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![Page 1: eeevcet.files.wordpress.com · Web viewUNIT IV-TUTORIAL – 3 PERIODS. A conductor of length 100cm moves at right angles to a uniform field of strength 10,000 lines per cm. 2. with](https://reader034.vdocuments.mx/reader034/viewer/2022052011/60267bc6ecc85c6cf46405ea/html5/thumbnails/1.jpg)
VELAMMAL COLLEGE OF ENGINEERING AND TECHNOLOGY, Madurai – 625 009
Department of Electrical and Electronics EngineeringAcademic Year 2016-2017
Name of the Staff : K.SnehaName of the Subject : Electromagnetic Theory/EE 8391Class : II Year EEE A & B SectionDate :
UNIT IV-TUTORIAL – 3 PERIODS
1. A conductor of length 100cm moves at right angles to a uniform field of strength 10,000 lines per cm2 with a velocity of 50 m/sec. Calculate the emf induced init. Find also the value of induced emf when the conductor moves at an angle of 300 to the field direction.Solution:
Length = 100 x 10-2 m ; velocity = v = 50 m/sec.
B = 10,000 lines /m2 = 1 Wb /m2 ; = 300
Case I :
= 900
e = Blv sin
= 1 x 100 x 10-2 x 50
= 50 volts.
Case II:
= 300
e = 1 x 100 x 10-2 x sin 300
= 25 volts.
2. A conducting cylinders of radius 5cm and height 20cm rotates at 600rev / sec in a radial field B = 0.5Tesla. Sliding contacts at the top and bottom are connected to voltmeter. Find the induced voltage?Solution:
Induced voltage, V = ∫ E⋅d l
E=v×B
Velocity, u = x B ; B = 0.5a Tesla
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= 2 N
u = 2 x 600 x 0.05 a x 0.5 a
E = 94.25 (-az)
As length is varying along z-axis
dl = dz az
V=∫0
0. 2
E (−az )⋅dz az
=−94 .25×0 . 2= −18 .85 volts .
3. The conduction current flowing through a wire with conductivity = 3 x 107 S/m and relative permittivity r = 1 is given by Ic = 3 sint (mA). If = 108 rad /sec. Find the Displacement current?Solution:
The displacement current density, JD = (dE / dt)
E is found from conduction current density as,
Jc = E
And Ic = E x A
E=I cσA
=3×10−3sinωt3×107×A
=1×10−10
Asinωt
∂E∂ t
=1×10−10
Aωcosωt
Jd=εω(1×10−10
A )cosωt = 8 .854×10−4
Acosωt
Displacement current, Id = ( Jd /A) = 8.854 x 10-14 cost Ampere.
4. Find JD:a) Next to your radio where local AM station provide E = 0.02 sin[0.01927(3 x 108t-z)]
ax
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b) In a good conductor with = 107 V/m and J = 107 sin 120t ax
Solution:
a) JD =
∂D∂ t
=ε0∂E∂ t
= 8.854 x10-12 [ 0.02 cos[0.01927(3 x108t-z)] x 0.01927 x 3 x 108]
= 1.02 cos[0.01927(3 x 108t-z)]
b) J = EE = sin 120 t
JD =
∂E∂ t =3.33 cos 120t.
5. Let µ=10-5 H/m, = 4 x 10-9 F/m, = 0 and V = 0. Find K, so that each of the following pairs of fields satisfies Maxwell’s equations,
(i) D = 6 ax – 2y ay + 2z az nC / m2
H = Kx ax + 10y ay – 25z az A/m
(ii) E = (20y – kt) ax V/m.
H = (y + 2 x 106t) az A/m.
Solution:
µ=10-5 H/m ; = 4 x 10-9 F/m ; = 0 ; V = 0
Case (i):
D = 6 ax – 2y ay + 2z az nC / m2
H = Kx ax + 10y ay – 25z az A/m
Maxwell’s equation is
B = µH = 10-5(y + 2 x106t) az
-
∂B∂ t = -20az -----(1)
∇×E=−∂B∂ t
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∇×E=¿|ax a y az ¿||∂∂ x
∂∂ x
∂∂ x ¿|¿
¿¿¿
-----(2)
From (1) and (2), equation is verified.
Similarly verify other equations.
Case (ii):
E = (20y – kt) ax V/m.
H = (y + 2 x 106t) az A/m.
