ee341 2 chapter 7

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ICE2341ICE2341 Electromagnetics Wave2009. Feb. 19

Chapter 7

Electromagnetics Wave and Antenna

Prof. Seong-Ook Park

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7. Time-varying fields and Maxwells Equations (contents)

7-1 Introduction

7-2 Faradays Law of Electromagnetic InductionFundamental postulate for electromagnetic induction

7-2.1 A Stationary Circuit in a Time-Varying Magnetic Fieldtransformer emf (electromotive force), Lenzs law

7-2.2 Transformersideal transformer, real transformer, coefficient of coupling, eddy current

7-2.3 A Moving Conductor in a Static Magnetic Fieldflux cutting emf or motional emf

7-2.4 A moving Circuit in a Time-Varying Magnetic Fieldgeneral forms of Faradays law

7-3 Maxwells EquationsDisplacement current, Maxwells Equation

7-4 Potential Functions

7-5 Electromagnetic Boundary Conditionsgeneral statements, interface between two lossless medias, interface between dielectric and conductor7-6 Wave Equations and Their Solutions

7-7 Time-Harmonic Fields

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Fundamental EM field quantitiesFundamental EM field quantities::

1. Electric field intensity E

2. Electric flux density D

3. Magnetic flux density B

4. Magnetic field intensity H

The principal objective of studying electromagnetism is to

understand the interaction between charges and currents at a

distance.

( )mV/

( )T

( )mA /

)mC( 2/

We have fundamental postulates relating E, D, B, H and the

source quantities J and r.

Maxwells Equation

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Electrostatic postulates

Magnetostatic postulates

0E =

v

JHvv

=r= Dv 0B =

v

Faradays law of Electromagnetic Induction:

t=B-E

vv

differential form

= sc B-E sddtd

ldvvvv

integral form

0Ec = ldvv

IHc

= ldvv 0B

s = sdvv

QsdDs

=vv

oror

oror

Constitutive relations (linear and isotropic media) :

D = E B = H

= r0 : permittivity = r0 : permeability

0 : 1/36 x 10-9 (F/m) 0 = 4 x 10

-7 (H/m)

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In the time-varying case, Maxwell initially considered the

following 4 postulates:

t=B

-Evv

(1) JHvv

= (2)

r= Dv

(3) 0B =v

(4)

Maxwell equation can express in integral form :

=

-=

s

sc

QdsD

dsdtBddlE

v

v

v

=

=

s

s

dsB

IdlH

0v

v(1) (2)

(3) (4)

I: Total current through S.

Q : Total charge contained in volume V bounded

by Surface S

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Fundamental Postulate for Electromagnetic Induction

where Sis open surface, ds is differential surface area vector normal to the surface(p.22,(2-34)), Cis contour bounding the surface, and dlis differential line vector

along the contour.

Eq.(7.2) is reduced from Eq.(7.1), applying Stokess theorem (p.59,(2-143))

Eq.(7.1) reduces to the equation in static case.

The postulate (7.2) reduces to Faradays law of electromagnetic induction.This is considered as a postulate because of experimental law.

(7.1)

(7.2)C S

BEt

B E dl d s

t

= -

= -

urur

urur r

g g

C

jSD

S

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Reduction of Postulate to Faradays Law

Eq.(7.6) is expression ofFaradays law of electromagnetic induction:

The electromagnetic force induced in a stationary closed circuit is equal to the

negative rate of increase of magnetic flux linking the circuit.

Lenzs law: The induced emf will cause a current to flow in the closed loop in such a

direction to oppose the change in the linking magnetic flux.

C S

d E dl B d s

dt= -

ur ur rg g (7-3)

C S

B E dl d s

t

= -

urur r

g g

(7-2)

CV E dl =

urg = emf induced in circuit with contour C(V) (7-4)

S

d

B d sdtF =

ur r

g = magnetic flux crossing surfaceS

(Wb) (7-5)

dV

dt

F= - = transformer emf (V) (7-6)

C

jSD

S

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Time-Varying Field and Wave

The electric field and magnetic field are related by Faradays law and the governing

equations come to be as follows.

Are they a complete set to represent the time-varying fields?

Fundamental relations Electric field Magnetic field

Governing equations

Constitutive relations

(linear and isotropic media)

(3-102)e=D E1

(6-80b)

m

=H B

(7-47b)

0 (7-47d)

=

=

H J

Bg

(7-47a)

(7-47c)

t

r

= -

=

BE

Dg

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Displacement Current Density (1)When conditions are changed from static to non-static, what changes arise on the

electromagnetic equations?

One of such changes is expressed by Faradays law.Another is displacement current to satisfy the equation of continuity.

Equation of continuity

Governing equation (7-47b) for magnetostatic model does not satisfy the equation ofcontinuity (7-48) as follows:

Something time-dependent termf(t) is missing in (7-47b).

Maxwell began by consdering these known laws and expressing them as differential eq. He then noticed that

there was something strange about eq. (7-47b). If one takes the divergence of this equation, the left-

hand slide will be zero, because the divergence of a curl is always zero. But if the divergence ofjis

zero, then the total flus of current out of any closed surface is also zero.

J (7-48)t

r = -

( )H J 7-47b =

( ) ( )H 0 J, 7-47b = =

( ) ?f t+

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Displacement Current Density (2)

In order to satisfy both null identity and the equation of continuity,

Putting Gausss law (7-47c) into (7-50), generalized Amperes circuital law (7-51) is

obtained and reduces to (7-52).

Displacement current density

introduces by J. C. Maxwell

( ) ( )H 0 J 7-50t

r = = +

( ) ( )D

H 0 J , 7-51t

= = +

( )D

H J , 7-52t

= +

The flux of current from a closed surfaceis the decrease of the charge inside the surface. This certainly cannot in general be zero because we know that

the charges can be moved from one place to another. Maxwell appreciated this difficulty and proposed that it could be avoided by adding the term to

the right-hand side of eq 7-52. If we take away the scaffolding he used to build it, we find the Maxwells beautiful edifice stands on its own. He brought

together all of the laws of the electricity and Magnetism and made one complete and beautiful theory.

t

r = -

Jg

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E

E

E

j

G

?B

j

Q

What is the magnetic field ofa spherically symmetric current?

J-v

=rt

: equation of continuity

How the new term works

As our first example, we consider that happens with a spherically symmetric

radical distribution of current. Suppose we imagine a little sphere with

Radioactive material on it.

This radioactive material is squirting out some charged particles.We would have a current that is everywhere radically outward. We will assume

that is has the same magnitude in all directions.

