MCQ 10.3.22 The directivity of a /2l long wire antenna is(A) 1.5 (B) 1.66

(C) 2 (D) 2

MCQ 10.3.23 The characteristic impedance of TV receiving antenna cable is 300W. If the conductors are made of copper separated by air and are 1 mm thick, what is the phase velocity and phase constant when receiving VHF channel 3 (63 MHz) and VHF 69 (803 MHz) ?(A) 1.32 rad/m and 17.82 rad/m

(B) 1.52 rad/m and 16.82 rad/m

(C) 1.52 rad/m and 17.82 rad/m

(D) 1.32 rad/m and 16.82 rad/m

MCQ 10.3.24 An antenna located on the surface of a flat earth transmits an average power of 200 kW. Assuming that all the power is radiated uniformly over the surface of a hemisphere with the antenna at the center, the time average poynting vector at 50 km is

(A) Zero (B) /W ma2r

2

p

(C) /W m40 2mp (D) /W ma40r

2mp

MCQ 10.3.25 An antenna can be modeled as an electric dipole of length 5 m at 3 MHz. Find the reduction resistance of the antenna assuming uniform current over the length.(A) 2W (B) 1W

(C) 4 W (D) 0.5W

GATE 1998

GATE 1998

GATE 1996

IES EC 2012

IES EC 2011

IES EC 2011

IES EC 2011



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For View Only Shop Online at www.nodia.co.inMCQ 10.3.26 A dipole with a length of 1.5 m operates at 100 MHz while the other has a length

of 15 m and operates at 10 MHz. The dipoles are fed with same current. The power radiated by the two antennas will be(A) the longer antenna will radiate 10 times more power than the shorter one.

(B) both antennas radiate same power.

(C) shorter antenna will radiate 10 times more power than the longer antenna

(D) longer antenna will radiate 10 times more power than the shorter antenna

MCQ 10.3.27 A short current element has length .l 0 03l= , where l is the wavelength. The radiation resistance for uniform current distribution is(A) 0.072 2p W (B) 80 2p W

(C) 72W (D) 80W

MCQ 10.3.28 In a three element Yagi antenna(A) All the three elements are of equal length

(B) The driven element and the director are of equal length but the reflector is longer than both of them

(C) The reflector is longer than the driven element which in turn is longer than the director

(D) The reflector is longer than the driven element which in turn is longer than the reflector

MCQ 10.3.29 Multiple member of antennas are arranged in arrays in order to enhance what property ?(A) Both directivity and bandwidth (B) Only directivity

(C) Only bandwidth (D) Neither directivity nor bandwidth

MCQ 10.3.30 If the total input power to an antenna is Wt , the radiated power is Wr , and the radiation intensity is f, then match List-I with List-II and select the correct answer using the code given below the lists:

List-I List-II

a. Power gain 1. /W Wr t

b. Directive gain 2. /W 4r p

c. Average power radiated 3. /W4 tpf

d. Efficiency of the antenna 4. /W4 rpfCodes : a b c d(A) 3 4 2 1(B) 4 3 2 1(C) 3 4 1 2(D) 4 3 1 2

IES EC 2011

IES EC 2010

IES EC 2010

IES EC 2009

IES EC 2008




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MCQ 10.3.31 Where does the maximum radiation for an end-fire array occur?(A) Perpendicular to the line of the array only(B) Along the line of the array(C) AT 45c to the line of the array(D) Both perpendicular to and along the line of the array

MCQ 10.3.32 As the aperture area of an antenna increases, its gain(A) increases

(B) decreases

(C) remains steady

(D) behaves unpredictably

MCQ 10.3.33 Which one of the following is correct ? Normal mode helical antenna has(A) low radiation efficiency and high directive gain

(B) high radiation efficiency and low directive gain

(C) low radiation efficiency and low directive gain

(D) high radiation efficiency and high directive gain

MCQ 10.3.34 For taking antenna far field pattern, what must be the distance R , between transmitting and receiving antennas ?

(A) R D2>2

l (B) R D3

4>2 2l

(C) R D2

> 2

2

l (D) R D2> 2

2

lMCQ 10.3.35 A transmitting antenna has a gain of 10. It is fed with a signal power of 1 W.

Assuming free-space propagation, what power would be captured by a receiving antenna of effective area 1 m2 in the bore sight direction at a distance of 1 m ?(A) 10 W (B) 1 W

(C) 2 W (D) 0.8 W

MCQ 10.3.36 The Fraunhofer region where the pattern measurement of transmitting antenna has

to be taken from a distance of D22

l , where D is the maximum aperture dimension and l is the free-space wavelength. What is the region generally known as ?(A) The near field

(B) The far field

(C) Quiet zone

(D) Induction field

MCQ 10.3.37 Match List I (Type of Antenna) with List II (Example) and select the correct answer using the code given below the lists :

IES EC 2008

IES EC 2008

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IES EC 2007

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For View Only Shop Online at www.nodia.co.inList-I List-II

a. Aperture antenna 1. Helical antenna

b. Circularly polarized 2. Point source antenna

c. Frequency independent 3. Log periodic antenna

d. Isotropic antenna 4. Microstrip antennaCodes :

a b c d(A) 3 2 4 1(B) 4 1 3 2(C) 3 1 4 2(D) 4 2 3 1

MCQ 10.3.38 A TEM wave impinges obliquely on a dielectric-dielectric boundary ( 2,r1e = 1)r2e = . The angle of incidence for total reflection is

(A) 30c (B) 45c

(C) 60c (D) 75c

MCQ 10.3.39 In a four element Yagi-Uda antenna(A) There is one driven element, one director and two reflectors

(A) There is one driven element, two directors and one reflector

(C) There are two driven elements, one director and two reflectors

(D) All the four elements are driven elements

MCQ 10.3.40 Assertion (A) : For extremely high frequency ranges or above, compared to linear antennas, aperture antennas are more useful.Reason (R) : The larger the effective area of an antenna, the sharper is the radiated beam.(A) Both A and R are true and R is the correct explanation of A




MCQ 10.3.41 The current distribution along a travelling wave antenna can be written in the form(A) Z e j z0= b- (B) sinZ z

0b=^ h

(C) Z0

=^ h (D) ( )cosZ t z0

w b= -^ h

MCQ 10.3.42 Following antenna is frequently used for local area transmission at UHF/VHF(A) Ground monopole (B) Turnstile antenna

(C) Slot antenna (D) Loop antenna

IES EC 2003

IES EC 2003

IES EC 2003

IES EC 2002

IES EC 2002




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MCQ 10.3.43 For frequencies up to 1650 kHz, the transmitting antenna used is a(A) parabolic dish (B) vertical antenna

(C) Yagi antenna (D) turnstile antenna

MCQ 10.3.44 The radiation field of an antenna at a distance r varies as(A) 1/r (B) /r1 2

(C) /r1 3 (D) /r1 4

MCQ 10.3.45 The wave radiated by a helical antenna is(A) linearly polarized

(B) right circularly polarized

(C) left circularly polarized

(D) elliptically polarized

MCQ 10.3.46 In a certain microstrip patch antenna, the unexcited patch is of length L , width W , thickness of the substrate being h and its relative permittivity re . Then, the capacitance of the unexcited patch is(A) /LW hre (B) /LW hr0e e