Consider the Maxwell’s equation,
Here,
∂D∂ t = 0 --- (3)
∇×H=¿|ax ay az ¿||∂∂ x
∂∂ x
∂∂ x ¿|¿
¿¿¿
Hence verified.
Similarly for other Maxwell’s equations.
6. (a) An area of 0.65 m2 in the z = 0 plane is enclosed by a filamentary conductor. Find the induced voltage, given that, B = 0.05 cos 103t (ay+az)/2.Solution:
v = −∫
S
∂B∂ t
⋅ds az
=∫S
50 sin 103 t (ay+az
√2 )⋅ds az
¿23 . 0 sin103 t volts .
∇×H=∂D∂ t
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(b) Find the induced voltage in the conductor along x-axis of length 0.2m from the origin where B = 0.04 ay Tesla and U = 2.5 sin 103t az.
Solution:
Em = U ¿ B
= 0.10 sin 103t (-ax) V/m
v = −∫
0
0 .2
0 . 10sin 103 t(−ax )⋅dx ax
= -0.02 sin103t volts.
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VELAMMAL COLLEGE OF ENGINEERING AND TECHNOLOGY, Madurai – 625 009
Department of Electrical and Electronics EngineeringAcademic Year 2014-2015
Name of the Staff : K.SnehaName of the Subject : Electromagnetic Theory/EE 8391Class : II Year EEE A & B SectionDate :
UNIT III-TUTORIAL – 3 PERIODS
1. Determine H at P1(0.4, 0.3, 0) in the field of an 8A filamentary current directed inward from infinity to the origin on the positive x-axis and then outward to infinity along y-axis.Solution:
Magnetic Field Intensity for a finite length current element is given as,
H x=I4 πρx
(sinα 2 x−sinα1 x ) aφ
H y=I4 πρ y
(sin α 2 y−sinα 1 y ) aφ
H x=84 π (0 .3 )
(sin 53 .1−sin(−90)) az
= 12π
az A /m .
H x=I4 πρx
( sin(36 . 9)−sin(90 ))(− az )
=−8π
az A/m .
H=H x+H y=−6 .37 az A /m .
2. Consider a Square loop of sides (1, 0, 0), (1, 2, 0), (3, 0, 0) and (3, 2, 0) in the z = 0 plane carrying 2mA in the field of an infinite filament on the y-axis. And a current of 5A flows towards the origin along the y-axis. Calculate the total force on the loop.Solution:
The Magnetic Field Intensity is given as,
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H=I2πx
az=152πx
az A /m
and B=μ0 H =3×10−6
xazTesla
Force equation is, F=−I∮B×dl
=−2×103×3×10−6 [∫1
3 az
x dx ax+∫0
2 az
3 dy ay+∫3
1 az
x dx ax+∫2
0 az
1 dy ay ]on solving,
F = -8 axNewton.
3. An Air-cored toriod of cross-sectional area 6cm2, mean radius of 15 cm with an sir gap of 2mm. The magnetic field intensity of steel is 200A-t and B =1T and find:
a. Vm,air
b. Vm,steel
c. The current required in a 1300 turn coil linking the left leg.Solution:
(a) Rair =
dair
μS= 2×10−3
4 π×10−7×6×10−4 =2. 65×106 A−t /Wb .
= BS = 1(6 x 10-4) = 6 x 10-4 Wb.
Vm, air = (6 x 10-4) (2.65 x 106) = 1590 A-t.
(b) Hsteel = 200 A-t.Vm, steel = Hsteel dsteel = 200 x 0.3 =188 A-t.
(c) Total mmf = 1778 A-tCoil Current = (total mmf / number of turns) = 3.56 A.
4. State Ampere’s circuital law and explain how it can be used for magnetic field calculations:Ans:
It states that the line integral of Magnetic Field Intensity about any closed path is exactly equal to the direct current enclosed by that path.
∮ H⋅dl=I enclosed
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1
2
(0, 4)(8, 4)
(8, 0)(0, 0) 10A
2m
4m
In Ampere’s law , to determine H there must be a degree of symmetry and the following conditions is to be met:
1. At each point of the closed path H is either tangential or normal to the path.2. H has the same value at all points of the path where H is tangential.
