Field Theory Gks Compiled

8/6/2019 Field Theory Gks Compiled

1/90

e notes for the lectures on 22-5-2006 and 23-5-2006

(under revision)

e-Notes by Prof.G.K.Suresh, SIT, Tumkur

UNIFORM PLANE WAVES

In free space ( source-less regions where 0J = = = ), the gauss law is

0 0

0 ________ (1)

D E E or

E

= = = =

=

g g gr

g

The wave equation for electric field, in free-space is,

22

2________ (2)

EE

t

=

r

The wave equation (2) is a composition of these equations, one each component wise,

ie,

2 2

2 2

2 2

2 2

2 2

2 2

_______(2)

_______(2)

_______(2)

Ex Eya

x t

Ey Ey by t

Ez Ez c

z t

=

=

=

Further, eqn. (1) may be written as

0 ________ (1) Ex Ey Ez

a x y z

+ + =

For the UPW, E is independent of two coordinate axes; x and y axes, as we have assumed.

0x y

==

Therefore eqn. (1) reduces to

0 ______ (3)zE

z

=

ie., there is no variation of Ez in the z direction.

1


2/90

Also we find from 2 (a) that

2

2

Ez

t

= 0 ____(4)

These two conditions (3) and (4) require that Ez can be

(i) Zero

(ii) Constant in time or

(iii) Increasing uniformly with time.

A field satisfying the last two of the above three conditions cannot be a part of wave motion.

Therefore Ez can be put equal to zero, (the first condition).

The uniform plane wave (traveling in z direction) does not have any field components ofE& H

in its direction of travel.

Therefore the UPWs are transverse., having field components (of E& H) only in directions

perpendicular to the direction of propagation does not have any field component only the

direction of travel.

RELATION BETWEEN E& H in a uniform plane wave.

We have, from our previous discussions that, for a UPW traveling in z direction, both E& H

are independent of x and y; and E& H have no z component. For such a UPW, we have,

( 0) ( 0) _____ (5)

( 0)

( 0) ( 0) _____ (6)

( 0)

y x

z z

y x

z z

i j k

E E E i j

x y z

Ex Ey Ez

i j k

H H H i j

x y z

Hx Hy Hz

= = = = +

=

= = = = +

=

r

r

Then Maxwells curl equations (1) and (2), using (5) and (6), (2) becomes,

2

Ez = 0


3/90

______ (7)

______ (8)

E Ex Ey Hy Hx H i j i j

t t t z z

and

H Hx Hy Ey Ex E i j i j

t t t z z

= = + = +

= = = +

r

rr

Thus, rewriting (7) and (8) we get

______ (7)

______(8)

Hy Hx Ex Eyi j i j

z z t t

Ey Ex Hx Hyi j i j

z z t t

+ = +

+ =

Equating i th and j th terms, we get

3


4/90

( )

( ) ( )

1 0 0

1 0 0 0 1

' '0 01

0

'

1

0

______9( )

______9( )

______9( )

______9( )

1; . ,

. .

.9( ), ,

Hy Exa

z t

Hx Eyb

z t

Ey Hx

i cz t

and

Ex Hyd

z t

Let

Ey f z t Then

EEy

f z t f t

Fromeqn c weget

Hx f f

t

Hx f dz

=

=

=

= =

==

= =

= .c+

( )' 0' '11 1

1z

Now

z tf f f

z z

fH Cz

==

=+

4


5/90

( )' 0' '11 1

11

Now

z tf f f

z z

fdz c f c

z

Hx Ey c

= =

= += +

= +

The constant C indicates that a field independent of Z could be present. Evidently this is not a

part of the wave motion and hence is rejected.

Thus the relation between HX and EY becomes,

__________(10)

x y

y

x

H E

E

H

=

=

Similarly it can be shown that

x

y

EH

=

_____________ (11)

In our UPW, x yE E i E j= +

5


6/90

2

22

2

0

_______ ( )

EE E

t t

E E E xi

t t

But E

E

= +

=

=

r r

r rr

rg

r

DERIVATION OF WAVE EQUATION FOR A CONDUCTING MEDIUM:

6


7/90

In a conducting medium, = 0, = 0. Surface charges and hence surface currents exist,

static fields or charges do not exist.

For the case of conduction media, the point form of maxwells equations are:

________ ( )

_________ ( )

0 _________ ( )

0 _________ ( )

( ),

D E H J E i

t t

B H E iit t

D E E iii

B H H iv

Taking curl onbothsides of equation i we get

E

H E t

= + = +

= =

= = =

= = =

= +

=

r r r

r rr

r r rg g g

r r rg g g

rr r

r

( )

( )

2

2

2

22

2

________ ( )

.( ) .( ),

_________ ( )

_________ ( )

.( )

_________(

E E vt

substituting eqn ii ineqn v we get

H H H vi

t t

But H H H vii

eqn vi becomes

H H H H vii

t t

+

= +

=

=

r

r rr

r r rg

r rr r

g

22

2

)

1 1

0 0

.( ) ,

0 ________ ( )

i

B But H B

eqn viii becomes

H H H ix

t t

= = = =

=

rr r

g g g g

r rr

This is the wave equation for the magnetic field H in a conducting medium.

