Page 115 - Electromagnetics

law states
∂E z (ρ, t) 1 ∂ ∂H(ρ, t)
ˆ
∇× E(ρ, t) =−φ + ˆ z [ρE φ (ρ, t)] =−µ . (2.344)
∂ρ ρ ∂ρ ∂t
Equating components we see that ∂ H ρ /∂t = 0, and because our interest lies in wave
solutions we take H ρ = 0. Ampere’s law in a homogeneous lossless region free from
impressed sources states in a similar manner
∂ H z (ρ, t) 1 ∂ ∂E(ρ, t)
ˆ
∇× H(ρ, t) =−φ + ˆ z [ρH φ (ρ, t)] =
. (2.345)
∂ρ ρ ∂ρ ∂t
Equating components we find that E ρ = 0. Since E ρ = H ρ = 0, both E and H are
perpendicular to the ρ-direction. Note that if there is only a z-component of E then
there is only a φ-component of H. This case, termed electric polarization, results in

∂E z (ρ, t) ∂ H φ (ρ, t)
= µ .
∂ρ ∂t
Similarly, if there is only a z-component of H then there is only a φ-component of E.
This case, termed magnetic polarization, results in

∂ H z (ρ, t) ∂E φ (ρ, t)
− =
.
∂ρ ∂t
ˆ
ˆ
Since E = φE φ + ˆ zE z and H = φH φ + ˆ zH z , we can always decompose a cylindrical
electromagnetic wave into cases of electric and magnetic polarization. In each case the
resulting field is TEM ρ since the vectors E, H, ˆρ are mutually orthogonal.
Wave equations for E z in the electric polarization case and for H z in the magnetic
polarization case can be found in the usual manner. Taking the curl of (2.344) and
substituting from (2.345) we find

1 ∂ ∂E z ∂ 1 ∂
ˆ
∇× (∇× E) =−ˆ z ρ − φ [ρE φ ]
ρ ∂ρ ∂ρ ∂ρ ρ ∂ρ
2
2
2
1 ∂ E 1 ∂ E z ∂ E φ
ˆ
=− 2 2 =− 2 ˆ z 2 + φ 2
v ∂t v ∂t ∂t
where v = 1/(µ
) 1/2 . Noting that E φ = 0 for the electric polarization case we obtain the
wave equation for E z . A similar set of steps beginning with the curl of (2.345) gives an
identical equation for H z .Thus
1 ∂ ∂ E z 1 ∂ 2 E z
ρ − = 0. (2.346)
2
ρ ∂ρ ∂ρ H z v ∂t 2 H z
We can obtain a solution for (2.346) in much the same way as we do for the wave
equations in § A.1. We begin by substituting for E z (ρ, t) in terms of its temporal Fourier
representation
1 ∞ jωt
˜
E z (ρ, t) = E z (ρ, ω)e dω
2π
−∞
to obtain
1 ∞ 1 ∂ ∂ ω 2 jωt
˜
˜
ρ E z (ρ, ω) + E z (ρ, ω) e dω = 0.
2π −∞ ρ ∂ρ ∂ρ v 2

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