(C) /LW hre (D) /LW hr0e e

MCQ 10.3.47 A radio communication link is to be established via the ionosphere. The virtual height at the mid-point of the path is 300 km and the critical frequency is 9 MHz. The maximum usable frequency for the link between the stations of distance 800 km assuming flat earth is(A) 11.25 MHz (B) 12 MHz

(C) 15 MHz (D) 25.5 MHz

MCQ 10.3.48 Assertion (A) : Programmes broadcast by radio stations operating in the medium wave band of 550 to 1650 kHz situated at long distance in excess of 500 km cannot be heard during day-time but may be heard during night time.Reason (R) : In the night-time, radio waves reflected from the F -layer suffer negligible attenuation since D -and E -layers are absent during the night-time.(A) Both A and B are true and R is the correct explanation of A

(B) Both A and R are true and but R is NOT the correct explanation of A



MCQ 10.3.49 Assertion (A) : For an end-fire array, the current in successive antennas must lag in phase.Reason (R) : Radiation of successive antennas will cancel along the axis.(A) Both A and R are true and R is the correct explanation of A

IES EC 2001

IES EC 2001

IES EC 2001

IES EC 2001

IES EC 2001

IES EC 2001

IES EC 2001



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For View Only Shop Online at www.nodia.co.in(B) Both A and R are true but R is NOT the correct explanation of A



MCQ 10.3.50 Assertion (A) : The radio horizon for space wave is more than the optical horizon.Reason (R) : The atmosphere has varying density.(A) Both A and R are true and R is the correct explanation of A




MCQ 10.3.51 What is the radiation resistance of a dipole antenna /20l long approximately equal to ?(A) 2 W (B) 40 W

(C) .0 6 W (D) 20 W

MCQ 10.3.52 Consider the following statements about the effective length of a half wave dipole (Elevation angle q is measured from the dipole axis) :1. Effective length is a function of q

2. Effective length is maximum for /2q p=

3. Maximum effective length is larger than physical length

4. Effective length is the same for the antenna in transmitting and receiving modes.


(C) 1, 2 and 3 (D) 1, 3 and 4

***********

IES EC 2001

IES EE 2008

IES EE 2004




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SOLUTIONS 10.1

SOL 10.1.1 Option (D) is correct.The boundary between near and far zone is defined by r r0= (distance from the antenna) as

r0 d2 2

l= where d is the length of dipole.

So, the near and far zones of the field are as following :Near zone for r r> 0 and far zone for r r> 0

Now, for the Hertzian dipole of length /50l , we have

r0 /502 2

ll

= ^ h 1250

l=

Since r 52l= r> 0

and r 5l= r> 0

So, both the positions are at far zone(radiation zone).

SOL 10.1.2 Option (C) is correct.Given, the operating frequency of the antenna is f 25 MHz=Since, the antenna is quarter wave monopole so, the length of the monopole antenna will be given as l /4l=where l is the operating wavelength of the antenna given as

l 12 mfc

25 103 10

6

8

#

#= = =

Thus, we get the length of antenna as

l 3 m412= =

SOL 10.1.3 Option (D) is correct.Given, the current fed to the antenna is ( )i t 83.3 cos mAtw=So, the magnitude of the current flowing in the antenna is I0 83.3 10 A3

#= -

and from the figure we get the location of point P as r 100 10Km m5= =



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For View Only Shop Online at www.nodia.co.inand q /2p=Therefore, the electric field strength at point P is given as

E sf sin

cos cos

r

I

220 0

p q

h p q=

a k

.

sin

cos cos

2 10 2

120 83 3 10 2 90

5

3# c

p p

p p=

-

^

^ ^ a

h

h h k

.

2 10 1120 83 3 10 1

5

3#

pp

=-

^ ^

^ ^ ^

h h

h h h 5 10 5

#= - 50 /V mm=

SOL 10.1.4 Option (A) is correct.Since, the amplitude of current decreases linearly toward zero at the top so, the current amplitude at a height z above the plane is given as

I z^ h I hz10= -a k

where I0 is amplitude of source current and h is the height of the antenna.Therefore, the effective length of the antenna is

le hz dz1

h

0= -a k#

25 mz hz h h h2 2 2

h2

0= - = - = =: D (given 50 mh = )

SOL 10.1.5 Option (B) is correct.Length of antenna, dl 7.5 m=Operating frequency, f 2 2 10MHz Hz6

#= =So, the operating wavelength of the antenna is

l 1.5 10fc

2 103 10

6

82

#

##= = =

Therefore, the radiation resistance of the antenna is given as

Rrad dl80 2

2p l= b l

..80

1 5 107 52

2

2

#p= b l 1.97 W=

SOL 10.1.6 Option (A) is correct.Since, the current has the step distribution and both the current levels are distributed for equal intervals so, the average current will be given as

Iavg .I I

I22 0 750

00=

+=

Since, the average current flowing in the antenna is .0 75 times the uniform current I0 therefore, the radiated power will be .0 75 2

^ h times of the value obtained for I0 and due to the same reason the radiation resistance will down to .0 75 2

^ h times the value for a uniform current.

i.e. Rrad . dl0 75 802 22

p l= ^ bh l; E . 80 .0 5625 0 032 2p= ^ h8 B 0.4 W=




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SOL 10.1.7 Option (B) is correct.Since, the dipole must match the line impedance.i.e. Rrad Z0= where Z0 is characteristic impedance so, we get Rrad 60W=

dl80 22

p lb l 60=

80/c fdl2

2p c m 60=

dl 8060

100 103 10 /

2 6

8 2 1 2

##

#p

= c m= G ( 100 MHzf = )

0.827 m=

SOL 10.1.8 Option (D) is correct.Operating frequency, f 0.2 0.2 10GHz Hz9

#= =Radiation resistance, Rrad 31.6W=So, the operating wavelength of antenna is

l .

1.5 mfc

0 2 103 10

9

8

#

#= = =

Now, the radiation resistance of the antenna is defined as

Rrad dl80 2

2p l= b l

So, putting all values we get

31.6 dl80 22

p l= b l

dll .0 2.

dl 5. l

i.e. Antenna is one fifth wave dipole.

SOL 10.1.9 Option (A) is correct.Current flowing in the antenna, ( )i t cosI t2 100

7#p= ^ h

Radius of the circular loop, b 30 30 10cm m2#= = -

So, we get the operating frequency of the antenna as

f 10 Hz22 107

7#p

p= =

The operating wavelength of the antenna is

l 30 mfc

103 10

7

8#= = =

Since, b>>l so, the radiation resistance of the antenna is given as

Rrad S320

4

4 2

lp= where S is area of the circular loop.

lb320

4

4 2 2#

lp

= ^ h (S b2p= )



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30

320 30 104

6 2 4# # #p

=-

^

^

h

h

0.003076 3.076 mW W= =

SOL 10.1.10 Option (A) is correct.Cross sectional radius of antenna, a 4 4 10cm m2

#= = -

Conductivity of the antenna, s 2.9 10 /S m7#=

Length of antenna, dl 30 m=Operating frequency f 0.5 0.5 10MHz Hz6

#= =So, the surface resistance of the antenna is

Rs ..f

2 9 100 5 10 4 10

70

6 7

#

# # # #s

p m p p= =-

2.61 10 4 W#= -

Therefore, the loss resistance of the antenna is given as

Rl R adl

2s p= b l .2 61 102 4 10

3042#

# #p= -

-^ bh l

0.031W=

SOL 10.1.11 Option (D) is correct.The radiation resistance of the antenna is defined as

Rrad dl80 2

2p l= b l

where dl is the length of the antenna and l is the operating wavelength. So, we get

Rrad 80/c f302

2p #= c m ( /c fl = )

30 .803 10

0 5 1028

6 2

##

# #p=^

eh

o .1 97=

Therefore, the radiation efficiency of the antenna is

rh . ..