Application of Ampere’s Law:
Infinite long filament: Infinitely long coaxial transmission line:
5. A rectangular loop carrying 10A current is placed on Z = 0 plane. Find the Field Intensity at (4, 2, 0).Solution:
(i) H1 at P due to Horizontal side of 8m is,
H1=1
4 πh (cosα 1+cos α2)
h = 2m ; cos1 = cos2 = 0.894.
H1 = (10 / (4(2)) x 2 x 0.984) az = 0.72 az AT / m.
(ii) H2 at P due to vertical side of 4m.h = 4m ; cos1 = cos2 = 0.447.
H2=104 π×4
(2×0 .447 )
=0. 178 az AT /m .
H = H1 + H2
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21
P
4m
= 2 x 0.172 + 2 x 0.178
= 1.779 az.
6. Find H at the center of an Equilateral triangular loop of side 4m carrying current of 5A?
Solution:
Magnetic Field Intensity at appoint due to finite length of conductor,
H1=1
4 πh (cosα 1+cos α2) az
tan 300 = (h / 2) ; h = 1.154 m.
cos 1 = cos2 = cos 300
H1=54 π (1 .154 )
(cos30+cos30 )az
= 0 . 597 az AT /m .
Magnetic Field due to all the three sides is, 3 x 0.597 = 1.79 AT/m.
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VELAMMAL COLLEGE OF ENGINEERING AND TECHNOLOGY, Madurai – 625 009
Department of Electrical and Electronics EngineeringAcademic Year 2014-2015
Name of the Staff : K.SnehaName of the Subject : Electromagnetic Theory/EE 8391Class : II Year EEE A & B SectionDate :
UNIT V-TUTORIAL – 3 PERIODS
1. A free Space- sliver interface has Eio = 100 V/m on the free space side. The frequency is 15MHz and the = 61.7 Ms/m. Determine Ero and Eto at the interface.Solution:
For a Free space conductor interface, τ = -1
Ei 0
Er 0=−1 ⇒ E r 0=−100 V /m .
τ=Et 0
Ei 0=
2η2
η1+η2
η2 is for sliver medium,
η2=√ωμσ
∠ 450
=√2 π×150×106×4 π×10−7×161 .7×10−6 = 1.38×10−3Ω
η1 at the free space is 120.
./451035.7
1038.13771038.12
040
3
3
0
0
mVE
EE
t
i
t
2. A Steel pipe is constructed of a material for which µr = 180 and = 4 x 106 S/m. The two radii are 5 and 7mm and length is 75m. If the total current I(t) carried by the pipe is 8 cos t , where = 1200 rad/s.Find (i) Skin Depth. (ii) Effective Resistance.
Solution:
µr = 180 and = 4 x 106 S/m.
a1 = 5mm ; a2 = 7mm ; L = 75m
I(t) = 8 cost Ampere ; = 1200 rad/sec.
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(a) Skin Depth:
=
1√πf μσ
= 1√πfμr μ0 σ
=1¿ √π×1200π
2 π×180×4 π×10−7×4×106 ¿
¿¿
= 0.766 mm.
(b) Effective Resistance:
R =
L2πa2σδ
=752 π×7×10−3×4×106×0 .766×10−3
= 0.557 ohms.
3. In a non – Magnetic medium, E = 4 sin (2 x 107t – 0.8x) az V/m. Find (a) r and . (b) The time average power carried by the wave.Solution:
E = 4 sin (2 x 107t – 0.8x) az V/m
= 0 ; = 0
The wave is traveling along x- axis, H along (-ay) and E along az direction.
= 2 x 107t rad/sec; E0 =4 V/m.
= 0.8 rad / m.
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β=ωμ
⇒ μ=2π×107
0 . 8=7 . 85×107m /sec
u=1√μ0 μr ε0ε r
=3×108
√ μr εrFor non−magentic material⇒μr=1
εr=3×108
7 . 85×107 =14 . 6
Intrinsic Im pedance⇒η=√μεη=√ μ0 μr
ε 0εr=120π √1
14 .6=98 . 7 Ω
Average Power , Pav=E0
2
2|η|e−2αx cosθη ax
As α=0 and θη=0
Pav=E0
2
2ηax=
42
2×98 . 7=81 ax mW /m2
4. The Electric Field Intensity associated with a plane wave traveling in a perfect dielectric medium is givn by,
Ex(z, t) = 10 cos(2 x 10-7t – 0.1z) V/m.
