Page 318 - Electromagnetics

Substituting this into (4.341) and remembering that Z TM = η = (µ/ ) 1/2 is real for
lossless media, we have

1 2
ˇ
S av = ˆρ |E z0 | [N 0 (kρ)J 1 (kρ) − J 0 (kρ)N 1 (kρ)].
32η
Bythe Wronskian relation (E.88) we have
ˇ 2
|E z0 |
S av = ˆρ .
16πkρη
The power densityis inverselyproportional to ρ. When we compute the total time-
average power per unit length passing through a cylinder of radius ρ, this factor cancels
with the ρ-dependence of the surface area to give a result independent of radius:
2π ˇ 2
|E z0 |
P av /l = S av · ˆρρ dφ = . (4.342)
0 8kη
For a lossless medium there is no mechanism to dissipate the power and so the wave prop-
agates unabated. A similar calculation for the case of magnetic polarization (Problem
??) gives

ˇ 2
η|H z0 |
S av = ˆρ
16πkρ
and
ˇ 2
η|H z0 |
P av /l = .
8k
For a lossymedium the expressions are more difficult to evaluate. In this case we expect
the total power passing through a cylinder to depend on the radius of the cylinder, since
the fields decayexponentiallywith distance and thus give up power as theypropagate.
If we assume that the observation point is far from the z-axis with |kρ| 1, then we can
use (4.335) and (4.336) for the electric polarization case to obtain

1 1 e −2αρ 2
ˇ ∗ ˆ
ˇ
ˇ
S av = Re{E z ˆ z × H φ}= Re ˆρ |E z0 | .
φ
2 2 8πρ|k|Z ∗
TM
Therefore
2π 1 e −2αρ
ˇ
P av /l = S av · ˆρρ dφ = Re |E z0 | 2 .
0 Z ∗ TM 8|k|
We note that for a lossless material Z TM = η and α = 0, and the expression reduces to
(4.342) as expected. Thus for lossymaterials the power depends on the radius of the
cylinder. In the case of magnetic polarization we use (4.337) and (4.338) to get
1 1 e −2αρ 2
ˇ ˆ
ˇ
ˇ ∗
∗
S av = Re{E φ φ × H ˆ z}= Re ˆρZ TE |H z0 |
z
2 2 8πρ|k|
and
e −2αρ
ˇ

2
P av /l = Re Z ∗ |H z0 | .
TE
8|k|
© 2001 by CRC Press LLC