7


8/90

Next we consider the second Maxwells curl equation (ii)

________ ( )H

E iit

=

r

Taking curl on both sides of equation (ii) we get

( )

( ) 2 ________ ( )

;

HHE x

t t

But E E E

= = =

rr

r r rg

Vector identity and substituting eqn. (1) in eqn (2), we get

( ) 2

2

2

0

_______ ( )

E E E E

t t

E E

xit t

But E

= +

=

=

r r rg

r r

rg

(Point form of Gauss law) However, in a conductor, = 0, since there is no net charge within a

conductor,

Therefore we get 0E=gTherefore eqn. (xi) becomes,

22

2

E EE

t t

=

r____________ (xii)

This is the wave equation for electric field E in a conducting medium.

Wave equations for a conducting medium:

8


9/90

Regions where conductivity is non-zero.

Conduction currents may exist.

For such regions, for time varying fields

The Maxwells eqn. Are:

_________ (1)

__________ (2)

: ( / )

EH J

t

HE

t

J E Conductivity m

= +

=

=

r r

rr

r r

= conduction current density.

Therefore eqn. (1) becomes,

_________ (3)EH Et

= +

r r

Taking curl of both sides of eqn. (2), we get

( )

( )

( )

2

2

2

22

2

________(4)

( )

sin .(4) ,

_______(5)

1tan ,

E Ht

E E

t t

But

E E E vectoridentity

u gthiseqn becomesvectoridentity

E EE E

t t

But D

iscons t E D

= =

=

=

=

=

r r

r r

r r r

g

r rr r

g

rg

r rQ g g

Since there is no net charge within a conductor the charge density is zero ( there can be charge

on the surface ), we get.

10E D==

r rg g

Therefore using this result in eqn. (5)

we get

9


10/90

22

20 ________(6)

E EE

t t

=

r

This is the wave eqn. For the electric field E in a conducting medium.

This is the wave eqn. forE. The wave eqn. forH is obtained in a similar manner.

Taking curl of both sides of (1), we get

2

2

________(7)

________ (2)

(1) ,

________(8)

EH E

t

H But E

t

becomes

H HH

t t

= +

=

=

r r

rr

r rr

As before, we make use of the vector identity.

( ) 2 H H H = gin eqn. (8) and get

( )2

2

2

22

2

________ (9)

1 10 0

.(9)

________ (10)

H HH H

t t

But

BH B

eqn becomes

H HH

t t

=

= = = =

=

r rg

rr rg g g g

r rr

g

This is the wave eqn. forH in a conducting medium.

Sinusoidal Time Variations:

In practice, most generators produce voltage and currents and hence electric and magnetic fields

which vary sinusoidally with time. Further, any periodic variation can be represented as a weight

sum of fundamental and harmonic frequencies.

Therefore we consider fields having sinusoidal time variations, for example,

E = Em cos t

E = Em sin t

Here, w = 2 f, f = frequency of the variation.10


11/90

Therefore every field or field component varies sinusoidally, mathematically by an additional

term. Representing sinusoidal variation. For example, the electric field E can be represented as

( )( ) ( )

, , ,

., , ; , ,

E x y z t as

ie E r t r x y z r r r%

Where E is the time varying field.The time varying electric field can be equivalently represented, in terms of corresponding phasor

quantity Er

(r) as

( ) ( ), ________ (11)j teE r t R E r e =

r%

The symbol tilda placed above the E vector represents that E is time varying quantity.

The phasor notation:

We consider only one component at a time, say Ex.

The phasor Ex is defined by

( ) { ( ) }, ________ (12) j t x e xE r t R E r e =r%

| Ex |

| Ex |

( )xE r denotes Ex as a function of space (x,y,z). In general ( )xE r is complex and hence can be

represented as a point in a complex and hence can be represented as a point in a complex plane.

(see fig) Multiplication by jwte results in a rotation through an angle wt measured from the angle

. At t increases, the point Ex jwte traces out a circle with center at the origin. Its projection on

the real axis varies sinusoidally with time & we get the time-harmonically varying electric field

Ex%(varying sinusoidally with time). We note that the phase of the sinusoid is determined by ,

the argument of the complex number Ex.

Therefore the time varying quantity may be expressed as

{ } ________ (13)

cos( ) ________ (14)

j j t

x e x

x

E R E e e

E t

=

= +

%

11

t


12/90

Maxwells eqn. in phasor notation:

In time harmonic form, the Maxwells first curl eqn. is:

_______ (15)DH Jt

= +

r%r r% %

using phasor notation, this eqn. becomes,

( ) ________ (16) j t j t j t e e e R He R De R Jet

= +

r r r

The diff. Operator & Re part operator may be interchanged to get,

( ) ( )

( ) 0

j t j t j t

e e e

j t j t

e e

j t

e

R He R De R Jet

R j D e R Je

R H j D J e

= + = +

=

r r r

r r

r r r

This relation is valid for all t. Thus we get

________(17)H J j D =+

This phasor form can be obtained from time-varying form by replacing each time derivative by

., jw ie is to be replaced byt

For the sinusoidal time variations, the Maxwells equation may be expressed in phasor form as:

( )(17)

(18)

(19)

(20) 0 0

LS

LS

V VSV

S

H J j D H dL J j D ds

E j B E dl j B ds

D D ds d

B B ds

= + = +

= =

= = = =

rg g

rr r r r rg g

r r rg g

r r rg g

The continuity eqn., contained within these is,

_______ (21)S

vol

J j J ds j dv = = r

g g

The constitutive eqn. retain their forms:

12


13/90

D E

B H

J E

=

=

=

r r

r r____ (22)

For sinusoidal time variations, the wave equations become

{ }

{ }

2 2

2 2

( )

( )

E E for electric field

H H for electric field

=

= r r_________ (23)

Vector Helmholtz eqn.