(i) What is velocity of propagation.(ii) Write down an expression for magnetic field intensity associated with the
wave µ = µ0.Solution:
Ex(z, t) = 10 cos(2 x 10-7t – 0.1z) V/m.
Here, = 2 x 10-7 ; = 0.1.
(I) Velocity of Propagation = ( / ) = (2 x 10-7 / 0.1) = 2 x 10-6 m / sec.(II) Hy = (Ex / )
= √ με=√ μ0
ε =√ 4 π×10−7
8 . 854×10−12=376 . 73
Hy = 0.027 cos(2 x 10-7t – 0.1z) A/m.
5. Find the values of phase shift constant, attenuation constant and intrinsic impedance of the medium with r = 2.25 and = 10-4 V/m for a wave of frequency 2.5 MHz.Solution:
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Propagation constant = √ j ωμ (σ+ jωε )
= 0.025 + j 0.0786
= 0.025 rad / m and = 0.0786 rad / m.
= √ jωμσ+ j ωε
= 240 .34∠17 .72 Ω
6. Compare the conduction and displacement current densities in copper ( = 0 , µ = µ0 and = 5.8 x 107 S/m) at a frequency of 1MHz. Repeat for Teflon which has = 2.10, µ = µ0 and = 3 x 10-8 S/m at 1MHz.Solution:
Assuming sinusoidal variation of the electric field in the material,
E = E0 sint V/m.
Where, = 2f and f = 106
Conduction current density, J = E
= E0 sint A/m2
Displacement current density, J =
∂D∂ t
=ε ∂E∂ t
=ωε E0cos ωt A /m2
| JJ d|= σ
ωε
For copper,
σωε
= 5.8×107
2π×106 136 π
×10−9=1012
(conduction current dominates the displacement current)
For Teflon,
σωε
= 3×10−8
2π×106×2 .1× 136π
×10−9=2. 57×10−4
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VELAMMAL COLLEGE OF ENGINEERING AND TECHNOLOGY, Madurai – 625 009
Department of Electrical and Electronics EngineeringAcademic Year 2014-2015
Name of the Staff : K.SnehaName of the Subject : Electromagnetic Theory/EE 8391Class : II Year EEE A & B SectionDate :
UNIT II-TUTORIAL – 3 PERIODS
1. Find the total electric Field intensity at point P(0, 6, 5)m due to a charge of 20 C located at (2, 0, 6)m and charge of 60C located at (0, -1, 2)m and charge of 100C located at (2, 3, 4)m.Solution:
Q1 = 20 C at (2, 0, 6)m;
Q2 = 60C at (0, -1, 2)m
Q3 = 100C at (2, 3, 4)m.
Electric Field Intensity at (0, 6, 5)
Eo=E1+ E2+ E3
Eo=Q1
4πε 0r1ar 1 +
Q2
4 πε0r2ar2 +
Q3
4 πε0r3ar 3
R1=−2 ax+6 a y−az
|R1|=√4+36+1=√101=6 . 4m
R2=7 a y+3 az
|R2|=√49+9=7. 615m
R3=−2 ax+3 a y+ az
|R1|=√4+9+1=3 .74 m
E=9×109¿[20×10−6
(6 . 4 )3(−2 ax+6 a y− az ) ¿]¿
¿
¿
E=−3781.15 { ax+64291.82 { a¿ y+20185 . 9 az V /m .¿
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2. A charge q1is located at each of the four corners of a square of side ‘a’ meter. Find the charge which is to be located at center of the square to stabilize the system.Solution:
FA = FBA + FDA + FCA + FEA
FBA=1
4 πεo
q12
R2 aBA = ka2 q1
2 aBA
FDA=1
4 πε o
q12
R2 aDA = ka2 q1
2 aDA
FCA=1
4 πεo
q12
(√2a2 )2aCA = k
2a2 q12 aCA
FEA=1
4 πε o
q12
(√2a2
2 )2 aEA = k
a2 /2q2
2 aEA
The system is stable (i.e.) FA= 0
0 = FBA + FDA + FCA + FEA
0=kq1
a2 [q1 aBA+q1 aDA+q1
2aCA+2q2 aEA ]
Here , aBA = cos 45o⋅aCA ; aDA=cos 45o⋅aCA; aEA= aCA
0=kq1
a2 [q1 cos 45o aCA+q1 cos45o aCA +q1
2aCA+2q2 aCA]
0=kq1
a2 [q1 (1√2 ) aCA+q1 (1√2 ) aCA +q1
2aCA+2q2 aCA ]
q2=−[√2q1+q1
22 ] = −
(√2+1/2)2
= −0 .957q1
3. An uniform line charge is lying along y = 3, z = 5 if L = 30nC/m. Find E at (a) the origin (b) PB (0, 6, 1) (c) PC (5, 6, 1).Solution:
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(a) E=
ρL
2 πεoRaR
=ρL
2πε oRR|R|
R=−3 a y−5 az |R|=√9+25=√34
E=30×10−9
2π×8 .854×10−12×34[−3a y−5az ]
E=−47 . 6 a y−79 . 31 { a z
V /m . ¿
(b) PB (0, 6, 1) :
R=3 ay−4 az |R|=√9+16=5
E=30×10−9
2π×8 .854×10−12×25[3ay−4az ]
E=64 . 8 ay−86 . 4 az V /m .