( ) 0( )

HE

jt

e = +

2( ) = - 4 ( )Q r

r j rt

p

DH j

t

= +

The electric field at the radius r must be , so long as the charge

is symmetrically distributed, as we assume. It is radial, and its rate of changes is then

2

0

1 ( )=

4

E Q r

t r tpe

2

2 2

0 0 0

( )E 1 1 -4 ( ) ( )

4 4

dQ rr j r j r dt

t r r

p

pe pe e

= = = -

2

0

( )=4

Q rErpe

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1GLOOP

I

I

B

r

( )a

1GLOOP

I

2GLOOP

B

eB

1S

E

2S

1S

( )b

+ + + + + + + + + +

- - - - - - - --

Fig. The magnetic field near a charging capacitor.

As our first example, we consider the magnetic field of a wire used to charge a parallel-plate condenser.

Suppose however, that we now slowly move the curve G downward. We get always the same result until wedraw even with the plates of the condenser. Then the current I goes to zero. Does the magnetic field disappear?

That would be quite strange.

From our discussion so far of Maxwells new term, you may have the impression that it doesnt add much

that it just fixes up the equations to agree with what we already expect. It is true that if we just consider

by itself, nothing particularly new comes out. The words by itself are, however, all important.

DH J

t

= +

H J, = 2 ,r H Ip =2

IH

rp=

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Maxwells re-analysis of Gausss and Amperes laws:

0area

Qd = AE Id 0

line

= sB

Note thatdt

dQIQ =

00area0area

1

QI

dt

dQd

dt

dQd ==

= AEAE

Maxwell-Ampere law:

( )

dtdI

ddt

dIIId

E

Q

F+

+=+=

000

area0000

line

AEsB

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EXAMPLE: A Van de Graaff generator with a spherical

bowl of radiusR is charged at a constant rate with a

currentI. find the magnetic induction field at a point

above the sphere. (Assume the current enters at the

bottom of the bowl.)

z

ad S

dl

I

z

dl

d S

I

Figure. The surface enclosed by the loopcan be taken to be either the flat surface onthe left, on the sph erical cap on the right.

Solution: As charge accumulates on the bowl,

the electric field will increase at the following rate:

2 2

0 0

E 1

4 4

rr

dQa

aIdt

t r rpe pe

= =

Drawing a loop of radius a about the z axis above the sphere,

we have

0 0

0

2

EB S

2 S

4

r

dl dt

I aaB d

rj

m e

mp

p

=

=

The surface integral can be evaluated easily either over the flat

surface included by the loop or alternatively over a spherical capconcentric with the sphere, as shown in Figure.

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(a) For the flat surface, we need thezcomponent

ofrin order to compute rdS. Calling the

cylindrical radial coordinate

we find

( ), rz

r =2 2cos , , z r r z q r= = +and

( )

( )

0

2 30

0

302 2

0

2 2

0

0

2 2

28

4

4

4

1

a

a

a

I z B z d a r

I zd

az

I

I z

a z a

z

a z

jm pr rp

mr

p

r

m

p r

m

pr

=

=+

-=

-

+

+

=

(b) The element of surface area for the spherical

cap on the right of Figure is

leading to

2S

rd a r d = W

22 sin ,ra r dp q q=

( )

( )( )

1

11

2 2

2sin0

2 20

sinsin0 0

00

0

2 2

2 sin

8

sin cos4 4

4

1

a rr r

aa r

z a

I a a r d Ba r

I Id

a

I z

a z a

a

jm p q q

p

m mq

m

q qp p

p

-

--

+

-

=

= = -

= +

When we added a new term to the equation for the curl of E,we found that a whole new class of phenomena was

described. We shall see that Maxwells little addition to the equation for also has far-reaching consequences. We

can touch on only a few of them.

z

a

d S

d l

I

z

d l

d S

I

t

+=D

JH

vvv

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Example 7-5 An a-c voltage source of amplitude V0 and angular frequency , vc =

V0

sint, is connected across a parallel-plate capacitor C1

, as shown in Fig. 7-7.

(a) Verify that the displacement current in the capacitor is the same as the

conduction current in the wires. (b) Determine the magnetic field intensity at a

distance r from the wire.

parallel plate capacitor with areaA,

separation d, and dielectric media ofe.

S1 is a planar disk surface crossing

the wire (no displacement current).

S2 is curved surface passing through

the dielectric medium (no conduction

current).

conduction current displacement current

Parallel-Plate Capacitor connected to ac Voltage Source

0 sincv V tw=

1

AC

de=

Generalized Ampere's circuital law

C Sdl d

t

= + D

H J sg g

1C

2S

1S

C

ci

cv

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Parallel-Plate Capacitor connected to ac Voltage Source (Trace the equations!)

a) The conduction current in the connecting wire is

For a parallel-plate capacitor with area A, plate separation d, and a dielectric

medium of permittivity the capacitance is

With a voltage vc appearing between the plates, the uniform electric field intensity

E in the dielectric is equal to (neglecting fringing effects) E=vc/d, whence

The displacement current is then

( )1 1 0 cos Acc dvi C C V t dt

w w= =

1 .A

Cd

e=

0 sin .V

D E t d

e e w= =

0

1 0

D cos

cos . Q.E.D.

DA

c

Ai ds V t t d

C V t i

e w w

w w

= = = =

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Parallel-Plate Capacitor connected to ac Voltage Source (Trace the equations!)

b) the magnetic field intensity at a distance r from the conducting wire can be found by

applying the generalized Amperes circuital law, Eq.(7-54b), to contour C in Fig. 7-7. Two

typical open surfaces with rim C may be chosen: (1) a planar disk surface S1, or (2) acurved surface S2 passing through the dielectrc medium. Symmetry around the wire

ensures a constant H along the con tour C. The line integral on the left side of Eq. (7-54b) is

For the surface S1, only the first term on the right side of Eq. (7-54b) is nonzero because oncharges are deposited along the wire and , consequently, D=0.

since the surface S2 passes through the dielectric medium, no conduction current flows

through S2. If the second surface integral were not there, the right side of Eq.(7-54b) would

be wero. This would result in a contradiction. The inclusion of the displacement-current term

by Maxwell eliminates this contradiction. As we have shown in part (a), id=ic. Hence we

obtain the same result whether surface S1 or surface S2 is chosen. Equating the weoprevious integrals, we find that

H 2C

d rHfp = l

11 0J cosc

Sds i C V t w w = =

( )1 0 cos /D

H J (7-52

4b)C S

C V H t A dl dsm

trf w w

p

= + = g

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Maxwells Equations

The above fourMaxwells equations, together with following equation of

continuity (7-48) and Lorentzs force equation (6-5) form the foundation of

electromagnetic theory.

