# Document 15607096 ```The Maxwell Equations
Farady’s law and the Electromagnetic “force” (EMF).
Faraday's law relates the electromotive force, , with the change of the magnetic flux with time,
 
1 d
c dt
The electromagnetic force around the circuit through which the flux is passed is
 


E

d
 



ˆ da


E
n

C
S
We note that






 1 B
1 d
1d
1


B

d
x

d


B

d


v
 
 nˆda
c dt
c dt S
c S
c S t




Hence when the loop is not moving, we have

  1 B
 E 
0
c t
The Maxwell equations in vacuum
 
 
  E  4
B 

0
  1 B
  1 E 4 
 E 
 0  B 

J
c t
c t
c
1
Implicit in those are the continuity equation,
  
J 
0
t
and the Lorentz force
 



vB

F  q E 
c


The Maxwell equations in macroscopic media
 
 
  D  4 f
B  0


  1 B
  1 D 4 
 E 
 0  H 

Jf
c t
c t
c
with
 

D  E  4P
 

H  B  4M
Boundary conditions
Let the fields in two neighboring media be denoted by 1 and 2. Let the surface charge density on the
surface between the two media be
boundary conditions are


( D2  D1 )  nˆ  4


( E2  E1 )  nˆ  0

and let the surface current density there be

K . Then the


( B2  B1 )  nˆ  0


4 
nˆ  ( H 2  H1 ) 
K
c
2
Scalar and vector potentials

A
Let us introduce the vector potential
such that
Then, from the Maxwell’s equations, we have
  
 
B    A . This ensures that   B  0 .

   1 A 
0
   E 
c t 

and therefore we choose


  1 A 
E 
  

c t 

These choices satisfy the two homogeneous Maxwell’s equations. The other two equations determine
the scalar and the vector potentials,
1  
 2 
  A  4
 c t
2
 1  A    1 
4 
2
 A  2 2  (  A 
)
J
c t
c t
c
We note that both the electric and the magnetic field are unchanged by the transformation

  
A  A'  A  
1 
  '   
c t
and therefore we can choose the gauge
3
  1 
 A
0
c t
to obtain the Maxwell’s equations in the form
1  2
   2 2  4
c t
 1 2 A
4  (Lorentz gauge)
2
 A 2 2  
J
c t
c
 
But of course, we could have stayed with   A  0 , and then to obtain
2
2  4
 1  2 A  1 
4 
2
 A  2 2  (
)
J
c t
c t
c
(Coulomb gauge)
Poynting theorem
The rate of work done by the electromagnetic fields on a moving charge q is
 
qv  E . For a system
of current density, that rate will be









1
D 


d
r
J

E

d
r
c
E

(


H
)

E


4  
t 
Using
  
  
  
  ( E  H )  H  (  E )  E  (  H )
we find
4












1
D
B 

 dr J  E   4  dr  c  ( E  H )  E  t  H  t 
Denoting the total energy density of the electromagnetic field by U,

1    
U
ED BH
8

we obtain the conservation law
 
U  
   S  J  E
t
total energy rate
mechanical work
with the Poynting vector

c  
S
EH
4
Exercise: A segment of a wire of length L and radius a, carrying the current I. A voltage drop V is
applied on the wire. Find the Poynting vector.
Solution: We us cylindrical coordinates, placing the wire along the z-direction. The magnetic field
produced by the current in the wire (on the surface of the wire) is
 2I
B  ̂
ca
5
and the electric field is
 V
E  zˆ ,
L
The Poynting vector at the surface of the wire is

c  
I V
S
EH 
ˆ
4
2a L ,
which is the power supplied by the battery,
2aL .
P  IV
, divided by the surface area of the wire,
Electromagnetic waves
In the absence of sources (i.e., no free charges or free currents) the Maxwell’s
 
 
  D  0
  B  0
  1 B
  1 D
 E 
 0  H 
0
c t
c t


Let us describe the medium by its dielectric constant, D  E , and magnetic permeability,


B  H . We then have
6
 
 
  E  0
  B  0
  1 B
   E
 E 
 0  B 
0
c t
c t
We see that each of the filed components satisfy the wave equation
1  2u
 u 2 2 0
v t
2
with the velocity
v
c

The solution is (in general)
 
 

ik  r it
ik  r  it
u(r , t )  Ae
 Be
with the dispersion relation relating the wave-vector with the frequency
k

v
 

c
Let us now solve for the electromagnetic fields. We write
 
 iknˆ r it
E (r , t )  E0e
 
 iknˆ r it
B(r , t )  B0e
7
with the convention that the physical fields are the real parts of the above expressions. From the
homogeneous Maxwell’s equations we have that

E0  nˆ  0

B0  nˆ  0
and therefore both the electric and the magnetic fields are perpendicular to the direction of the
wave propagation. Such a wave is called &quot;transverse wave&quot;. From the other two Maxwell’s equations
we have


