Page 411 - Engineering Electromagnetics, 8th Edition

CHAPTER 11 The Uniform Plane Wave 393

crossing the conductor surface within this area,
b L 1 1
2
2
P L = S z da = σδE e −2z/δ dx dy = 4 σδbLE x0
x0
area 0 0 4 z=0
In terms of the current density J x0 at the surface,
J x0 = σ E x0
we have
1 2
P L = δbL J x0 (88)
4σ
Now let us see what power loss would result if the total current in a width b were
distributed uniformly in one skin depth. To find the total current, we integrate the
current density over the infinite depth of the conductor,
∞
b
I = J x dy dz
0 0
where
z
J x = J x0 e −z/δ cos ωt −
δ
or in complex exponential notation to simplify the integration,
J xs = J x0 e −z/δ − jz/δ
e
= J x0 e −(1+ j)z/δ
Therefore,

∞
b
I s = J x0 e −(1+ j)z/δ dy dz
0 0

∞
= J x0 be −(1+ j)z/δ −δ
1 + j 0
J x0 bδ
=
1 + j
and
J x0 bδ π
I = √ cos ωt −
2 4
If this current is distributed with a uniform density J throughout the cross section

0 < y < b,0 < z <δ, then
J x0 π

J = √ cos ωt −
2 4
The ohmic power loss per unit volume is J · E, and thus the total instantaneous power
dissipated in the volume under consideration is
1 J 2 π
2
P Li (t) = (J ) bLδ = x0 bLδ cos 2 ωt −
σ 2σ 4