R RR

1 97 0 0311 97

rad l

rad= + = + . %98 6=

SOL 10.1.12 Option (B) is correct.Given,Operating frequency, f 100 10MHz Hz8= =Radius of circular loop, b 20 20 10cm m2

#= = -

Cross sectional radius of wire, a 5 5 10mm m3#= = -

Conductivity of copper, s 5.8 10 /S m7#=

The surface resistance of antenna is given as

Rs f 0

sp m=

.5 8 1010 4 10

7

8 7

#

# # #p p=-

2.61 10 3 W#= -

So, the loss resistance of the antenna is

R ab Rl s= b l .

5 1020 10 2 61 103

23

#

## #= -

-- 0.104 W=




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SOL 10.1.13 Option (D) is correct.Given that the quarter wave monopole antenna is connected to transmission line. So, the load impedance of transmission line will be the input impedance of monopole antenna.i.e. ZL Zin=Since, the input impedance of quarter wave monopole antenna is Zin . .j36 5 21 25 W= +^ h

So the reflection coefficient of transmission line is given as

G Z ZZ Z0

L

L

0= +

- . .. .

jj

36 5 21 5 7536 5 21 25 75

=+ ++ -

^

^

h

h

. .0 3874 140 3< c=Therefore, the standing wave ratio along the transmission line is

S .11

2 265GG

=-+

=

SOL 10.1.14 Option (D) is correct.Radiated power of an antenna is defined as

Prad I dl12

2 2

02

p h b= ^ h (1)

where I is the current in the antenna, dl is the length of the antenna and b is the phase constant.Now, the maximum electric field intensity at a distance R from the antenna is defined as

E maxq IdlR40

ph b= b l (2)

So, comparing equation (1) and (2), we get

E maxq R P1 90 rad=

.10 10

1 90 0 4 1033

## #= ( 0.4 kWPrad = )

19 /mV m=

SOL 10.1.15 Option (D) is correct.Given, current flowing in the antenna is ( )i t 41.7 cos mAtw=So, the magnitude of the current flowing in the antenna is I0 41.7 mA=Now, for a quarter wave monopole antenna, radiation resistance is Rrad 73. WSo, the average power radiated by the antenna is given as

Prad I R21

rad02= . 732

1 41 7 10 3 2## #= -

^ h

63.5 mW=



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Total radiated power, Prad 250 mW= 0.25 W=Length of antenna, dl .0 01l=Now, the radiated power of an antenna in terms of current I0 flowing in the antenna is defined as

Prad I R21

rad02= ^ h (1)

where Rrad is the radiation resistance of the antenna. Since, the current is linearly distributed over the antenna So, we get the average current in the antenna as

Iavg I20=

Since, the average current flowing in the antenna is half of the uniform current I0 therefore, the radiated power will be th4

1 of the value obtained for uniform current in equation (1)

i.e. Prad 21 I R4

102rad= b l

.0 25 80I dl81

02 2

2p l# #= b l

.0 25 .I 10 0 0102 2 2p= ^ ^h h

I02 .25 33=or, I0 5.03 A=

SOL 10.1.17 Option (D) is correct.Length of antenna, dl .0 02l=Total radiated power, Prad 4 W=Since, the monopole antenna is extending over the conducting plane so, the power will be radiated only over the upper half space and therefore, the radiation resistance of the antenna will reduces to half of its value

i.e. Rrad dl

21 80 2

2p l= b l; E

dl40 22

p l= b l

As the current is distributed linearly. So, the average current in the antenna is

Iavg I20=


1 of the value obtained for I0.

i.e. Prad I dl

21

2 4002

22

p l= b bl l; E

or, I0 dlP

10

2 /rad

22

1 2

p l

=b l

> H

.10 0 022 4 /

2 2

1 2#

p=

^ h= G 14.2 A=

SOL 10.1.18 Option (A) is correct.Operating frequency, f 0.2 GHz=So, the operating wavelength of the Hertzian dipole is




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l .

1.5 mfc

0 2 103 10

9

8

#

#= = =

Now, the effective area of the dipole is defied as

Ae G4 d

2

pl=

where Gd is the directive gain and since the directive gain of Hertzian dipole is . sin1 5 2q so, putting this value, we get

Ae .

. .sin sin41 5

1 5 0 272

2 2

p q q= =^^

hh

Therefore, the maximum effective area of the dipole is A ,maxe 0.27 m2= (maximum value of sinq is 1)

SOL 10.1.19 Option (C) is correct.The time average power density of the incident wave is defined in terms of received power as

Pave APe

r=

where, Pr is the received power and Ae is the effective aperture area and as calculated in the previous question, the maximum effective area of the Hertzian dipole is Ae 0.27 m2=So, we get the average power density of the incident wave as

Pave ..

0 271 5 10 6

#=-

5.56 /W m2m=

SOL 10.1.20 Option (D) is correct.For a quarter-wave monopole antenna pattern function is

f q^ h /

sin

cos cos2

qp q

=^ h8 B

So, the normalized radiation intensity of the quarter wave monopole antenna is given as ,U q f^ h f 2 q= ^ h

/

sin

cos cos22

2

q

p q=

^ h8 B

Therefore, the maximum radiation intensity is Umax 1=Now, the power radiated by the quarter wave monopole antenna is evaluated as

Prad , sinU d dq f q q f= ^ h# "- ,#

sin

cos cossin d d2/

2

2

0

2

0

2

q

p qq q f=

pp 9 C

## .2 0 609p= ^ ^h h

Therefore, the directivity of quarter wave monopole antenna is

D PU4 max

rad

p=



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..

2 0 6094 3 28

pp= =

^ ^h hSOL 10.1.21 Option (B) is correct.

As the radiation intensity in all directions are same so, ,U q f^ h Uave=where, ,U q f^ h is radiation intensity in a particular direction and Uav is the average radiation intensity. So, the directive gain in a particular direction is

,Gd q f^ h ,UU

UU 1

ave ave

aveq f= = =^ h

Therefore the directivity of the antenna is D G 1,maxd= =

SOL 10.1.22 Option (C) is correct.Maximum radiation intensity, Umax 1 /W Sr=Efficiency of antenna, h %95=Input power of antenna, Pin 0.8 Watt=So, the output radiated power is given as Prad Pinh= . .0 95 0 8#= ^ ^h h 0.76 Watt=Therefore, the directivity of antenna is evaluated as

, D PU4 max

rad

p= .0 764 1#p=

.16 53=

SOL 10.1.23 Option (A) is correct.Maximum radiation intensity, Umax 1.5 /W Sr=Directivity of antenna, D .20 94=Since, the directivity of antenna is defined as

D PU4 max

rad

p=

So, radiated power of the antenna is given as

Prad ..

