Apr 2, 2001

Notes for Lecture Notes for Lecture #30

Electromagnetic wave guides

In order to understand the operation of a wave guide, we must first learn how electromagnetic waves behave in a dissipative medium. A plane wave solution to Maxwell's equations of the form:

E = E0eik [^(k)] ·r - i wt       and      B = k
w
^
k
 
×E0 eik [^(k)] ·r - i wt
(1)
for the electric and magnetic fields, with the wave vector k satisfying the relation:
k2 = w2 me º R + i I.
(2)
We can determine the complex wavevector kr + i ki according to
kr = æ
ç
ç
ç
è
Ö
R2+I2
 
+R

2
ö
÷
÷
÷
ø
1/2


 
    and    ki = æ
ç
ç
ç
è
Ö
R2+I2
 
-R

2
ö
÷
÷
÷
ø
1/2


 
(3)
The form of the frequency dependent constants R and I depend on the materials. For the Drude model at low frequency (Eq. 7.56), R = w2 meb and I = wms, for example. The value of ki determines the rate of decay of the field amplitudes in the vicinity of the surface, with the skin depth given by d º 1/ki. In the limit that I >> R, as in the case of a good conductor at low frequency, d » (2/(wms) )1/2.

For an ïdeal" conductor I ® ¥, so that the fields are confined to the surface. Because of the field continuity conditions at the surface of the conductor, this means that, Btangential ¹ 0 (because there can be a surface current), Enormal ¹ 0 (because there can be a surface charge), and Bnormal = 0.

Suppose we construct a wave guide from an ïdeal" conductor, designating [^(z)] as the propagation direction. We will assume that the fields take the form:

E = E(x,y) eikz-iwt         and          B = B(x,y) eikz-iwt
(4)
inside the pipe, where now k and e are assumed to be real. Assuming that there are no sources inside the pipe, the fields there must satisfy Maxwell's equations (8.16) which expand to the following :
Bx
x
+ By
y
+ i k Bz = 0.
(5)
Ex
x
+ Ey
y
+ i k Ez = 0.
(6)
Ez
y
- i k Ey = i wBx.
(7)
i k Ex - Ez
x
= i wBy.
(8)
Ey
x
- Ex
y
= i wBz.
(9)
Bz
y
- i k By = -i mew Ex.
(10)
i k Bx - Bz
x
= -i mew Ey.
(11)
By
x
- Bx
y
= -i mewEz.
(12)

Combining Faraday's Law and Ampere's Law, we find that each field component must satisfy a two-dimensional Helmholz equation:

æ
ç
è
2
x2
+ 2
y2
- k2 + mew2 ö
÷
ø
Ex(x,y) = 0,
(13)
with similar expressions for each of the other field components. For the rectangular wave guide discussed in Section 8.4 of your text a solution for a TE mode can have:
Ez(x,y) º 0         and        Bz(x,y) = B0 cos æ
ç
è
m px
a
ö
÷
ø
cos æ
ç
è
n p y
b
ö
÷
ø
,
(14)
with k2 º k2mn = mew2 -[([(m p)/ a])2 + ([(n p)/ b])2] . From this result and Maxwell's equations, we can determine the other field components. For example Eqs. (7-8) simplify to
Bx = - k
w
Ey       and       By = k
w
Ex.
(15)
These results can be used in Eqs. (10-11) to solve for the fields Ex and Ey and Bx and By:
Ex = w
k
By = -i w
k2-mew2
Bz
y
= -iw
é
ê
ë
æ
ç
è
m p
a
ö
÷
ø
2

 
+ æ
ç
è
n p
b
ö
÷
ø
2

 
ù
ú
û
n p
b
B0 cos æ
ç
è
m px
a
ö
÷
ø
sin æ
ç
è
n p y
b
ö
÷
ø
,
(16)
and
Ey = - w
k
Bx = i w
k2-mew2
Bz
x
= i w
é
ê
ë
æ
ç
è
m p
a
ö
÷
ø
2

 
+ æ
ç
è
n p
b
ö
÷
ø
2

 
ù
ú
û
n p
a
B0 sin æ
ç
è
m px
a
ö
÷
ø
cos æ
ç
è
n p y
b
ö
÷
ø
.
(17)
One can check this result to show that these results satisfy the boundary conditions. For example, Etangential = 0 is satisfied since Ex(x,0) = Ex(x,b) = 0 and Ey(0,y) = Ey(a,y) = 0. This was made possible choosing ÑBzûsurface·[^(n)] = 0, where [^(n)] denotes a unit normal vector pointing out of the wave guide surface.


File translated from TEX by TTH, version 2.20.
On 2 Apr 2001, 10:30.