Pairs of E, D and B, H are related by constitutive relations:

Differential Form Integral Form Significance

Faradays law

Amperes circuital law

Gausss law

No isolated magneticcharge

BE

t = -

D r =

DH J

t

= +

B 0 =

Ec

dd

dtF = - l

DH

c Sd I ds

t

= +

l

DS

ds Q =B 0

Sds =

( ) ( ) ( )J 7-48 F E B 6-5q utr

= - = +

D E and H B/e m= =

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Maxwells Equations: Comparison with Static Models

Maxwells equations Time-varying Static

Faradays law

Amperes circuital law

Gausss law

No isolated magnetic charge

Constitutive relations

Equation of continuity

Lorentzs force equation

( )( ),

,t

tt

= -

B RE R ( ) 0 =E R

( ) ( )( ),

, ,t

t tt

= +

D RH R J R ( ) ( ) =H R J R

( ) ( ), ,t tr =D R Rg ( ) ( )r =D R Rg

( ), 0t =B Rg ( ) 0 =B Rg

( ) ( ), ,t te=D R E R

( ) ( )1

, ,t tm

=H R B R

( ) ( )e=D R E R

( ) ( )1

m=H R B R

( )( ),

,t

tt

r = -

RJ Rg ( )

( ),

t

t

r = -

RJ Rg

( ) ( ) ( )( ), , ,t q t u t = + F R E R B R ( ) ( ) ( )( )q u= + F R E R B R

(Radius vector to represent the position) x y z x y z = + +R a a a

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Maxwells Equations: All Independent?

Maxwells equations (43=12 equations)

Constitutive relations (23=6 equations)

Equation of continuity (3 equations)

Total number of scalar equations is 21 equations.

Total number of unknowns is 43= 12.

Considering the constitutive relation, number of unknown is 6.

The two curl equations are sufficient to determine the E and H. The two divergence equations of Maxwells can be derived from the two curl

equations by making use of the equation of continuity. (Prob. P.7-11)

, , , 0t t

r = - = + = =

B DE H J D Bg g

1,e

m= =D E H B

t

r = - Jg

, , ,E B H D

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Potential Functions

Magnetic flux density B and electric field intensity E can be expressed by a

magnetic potential A and the electric potential V:

due to charge distribution due to time-varying current J

(this term is zero in static case)

Eq.(7-57) can be reduced from (7-55) and Faradays law (7-1):

To be consistent with the definition of the scalar electric potential Vforelectrostatics, we write

( ) ( ) ( ) ( )A

B A T . 7-55 E V/m . 7-57Vt

= = - -

( )

( )

BE . 7-1

AE A or E 0

t

t t

= -

= - + =

AE V

t

+ = -

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Potential Functions: Comparison with Static Field

In the static case, B and E can be expressed independently by a A and V:

From the Gausss law and (3-43),

Poissons Equation => solution

From the Amperes law and (6-15),

vector Poissons Equation

=> solution

A is related only to B which is purely rotational

and its divergence is zero.But in time-varying case,A is related to both ofB and E, and is not purely

rotational.

B A (6-15)= AE (3-43)V t= - -

( ) 2 2B A A A A Jm = = - = - =

2E V Vr

e = - = - =

'

1'

4 VV dv

R

r

pe=

''

4 Vdv

R

m

p=

JA

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Nonhomogeneous Wave Equation forA

The wave equation for vector potential A is obtained by substituting B and Eexpressed by potentials into one of Maxwell equations, and by applying Lorentz

condition (Lorentz gauge) for potentials (7-62):

In order to define a vector, specifications are necessary both for curl and divergence.

The curl of A is specified by (7-55) and there remains liberality of divergence. The

Lorentz condition is specification of the divergence. The Lorentz condition is

consistent with static condition A=0 and the equation of continuity. (Prob. P.7-12)

( )2

2

2

AA J 7-63

tme m

- = -

DH J

t

= +

A

A J Vt t

m me = + - -

( )2

2

2AA A J V

t tm me me - = - -

2

2

2AA J A V

t tme m me - = - + +

( )A 0, 7-62V

tme

+ =

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Nonhomogeneous Wave Equation forV

The wave equation for scalar potential Vis obtained by substituting E expressed bypotentials into Gausss law, and by applying Lorentz condition for potentials (7-62)

as follows:

(Gausss law)

The Lorentz condition uncouples the wave equations forA and V.

(Each wave equation includes only A orV)

The inhomogeneous wave equations reduce to Poissons equations in static

cases. Nonhomogeneous indicates a type of differential equation, and is also used to

indicate a type of a medium. Confusing!

( )2

2

2

VV 7-65

t

rme

e

- = -

D r =

( )2A

, AV Vt t

re r

e

- + = + = -

( )A 0, 7-62V

tme

+ =

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from the identity we may define a scalar V by0,)V(

V-A

E =

+ t

vv

(this definition is consistent with the electrostatic case)

Hence, we have

t

-=

AV-E

vv

Wave equations

t

+=D

JH

vvv

te+=m EJB1

vvv

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-

me+m=t

AV-

tJ)A(

vvv

2

2

2 AVJA)A(tt meme-m=-

vvvv

me++m=

me-tt

VAJ-

AA

2

22

vvv

v(*)

To simply (*), we choose

t

me=V

-Av

Lorentz condition (Lorentz

gauge) for potentials

- Relation between A and V

(note: a vector is well defined if both its curl

and divergence are specified)

Av

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As a result, we have

J-

A

A 2

22

vv

v

m=

me- t

non-homogeneous wave equation for vector potentialAv

Notes :

1. Wave equation : wave traveling with a velocity equal to

2. In static case, Lorentz gauge reduces to

me

1

0= Av

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Electromagnetic Boundary Conditions: Medium

Medium (Oxford Dictionary)

1. a way of communicating information, etc. to people

2. something that is used for a particular purpose3. the material or the form that an artist, a writer or musician uses

4. (biology) a substance that something exists or grow in or it travels through

5. a person who claims to be able to communicate with the spirits of dead people.

Medium for an electromagnetic wave is expressed by e, m, and s.

simple medium (linear, isotropic, and homogeneous)

linear medium nonlinear medium

isotropic medium anisotropic medium

homogeneous medium inhomogeneous medium

nonconductive medium conductive medium

lossless (good) dielectric perfect (good) conductor => Appendix B

contiguous: touching or next to something (Oxford Dictionary)

continuous: extending without abrupt change (My opinion)

( )Ee

x y z e e e

( ), ,x y ze' "je e e= -

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Electromagnetic Boundary Conditions

Boundary conditions are derived applying the integral form of Maxwells equations to

small region at an interface of two media in manners similar to those used in

obtaining the boundary conditions for static fields.As the result, they are exactly the same as those for static fields.