In a conducting medium, these become

( )

( )

2 2

2 2

0

0

E j E

H j H

+ =

+ =r r ________ (24)

Wave propagation in a loss less medium:

In phasor form, the wave eqn. for VPW is

222

22

2

2

1 2

; _______ (25)

_______ (26)

y

y

j x j x

y

EEE

Exx

E

E C e C e

= = = = +

r

r

C1 & C2 are arbitrary constants.

The corresponding time varying field is

( ) ( )

( ) ( )

( ) ( )

1 2

1 2

,

______ (27)

cos cos ______ (28)

j t

y e y

t z t z j j

e

E x t R E x e

R C e C e

C t z C t z

+

= = +

= + +

%

When C1 and C2 are real.

Therefore we note that, in a homogeneous, lossless medium, the assumption of sinusoidal time

variations results in a space variation which is also sinusoidal.

Eqn. (27) and (28) represent sum of two waves traveling in opposite directions.

If C1 = C2 , the two traveling waves combine to form a simple standing wave which does not

progress.

If we rewrite eqn. (28) with Ey as a fn of (x- t),

we get =

Let us identify some point in the waveform and observe its velocity; this point is ( )t x a =

constant

13


14/90

Then

' 'a t

dx x

dt t

= = = =

Q

This velocity is called phase velocity, the velocity of a phase point in the wave.

is called the phase shift constant of the wave.

Wavelength: These distance over which the sinusoidal waveform passes through a full cycle of

2 radians

ie.,

14


15/90

0

2

2 2

2

;

1:

Z

or

But

f

or

f f inH

=

= =

===

=

==

Wave propagation in a conducting medium

We have,

Where

( )

2 2

2 2

0E E

j

j j

=

= += +

is called the propagation constant is, in general, complex.

Therefore, = + j

= Attenuation constant

= phase shift constant.

The eqn. for UPW of electric field strength is

22

2

EE

x

=

r

One possible solution is

( ) 0xE x E e =

Therefore in time varying form, we get

( )

0

, x j te

x jwt

e

E x t R E e e

e R E e

= =

r%

This eqn. shown that a up wave traveling in the +x direction and attenuated by a factor xe .

The phase shift factor

15


16/90

2

and velocity f

=

= =

= Real part of = RP ( ) j j t +

=

2

2 2

2

2 2

1 12

1 12

+

= + +

Conductors and dielectrics:

We have the phasor form of the 1st Maxwells curl eqn.

c dispH E j E J J = + = +

where cJ E= = conduction current density ( A/m2 )

disp J j E = = displacement current density ( A/m2 )

cond

disp

J

J

=

We can choose a demarcation between dielectrics and conductors;

1

=

*1

> is conductor. Cu: 3.5*108

@ 30 GHz

* 1

t2 ) we get another function of z namely ( )1 0 2 f z t . This is

nothing but time shifted version of ( )1 0 1 f z t , shifted along + z axis by a distance z = ( )0 2 1t t .

This means that the function f1 ( )0x t has traveled along + z axis with a velocity 0 . This iscalled a traveling wave.

On the other hand ( )2 0 f z t + represents a wave traveling along z axis with a velocity 0 and

is called a reflected wave, as we shall further seen in the next semester, in the topic transmission

line.

This shows that the wave equation has two solutions ( as expected, since the wave eqn. is a

second order PDE ) a traveling wave ( or forward wave ) along + z direction represented by

f1 ( )0z t and the other a reverse traveling wave ( reflected wave ) along z axis. If there is no

reflecting surface, the second term of eqn. (8) is zero, resulting is

E = f1 ( )0z t _________(9)

Remember that eqn. (9) is a solution of the wave equation and is only for the particular case

where the electric field E is independent of x and y directions; and is a function of z and t only.

29


30/90

Such a wave is called also the equation does not indicate the specific shape of the wave

(amplitude variation) and hence is applicable to any arbitrary waveform.

UNIFORM PLANE WAVES:

In free space ( source-less regions where 0J = = = ), the gauss law is

0 0

0 ________ (1)

D E E or

D

= = = =

=

g g gr

g

The wave equation for electric field, in free-space is,

22

2________ (2)

EE

t

=

r

30


31/90

The wave equation (2) is a composition of these equations, one each component wise,

ie,

2 2

2 2

2 2

2 2

2 2

2 2

_______(2)

_______(2)

_______(2)

Ex Eya

x t

Ey Eyb

y t

Ez Ez c

z t

=

=

=

Further, eqn. (1) may be written as

0 ________ (1) Ex Ey Ez

a x y z

+ + =

For the UPW, E is independent of two coordinate axes; x and y axes, as we have assumed.