20 944 1 5p

= ^ h

0.9 Watt=

SOL 10.1.24 Option (B) is correct.From the given value of radiation intensity, we get maximum radiation intensity of the antenna as Umax 1=So, the radiated power of the antenna is evaluated as

Prad , sinU d dq f q q f= ^ h# sin sin d d/

0

2

0

2

q q q f=f

p

q

p

==^ ^h h## 2

2p=

Therefore, the directivity of antenna is

D /

.PU4

24 1

2 546max

rad2

ppp

= = =^ h




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SOL 10.1.25 Option (C) is correct.Given, the field pattern of antenna,

U q^ h /

/

4 0 3

0 3

< << <q p

p q p= *

So, the total radiated power of the antenna is given as

Prad sinU d d/

0

3

0

2q q q f=

q

p

f

p

==^ h##

cos4 2 /0

3# p q= - p

6 @

4p=Therefore, the directivity of the antenna is

D 4

PU4

44

4max

rad

pp

p= = =^ h

SOL 10.1.26 Option (D) is correct.The three element antenna array has the current ratio : :1 2 1

We can split the middle element to two elements each of them carrying current0I0 c as shown below.

Now all the four elements are carrying current 0I0 c and separation between them are /d 2l= . So, this array can be replaced by two array antenna with two elements as shown below :

Since the currents are in same phase, so the phase difference between the currents will be zero.i.e. a 0= and separation between the antennas as obtained from the above shown figure is d /2l=

SOL 10.1.27 Option (B) is correct.As shown below the three element array displaced by /2l .

Now we split all the elements with current I0 as shown below :



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The three current elements I0 located at the same position can be treated as the single element carrying current I3 0 as shown below :

Thus, the current ratio will be 1 : 3 : 3 : 1 of the four element array.

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SOLUTIONS 10.2

SOL 10.2.1 Option (B) is correct.Current in the dipole, ( )i t 0.5 10sin At8=Length of the dipole, dl /100l=So, the magnitude of the current flowing in dipole is I0 0.5=and from the shown figure, we get r 100l= and 60cq =Now, the magnetic field components at any point , ,r q f^ h due to hertizian dipole located at origin are defined as H sq 0Hrs= =

and H sf sinrjI dl e4

j r0

pb q= b-

where I0 is the magnitude of current flowing in Hertzian dipole, dl is the length of dipole and b is phase constant. So, putting all the given values, we get

H sf .

sinj

e4 100

0 5 2100

60 j 2 100cp l

lp l

= lp l-

^

^ b bb ^

h

h l ll h

j4 10 2

34

##l

=^ h

As, l 2c2103 10 68

8#

wp p p#= = = ( 10 /rad s8w = )

Therefore, H sf j24 10 2

34

##p

=

1.1486 10 /A mej6 90#= c-

Thus, the net magnetic field intensity at point P will be H Im H e as

j t= fw

f^ h . sin t1 1486 10 906# cw= +-

^ h

1.15 /sin A mt10 908 c m= +^ h

SOL 10.2.2 Option (B) is correct.The field intensities of the Hertzian monopole are defined as

E sq sinrj I dl e4

j r0

ph b q= b-

and H sf sinrjI dl e4

j r0

pb q= b-

So, the time average power of the Hertzian monopole is



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Pave Re E H21 *

s s#= " , sinI dlr2

140

2

20

22

ph b q= b l

Now, the radiation intensity of the antenna is given as

,U q f^ h r Pave2= sin

I dl32 20

2

02 2

ph b q= ^ h

So, the maximum radiation intensity is

Umax I dl32 20

2

02

ph b= ^ h

(maximum value of sin 1q = )

As the radiated power of an antenna is given as

Prad , sinU d dq f q q f= ^ ^h h#where, the integral is taken in the range 0 < < 2q p , 0 2< <f p for Hertzian monopole. So, we get

Prad sin sinI dl

d d32

/

20

2

0

2

02 2

0

2

ph b q q q f=

pp ^ h##

sinI dl

d d32

/

20

2

02 3

0

2

0

2

ph b q q f=

p p^d c

hn m= G# #

I dl32 3

42

02

02

ph b p= ^

bh

l

Since, the directivity of an antenna is defined as

D PU4 max

rad

p=

So, putting the values obtained above we get the directivity of Hertzian monopole antenna as

D /4 3

4 1pp

=^

^

h

h 3=

SOL 10.2.3 Option (B) is correct.The field intensities of Hertzian dipole antenna are defined as

H sf sinrjI dl e4

j r0

pb q= b-

E sq H sh= f

So, average radiated power of the antenna is given as

Pave Re E H21 *

s s#= sinI dlr2

140

2

20

22

ph b q= b l

The radiation intensity of the antenna is defined as ,U q f^ h r Pave

2=

sinI dl32 20

2

02 2

ph b q= ^ h

So, the total radiated power of the antenna is

Prad , sinU d dq f q q f= ^ ^h h#

sinI dl

d d32 20

2

02

0

2

0

3

ph b q q f=

pp ^ h## I dl32 3

82

02

02

ph b p= ^

bh

l

Since, the directive gain of the antenna is defined as




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Gd ,

PU4rad

p q f= ^ h

Therefore, we get the directive gain of the Hertzian dipole antenna as

Gd /sin

8 34 2

pp q

= ^ h sin2

3 2q=

SOL 10.2.4 Option (B) is correct.Current in 1st antenna, I s1 I 00 c=Current in the 2nd antenna I s2 I 1800 c=Separation between two antennas d /2l=So, the phase difference between the two currents is a 180c= radp= .The unit pattern function of a Hertzian dipole antenna (i.e., the unit pattern function of both the antenna) is f1 q^ h cosq= where q is angle with z -axisThe field pattern of f1 q^ h has been plotted below :

Now, the group pattern function of the two antenna is defined as

f2 q^ h cos cosd21 b q a= +^ h: D

where a is the phase difference, b is phase constant and d is the separation between two antennas. So, we get

f2 q^ h cos cos21 2

2lp l q p= +b l; E cos cos2

1 p q p= +^ h: D

This field pattern is plotted as below :



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For View Only Shop Online at www.nodia.co.inTherefore, the resultant pattern f q^ h of the antenna array will be drawn by just multiplying these two patternsi.e. f q^ h f f1 2#q q= ^ ^h h6 6@ @

Thus, the obtained plot for the antenna array has been shown below :

SOL 10.2.5 Option (D) is correct.Separation between the two antennas, d /4l=Phase difference between the currents, a /2p=-The unit pattern function of a Hertzian dipole antenna (i.e., the unit pattern function of both the antenna) is f1 q^ h cosq= where q is angle with z -axisThis field pattern has been plotted below :

Now, the group pattern function of the two antenna is defined as

f2 q^ h cos cosd21 b q a= +^ h: D

where a is the phase difference between the currents in the dipole, b is phase constant and d is the separation between two antennas. So, we get

f2 q^ h cos cos21 2

4 2lp l q p= -b l; E

cos cos21

2 2p q p= -c m< F

It’s null (zero) will be at q p= and maxima will be at 0cq = . So, the field pattern f2 q^ h is as plotted below




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Therefore, the resultant pattern f q^ h of the antenna array will be drawn by just multiplying these two patternsi.e. f q^ h f f1 2#q q= ^ ^h h6 6@ @

