Page 119 - Electromagnetics

From this we see that when E φ = 0 the component E θ must obey
f E (r, t)
E θ (r,θ, t) = .
sin θ
By (2.350) there is only a φ-component of magnetic field, and it must obey H φ (r,θ, t) =
f H (r, t)/ sin θ where
∂ 1 ∂
− µ f H (r, t) = [rf E (r, t)]. (2.351)
∂t r ∂r
Thus the spherical wave has the property E ⊥ H ⊥ r, and is TEM to the r-direction.
We can obtain a wave equation for E θ by taking the curl of (2.350) and substituting
from Ampere’s law:
1 ∂ 2 ∂ ∂ ∂
∇× (∇× E) =−θ ˆ [rE θ ] =∇ × −µ H =−µ σE +
E .
r ∂r 2 ∂t ∂t ∂t
This gives
∂ 2 ∂ ∂ 2
[rf E (r, t)] − µσ [rf E (r, t)] − µ
[rf E (r, t)] = 0, (2.352)
∂r 2 ∂t ∂t 2
which is the desired wave equation for E. Proceeding similarly we find that H φ obeys

∂ 2 ∂ ∂ 2
[rf H (r, t)] − µσ [rf H (r, t)] − µ
[rf H (r, t)] = 0. (2.353)
∂r 2 ∂t ∂t 2
We see that the wave equation for rf E is identical to that for the plane wave field E z
(2.331). Thus, we can use the solution obtained in § A.1, as we did with the plane wave,
with a few subtle differences. First, we cannot have r < 0. Second, we do not anticipate
a solution representing a wave traveling in the −r-direction — i.e., a wave converging
toward the origin. (In other situations we might need such a solution in order to form a
standing wave between two spherical boundary surfaces, but here we are only interested
in the basic propagating behavior of spherical waves.) Thus, we choose as our solution
the term (A.45) and find for a lossless medium where = 0
1 r
E θ (r,θ, t) = A t − . (2.354)
r sin θ v
From (2.351) we see that
1 1 r
H φ = A t − . (2.355)
µv r sin θ v
Since µv = (µ/
) 1/2 = η, we can also write this as
ˆ r × E
H = .
η
We note that our solution is not appropriate for unbounded space since the fields have
a singularity at θ = 0. Thus we must exclude the z-axis. This can be accomplished
ˆ
by using PEC cones of angles θ 1 and θ 2 , θ 2 >θ 1 . Because the electric field E = θE θ is
normal to these cones, the boundary condition that tangential E vanishes is satisfied.
It is informative to see how the terms in the Poynting power balance theorem relate for
a spherical wave. Consider the region between the spherical surfaces r = r 1 and r = r 2 ,
r 2 > r 1 . Since there is no current within the volume region, Poynting’s theorem (2.299)
becomes
1 ∂
(
E · E + µH · H) dV =− (E × H) · dS. (2.356)
2 ∂t V S

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