0x y

==

Therefore eqn. (1) reduces to

0 ______ (3)zE

z

=

ie., there is no variation of Ez in the z direction.

Also we find from 2 (a) that2

2

Ez

t

= 0 ____(4)

These two conditions (3) and (4) require that Ez can be

(iv) Zero

(v) Constant in time or

(vi) Increasing uniformly with time.

A field satisfying the last two of the above three conditions cannot be a part of wave motion.

Therefore Ez can be put equal to zero, (the first condition).

The uniform plane wave (traveling in z direction) does not have any field components ofE& H

in its direction of travel.

Therefore the UPWs are transverse., having field components (of E& H) only in directions

perpendicular to the direction of propagation does not have any field component only the

direction of travel.

31

Ez = 0


32/90

RELATION BETWEEN E& H in a uniform plane wave.

We have, from our previous discussions that, for a UPW traveling in z direction, both E& H

are independent of x and y; and E& H have no z component. For such a UPW, we have,

( 0) ( 0) _____ (5)

( 0)

( 0) ( 0) _____ (6)

( 0)

y x

z z

y x

z z

i j k

E E E i j

x y z

Ex Ey Ez

i j k

H H H i j

x y z

Hx Hy Hz

= = = = +

=

= = = = +

=

r

r

Then Maxwells curl equations (1) and (2), using (5) and (6), (2) becomes,

______ (7)

______ (8)

E Ex Ey Hy Hx H i j i j

t t t z z

and

H Hx Hy Ey Ex E i j i j

t t t z z

= = + = +

= = = +

r

rr

Thus, rewriting (7) and (8) we get

______ (7)

______(8)

Hy Hx Ex Eyi j i j z z t t

Ey Ex Hx Hyi j i j

z z t t

+ = + + =

Equating i th and j th terms, we get

32


33/90

( )

( ) ( )

1 0 0

1 0 0 0 1

' '0 01

0

'

1

0

______9( )

______9( )

______9( )

______9( )

1; . ,

. .

.9( ), ,

Hy Exa

z t

Hx Eyb

z t

Ey Hx

i cz t

and

Ex Hyd

z t

Let

Ey f z t Then

EEy

f z t f t

Fromeqn c weget

Hx f f

t

Hx f dz

=

=

=

= =

==

= =

= .c+

( )' 0' '11 1

1z

Now

z tf f f

z z

fHz

==

=

33


34/90

( )' 0' '11 1

11

Now

z tf f f

z z

fdz c f c

z

Hx Ey c

= =

= += +

= +

The constant C indicates that a field independent of Z could be present. Evidently this is not a

part of the wave motion and hence is reflected.

Thus the relation between HX and EY becomes,

__________(10)

x y

y

x

H E

E

H

=

=

Similarly it can be shown that

x

y

EH

=

_____________ (11)

In our UPW, x yE E i E j= +

34


35/90

2

22

2

0

_______ ( )

EE E

t t

E E E xi

t t

But E

E

= +

=

=

r r

r rr

rg

r

DERIVATION OF WAVE EQUATION FOR A CONDUCTING MEDIUM:

35


36/90

In a conducting medium, = 0, = 0. Surface charges and hence surface currents exist,

static fields or charges do not exist.

For the case of conduction media, the point form of maxwells equations are:

________ ( )

_________ ( )

0 _________ ( )

0 _________ ( )

( ),

D E H J E i

t t

B H E iit t

D E E iii

B H H iv

Taking curl onbothsides of equation i we get

E

H E t

= + = +

= =

= = =

= = =

= +

=

r r r

r rr

r r rg g g

r r rg g g

rr r

r

( )

( )

2

2

2

22

2

________ ( )

.( ) .( ),

_________ ( )

_________ ( )

.( )

_________(

E E vt

substituting eqn ii ineqn v we get

H H H vi

t t

But H H H vii

eqn vi becomes

H H H H vii

t t

+

= +

=

=

r

r rr

r r rg

r rr r

g

22

2

)

1 1

0 0

.( ) ,

0 ________ ( )

i

B But H B

eqn viii becomes

H H H ix

t t

= = = =

=

rr r

g g g g

r rr

This is the wave equation for the magnetic field H in a conducting medium.

36


37/90

Next we consider the second Maxwells curl equation (ii)

________ ( )H

E iit

=

r

Taking curl on both sides of equation (ii) we get

( )

( ) 2 ________ ( )

;

HHE x

t t

But E E E

= = =

rr

r r rg

Vector identity and substituting eqn. (1) in eqn (2), we get

( ) 2

2

2

0

_______ ( )

E E E E

t t

E E

xit t

But E

= +

=

=

r r rg

r r

rg

(Point form of Gauss law) However, in a conductor, = 0, since there is no net charge within a

conductor,

Therefore we get 0E=gTherefore eqn. (xi) becomes,

22

2

E EE

t t

=

r____________ (xii)

This is the wave equation for electric field E in a conducting medium.

Wave equations for a conducting medium:

37


38/90

Regions where conductivity is non-zero.