Thus, the obtained pattern for the antenna array has been shown below :

SOL 10.2.6 Option (B) is correct.The normalized array factor for the antenna is given as

AF n^ h ! ....NeN N

e e1 1 21j j j N2 1

/= + +

-+y y y-^

^h

h

where y cosdb q a= +^ h

and / ! ! ....NN N N N N

1 21

31 2

= + +-

+- -

+^ ^ ^h h h

1 1 2N N1 1= + =- -^ h

So, AF n^ h e2

1 1Nj N

11= + y

-- e e e

21 / / /N

j N j j N1

2 1 2 2 1= +y y y-

- - -

/cos2

1 2NN

11y= -

-

Therefore, the group pattern function of the array is

f q^ h cos cosd2

N 1b q a= + -

b l

SOL 10.2.7 Option (A) is correct.Maximum electric field, Emax 6 / 6 10 /mV m V m3

#= = -

Location of point of field maxima, r 40 40 10km m3#= =

Total radiated power is Prad 100 10kW W5= =The average radiated power of an antenna is defined as

Pave Re E H21 *

s s#= " ,



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For View Only Shop Online at www.nodia.co.inSo, the radiation intensity of the antenna is given as ,U q f^ h r Pave

2=

Rer E H2*

s s

2

#= " ,

Therefore, the maximum radiation intensity of the antenna is

Umax Rer E H2*

s s

2

#= " , r E22 2

h= H Eh=b l

r E2 max

22

h= ^ h 2 12040 10

6 103 2

3 2

#

## #p= -^

^h

h 1200h p=^ h

Since, the directivity of an antenna is defined as

D PU4 max

rad

p=

So, we get D 2 120 10

4 40 10 6 105

3 2 3 2

# #

# # # #

pp

=-

^ ^h h .0 0096=

Therefore, in decibel the directivity is given as 10 log D10 20.18 dB=-

SOL 10.2.8 Option (B) is correct.Consider the maximum power gain is Gp and directive gain is Gd so, the radiation efficiency is defined as

rh GGd

p=

or, Gp Gr dh= . G0 95 d= ^ h ( 95%rh = )Therefore, the maximum power gain is G ,maxp . G0 95 ,maxd= ^ h . D0 95= ^ h D G ,maxd=^ h

. .0 95 0 0096#= ^ h .0 00912= .9 12 10 3#= -

SOL 10.2.9 Option (B) is correct.Minimum detectable power, Pmin 0.13 mW=Transmitted power, Prad 30 30 10kW W3

#= =Operating frequency, f 3 3 10GHz Hz9

#= =Target cross section, s 1.25 m2=Radius of antenna, a 1.8 m=Since, the effective area of the antenna is %70 of it’s actual area so, the effective area of the antenna is

Ae a10070 2

# p= ^ h . .0 7 1 8 2# #p= ^ ^^h h h 7.125 m2=

As the maximum range is the point where the received power is equal to the minimum detectable power. So, the received power by the target located at its maximum range is Pr Pmin= 0.13 0.13 10mW W3

#= = -

Now, the operating wavelength of the antenna is

l fc= 0.1 m

3 103 10

9

8

#

#= =




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So, the directive gain of the antenna is given as

Gd A4 e2l

p= .

.0 1

4 7 1252

#p=

^

^

h

h

2850p=Since, the maximum detectable range of the antenna is defined as

rmax G

PP

4

/d

r

rad3

2 2 1 4

pl s=^ h

= G

where Pr is the received power by the target located at its maximum range. So, putting all the values in the above expression, we get

rmax . .

.40 1 2850 1 25

0 13 1030 10

/

3

2 2

3

3 1 4

#

#pp

= -^

^ ^ ^

h

h h h> H

584.27 m=

SOL 10.2.10 Option (A) is correct.As calculated in previous question, the maximum detectable range of radar is rmax 584.3 m=So, half of the range will be at the position

r r21

max= 292.2 m=

Therefore, the time average power density at half of the range of the radar is

Pave rG P4d rad

2p=

.4 292 22850 30 10

2

3# #

pp

=^

^

h

h

250.35 /W m2=

SOL 10.2.11 Option (D) is correct.Current amplitude I0 50 A=Operating frequency, f 180 180 10kHz Hz3

#= =Effective length, l 20 m=Location of the observation point, R km m80 8 104

#= =So, the maximum field intensity at the observation point is given as

E maxq RI dl4

00p h b= R

I l4 2 e00 #p h b=

As, the operating wavelength is

fcl =

180 103 10

610

3

8 4

#

#= =

and so the phase constant is

b 2102 6 12 104

4# #l

p p p= = = -

Therefore, the maximum field intensity at the observation point is

E maxq 4 8 10

50 120 12 10 2 2044

# ## # # # #p

p p= -

^^

hh

.0 002827= 2.83 /mV m=

SOL 10.2.12 Option (B) is correct.The time average power density of antenna is defined as



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For View Only Shop Online at www.nodia.co.in Pave Re E H2

1 *s s#= " ,

So, the time average radiated power is given as

Prad dSPaveS

:= # sinRE

d d21 /

2

0

2

0

2

0

2

h q q f= qpp

##

sinRRI l d d2

14 2

/

e0

2

0

2

0

20

0

2

#h p h b q q f=pp

b l##

sinI l d d21

42/e

00

0

2

0

2 2#h p

b q q f=pp

b l##

sinI l d21 2 4

2 /e

00

2

0

2

# ##p h p

b q q=p

b l #

120 450 12 10 2 20 1

4 2

# ## # # #

#p p pp=

-

b l

426.37 0.43W kW= =

SOL 10.2.13 Option (D) is correct.As calculated in previous question the time average radiated power is Prad 0.43 kW=Amplitude of the current in the antenna is I0 50 A=So, the radiation resistance of the antenna is given as

Rrad .

IP2

502 0 43 10rad

02 2

3# #= =

^ h

0.34 W=

SOL 10.2.14 Option (D) is correct.Cross sectional radius of wire a 6 6 10mm m3

#= = -

Radius of the circular loop, b 1 m=Operating frequency, f 0.5 0.5 10MHz Hz6

#= =No. of turns, N 10=So, the operating wavelength of the antenna is

l .

6 10 mfc

0 5 103 10

6

82

#

##= = =


Rrad 320N b2 64

p l#= b l 10 3206 10

12 62

4

##

p#= b l 2.37 10 4 W#= -

SOL 10.2.15 Option (A) is correct.As calculated in the previous question, radiation resistance of the antenna is Rrad 2.37 10 4 W#= -

So, the surface resistance of the antenna is given as

Rs f 0

sp m=

..

