Page 407 - Electromagnetics

˜
˜
˜
we see that E ∼∇ × (∇× A e ) has all three components. This choice of A e produces a
˜
field TM to the r-direction. We need only choose ∇· A e so that the resulting differential
equation is convenient to solve. Substituting the above expressions into (5.162) we find
that
ˆ
ˆ
2 ˜
˜
2 ˜
2 ˜
ˆ r 1 ∂
∂ A e 1 ∂ A e θ ∂ A e φ ∂ A e 2 ˜
− sin θ + + + − ˆ rk A e =
r sin θ r ∂θ ∂θ r sin θ ∂φ 2 r ∂r∂θ r sin θ ∂r∂φ
2 ˜ ˆ 2 ˜ ˆ 2 ˜
k ∂φ e θ k ∂φ e φ k ∂φ e
ˆ r + + . (5.164)
jω ∂r r jω ∂θ r sin θ jω ∂φ
˜ ˜ ˜
Since ∇· A e only involves the derivatives of A e with respect to r, we may specify ∇· A e
indirectly through
˜
˜ jω ∂ A e
φ e = .
k 2 ∂r
With this (5.164) becomes
2 ˜
2 ˜
˜
1 1 ∂
∂ A e 1 ∂ A e 2 ˜ ∂ A e
sin θ + + k A e + = 0.
r sin θ r ∂θ ∂θ r sin θ ∂φ 2 ∂r 2
Using
˜
2 ˜
1 ∂ 2 ∂
A e ∂ A e
r =
r ∂r ∂r r ∂r 2
we can write the differential equation as
˜
2 ˜
˜
˜
1 ∂ 2 ∂(A e /r) 1 ∂ ∂(A e /r) 1 ∂ (A e /r) 2 A e
r + sin θ + + k = 0.
2
2
2
2
r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ 2 r
˜
The first three terms of this expression are precisely the Laplacian of A e /r.Thus we
have
˜ A e

2
2
(∇ + k ) = 0 (5.165)
r
˜
and the quantity A e /r satisfies the homogeneous Helmholtz equation.
˜
˜
The TM fields generated by the vector potential A e = ˆ rA e may be found by using
(5.160) and (5.161). From (5.160) we have the electric field
˜

˜ ˜ ˜ ˜ jω ∂ A e
E =− jωA e −∇φ e =− jωˆ rA e −∇ .
k 2 ∂r
Expanding the gradient we have the field components
1
∂ 2
˜ 2 ˜
E r = + k A e , (5.166)
jω ˜µ˜ c ∂r 2
2 ˜
1 1 ∂ A e
˜
E θ = , (5.167)
c
jω ˜µ˜ r ∂r∂θ
2 ˜
1 1 ∂ A e
˜
E φ = . (5.168)
c
jω ˜µ˜ r sin θ ∂r∂φ
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