1. The tangential component of an E field is continuous across an interface.

2. The tangential component of an H field is discontinuous across an interface

where a surface current exists, the amount of discontinuity being determined by

Eq. (7-66b)

3. The normal component of a D field is discontinuous across an interface where a

surface charge exists, the amount of discontinuity being determined cy Eq. (7-

66c).

4. The normal component of a B field is continuous across an interface.

( ) ( ) ( ) ( ) ( )1 2 n2 1 2 3V/m ; 7 66 a H H J A/m . 7 66t t E E a b= - - = -

( ) ( ) ( ) ( ) ( )2n2 1 2 3 1 2a D D C/m . 7 66 T . 7 66n nc B B d r - = - = -

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Boundary Conditions

For problem involving contiguous regions of different m &

e, we need to know the boundary conditions:

From the integral forms of Maxwells equations, we get

For tangential components,

tt 21 EE =( ) s212 JHHa

vvv=-n

For normal components

nn 21 BB =( ) s212 DDa r=-

vvn

medium 1 2na

medium 2

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From Maxwells equations: (1) Faradays Law

--

-=-

-=

-

-

-

-=

-

=

l

EE

t

BwEE

t

B

w

EE

l

EE

t

B

y

E

x

E

BE

z

z

zxy

3412

2134

t-

vvvvv

vvvvv

vvv

vv

When w tends to 0, E1t=E2t

Tangential E-field is continuous across an interface

E1

E2

E3 E4w

l

medium 1

medium 2

y

xz

11,em

22 ,em

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From Maxwells equations: (2) Amperes Law

wl

HHDwJHH

DJ

w

HH

l

HH

DJ

y

H

x

H

DJH

zz

zz

zz

xy

zz

-+

+=-

+=-

--

+=

-

+=

4312

2134

t

t

t

t

vvvvvv

vv

vvvv

vv

vv

vvv

H1

H2

H3 H4w

l

medium 1

medium 2

y

xz

When w tends to 0, Jzw Js , Ht2 Ht1 = Js

Tangential H-field is discontinuous across an interface

where a free surface current exists

( )2 1 2n sa H H J \ - =v v v

11,em

22 ,em

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From Maxwells equations: (3) Gausss Law

( )( ) SSDDa

aDaDdSD

D

sn

nnS

D=D-

+=

=

r

r

212

1221

vv

vvv

v

na1

na2

Ss Dr

( )

( ) snnsn

DD

DDa

r

r

=-

=-\

21

212

vv

vv

Normal component of D field is discontinuous across

an interface where a surface charge exists.

The amount of discontinuity being equal to the surfacecharge density

D1

D2

medium 1

medium 2

11,em

22 ,em

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From Maxwells equations: (4)

( ) 0

0

0

212 =D-

=

=

SBBa

dSB

B

n

Svv

v

v

na1

na2

( )

nn

nn

n

HH

BB

BBa

2211

21

212 0

mm ==

=-vv

vv

Normal component of B field is continuous across an

interface

B1

B2

medium 1

medium 2

11,em

22 ,em

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* To solve a practical problem, only the tangential

components of E and H are sufficient in considering the

boundary conditions.

Special Cases

(1) Interface between 2 lossless linear media,

No free charges and no surface currents at interfacebetween two lossless media

nn

t

t

EEDD

D

DEE

2211n2n1

2

1

2

1t2t1

vvvv

v

vvv

ee

e

e

==

==

nn

t

t

HHBB

B

BHH

2211n2n1

2

1

2

1t2t1

vvvv

v

vvv

mm

m

m

==

==

(2) Interface between a dielectric and a perfect conductor

0,0 ==\ ssJ r

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In solving field problems, good conductors are often consideredas perfect conductors in regard to boundary conditions.

* the charges can only reside on the surface.

Conductivities of materials (s)silver iron

copper seawater

gold distilled water

aluminum transformer oil

brass

710176 =s .710805 =s .710104 =s .

710543 =s .1110-=s

4102 -=s

4=s

710=s

710571 =s .

Interior of a perfect conductor,

E=0 ( otherwise J = sE ). Therefore, D=0

Interrelationship between (E, D) and (B, H) fromMaxwells, B=H=0

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Medium 1 Medium 2

0E t1 =s12n JHa

vv=

s12n Da r=v

0B n1 =

0E t2 =

0B n2 =

0H t2 =0D n2 =

1

sn1E e

r=

1

2

2na

t1H

sJ

st1 JH =

2 perfect

1 dielectric

2na

conductor

n1E

1

2

+ + + + + + + + + +sr

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Interface between Two Lossless Linear Media

A lossless linear media can be specified by a permittivity e and permeability m with

s = 0. And then, rs = 0 and Js = 0.

Table 7-3

Boundary Conditions between

Two Lossless Media

( ) ( )1 2 2 1 2 2 1 2 1 2, 0, 0,t t n s n s n n E E B Br= - = = - = = =a H H J a D Dg

1 11 2

2 2

tt tt

DE ED

ee= =

11 2

2

tt tt

BH H

B

m

m

1

2

= =

1 2 1 1 2 2n n n n D D E E e e= =

1 2 1 1 2 2n n n n B B H H m m= =

( )7 67a-

( )7 67b-

( )7 67c-

( )7 67d-

I f b Di l i d P f C d

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Interface between Dielectric and Perfect Conductor

A perfect conductor is one with an infinite conductivity. Good conductors such as

silver, copper, gold and aluminum with conductivities of the order of 107 (S/m)

are approximately perfect one.Table 7-4

Boundary Conditions between a Dielectric (Medium 1) and

a Perfect Conductor (Medium 2) (Time-varying Case)

On the Side of Medium 1 On the side of Medium 2

1 0tE = 2 0tE =

2 0tH =2 1 3a H Jn =

2 1 3a Dn r = 2D 0n =

20nB =1 0nB =

( )7 68a-

( )7 68b-( )7 68c-

( )7 68d-

31 1

1

E nEr

e= = ( )7 69-

1 1 3H H Jt= = ( )7 70-

Medium 1

(dielectric) Medium 1(dielectric) Medium 1(dielectric)

Medium 2

(perfect conductor)Medium 2

(perfect conductor)Medium 2

(perfect conductor)

1E

1E

1H3

Jn2

a

Additional Boundary Condition

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If medium #1 is a very good condition

( ) 1s we ?

2

sJ n H;

12 2

1

(1 )2

t ss Z J n jwm

sE = = H +

1

1 1

s s

jZ

j

wmh

s we= =

+2( ) 1s we ?

1

1

(1 )2

sZ jwm

s+;


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1 1( )s we ?