2 9 100 5 10 4 10

7

6 7

#

# # # #p p=-

2.61 10 4 W#= -




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Therefore, the loss resistance of the antenna is

Rl N ab Rs#= b l .10

6 101 2 61 103

4#

## #= -

-b l

0.435W=Thus, the radiation efficiency of the antenna is

radh 0.055%R RRrad l

rad= + =

SOL 10.2.16 Option (B) is correct.Radiation function of the dipole antenna of height h is defined as

F q^ h sincos cos cosh h

qb q b

=-^ h

Since, the height of dipole antenna is /h 8l= . So, we get

F q^ h . .

sincos cos cos1 25 1 25

qq p

=-^ ^h h

This function has been drawn as to obtain the pattern shown below :

SOL 10.2.17 Option (D) is correct.Since, the point , ,P 0 0 1000^ h lies along the axial direction of antenna carrying current in az direction, so it’s contribution to the field will be zero. Now for the antenna carrying current along ax direction, we haveAmplitude of the current in antenna, I0 4 A= ( ( ) 4 cos Ai t tw= )Length of the antenna, dl 0.1 m=The position of point P is r 1000= and 90cq = as shown in the figure below :

So, the electric field component in free space is defined as E sq H s0h= f



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sinrjI dl e4

j r0

0h pb q= b-

b l

.

sinj

e4 1000

120 4 22 0 1

90 j22 1000c

p

p pp

= pp-

^

^ ^ b ^^

h

h h l hh 2b l

p=b l

1.2 10 /V mj e j2 1000#= - -

Since, rest of the components of field will be zero so, we get the net electric field as Es E as= q q .j e a1 2 10 j

x2 1000

#= -- -^ ^h h

. /V mj e a1 2 10 jx

2 1000#=- - -

^ h

Similarly, at point , ,Q 1000 0 0^ h the contribution due to antenna carrying current along x -axis will be zero while the electric field due to antenna along az will be Es . /V mj e a1 2 10 j

z2 1000

#=- - -^ h

SOL 10.2.18 Option (C) is correct.Since, the antenna are carrying current along ax and az while the point is located at y -axis so, both the antenna will contribute to the field. Therefore, summing the fields obtained due to the two antennas in previous question, we get, Es .j e a a1 2 10 j

x z2 1000

#=- +- -^ ^h h

So, in the time domain tE^ h Re eE cot

sj= ^ h

. /sin V mt a a1 2 10 1000 x z2

# w= - +-^ ^ ^h h h

Thus, the field at t 0= at point , ,0 1000 0^ h is E . /V ma a9 92 10 x z

3#=- +-

^ ^h h

SOL 10.2.19 Option (D) is correct.The field component due to the current element is given as

E sq sin e10 j r10

p q= p-

So, at point P (r 100= , /2q p= , /6f p= )

E sq sin e10010

2j10 100p= p-

a^

kh

0.1 /V me j1000= p-

SOL 10.2.20 Option (C) is correct.Since, the vertical element is shifted from origin to a point .y 0 1= on the y -axis the distance of point P from the two locations of antenna is approximately same and therefore the magnitude of field component, E sq will be same in both cases but the phase angle will change due to the change in location of current element.So, the field intensity at point P due to the new location of vertical element is given as E s1q E es j r l10= q

p- -^ h (1)where l is the difference between the length of point P from two locations as shown in figure below :




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l . cos0 1 3p= a k .0 05=

Putting the value in equation (1), we get the field component as E s1q . e e0 1 j j l1000 10= p p- ^ h

0.1 /V me e .j j1000 0 5= p p-

SOL 10.2.21 Option (B) is correct.Radiation resistance of a short circuit current element is determined as

Rrad l80 2

2p l= b l where l is the length of dipole

Rrad .80 0 062

2p l

l= b l 2.84 W=

But, as the current is not uniform so, we determine the average current through the element. Now, from the given expression of current in the element, we get

I z1^ h I ll z2

0= +b l for l z2 0# #

and I z2^ h I ll z2

0= -b l for z l0 2# #

Therefore, the average current in the element is given as

Iavg I z I z

21 2=

+^ ^h h I ll z I l

l z

2

2 20 0

=

+ + -b bl l

I20=


1 of the value obtained for I0 and due to the same reason the radiation resistance will down to th4

1 of its value.

i.e. Rrad net6 @ R41

rad= .41 2 84#= 0.71W=

SOL 10.2.22 Option (B) is correct.Length of wire, dl 1 0.01cm m= =Operating frequency, f 0.3 0.3 10GHz Hz9

#= =Cross section radius, r 1 10mm m3= = -

So, radiation resistance is given as

Rrad 80 dl22

p l= b l



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For View Only Shop Online at www.nodia.co.in 80

.

.

0 3 103 10

0 012

9

8

2

#

#p=

cf m p 0.07895W= f

cl =b l

Now, the ohmic resistance of the wire is defined as

Rl aL

2s p d=where s " Conductivity a " Radius of the cross section d " Skin depth L " Length of the wireSince, the skin depth of the wire is given as

d f1p ms

= . .0 3 10 4 10 5 8 10

19 7 7

# # #p p=

-^ ^ ^h h h

.3 82 10 6#= -

So, we get Rl . ..

5 8 10 2 10 3 82 100 01

7 3 6# # #p

= - -^ ^ ^h h h

0.0072W=Therefore, the ratio of the radiation resistance to the ohmic resistance of wire will be

RRl

rad .10 977 11.=

SOL 10.2.23 Option (D) is correct.Length of antenna, dl 2 0.02cm m= =Radiated power, Prad 2 W=Operating frequency, f 0.6 0.6 10GHz Hz9

#= =So, operating wavelength of antenna is

l .

0.5 mfc

0 6 103 10

9

8

#

#= = =


Rrad dl80 2

2p l= b l .

.80 0 50 02

125162

2 2

p p= =b l

As the radiated power of the antenna is defined as

Prad I R21

rad02= ^ h

I R. . .r m s rad2= ^ h ( /I I 2. .r m s 0= )

So, the rms current in the antenna is

I . . .r m s /1.26 AR

P16 125

2rad

rad2p

= = =^ h

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SOLUTIONS 10.3

SOL 10.3.1 Option (B) is correct.The directivity of an antenna is defined as

D UUmax

ave=

where Umax is the maximum radiation intensity of the antenna and Uave is the average radiation intensity. Since, the given antenna has the radiation pattern ( )U q cos4q= (0 /2# #q p )So, the maximum radiation intensity is Umax 1=The average radiation intensity is

Uave ( , )F d41p q f W= # ( , )sinF d d4

10

2

0

2

p q f q q f=pp

; E##

cos sin d d41 /

4

0

2

0

2

p q q q f=pp

; E## cos41 2 5

/5

0

2

p p q= -p

b l; E

41 2 0 5

1#p p= - +: D 4

152

101

#pp= =

Therefore, the directivity of the antenna is

D 101 10= =

or, ( )in dBD 10 10 10log dB= =

SOL 10.3.2 Option (D) is correct.The beam-width of Hertzian dipole is 180c so, its half power beam-width is 90c.

SOL 10.3.3 Option (A) is correct.The operating wavelength of the antenna is

l fc

20 103 10

2003

9

8

#

#= = = ( 20 GHzf = )

Therefore, the gain of parabolic antenna is given as

Gp D2

2hp l= b l

0.7 30705.412

1003

2p#= =c m (efficiency, %70h = )

or, log G10 p10 .44 87= dB

SOL 10.3.4 Option (C) is correct.Using the method of images, the configuration is as shown below



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Here ,d l= ,a p= thus, d 2b p=

So, the array factor of the antenna is given as

. .A F cos cosd2

b y a= +: D

cos cos2

2p y p= +: D ( )sin cosp y=

SOL 10.3.5 Option (B) is correct.Since, the antenna is installed at conducting ground. So, the power will be radiated only on the half side of the antenna and therefore, the radiation resistance of the antenna will be half of its actual value and given as

Rrad 80 dl21 2 2

pl

= ` j; E 0.