(c) PC (5, 6, 1):
R=3 ay−4 az |R|=√9+16=5
E=30×10−9
2π×8 .854×10−12×25[3ay−4az ]
E=64 . 8 ay−86 . 4 az V /m .
4. A positive point charge of 100C is located in air at x = 0, y = 0.1m and another such charge at x = 0, y = -0.1m. What is the magnitude and direction of E and what is absolute potential V at x = 0.2m and y = 0.Solution:
Q = 100C at (0, 0.1) and (0, -0.1)
E1=Q4πε or1
2ar 1
ar1 =0.2 ax−0 .1 a y
0.2236
Er 1=100 ×10−12
4 πε0 (0 . 2236 )2 [0 .2 ax−0 .1 a y
0 .2236 ] = 16 .1 ax−8 .05 { ay
¿
Similarly,
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E2=Q4πε or2
2ar 2
ar 2 =0 .2 ax+0 . 1 ay
0 .2236
Er 2=100 ×10−12
4 πε 0(0 . 2236 )2 [0 . 2ax+0 .1 a y
0 . 2236 ] = 16 .1 ax+8 . 05 { ay
¿
The Electric Field Intensity at (0.2, 0) is,
E=E1+E2=32 . 2 ax V/m.
Absolute Potential is,
V 1=Q4 πεo r1
Er 2=100 ×10−12
4 πε 0(0 .2236 )= 4 .025 V
V 2=Q4 πε or2
Er 2=100 ×10−12
4 πε 0(0 .2236 )= 4 .025 V
Absolute Potential at (0.2, 0) is,
V = V1+ V2
= 8.05 V.
5. Conducting spherical shells with radii a = 10cm and b = 30cm are maintained at a potential difference of 100V such that V(r = b) = 0 and V(r = a) = 100V. Determine V and E in the region between the shells:Solution:
a =10 cm ; b = 30cm ; v =100 V ;
Boundary conditions are,
r = b at v = 0V.
r = a at v = 100V
Laplace equation in spherical co-ordinates is,
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∇2V= 1r 2
∂∂r (r2 ∂V
∂r )+ 1r 2 sinθ
∂∂r (sin θ ∂V
∂θ )+ 1r2 sin2θ (∂
2V∂ ϕ2 )
V varies only with respect to r,
1r2
∂∂ r (r2 ∂V
∂r ) = 0
V = -(A/r) + B ----- (1)
A and B are constants.
Applying the boundary conditions,
r = b at v = 0V.
0 = - (A / b) + B
B = A / b
(1) ----- V = -(A / r) + (A / b)
r = a at v = 100V
V0 = - (A / a) + (A / b)
A =
V 0
[ 1b−1
a ]
V=−V 0
[1b−1a ]
1r+
V 0
[1b− 1a ]
1b
= V 0 [ 1r−1
b1a−1
b ] = 15[1r−103 ] volts
E=−∇V =− dVdr
ar =− Ar2 ar =
V 0
r2 [1a−1b ]
= 15r2 ar V /m .
6. As shown in figure below, charge distributed along the z-axis between Z = 5m with a uniform density, L = 20nC/m. Determine E at (2, 0, 0)m in Cartesian co-ordinates. Also explain the answer in cylindrical co-ordinates.
