Page 112 - Electromagnetics

With H 0 = E 0 /η we have
E 2 0 2
t = ˆ z µf (t). (2.341)
2η 2
As a numerical example, consider a high-altitude nuclear electromagnetic pulse (HEMP)
generated by the explosion of a large nuclear weapon in the upper atmosphere. Such
an explosion could generate a transient electromagnetic wave of short (sub-microsecond)
duration with an electric field amplitude of 50, 000 V/m in air [200]. Using (2.341),
we find that the wave would exert a peak pressure of P =|t|= .011 Pa = 1.6 × 10 −6
2
lb/in if reflected from a perfect conductor at normal incidence. Obviously, even for this
extreme field level the pressure produced by a transient electromagnetic wave is quite
small. However, from (2.340) we find that the current induced in the conductor would
have a peak value of 133 A/m. Even a small portion of this current could destroy a
sensitive electronic circuit if it were to leak through an opening in the conductor. This is
an important concern for engineers designing circuitry to be used in high-field environ-
ments, and demonstrates why the concepts of current and voltage can often supersede
the concept of force in terms of importance.
Finally, let us see how the terms in the Poynting power balance theorem relate. Con-
sider a cubic region V bounded by the planes z = z 1 and z = z 2 , z 2 > z 1 . We choose
the field waveform f (t) and locate the planes so that we can isolate either the forward
or backward traveling wave. Since there is no current in V , Poynting’s theorem (2.299)
becomes
1 ∂
(
E · E + µH · H) dV =− (E × H) · dS.
2 ∂t V S
Consider the wave traveling in the −z-direction. Substitution from (2.336) and (2.337)
gives the time-rate of change of stored energy as
1 ∂ 2 2
S (t) =
E (z, t) + µH (z, t) dV
cube
2 ∂t V
2
1 ∂ z 2 (vµ) H 0 2 2 z H 0 2 2 z
= dx dy
f t + + µ f t + dz
2 ∂t x y z 1 4 v 4 v
1 ∂ H 0 2 z 2 2 z
= µ dx dy f t + dz.
2 ∂t 2 x y z 1 v
Integration over x and y gives the area A of the cube face. Putting u = t + z/v we see
that
2
H ∂ t+z 2 /v 2
0
S = Aµ f (u)v du.
4 ∂t t+z 1 /v
Leibnitz’ rule for differentiation (A.30) then gives
µvH 0 2 2 z 2 2 z 1
S cube (t) = A f t + − f t + . (2.342)
4 v v
Again substituting for E(t + z/v) and H(t + z/v) we can write

S (t) =− (E × H) · dS
cube
S
2
vµH 0 2 z 1
=− f t + (−ˆ p × ˆ q) · (−ˆ z) dx dy −
x y 4 v
vµH 0 2 z 2
2
− f t + (−ˆ p × ˆ q) · (ˆ z) dx dy.
x y 4 v

© 2001 by CRC Press LLC