40 5 10

5023

2

#p= c m

52 2p W=

SOL 10.3.6 Option (B) is correct.The array factor of the antenna is defined as

. .A F cos sind2

b q a= +b l

Here, d 4l=

and a 90c=

Thus, . .A F cossin

2

24 2q

=+l

p l p

c m cos sin4 2p q p= +a k

2b lp=b l

The option (A) satisfy this equation.

SOL 10.3.7 Option (B) is correct.The directive gain of an antenna at a particular direction ,q f^ h is defined as

( , )Gd q f ( , )PU4rad

p q f= (1)

Since, for lossless antenna Prad Pin=So, we get Prad P 1in= = mWAgain the directive gain of the antenna is given 10 ( , )logGd q f 6= dBSo, ( , )Gd q f .3 98=Putting it in equation (1) we get the total power radiated by antenna as 4 ( , )Up q f ( , )P Grad d q f= 1 3.98 3.98m#= = mW

SOL 10.3.8 Option (A) is correct.Normalized array factor is given as




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. .A F 2 cos 2y=

where, y sin cosdb q f d= + q 90c= , d s2= , f 45c= , d 180c=

So, . .A F 2 cos 2y= 2 cos sin cosd

2b q f d= +; E

2 cos coss22 2 45 2

180c clp= +: D

2 cos s 90clp= +9 C 2 sin s

lp= a k

SOL 10.3.9 Option (C) is correct.The signal strength (power) at a distance r from an antenna is inversely proportional to the distance r .

i.e. P r12\

So, PP

2

1 rr

1222

= (1)

Since, 3 dB decrease " Strength is halved (10 10 2/ .3 10 0 3= = )

Therefore, PP

2

1 2=

Substituting it in equation (1), we get

2 r52

22

= ( 5 kmr1 = )

or r2 5 2= km 7071= mThus, the required distance to move is d r r2 1= - 7071 5000 2071= - = m

SOL 10.3.10 Option (C) is correct.We have l 492= m

and height of antenna, dl 124= m 4. l

So, it is a quarter wave monopole antenna and radiation resistance of a quarter wave monopole antenna is 36.5W.

SOL 10.3.11 Option (D) is correct.We have y cosdb q d= + (1)

where d 4l= Distance between elements

y 0= Because of end fire q 60c=Putting all the values in equation (1) we get

0 60cos24 cl

p l d#= + 2 21p d#= +



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p=-

SOL 10.3.12 Option (C) is correct.For a dipole antenna we have

BW 1(Diameter)

\

So, as diameter increases Bandwidth decreases.

SOL 10.3.13 Option (A) is correct.Far field region for an antenna is defined for the distance r from the antenna as

r 2d>2

lwhere d is the largest dimension of the antenna and l is the operating wavelength. Now, the operating wavelength of the antenna is given as

l fc

4 103 10

403

9

8

#

#= = = m

So, for the closest far field we have

r d2 2

l= (2.4)2340

2#= e o ( . )

150380 2 4 2

# .= m

SOL 10.3.14 Option (D) is correct.We know that for a monopole its electric field varies inversely with r 2 while its potential varies inversely with r . Similarly, for a dipole its electric field varies inversely as r 3 and potential varies inversely as r 2.In the given expression both the terms r r

1 11 2+- -_ i are present, so, this potential is

due to both monopole and dipole.

SOL 10.3.15 Option (A) is correct.Power received by an antenna is defined as

Pr rP A

4t

e2p #=

where Pt is the power radiated by the transmitting antenna, r is the distance between transmitter and receiver and Ae is the effective aperture area of the receiving antenna. So, we get

Pr ( )4 100251 500 10

2

4

# #

# #p

=-

( , 100 , 251cm m WA r P500e t2= = = )

100 Wm=

SOL 10.3.16 Option (D) is correct.Magnetic field intensity in terms of vector potential is defined as

H A1 dm #=

where A is auxiliary potential function.So, H:d ( ) 0A: #d d= =and H#d ( ) 0A# #d d= =Y




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SOL 10.3.17 Option (A) is correct.Radiation resistance of a circular loop is given as

Rrad N S

38 3

2Thpl

= ; E

Rrad N 2\

where N is number of turns. Since, the radiation resistance of a circular loop is .0 01W.

i.e. Rr1 0.01W=So, we get the net radiation resistance of the five turns of such loop as Rr2 N Rr2

1#= ( ) .5 0 012#= 0.25W= (N 5= )

SOL 10.3.18 Option (D) is correct.Aperture area of a receiving antenna is defined in terms of received power as

Aperture Area RePoynting vector of incident wave

Power ceived=

Ae PPt

r=

Since, Pt E

0

2

h= ( 1200h p= is intrinsic impedance of space)

So, Ae ( )120 3.142 10

20 102 10

E

6

3 2

6

0

2#

#

## #= =

h

-

-

-

_ i

.400 10

2 10 12 3 146

6

#

# # #= -

-1.884 m2=

SOL 10.3.19 Option (C) is correct.Maximum usable frequency( fmax) in terms of incidence angle (i ) is defined as

fmax sin ifo=

where f0 is critical frequency. So, we get

fmax sinMHz

608

c=

23

8=c m

MHz3

16=


Far field r1

\

SOL 10.3.21 Option (C) is correct.The maximum usable frequency is given as

fm sin if0=

where i is launching angle and f0 is critical frequency so, we get

20 106# sin i

10 106#=

or, sin i 21=

or, i 30c=




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For View Only Shop Online at www.nodia.co.inThe directive gain of half wave /2l^ h dipole antenna is given as

Gd .sin

cos cos1 66 2

2

2

q

p q=

a k

So, the directivity of the antenna is D G ,maxd=

Since, the maximum value of the function sin

cos cos2

22

qqp

^ h is 1. So, the directivity of

/2l long wire antenna is D .1 66 1= ^ h .1 66=

SOL 10.3.23 Option (A) is correct.Since, the EM waves are travelling in free space, So the phase velocity of the wave will be equal to the velocity of light in free space.i.e. vp c=So, at frequency, f 63 MHz= (Channel 3)

wavelength, l 4.76 mfc

63 103 10

6

8

#

#= = =

So, phase constant, b 1.32 /rad m2lp= =

and at frequency, f 803 MHz= (channel 69)

wavelength, l 0.374 mfc

803 103 10

6

8

#

#= = =

So, phase constant, b 16.82 /rad m2lp= =

SOL 10.3.24 Option (A) is correct.Since, the antenna is located at earth so, power radiated to the hemisphere will be half of the transmitted value.

i.e. Pr P2t= 200 kW

2= 100 kW=

Now, the average poynting vector (power radiated per unit area) at a distance r from the antenna is given as

Pave rP ar r2p

=

where ar denotes the direction of Poynting vector. So, for 50 kmr = , we have

Pave a50 10

100 10r3 2

3

#

#

p=

^ h /W ma40

r2mp=

SOL 10.3.25 Option (B) is correct.Radiation resistance of a dipole antenna is defined as

Rrad dl80 2

2p l= b l ...(1)

Given,The length of dipole, dl 5 m=operating frequency, f 3 MHz= 3 10 Hz6

#=So, the operating wave length of the antenna is given as




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l 100 mfc

3 103 10

6

8

#

#= = =

Putting these values in equation (1) we get

Rrad 80 10052

2p= b l 400

80 2p= 1.97 2. W=

SOL 10.3.26 Option (C) is correct.The radiated power of an antenna is defined as

Prad I R21

rad02=

i.e. Prad Rrad\ ...(i)Now, the radiation resistance of the antenna is given as

Rrad dl80 2

2p l= b l

i.e. Rrad dl

2

2

\l^ h

Since l fc=

So, we get Rrad f dl2 2\ ^ h ...(ii)

Combining eq(1) and (2) we conclude that Prad dl f2 2

\ ^ ^h h

Now, for the 1st antenna we have dl f^ ^h h . .1 5 100 10 1 5 106 8

# #= =^ ^h h

for 2nd antenna dl f^ ^h h .15 10 10 1 5 106 6

# #= =^ ^h h

Since, the product of length and frequency are same for both the antenna So, the power radiated by both the antennas will be same.