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R
(2, 0, 0)Y
+5
-5
1cm
Glass
Air
Solution:
L = 20nC /m.
Field Point (2, 0, 0)m
Electric Field Intensity at a point due to a line charge (for finite line charge)
E=ρl
2πε0 lρ√ρ2+ l2
= 2×10−9
2×π×8 . 854×10−12×25√52+22
= 166 . 898 ax V /m .
7. A parallel plate capacitor with a separation d = 0.1cm has 29000V applied when free space is the only dielectric. Assume that air has a dielectric strength of 30,000 V/cm. Show why 290,000V/cm and thickness d2 = 0.2cm is inserted as sown in figure:
Solution:
d = 1 cm ; v =29000 V ; E1 = 30 KV / cm
E2 = 290 KV /cm ; d2 = 0.2 cm;
V = potential across the capacitor = 29KV.
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When glass dielectric is inserted,
V = V1 + V2
= E1 d1 + E2 d2
= 0.8 E1 + 0.2 E2
V 1
V 2=d1
ε0
ε0ε r
d2⇒ V 1=26 V 2
V 1+V 2=29000026V 1+V 2=29000V 1=27925 .925 volts . V 2= 1074 .074 volts .
Then , E1=V 1
d1= 34 .907 KV /cm
8. Find the work done in moving a point charge Q = 5C from the origin to (2m, /4, /2) spherical co-ordinates, in the field,
E = 5e-r/4 ar + (10 / r sin)a V / m.
Solution:
Q = 5C
Work done = ∮SE⋅dl .
dl = dr ar+ r d a + r sin d a
E = 5e-r/4 ar + (10 / r sin) a V/m.
Work done = ∫0
2
5e−r /4dr+∫0
π /210r sin θ
r sin θ dφ
= [ 5e−r/4
−(1/4 ) ]02
+10 [φ ]0π /2
= - 20 [ 0.6065 -1] +5
= 7.869 + 5.
Workdone = 23.5769.
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9. A hallow sphere is charged to 12C of electricity. Find the potential (a) at its surface (b) inside the sphere (c) at a distance 0.3m from the surface. The radius of sphere is 0.1m.Solution:
(a) Potential at the surface:
V =
Q4 πε0 r
If r = a and so V =
12×10−6
4 π×8 .854×10−12×0. 1=1 .08×10Volts .
(b) Potential Inside:E = -V
E = 0 and so V = constant
(c) Outside:r = 0.3m (r > a)
V =
Q4 πε0 r
r = 0.4m
V =
12×10−6
4 π×8 .854×10−12×0. 4=2 .7×106Volts .
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VELAMMAL COLLEGE OF ENGINEERING AND TECHNOLOGY, Madurai – 625 009
Department of Electrical and Electronics EngineeringAcademic Year 2014-2015
Name of the Staff : K.SnehaName of the Subject : Electromagnetic Theory/EE 8391Class : II Year EEE A & B SectionDate :
UNIT I-TUTORIAL 3 PERIODS
1. Given A = 2ax + 4ay and B = 6ay – 4az. Find the smaller angle between them using, (a) the cross product (b) the dot product.Solution:
(a) the cross product:
A x B =
|ax a y az ¿||2 4 0 ¿|¿¿
¿¿
|A|=√(2 )2+( 4 )2+(0)2=4 . 47|B|=√(0 )2+(6 )2+(−4 )2=7 .21|A×B|=√(−16 )2+(8)2+(12)2=21 .54
Then, |A×B|=|A||B|sinθ
Sin = 21.54 / ((4.47)(7.21)) = 0.688.
= 41.9o
(b) the dot product:
A. B =(2)(0) +(4)(6)+(0)(-4) = 24.
|A⋅B|=|A||B|cos θ
cos = 24/(4.47)(7.21) = 0.745
= 41.9o
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2. Use the spherical co-ordinate system to find the area of the strip on the spherical shell of radius ‘a’. What results when =0 and = ?Solution:
The differential surface element (in spherical co-ordinate system) is,
dS = r2sin d d
Then,
A = ∫0
2 π
∫α
β
a2 sinθ dθ dφ
= 2a2 (cos - cos)
here, =0 and = then,
A = 2a2 (cos 0o – cos )
= 4a2.
This is the surface area of the entire sphere.