Conduction currents may exist.

For such regions, for time varying fields

The Maxwells eqn. Are:

_________ (1)

__________ (2)

: ( / )

EH J

t

HE

t

J E Conductivity m

= +

=

=

r r

rr

r r

= conduction current density.

Therefore eqn. (1) becomes,

_________ (3)EH Et

= +

r r

Taking curl of both sides of eqn. (2), we get

( )

( )

( )

2

2

2

22

2

________(4)

( )

sin .(4) ,

_______(5)

1tan ,

E Ht

E E

t t

But

E E E vectoridentity

u gthiseqn becomesvectoridentity

E EE E

t t

But D

iscons t E D

= =

=

=

=

=

r r

r r

r r r

g

r rr r

g

rg

r rQ g g

Since there is no net charge within a conductor the charge density is zero ( there can be charge

on the surface ), we get.

10E D==

r rg g

Therefore using this result in eqn. (5)

we get

38


39/90

22

20 ________(6)

E EE

t t

=

r

This is the wave eqn. For the electric field E in a conducting medium.

This is the wave eqn. forE. The wave eqn. forH is obtained in a similar manner.

Taking curl of both sides of (1), we get

2

2

________(7)

________ (2)

(1) ,

________(8)

EH E

t

H But E

t

becomes

H HH

t t

= +

=

=

r r

rr

r rr

As before, we make use of the vector identity.

( ) 2 H H H = gin eqn. (8) and get

( )2

2

2

22

2

________ (9)

1 10 0

.(9)

________ (10)

H HH H

t t

But

BH B

eqn becomes

H HH

t t

=

= = = =

=

r rg

rr rg g g g

r rr

g

This is the wave eqn. forH in a conducting medium.

Sinusoidal Time Variations:

In practice, most generators produce voltage and currents and hence electric and magnetic fields

which vary sinusoidally with time. Further, any periodic variation can be represented as a weight

sum of fundamental and harmonic frequencies.

Therefore we consider fields having sinusoidal time variations, for example,

E = Em cos t

E = Em sin t

Here, w = 2 f, f = frequency of the variation.39


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Therefore every field or field component varies sinusoidally, mathematically by an additional

term. Representing sinusoidal variation. For example, the electric field E can be represented as

( )( ) ( )

, , ,

., , ; , ,

E x y z t as

ie E r t r x y z r r r%

Where E is the time varying field.The time varying electric field can be equivalently represented, in terms of corresponding phasor

quantity Er

(r) as

( ) ( ), ________ (11)j teE r t R E r e =

r%

The symbol tilda placed above the E vector represents that E is time varying quantity.

The phasor notation:

We consider only one component at a time, say Ex.

The phasor Ex is defined by

( ) { ( ) }, ________ (12) j t x e xE r t R E r e =r%

( )xE r denotes Ex as a function of space (x,y,z). In general ( )xE r is complex and hence can be

represented as a point in a complex and hence can be represented as a point in a complex plane.

(see fig) Multiplication by jwt

eresults in a rotation through an angle wt measured from the angle

. At t increases, the point Ex jwte traces out a circle with center at the origin. Its projection on

the real axis varies sinusoidally with time & we get the time-harmonically varying electric field

Ex%(varying sinusoidally with time). We note that the phase of the sinusoid is determined by ,

the argument of the complex number Ex.

Therefore the time varying quantity may be expressed as

{ } ________ (13)

cos( ) ________ (14)

j j t

x e x

x

E R E e e

E t

=

= +

%

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Maxwells eqn. in phasor notation:

In time harmonic form, the Maxwells first curl eqn. is:

_______ (15)DH Jt

= +

r%r r% %

using phasor notation, this eqn. becomes,

( ) ________ (16) j t j t j t e e e R He R De R Jet

= +

r r r

The diff. Operator & Re part operator may be interchanged to get,

( ) ( )

( ) 0

j t j t j t

e e e

j t j t

e e

j t

e

R He R De R Jet

R j D e R Je

R H j D J e

= + = +

=

r r r

r r

r r r

This relation is valid for all t. Thus we get

________(17)H J j D =+

This phasor form can be obtained from time-varying form by replacing each time derivative by

., jw ie is to be replaced byt

For the sinusoidal time variations, the Maxwells equation may be expressed in phasor form as:

( )(17)

(18)

(19)

(20) 0 0

LS

LS

V VSV

S

H J j D H dL J j D ds

E j B E dl j B ds

D D ds d

B B ds

= + = +

= =

= = = =

rg g

rr r r r rg g

r r rg g

r r rg g

The continuity eqn., contained within these is,

_______ (21)S

vol

J j J ds j dv = = r

g g

The constitutive eqn. retain their forms:

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43/90

Then

' 'a t

dx x

dt t

= = = =

Q

This velocity is called phase velocity, the velocity of a phase point in the wave.

is called the phase shift constant of the wave.

Wavelength: These distance over which the sinusoidal waveform passes through a full cycle of

2 radians

ie.,

0

2

2 2

2

;

1:

Z

or

But

f

or

f f inH

=

= =

===

=

==

Wave propagation in a conducting medium

We have,

43


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Where

( )

2 2

2 2

0E E

j

j j

=

= += +

is called the propagation constant is, in general, complex.