2

sn H J =

12 2

1

(1 )2

t ss E Z J n H jwm

s= = +

:s1 =

2

s J n H ;

2E

2H2

D2

B

s2m

2e2

(2)1E

1H

1D

1B

s1m

1e1(1)

n S S S Z R jX = +x

z

y

( large)s 1

Boundary Condition: Time-Harmonic Fields

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Boundary Condition: Time-Harmonic Fields

x x z z a aE = E + EExample based on Fig.1-6

. , , , .x x z zx zand are continuous Also are continuous x z x z

E E E EE E

( ) 0 (0) ( ) (0)

0

x y z

x z x x z z x y z

z z

a a a

a a a a a

x z z x

E E E = E + E = = + - +

E E

(0) ( ) (0 () )x z x x y y z z x y z a j j ax

a aa az

wm wm = - H = - H +E E

E = + +- +

H H

( 1) ( )yxz z

yx

z x z xj

jwm

wm- H = HE EE E- -

-

=

Significance of Boundary Conditions

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Significance of Boundary Conditions

In this section we have discussed the relations that field vectors must satisfy at

an interface between different media. Boundary conditions are of basic

importance in the solution of electromagnetic problems because generalsolutions of Maxwells equations carry little meaning until they are adapted to

physical problems each with a given region and associated boundary conditions.

Maxwells equations are partial differential equations. Their solutions will contain

integration constants that are determined from the additional information

supplied by boundary conditions so that each solution will be unique for each

given problem.

particular solution

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GAUSSS LAW - EXAMPLES

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point

charge: q

volume

charge

density: o(coulombs/m3)

surface

charge

density: s(coulombs/m3)

There is no free magnetic charge of any kind, so$

s

B nda 0 = always

$S VD nda dv =

$ q f =

$ 2

2r

0 0

2r r 2

E r nr sind d q

qE 4r =q E =

4r

$p p

f f

= = =

= =

$2 r 2

2 2

r o

0 0 0 0 0

3

2 r oo r o o

3

o or o2

E r nr sindd = r sindrdd

r r4r E 4 E (for r < R ) ( D )3 3

R E (for r > R ) ( D 0)

3r

$

s

o s

S V S

sn s n

2 E nda= dv= da

12 E A= A E =

2

z

r

sinqr

sinq fr d

qrd

fdq

nE

ro o

R

da

En

e

e

E

E

ss

nA

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FIELDS PERPENDICULAR TO BOUNDARIES

Using Gausss Law $ =( D nda dv) :S V

= d

-

-

S V1n 2n S

1 1n 2 2n S

D nda dv (lim 0)

(D D )A= A

E E =

More formally: $

$ - = - =

1 2 S

1 2 2 S

n (D D )

n (E E )

The jump in normal D

is equal to the free

surface charge density

Likewise: $

$

- =

m - m =

1 2

1 2

1 2

n (B B ) 0

n ( H H ) 0

Normal B is continuous

$ =

- =

- =

S1n 2n

1n 2n

B nda 0

(B B )A 0

(B B ) 0

An

ssurface charge density ssurface S

d

1D

2D

1e

2e

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AMPERES LAW - EXAMPLES

line current: I(amps)

currentdensity: J(amps/m2)

surface current

density: Js(amps/m)

f

f

=

=

H 2r I

IH

2r

p

j

f

f

p = f = p

=

p= =

p

r 2 2

o o

0 0

o o o

2o o

o

rH 2 r J zrdrd J 2

2

JH r (for r < R ) ( H=J )2

J RH (for r > R ) ( H 0)

2 r

=

=

=

c ss

s

H ds J nda

H 2L J L

JH

2

$ $c sH ds J nda=

Hf

H

H

sJ

oJ

r

oR

Hf

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BOUNDARY CONDITIONS FOR PARALLEL FIELDS

Using Faradays Law: = - g g c Ad

E ds B ndadt

= -

- = d

=

P P

P P

c A

1 2

1 2

d E ds B ndadt

(E E )L 0 (A 0 as 0)

Therefore E E and

- =1 2n (E E ) 0

Using Amperes Law: = + g g gc A Ad

H ds J da D dadt

= +

- = d

P P

g g gc A A

1 2 s a

dH ds J da D da

dt

(H H )L (J n )L (lim 0)

Therefore: - =1 2 sn (H H ) J

(There is no sheet

displacement current.)

n

2E P

1E P

1E

2E

1e

2e

an

d

ds

A

L

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FIELDS INSIDE PERFECT CONDUCTORS

Electric Fields Inside Perfect conductors

Magnetic Fields Inside Perfect Conductors

s

=

m

If and E 0 :

But if J :

But if H :

But w cannot :

Since E 0 inside :

= s

= +

= m 2m m

Then J E

Then H since H J D t

Then W H 2 and w

Therefore

Therefore

E 0 inside

0 inside

=

r =

perfect conductors

since E e = r

And therefore:

Since E 0= and E - B t, = B t 0 =therefore

B 0 inside=perfect conductors (unless constant and there

since the beginning of time)

BOUNDARY COMDITIONS, PERFECT CONDUCTORS

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General Boundary Condition:

1 2 s

1 2

n (D D )n (B B ) 0

- = s - =

1 2n (E E ) 0 - =

1 2 sn (H H ) J - =

2 2 2D B E 0= = inside = :

1 s

1

1

1 s

n D

n B 0

n E 0

n H J

= s

=

= =

H is parallel to perfect conductors

E is perpendicular to perfect conductors

Only surface charges and surface currents

1 1E ,D

2 2E ,D

1 1H ,B

2 2H ,B

n

ss

s

sJ

PLANE WAVE ELECTROMAGNETICS

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0EkE 2o2 =+

vv

oook emw=

In Cartesian coordinates:

0EkE x2

ox

222

=+

+

+

222 zyx

Consider a uniform plane wave which is independent ofx and

y, i.e.,0

EE x2

x

2

=

=

22 yx

we have

0Ek

Ex

2

ox

=+22

dz

d

( ) zjk-ozjk-

oxoo eEeEzE +=\ +

Plane wave in lossless media :

Free-space wavenumber :

PLANE WAVE ELECTROMAGNETICS

Consider the term

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Consider the term

( ) zox eEzE o-jk++ =

in the real situation( ) ( ) ]ezE[tzE tjxx

w++ = Re,( )zktE oo -w=

+ cos

which represents a wave propagating to the +z-direction.

For a particular point, we may calculate the phase velocity

(velocity of propagation of an equip-hase front) as follows:

constantzkt o =-w

ck

z

ooo=emw

w=

w== dt

dup c: velocity of light

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( ) ( ) ]ezE[tzE tjxxw++ = Re,

( )zktE oo -w=+ cos

constantzkt o =-w

2f2 ppw

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note:

o

ooo

2

c

f2

ck

lp

=p

=w

=emw=

k

koo

pl

pl

2

2

=

=\ l0 = free space wavelength

l = wavelength traveling in lossless media

zjkeE 00-v Co-sinusoidal wave traveling in z direction

with same velocity c.