3. Transform,
A = yax + xay +
x2
√x2+ y2az from Cartesian to cylindrical co-ordinates:
Solution:
We know that,
x = r cos ; y = r sin ; z = z ; r = √ x2+ y2
Hence, A = r sin ax + r cos ay + r cos2 az
Now, the projections of the Cartesian unit vectors on ar, a, and az are,
Dot (or) a a az
ax cos -sin 0
ay sin cos 0
az 0 0 1
Therefore,
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ax=cosφ ar−sinφ aφa y=sinφ ar+cosφ aφaz= az
and A = 2r sin cos ar + (r cos2 - r sin2)a + rcos2az.
4. Consider A = 5r ar + 2 sin a + 2 cos ain spherical co-ordinates. Transform the above vector to Cartesian co-ordinates.Solution:
The variables r, , , can be written interms of Cartesian variables as,
r=√x2+ y2+z2 ; cosθ= z√ x2+ y2+z2
; tanφ= yx
The vector can be written as,
A=5√x2+ y2+z2 ar+2 y
√x2+ y2aθ+
2 z√ x2+ y2+z2
aφ
The unit vectors ar , a, a can also transformed into Cartesian equivalents,
ar=x√x2+ y2+z2
ax+y√x2+ y2+z2
a y+z√ x2+ y2+z2
az
aθ=xz√x2+ y2+z2√x2+ y2
ax+yz√ x2+ y2+z2√ x2+ y2
ay−√ x2+ y2
√ x2+ y2+z2az
ar=− y√x2+ y2+z2
ax+x√x2+ y2
a yOn
combining the transformed component vector is,
A=(5 x+2 xyz√ x2+ y2+z2 (x2+ y2)
−2 yz√ x2+ y2+z2√ x2+ y2 )ax
+(5 y+2 y2 z( x2+ y2 )√x2+ y2+z2
+2 xy√ x2+ y2+z2 √x2+ y2 )a y+(5 z−2 y
√ x2+ y2+z2 )az
5. Prove that div(curl F) = 0 where, F = Fx ax + Fy ay + Fz az . Solution:
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div(curl F) = ∇⋅(∇×F )
curl F=∇×F=¿|ax ay az ¿|¿¿
¿¿
¿¿
div(curl F) = ∇⋅(∇×F )
= ( ∂∂ x
ax+∂∂ y
ay+∂∂ z
az)⋅(∇×F )
=( ∂∂ x
ax+∂∂ y
ay+∂∂ z
az)⋅¿ ¿
(∂F z
∂ y−∂F y
∂ z )ax−(∂ F z
∂ x−∂F x
∂ z )a y+(∂ F y
∂ x−∂Fx
∂ y )az
=0.
Hence proved.
6. Given that D = (10 3/4) a in cylindrical co-ordinates, evaluate both sides of divergence theorem for the volume enclosed by = 2, z = 0 and z =10.Solution:
Divergence theorem,
∮SD⋅ds=∫
V(∇⋅D )dv
L.H.S:
∮SD⋅ds=∫
z=0
10
∫φ=0
2 π10 ρ3
4aρ⋅ρ dφ dz aφ
we have, = 2,
∮SD⋅ds=∫
z=0
10
∫φ=0
2 π10×23
4aρ⋅(2)dφ dz aφ
= 40 x 2 x 10 = 800.
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R.H.S:
∫v(∇⋅D)dv= ∫
z=0
10
∫φ=0
2 π
∫ρ=0
2
(∇⋅D) ρ dρ dφ dz
∇⋅D=1ρ
∂∂ ρ
( ρ Δρ )
=1ρ
∂∂ ρ
(10 ρ3
4) = 1
ρ(10 ρ3) = 10 ρ2
∫v(∇⋅D)dv= ∫
z=0
10
∫φ=0
2 π
∫ρ=0
2
(∇⋅D) ρ dρ dφ dz
=∫0
10
∫0
2 π
(10 ρ4 /4 )02 dφ dz
= 40 x 2π x 10 = 800π.
Hence verified.
7. Show that A = 4ax–2 ay–az and B = ax+ 4ay–4az are perpendicular. To Prove:
* A and B are Perpendicular. (i.e) AB = 0
Proof:
AB = (4ax–2ay–az) (ax+ 4ay–4az)
= (4)(1) + (-2)(4) + (-1)(-4)
= 0.
Hence proved.