Therefore, = + j

= Attenuation constant

= phase shift constant.

The eqn. for UPW of electric field strength is

22

2

EE

x

=

r

One possible solution is

( ) 0xE x E e =

Therefore in time varying form, we get

( )

0

,x j t

e

x jwt

e

E x t R E e e

e R E e

= =

r%

This eqn. shown that a up wave traveling in the +x direction and attenuated by a factor xe .

The phase shift factor

2

and velocity f

=

= =

= Real part of = RP ( ) j j t +

=

2

2 2

2

2 2

1 12

1 12

+

= + +

Conductors and dielectrics:

44


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We have the phasor form of the 1st Maxwells curl eqn.

c dispH E j E J J = + = +

where cJ E= = conduction current density ( A/m2 )

disp J j E = = displacement current density ( A/m2 )

cond

disp

JJ

=

We can choose a demarcation between dielectrics and conductors;

1

=

* 1

>

is conductor. Cu: 3.5*108 @ 30 GHz

* 1

, then, both the reflection coefficients given by equations,

1

2

1

21

1

2

1

21

sincos

sincos

+

=

i

r

E

E( perpendicular polarization )

and

1

2

1

2112

1

2

1

2112

sincos/

sincos/

+

=i

r

E

E( parallel polarization )

become complex numbers when,1

21sin

>

Both coefficients take the form

+

jba

jbaand thus have a unit magnitude. In other words, the

reflection is total provided that 1 is great enough and also provided that medium (1) is denser

than medium. (2) but total reflection does not imply that there is no field in medium (2). In

medium (2), the fields have the form, ( )222 cossin Zyje

Snells law gives the y variation as, 212 / yje

And the Z variation as,

( )

1sin

1sin

sin1cos

12

2

12

12

2

12

22

222

+

=

=

=

Z

jZj

ZjZj

e

e

ee

In the above expression, the lower sign must be chosen such that the fields decrease

exponentially as Z becomes increasingly negative.

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ie.,2

21

2

2

11

2

2

12 sin1sincos

=

= jj

Therefore under the condition of TIR, a field does exist in the rarer medium. However, this field

has a phase progression along the boundary and decreases exponentially away from it. If is thus

the example of a non-uniform plane wave.

The phase velocity along the interface is given by ,

1

2

1

2

sin

Which, under the conditions of TIR is less than the phase velocity2

of a UPW in medium (2).

Consequently, the non-uniform plane wave in medium (2) is a slow wave. Also, since some kind

of a surface between two media is necessary to support this wave, it is called a surface wave.

.

Maxwells Equations

In static electric and magnetic fields, the Maxwells equations obtained so far are:

Differential form Controlling principle Integral form

0E = Potential around a closed path is zero 0LdE = (1)

D = Gausss Law dvSdD (2)

JH = Amperes Circuital Law SdJLdH (3)B Non-existence of isolated magnetic poles =0SdB (4)

Contained in the above equations is the equation of continuity for steady currents,

= 0dSJ0J (5)Modification of Maxwells equations for the case of time varying fields

First Modification of the first Maxwells equation [ ]0LdE;0E = ; ----(1)

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To discuss magnetic induction and energy, it is necessary to include time-varying fields, but

only to the extent of introducing the Faradays law.

Faradays law states that the voltage around a closed path can be generated by

a time changing magnetic flux through a fixed path ( transformer action) or

by a time-varying path in a steady magnetic field (electric generator action)

Faradays law: The electromotive force around a closed path is equal to the negative ofthe time rate of change of magnetic flux enclosed by the path.

S

SdBtt

LdE

In our study of electromagnetics, interest centers on the relation between the time

changing electric and magnetic fields and a fixed path of integration.

For this case the Faradays law reduces to,

S

SdBt

LdE

The partial derivative w.r.t time indicates that only variations of magnetic flux through a fixed

closed path or a fixed region in space are being considered.

Thus, for time varying fields, equation (1) gets modified to

t

BE

Sdt

BLdE

S

(6)

Second Modification: Modification of the Continuity equation for time varying fields:

Current is charges in motion. The total current flowing out of some volume must be equal to the rate of decrease of charge within the volume[charge cannot be created or destroyed- law of conservation of charges]. This concept is needed in order for understanding why current flows

between the capacitor plates. The explanation is that the current flow is accompanied by charge build up on the plates. In the form of equation,

the law of conservation of charge is

= dvdtd-SdJIf the region of integration is stationary, this relation becomes,

= dvt-SdJ ----(7)Applying divergence theorem to this equation, we get,

= dvt)dvJ( If this relation is to hold for any arbitrary volume, then, it must be true that,

t

J

---- ( 8)

This is the time-varying form of the equation of continuity that replaces equation (5).

Third Modification: Modification of the Maxwells equation for the Amperes Law:

81


82/90

Taking the divergence of equation (3) we getthe equation of continuity as,

0JH = (Divergence of curl is zero- vector identity).Thus Amperes law is inconsistent with the time varying fields for which the equation of

continuity ist

J

. To resolve this inconsistency, James Clerk Maxwell in the mid1860s suggested modification of the Amperes law to include the validity for time varying

fields also; He suggested substitution of Gausss law (2) into the equation of continuity (8)giving,

t

J

.But we know that D = .