* E- = 0 if concerned only wave traveling in

+z direction only

* In discontinuities medium, reflected wavesin opposite direction must considered.

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For the magnetic field:

H-jE ovv

wm=

( )

( )+++

+

++wm=

zyxo

x

HzHyHx-j

00zEz

00

zyx

( ) ( )

0

0

ojk-

=

=

=

=

+

++

+

+

z

z

o

o

x

o

y

x

H

eEz

j

z

zEjH

H

wmwm

eEk

H zoo

y

++ =wm

ojk-

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( )

( ) ( )zEzH

zE

x

o

y

x

o

o

o

++

+

=\

=

h

em

wm

1

1

( )Wp

e

m=h 377120

o

o

o

oh : intrinsic impedance of the free space

Instantaneous expression for :Hv

( ) ( ) ][Re, tjy ezHytzHw+=

v

( )zktEy oo

o -wh

=+

cos (A/m)

A uniform plane wave is an electromagnetic wave in which the electric and

ti fi ld d th di ti f ti t ll th l d

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magnetic fields and the direction of propagation are mutually orthogonal, and

their amplitudes and phases are constant over planes perpendicular to the

direction of propagation

( ) ( ) ]ezE[tzE tjxxw++ = Re,

( )zktE oo -w=+ cos

Let us examine a possible plane wave solution given

by

note: HEvv

^

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note: HE^

Er

Hv

and are transverse to the propagation direction .z

\ the uniform plane wave is a transverse electromagneticwave or TEM wave.

General form of TEM wave

( )zjkyjkxjk-

ozyxeEzyxE

--=

vv,,

with 2o2z

2y

2x kkkk =++

Define: wavenumber vector

zzyyxxr

nkkzkykxk zyx

++=

=++=v

v

Hv

Ev

n

I i i i d

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( )

( ) ( ) ( )rnjk-

rn-jkrk-j

11 v

vvv

vvvv

vvvv

=-=

==\

eEnrEjrH

eEeErE

o

oo

hwm e

mwm

h == k

Intrinsic impedance

of the medium :

Similarly,

( )

( ) ( )

( ) ( )

( )rHn

rHnjkj

rHj

rE

eHeHrH oo

v

v

vvv

vvvv vvv

-=

-=

=

==

1

1

rn-jkrk-j

h

we

we

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Uniform Plane Wave Solutions

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Strictly speaking, uniform plane waves can be producedonly by sources of infinite extent.

However, point sources create spherical waves.

Locally, a spherical wave looks like a plane wave.

Thus, an understanding of plane waves is very important inthe study of electromagnetics.

7-6 Wave Equations and Their Solutions

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Maxwells equations give a complete description of the relation between

electromagnetic fields and charge and current distributions.

Their solutions provide the answer to all electromagnetic problems, even though in

some cases the solution are difficult to obtain.

Special analytical and numerical techniques may be devised to aid in the solution

procedure; but they do not add to or refine the fundamental structure. Such is the

importance of Maxwells equations.

For a given charge and current distributions, r and J, we first solve thenonhomogeneous wave equations, (7-63) and (7-65), for potentials A and V. With A

and V determined, E and B can be found from Eqs. (7-57) and (7-55), respectively,

by differentiations.

2 22 2

2 2

AA J (7-63), (7-65)

AE (7-57), B A (7-55)

VV

t t

Vt

rme m me

e

- = - - = -

= - - =

Solutions for Wave Equations for Potentials

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At first, the solution ofV for an elemental point charge at time t, r(t)Dv,located at the origin of the coordinates is obtained and then summing the effects of

all the charge element in a given region.Spherical coordinates are convenient for the spherically symmetric system.

The scalar potential V depends only on radial distance Rand time t.

Except at the origin, V satisfies the following homogenous equations:

( )2

2

2 2

10 7-71

V VR

R R R t me

- =

( ) ( ) ( )1

R, R, , 7-72V t U t R

=

( )2 2

2 20, 7-73

U U

R tme

- =

( ) ( ) ( ), , 7-74U R t f t R me= -

( ) ( ) ( )1, / , 7-75V R t f t R uR

= -

introduction of new

variable

( )

Any twice differential function

of is solution of (7-73).t R me

22

20

VV

tme

- =

P.7-20 Proof of Solutions for Homogeneous Equation

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2 2

2 20 (7-73)

U U

R tme

- =

( )( )

( )( ) ( )

( ) ( )( )

( )

2

2' '

' " "

U t R t RU t R U t R

R R R R

t RU t R U t R U t R

R R

me me me me me

meme me me me me me

= = -

= = =

( ) ( ) ( ) ( )2

2' 1 ' "U t R U t R U t R U t R

t t Rme me me me = = =

( )( )

( )( )2

2where ' , "

dU x d U xU x U x

dx dx= =

2 2

2 2" " 0

U UU U

R tme me me

\ - = - =

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Homogeneous Vector Wave Equations

I f i h d J b th d ith i l

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( ) ( )2 2

2 2

2 2 2 2

1 E 1 HE 0. 7-81 H 0. 7-82 with 1/ ,u

u t u t me

- = - = =

B

Et

= -

D r =

DH Jt

= +

B 0 =

HE ,

tm

= -

E 0 =

EH ,t

e =

H 0 =

( )

2

2E Ht tm me

E

= - = -

( ) 2 2E E E E = - = -

22

2

EE 0;

tme

- =

In source free region where rand J are both zero, and with a simple

(linear, isotropic, and homogeneous) nonconducting medium, the wave

equations are given directly to E and H.

In this region with medium, Maxwellequations reduce to

Vector identity:

Time Harmonic Fields

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Maxwell's equations and all the equations derived from them so far in this chapter

hold for electromagnetic quantities with an arbitrary time-dependence. The actual

type of time functions that the Held quantities assume depends on the source

functions and J. In engineering, sinusoidal time functions occupy a unique position.

They are easy to generate; arbitrary periodic time functions can be expanded into

Fourier series of harmonic sinusoidal components; and transient nonperiodic

functions can be expressed as Fourier integrals. Since Maxwell's equations are

linear differential equations, sinusoidal time variations of source functions of a given

frequency will produce sinusoidal variations of E and H with the same frequency in

the steady state. For source functions with an arbitrary time dependence,electrodynamic fields can be determined in terms of those caused by the various

frequency components of the source functions. The application of the principle of

superposition will give us the total fields. In this section we examine time-harmonic

(steady-state sinusoidal) field relationships.