SOL 10.3.27 Option (B) is correct.Given the length of current element, .l 0 03l= .So, the radiation resistance of the system is given as

Rrad l80 2

2p l= b l .80 0 032

2p l

l= b l 0.072 2l W=

SOL 10.3.28 Option (D) is correct.In a three element Yagi antenna there are one reflector, one folded dipole (driven element) and one director. The length of reflector is greater than driven element which in turn is longer than the director.

SOL 10.3.29 Option (C) is correct.Antenna arrays are formed to produce a greater directivity i.e. more energy radiated in some particular direction and less in other directions.

SOL 10.3.30 Option (B) is correct.Input power Wt=Radiated power Wr=



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For View Only Shop Online at www.nodia.co.inRadiation intensity f=So,the power gain of the antenna is

Gp W4t

pf= 3a "^ h

Directive gain of antenna is

Gd W4r

pf= 4b "^ h

Average radiated power of the antenna is

Pr W4r

p= 2c "^ h

The efficiency of antenna is

h WWt

r= 1d "^ h

SOL 10.3.31 Option (C) is correct.Maximum radiation for an end fire array occurs along the line of the array.

SOL 10.3.32 Option (B) is correct.The gain of antenna is directly proportional to the aperture area. So, with increase of aperture area, received power increases and therefore the gain increases.

SOL 10.3.33 Option (B) is correct.In the helical antenna, normal mode of operation is very narrow in bandwidth and therefore the directivity is high. While the radiation efficiency is low.

SOL 10.3.34 Option (B) is correct.For an antenna near and far zone are specified by a boundary defined as

R d2 2

l=

where, R is the distance from antenna, d is the largest dimension of antenna and l is the operating wavelength of antenna. So, any target located at a distance R > 2D2

l from antenna is in the far zone for the antenna and any target located at a distance R < 2D2

l is in the near zone.

SOL 10.3.35 Option (B) is correct.Gain of transmitting antenna, Gdt 10=Transmitted power, Pt 1W=Effective area of receiving antenna, Aer 1 m2=Distance between transmitter and receiver, r 1 m=So, total received power by the receiving antenna is

Pr rP G A

4t

dt er2p=

4 11 10 12 # #p

=^

^ ^h

h h 0.79 W=


Since, the region r D2>2

l is called far zone for the antenna and as it is given that

in the Fraunhofer region measurement to be taken from a distance of D22

l from antenna so, the defined region is far zone or far field.




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SOL 10.3.37 Option (C) is correct.Helical antenna is used to provide circularly polarized wave and the log periodic antenna is frequency independent.

SOL 10.3.38 Option (C) is correct.For a wave travelling from medium 1 to medium 2, the incidence angle cq of the wave for which it is totally reflected by medium 2 is given as sinn c1 q sinn 902 c= (1)where n1 and n2 are the refractive index of medium 1 and medium 2 respectively.Since, refracting index of a medium having permittivity e and permeability m is defined as n me=So, putting it in equation (1), we get sin c0 1m e q 0 2m e=

sin cq 212

1ee= =

cq sin2

1 451 c= =-c m


Yagi-Uda antenna must have one reflector and one driven element while it can have any number of directors. So, the four element Yagi-Uda antenna will have 2 directors, one reflector, and one driven element.

SOL 10.3.40 Option (B) is correct.Directivity of an antenna is directly proportional to the effective area and therefore larger the effective area, sharper the radiated beam.This is the reason for using an aperture antenna instead a linear antenna for extremely high frequency ranges.

SOL 10.3.41 Option (B) is correct.Current distortion along a travelling wave antenna in general is defined as ,I z t^ h cos cotI z0 b= -^ h

but when we eliminate t by taking it’s phasor form, the current can be written as I z^ h I e j z0= b-



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Turnstile antenna is generally used at UHF/VHF for local area transmission.

SOL 10.3.43 Option (C) is correct.The frequencies upto 1650 kHz is in the range of medium frequency. Vertical radiators ranging from /6l to /5l used for broadcasting the medium frequencies as the operating conditions and economic consideration.

SOL 10.3.44 Option (B) is correct.The radiation field intensity of an antenna at a distance r is defined as

E sq sinrj I dl e4

j r0

ph b q= b-

or E sq sinr

I dlr410

\ph b q=

SOL 10.3.45 Option (A) is correct.The resultant field in a helical antenna is either circularly polarized or elliptically polarized depending on the pitch angle a.The radiated wave by a helical antenna is circularly polarized only when,

a tan c2

1

l= -a k

else it is elliptically polarized. In conclusion for a general term we can say the wave radiated by a helical antenna is elliptically polarized.

SOL 10.3.46 Option (A) is correct.The unexcited patch is shown below

The capacitance between the plates is given as

C separatoin between plates

Area of platese=

^

^

h

h h

LWr0e e=

SOL 10.3.47 Option (D) is correct.Maximum usable frequency between two stations of distance D is defined as

fMUF f hD1 2c

2= + b l

where h vertical height at the mid point of path and fc is critical frequency we put all the values to get,




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fMUF 9 10 1 2 3008006

2

##

= + b l 1.5 10 Hz7#= 15 MHz=

SOL 10.3.48 Option (B) is correct.D -layer is the lower most region of ionosphere which is present only during the day light hours and disappears at night because recombination rate is highest and also E -region is weekly lonised during night hour hence radiowave suffer negligible attenuation in night hour. This is the reason that the wave band which can’t be heard during day time but may be heard during night time.


SOL 10.3.50 Option (B) is correct.The atmosphere has varying density (refractive index) with the height from earth

given as dhdm . Radius of curvature of the wave path is

R ddhm=-

Solving it, we get the effective earth radius (Radio horizon)

34= actual earth radius (optical horizon)

SOL 10.3.51 Option (B) is correct.The radiation resistance of a dipole antenna is defined as

Rrad dl80 2

2p l= b l

Since, dl 20l=

So, Rrad /

80202

2

p ll= c m 2W=

SOL 10.3.52 Option (B) is correct.Effective length of a half wave dipole antenna is

le q^ h sin

cos cos2 2b q

p q=

a k> H

i.e. le is function of q .The maximum value of le is at /2q p= .

l 2ep

a k 2

2<b pl l= =

i.e. maximum value of le is less than its actual value 2l .

The effective length is the same for the antenna in transmitting and receiving modes.So, statements, 1, 2 and 4 and correct while statement 3 is incorrect.

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