Thereforewe get ,t

D

t

)D(J

, oninterchanging

the time and space differentiation.

Therefore 0)t

DJ(0

t

DJ =

+ or ----- (9)

This equation may be put into integral form by integrating over a volume and then applyingthe divergence theorem:

=+ 0Sd)tDJ( ------ (10)Equations 9 and 10 suggest that )

t

DJ(

+ may be regarded as a total current density fortime varying fields. Since D is the displacement density,

t

D

isknown as the

displacement current density.

Consider now a capacitor connected to an ac source as shown in figure.

I

When applied as shown in figure to a surface enclosing one plate of a two-plate capacitor,

equation ( 10 ) shows that during charge or discharge, the conduction current in the wire

attached to the plates is equal to the displacement current passing between the plates.

Maxwell reasoned that the total current density should replace J in Amperes law with the

result that

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t

DJH

+ ----- (11)Taking the divergence of this equation gives equation (9) and thus the inconsistency has

been removed. Note that the Equation (11) has not been derived from the preceding

equations but rather suggested by them. Therefore when Maxwell proposed it, it was a

postulate whose validity had to be established by experiment.

Integrating equation (11) over a surface and application of Stokes theorem gives theintegral form of the equation:

Sd)t

DJ(LdH

+ ---- (12)

This equation states that the mmf around a closed path is equal to the total current

enclosed by the path. Thus equations 11 and 12 replace the static form of Amperes law

(3).

Maxwells equations:

In summary, the Maxwells equations are as follows:

Differential form Controlling principle Integral form

JDH + Amperes Circuital Law SdJDLdH

(I)

BE Potential around a closed path is zero SdBLdE

(II)D = Gausss Law dvSdD

(III)

B Non-existence of isolated magnetic poles =0SdB (IV)Contained in the above equations is the equation of continuity,

t

J

= dvt-SdJ In all the cases the region of integration is assumed to be stationary.

WORD STATEMENT FORM OF FIELD EQUATIONS:

The word statements of the field equations may readily be obtained from the integral form of

the Maxwells equations:

I. The mmf around a closed path is equal to the conduction current

plus the time derivative of the electric displacement through any surface

bounded by the path.

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II. The emf around a closed path is equal to the time derivative of the

magnetic displacement through any surface bounded by the path.

III. The total electric displacement through the surface enclosing a

volume is equal to the total charge within the volume.

IV. The net magnetic flux emerging through any closed surface is zero.

Alternate way of stating the first two equations:

1. The magnetic voltage around a closed path is equal to the electric current

through the path.

2. The electric voltage around a closed path is equal to the magnetic current

through the path

Boundary Conditions using Maxwells equations:

The integral form of Maxwells equations can be used to determine what happens at theboundary surface between two different media.( Find out why not the differential form?)

The boundary conditions for the electric and magnetic fields at any surface of discontinuityare:

1. The tangential component of E is continuous at the surface. i.e., it is the samejustoutside

thesurface as it is at the inside the surface.

2. The tangential component of H is discontinuous across the surface except at thesurface of a perfect conductor. At the surface of a perfect conductor, the tangential

component of H is discontinuous by an amount equal to the surface current per unit

width.

3. The normal component of B is continuous at the surface of discontinuity.

4. The normal component of D is continuous if there is no surface charge density.

Otherwise D is discontinuous by an amount equal to the surface charge density.

y

Proof: XEx1 Ex2

1, 1 , 1 2, 2, 2( medium 1) ( medium 2)

EY1 EY2

84

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85/90

Ex3 Ex4

X

Fig 2 A boundary between two media

Let the surface of discontinuity be the plane x=0 as shown in fig 2.

Conditions on the tangential components ofE and H

1. Condition for Etan at the surface of the boundary:

Consider a small rectangle of width x and length y enclosing a small portion of each media(1) and (2).

The integral form of the second Maxwell equation ( II ) is,

S

SdBLdE

For the elemental rectangle of fig 2, we apply this equation and get

yxB2

xE

2

xEyE

2

xE

2

xEyE zx4x3y1x1x2y2

----(13)

where Bz is the average magnetic flux density through the rectangle yx . Now, as this area

of the rectangle is made to approach to zero, always keeping the surface of discontinuity between

the sides of the triangle. If Bz is finite, then as x o, the RHS of equation 13 will approachzero. If E is also assumed to be finite everywhere, then, x/2 terms of the LHS of equation 13will reduce to zero, leaving

Ey2 y - Ey1 y = 0for x = 0. Therefore

Ey2 = Ey1That is, the tangential component of E is continuous.

2. Condition for Htan at the surface of the boundary:

Now the integral form of the first Maxwells equation ( I ) is

Sd)JD(LdHS

For the elemental rectangle this equation becomes,

yx)JD(2

xH

2

xHyH

2

xH

2

xHyH zzx4x3y1x1x2y2 + ----(14)

If the rate of change of electric displacement D and the current density J are both considered to

be finite, then, as before, equation (14) reduces to

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Hy2 y - Hy1 y = 0for x = 0. Therefore

Hy2 = Hy1

That is, the tangential component of H is continuous (for finite current densities i.e., for any

actual case).