Electromagnetic wave is usually considered

in Frequency Domain

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( )

-

-

--

-

-

+

=

=

yx

yjkxjk

yx

yjkxjk

yx

dkdkezkkzyx

dydxezyxzkk

yx

yx

),,(2

1),,(

),,(),,(

2EE

EE

p

, yx kk

),,( zkkE yx

: , yx wave number

: Angular spectrum of E(x,y,z)

Arbitrary periodic time functions can be expanded into Fourier series of harmonicsinusoidal components; and transient non-periodic functions can be expressed as

Fourier integrals.We have seen that there are solutions of the wave equationwhich correspond to plane wave, and that any electromagnetic wave can bedescribed as a superposition of many plane waves.

q y

Equi-phasePlane

Propagation Direction

Fig. Plane wave with the direction of k

Use of Phasor

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Obtain in a series RLC circuit

with applied voltage

loop equation for a series RLC circuit

Instantaneous expression of loop equation for the circuit

phasors

Phasor expression of the loop equation

(much simpler)

( ) ( )cos ,i t I t w f= +( ) cose t E t w=

( )1 .di L Ri idt e t dt C

+ + =

( ) ( ) ( )1

sin cos sin cos . I L t R t t E t C

w w f w f w f w w

- + + + + + =

( ) ( )0

cos Re

j j t

e t E t Ee e

w

w

= = ( )Re j tsE e w=

( ) ( )Re j j ti t Ie ef w =

( )Re j tsI e w=

( )Re ,j tsdi

j I edt

ww=

Re .j tsIidt e

j

w

w

=

0j

s E Ee E = =

j

s I Ief=

,

1s s R j L I E

cw

w + - =

Convenience of Phasor UsageEXAMPLE 7-6 Express as first (a) and then (b) Determine3cos 4sint tw w- ( )1 1cos ,A tw q+ ( )2 2sin .A tw q+

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Solution We can conveniently use phasors to solve this problem

a) To express as we use as the reference and consider the sum of thetwo phasors 3 and since lags behind

Taking the real part of the product of this phasors and , we have

b) To express as we use as the reference and consider the sum of

the two phasors and -4

(The reader should note that the angle above is 143.1, not -36.9.) Now we take the imaginary part of theproduct of the phasor above and to obtain the desired answer:

( )1 1 ( )2 21 1 2 2, , , and .A Aq q

3cos 4sint tw w- ( )1 1cos ,A tw q+ cos tw( )/ 24 4 ,je jp- = sin tw = ( )cos / 2tw p- cos by / 2 rad:tw p( )1tan 4 /3 53.13 4 5 5

j j j e e

- + = =j te w

( )53.13cos 4sin Re 5 j j tt t e e ww w - = ( )5cos 53.1 .tw= +

( )1 1So, 5, and 53.1 0.927 rad .A q= = =3cos 4sint tw w- ( )1 1cos ,A tw q+ sin tw

( )/ 23 3je jp =( )1tan 3/ 4 143.13 4 5 5

j j j e e

- - - = =

j tew

( )

143.13cos 4sin Im 5 j j tt t e eww w - =

( )5sin 143.1 .tw= +

( )2 2Hance, 5, and 143.1 2.50 rad .A q= = =

( )7 92a-

( )7 92b-

Time Harmonic Maxwells Equations

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Instantaneous expression Phasor expression

BE

t

= -

D r =

DH J

t

= +

B 0 =

E H,jwm = -

E r e = /

H J E,jwe = +

H 0 =

( )7 94a-

( )7 94b-

( )7 94c-

( )7 94d-

( ) ( )E , , , Re , , ,j tx y z t E x y z e w =

Vector of real number

that depends on

location and time.

Vector phasor(vector of complex number)

that contains information on direction, magnitude,

and phase, and is independent of time.

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Procedure for Determining Electromagnetic Fields

Th f l d f d t i i th l t i d ti fi ld t ti

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The formal procedure for determining the electric and magnetic fields sue to time-

harmonic charge and current distributions is as follows:

1. Find phasors V(R) and A(R) from Eqs. (7-99) and (7-100).

2. Find phasors and

3. Find instantaneous for a

cosine reference.

The degree of difficulty of a problem depends on how difficult it is to perform the

integrations in Step 1.

( )E A R V jw= - - ( )B A.R =

( ) ( ) ( ) ( )E , Re and B , Re B j t j t R t R e R t R ew w = E =

Source Free Fields in Simple Media


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Wave equations Homogeneous vector Helmholtzs equations

HE

t

m

= -

E 0 =

EH

te

=

H 0 =

E H,jwm = -

E 0 =

H E,jwe =

H 0 =

( )7 104a-

( )7 104b-

( )7 104c-

( )7 104d-

22

2 2

1 EE 0.

u t

- =

( )7 105-

( )7 106-

kuww me= =

22

2 2

1 HH 0.

u t

- =

2 2E E 0k + =

2 2H H 0.k + =

1/ ,u me=

Principle of Duality

EXAMPLE 7-7 Show that if (E H) are solutions of source-free Maxwells equations in a simple medium

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EXAMPLE 7 7 Show that if (E,H) are solutions of source free Maxwell s equations in a simple medium

characterized by and , then so also are (E', H'), where

In the, above equation, is called the intrinsic impedance of the medium

Solution We prove the statement by taking the curl and the divergence of E and H and using Eqs. (7-104 a,

b, c, and d):

Equations ( 7-108 a, b, c, and d) are source-free Maxwells equations in E and H

E' Hh=E

H' .h

= -

( )7 107a-

( )7 107b-

h m e= /

( ) ( )E ' H Ejh h we = =

2 E H 'j jweh we h

= - - = -

( )2

1H E 'j jwm h we

h= =

( ) ( )1 1

H ' E Hjweh h

= - = - -

( )E ' H 0h = =

( )1H ' E 0h

= - =

( )7 108a-

( )7 108b-

( )7 108c-

( )7 108d-

Duality Theorem

Duality

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Electric Sources (J0, M=0) Magnetic Sources (J=0,M0)

A F FA E H H Ejj wwm e = = -

A F FA H J HE ME jjwe wm- = + = +

A J ' F M '4

4

jBR

V

jBR

V

edv

R

edv

R

em

pp

- -

==

2 2 2 2A A J F F Mb m b e + = - + = -

( ) ( )FA 11E A A H F Fj j j jwwm wmee

w= - - = - -

FA

1

1EA FH

m e = - =

Duality Theorem

D l titi

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Dual quantities

Eelectric Sources (J0, M=0) Magnetic Sources (J=0,M0)

FA E H

FA H E -

A F

J M

e m

m e

b b

1 /h h1/h h

Lossy Media

If the simple medium is conducting ( 0 ), a current J=E will flow, and Eq. (7-104b) should be changed

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p g ( ), , q ( ) g

to

with

complex permittivity

equivalent conductivity

loss tangent, loss angle

complex permeability

complex wavenumber

good conductor if

good insulator if

( )H E Ej j

j

ss we w e

w

= + = +

Ecjwe=

( )F/m .c js

e ew

= -

( )' '' F/m .c je e e= -

( )'' S/m .s we= ''tan .