B0  nˆ  E0
The Poynting vector, averaged over a period, is given by the real part of

c  *
S  
EH
8
Explanation:

c 1  * 1   *
c   *  *   * * 
S  
 ( E  E )  ( H  H ) 
E  H  E  H  E  H  E  H 
4 2
2
16
c   * * 

E  H  E  H 
16
In our case

c
S  
8
  2
| E0 | nˆ

The energy density, averaged over a period is
8
1
u  
16
   * 1  * 
 E  E  B  B 



and in our case
u  
  2
| E0 |
8
Electromagnetic waves in a conducting medium
( We consider the charge-free case.)
The Maxwell equations are
 
 
 
  D    E

0

 B  0

  1 B
  1 D 4 
 E 
 0  H 

Jf
c t
c t
c
Using Ohm’s law, relating the current density to the electric field via the conductivity


J  E
the fourth equation becomes

1    E 4 
 B 

E

c t
c
Therefore we now find
9



2
 E  4   E     E 
  2 2    0
 2    
2
c t  B  c t  B 
 B
Using again a plane-wave solution, the dispersion law (relating the wave
vector to the frequency) is now
k  
2
2 
4 
1

i


c2 
 
and the wave is damped.
Approximately, we find
2 

c
c 
2
k  (1  i )
c
k  

i
4
 1

4
 1

In a good conductor (high conductivity, law frequency), the electromagnetic
wave penetrates only over a length given by

c
2
which is called &quot;skin-depth&quot;.

Exercise: A wave of frequency
, polarized in the x-direction, propagates
along the z-direction in a good conductor which occupies the z&gt;0 space. Find the current in the
conductor.
10
Solution: Let us that
 ,   1.
The wave is then given by

E  E0 xˆeit e z (1i ) ,
and the current density established in the conductor is

J  E0 xˆeit e z (1i ) / 
Let us consider a slab of dimensions
direction) is
  h  dz . The current flowing in the slab (along the x-
dI  E0e  it e  z (1 i ) /  hdz
and the total current becomes

I   E0eit e z (1i ) /  hdz  E0
0
h
1 i
eit
Taking the real part to obtain the `true’ current, we find
I  E0h
cos(t 

4
)
2
We see that the current is not in-phase with the field.
11
Waveguides
Consider a rectangular waveguide of cross section of dimensions a and b and axis along the zdirection. In the absence of charge and current distributions, the electric and magnetic fields
satisfy


2



1  E
2 E


     2 2     0
 H  c t  H 
When the wave propagates along the z-direction, and the electric field is only in the x-y plane, then

E  xˆEx  yˆE y

H  xˆH x  yˆH y  zˆH z
and all the z-dependence of the fields is in the form
ik g z
e
. Denoting
  ck
we have
2H z 2H z

 (k 2  k g2 ) H z  0
2
2
x
y

  1 B
 0 we then find
From the Maxwell’s equation   E 
c t
k
k
Ex  H y , E y   H x
kg
kg
12
and from his other equation
ikEx 

  1 D
 H 
0
c t
we have
H z
H z
 ik g H y  0, ikEy 
 ik g H x  0
y
x
Therefore,
  k 2 
H z
ik g 1     H x 
 k  
x
  g 
  k 2 
H z
ik g 1     H y 
 k  
y
  g 
and it is sufficient to find
Hz
to find all fields. Now we solve for that component by variables
separation. Writing
H z ( x, y)  h1 ( x)h2 ( y)
we have
d 2 h1
2

k
h1  0,
x
2
dx
d 2h2
2

k
h2  0
y
2
dy
with
k x2  k y2  k 2  k g2  0
13
We now have the following solution
Hz  e
ikg z
E x  i
Ey  i
( A1 sin k x x sin k y y  A2 sin k x x cos k y y  A3 cos k x x sin k y y  A4 cos k x x cos k y y)
kk y
k
2
g
(1  [
k 2 1 ikg z
] ) e ( A1 sin k x x cos k y y  A2 sin k x x sin k y y  A3 cos k x x cos k y y  A4 cos k x x sin k y y)
kg
kkx
k
ik z
(1  [ ]2 ) 1 e g ( A1 cos k x x sin k y y  A2 cos k x x cos k y y  A3 sin k x x sin k y y  A4 sin k x x cos k y y)
2
kg
kg
Now we need to apply boundary conditions. Taking the surface of the waveguide as
a conductor then
Ey  0
at x=0 and x=a gives
A1  A2  0
and
k x  m / a
where m is an integer. Taking
Ex  0
at y=0 and y=b gives
A1  A3  0
and
k y  n / b
where n is an integer. Therefore the final solution is
Hz  e
ik g z
 Amn cos(
nm
mx
ny
) cos(
)
a
b
with
14
 m   n 
k  k 
 

a

  b 
2
2
g
2
2
limiting the allowed values of m and n!
15
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