In case of a perfect conductor: A perfect conductor has infinite conductivity. In such aconductor, the electric field strength E is zero for any finite current density. However, the actual

conductivity may be very large and for many practical applications, it is useful to assume it to be

infinite. Such an approximation will lead to difficulties because of indeterminacy in formulatingthe boundary conditions unless care is taken in setting them up. The depth of penetration of an ac

field into a conductor decreases as the conductivity increases. Thus in a good conductor a hf

current will flow in a thin sheet near the surface, the depth of the sheet approaching zero as theconductivity approaches infinity. This gives to the useful concept of a current sheet. In a current

sheet a finite current per unit width, Jz amperes per meter flows in a sheet of vanishingly small

depth x, but with the required infinitely large current density J, such thatlim x 0, J x = Jz

Now consider again the above example the mmf around the small rectangle. If the current

density Jzbecomes infinite as x0, the RHS of equation 14 will not become zero. Let Jzamperes per metre be the actual current per unit width flowing along the surface. Then as

x0, equation 14 for H becomes,

Hy2 y - Hy1 y = Jsz y.

Hence

Hy1 = Hy2 - Jsz ---- (15)

Now if the electric field is zero inside a perfect conductor, the magnetic field must also be zero,for alternating fields, as the second Maxwells equation shows. Then, in equation 15, Hy2 must be

zero.

So,

Hy1 = - Jsz ----(16)

This equation states that the current per unit width along the surface of a perfect conductor is

equal to the magnetic field strength H just outside the conductor. The magnetic field and the

surface current will be parallel to the surface, but perpendicular to each other. In vector notation,this is written as,

HnJ where n is the unit vector along the outward normal to the surface.

Conditions on the normal component ofB and D

3.Condition on the normal component of D (Dnor):

The integral form of the third Maxwells equation is

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S vol

dVSdD ----( III )

Consider the pill-box structure shown in fig 3. Applying the third Maxwells equation to this pill-

box structure, we getdSxdSDdSD

edgen2n1 ----(17)

DN1dS

1 1 1 X

2 2 2

DN2

Fig 3 A pill-box volume enclosing a portion of a boundary surface

In this expression, dS is the area of each of the flat surfaces of the pillbox, x is their separation,and is the average charge density within the volume x dS. edge is the outward electric fluxthrough the curved edge surface of the pillbox. As x 0, that is, as the flat surfaces of thepillbox are squeezed together, always keeping the boundary surface between them, edge0, forfinite values of displacement density. Also, for finite values of average density , the RHS ofequation (17) reduces to

0dSDdSD n2n1=

for x = 0.Then for the case of no surface charge condition on the normal components ofD

Dn1 = Dn2 ---- (18)

That is, if there is no surface charge, the normal component of D is continuous across the

surface.

In the case of a metallic surface: In the case of a metallic surface, the charge is considered to

reside on the surface. If this layer of surface charge has a surface charge density S Coulombsper square meter, the charge density of the surface layer is given by

3S C/mx

where x is the thickness of the surface layer. As x approaches zero, the charge densityapproaches infinity in such a manner that

S0x xlim =Then in fig 3, if the surface charge is always kept between the two flat surfaces as the seperation

between them is decreased, the RHS of equation (17) approaches S dS as x approaches zero.Equation 17 then reduces to

Sn2n1 DD = ----( 19 )87


88/90

When there is a surface charge density S, the normal component of displacement densityis discontinuous across the surface by the amount of the surface charge density.

For any metallic conductor the displacement density, D = E within the conductor will be asmall quantity( it will be zero in the electrostatic case, or in the case of a perfect conductor).

Then if the medium (2) is a metallic conductor, Dn2 = 0 ; and the equation (19) becomes

Dn1 = S ----(20)The normal component of the displacement density in the dielectric is equal to the surface

charge density on the conductor.

4 Condition on the normal component of B (Bnor):

The integral form of the fourth Maxwells equation is

=S

0SdB ----( IV )

The pill-box structure is again shown in fig 4 for magnetic flux density. Applying the fourth

Maxwells equation to this pill-box structure, we get

0dSBdSB edgen2n1 = ----(21)BN1

dS

1 1 1 X

2 2 2

BN2

Fig 4 A pill-box volume enclosing a portion of a boundary surface

In this expression, dS is the area of each of the flat surfaces of the pillbox, x is their separation,and is the average charge density within the volume x dS. edge is the outward electric fluxthrough the curved edge surface of the pillbox. As x 0, that is, as the flat surfaces of thepillbox are squeezed together, always keeping the boundary surface between them, edge0, forfinite values of magnetic flux density. The RHS of equation (21) reduces to

0dSBdSB n2n1 =for x = 0.Then the condition on the normal components of B since there are no isolatedmagnetic charges,

Bn1 = Bn2 ---- (22)

i.e., The normal component of magnetic flux density is always continuous across the

boundary.

http://www.ece.mcmaster.ca/faculty/nikolova/EM_downloads/LectureNotes/

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