'c

e sd

e we= @

' ''.jm m m= -

( )' ''c ck jw me w m e e = = -

,s we>>

,we s>>

Lossy Media

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2 2

c 0J E z j zb

xa T E e ea bs s - -= =

123

2 2

c 0J E z j zxa e eJ

a bs - -= =

2 2 2

0

2

0cJ A/m z j z z xa e eJ e J

a b a- - -= =

2 2

c0

0 0

/

0 0Js

z ze dz eJ d J dzJz a d- -

== =

( ) 22/2 2/

000 0

2

z zdzeJ eJd dd dd

-- = - - -

=

[ ]2 20 00 1 A/ms J JJ d d= - - =

0J

/

0 0

z z J e J ea d- -=

d

z

)0,( =+ tzEx

0

A

A-

ze a-

d

AAe 368.01 -

zbp p2 p3 p4mspa

df

11==

Power Dissipation in Lossy Media

Example 7-8 A sinusoidal electric intensity of amplitude 250 (V/m) and frequency 1

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p y p ( ) q y

(GHz) exists in a lossy dielectric medium that has a relative permittivity of 2.5 and a

loss tangent of 0.001. Find the average power dissipated in the medium per cubic

mater.Solution First we must find the effective conductivity of the lossy medium:

The average power dissipated per unit volume is

0

tan 0.001 ,cr

sd

we e= =

( ) ( )9

9 100.001 2 10 2.536

s pp

- =

( )41.39 10 S/m .-=

21 1

2 2 p JE E s= =

( ) ( )4 2 31

1.39 10 250 4.34 W/m .2

-= =

Electromagnetic Spectrum (1)

Wavelength Frequency Application and Photon Energy

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g

(m)

q y

f (Hz)

pp

Classificationgy

hf (eV)

2410

2110

1810

1510

12

10

1510-

1210-

910

-

10

10

-

610

-

310-

210-

(A)o

( )nm

( )mm

( )mm

( )cm

( )EHz

( )PHz

( )THz

Ultraviolet

Infrared

mm wave( )EHF 30-300 Ghz

X-ray

-rayg

910

610

310

310-

1

( )GeV

( )MeV

( )keV

( )eV

( )meV

Electromagnetic Spectrum (2)

Wavelength Frequency Application and Photon Energy

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(m) f (Hz) Classificationgy

hf (eV)

1210

910

610

310

60

1

310-

210-( )mm

( )cm110-

1

102

103104

105106

107108

10

( )m

( )km

( )Mm

( )Hz

( )Hz( )kHz

( )MHz

( )GHz

( )THz mm wave

( )EHF 30-300 GHz( )SHF 3-30 GHz( )UHF 300-3000 MHz

( )VHF 30-300 MHz

( )HF 3-30 MHz( )MF 300-3000 kHz( )LF 30-300 kHz

( )VLF 3-30 kHz( )ULF 300-3000 Hz

( )SLF 30-300 Hz

( )ELF 3-30 Hz

310

-( )meV

Review Questions

R.7-28 What do we mean by a retarded potential?

R 7 29 I h t d th t d ti ti d th l it f ti d d th tit ti

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R.7-29 In what ways do the retardation time and the velocity of wave propagation depend on the constitutive

parameters of the medium?

R.7-30 Write the source-free wave equations for E and H in free space.R.7-31 What is a phasor? Is a phasor a function of t? A function of ?

R.7-32 What is the difference between a phasor and a vector?

R.7-33 Discuss the advantages of using phasors in electromagnetics.

R.7-34Are conduction and displacement currents in phase for time-harmonic fields? Explain.

R.7-35 Write in terms of phasors the time-harmonic Maxwells equations for a simple medium.

R.7-36 Define wavenumber.

R.7-37 Write the expressions for time-harmonic retarded scalar and vector potentials in terms of charge and

current distributions.R.7-38 Write the homogeneous vector Helmholtzs equation for E in a simple, non-conducting, source-free

medium.

R.7-39 Write the expression for the wavenumber of a lossy medium in terms of its permittivity and

permeability.

R.7-40 What is meant by the loss tangent of a medium?

R.7-41 In a time-varying situation how do we define a good conductor?A lossy dielectric?

R.7-42 What is the velocity of propagation of electromagnetic waves? is it the same in air an in vacuum?

Explain.R.7-43 What is the wavelength range of visible light?

R.7-44 Why are frequencies below the VLF range rarely used for wireless transmission?

ProblemsP.7-24 Derive the general wave equations for E and H in a non-conducting simple medium where a charge

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distribution and a current distribution J exist. Convert the wave equations to Helmholtzs equations for

sinusoidal time dependence. Write the general solutions for E(R, t) and H(R, t) in terms of and J.

P.7-25 Given that

in air, find H and .

P.7-26 Given that

in air, find E and .

P.7-27 It is known that the electric field intensity of a spherical wave in free space is

Determine the magnetic field intensity H and the value of k.

P.7-28 In Section 7-4 we indicated that E and B can be determined from the potentials V and A. which are

related by the Lorentz condition, Eq. (7-98), in the time-harmonic case. The vector potential A was introduced

through the relation because of the solenoidal nature of B. In a source-free region. we

can define another type of vector potential Ar, such Assuming harmonic time dependence:

a) Express H in terms of Ar.

b) Show that Ac is a solution of a homogeneous Helmholzs equation.

( ) ( ) ( )9E a 0.1sin 10 cos 6 10 V/my x t zp p b= -

( ) ( ) ( )9H a 2 cos 15 sin 6 10 A/my x t zp p b= -

( )0E a sin cos .E t kRRq q w= -

B A= E 0, =cE A=

ProblemsP7-29 For a source-free polarized medium where =0, J=0, =0, but where there is a volume density ofpolarization P a single vector potential may be defined such that

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polarization P, a single vector potential c may be defined such that

a) Express electric field intensity E in terms of c and P.

b) Show that c satisfies the non-homogeneous Helmholtzs equation

The quantity c

is known as the electric hertz potential.

P.7-30 Calculations concerning the electromagnetic effect of currents in a good conductor usually neglect the

displacement current even at microwave frequencies.

a) Assuming r=1 and =5.70107 (S/m) for copper, cpmpare the magnitude of the displacement current

density with that of the conduction current density at 100(GHz)

b) Write the governing differential equation for magnetic field intensity H in a source-free good conductor.

0H .cjwe p=

2 2

0 0

0

c

Pkp p

e + = -

( )7 118-

( )7 119-

ee341 2 chapter 7

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