Page 410 - Electromagnetics

˜
We note immediately that A e /r satisfies the Helmholtz equation (5.165) since it has the
form of the separation of variables solution (D.113).
˜
˜
We may find the vector potential A h = ˆ rA h in the same manner. Noting that
˜ ˜
˜ E 0 − jkr cos θ E 0 sin φ ∂ − jkr cos θ
H r = sin θ sin φe = e
η ηjkr ∂θ
1
∂ 2 2
˜
= + k A h ,
jω ˜µ˜ c ∂r 2
we have the potential
˜ ˜ ∞ −n
A h E 0 k " j (2n + 1) 1
= sin φ j n (kr)P (cos θ). (5.181)
n
r ηω n(n + 1)
n=1
We may nowcompute the transverse components of the TM field using (5.167)–(5.170).
ˆ
For convenience, let us define a newfunction J n by
ˆ
J n (x) = xj n (x).
Then we may write
˜
∞
j E 0 cos φ "
˜ −n ˆ 1
E r =− j (2n + 1)J n (kr)P (cos θ), (5.182)
n
(kr) 2
n=1
˜ "
∞
j E 0
˜ ˆ 1
E θ = sin θ cos φ a n J (kr)P (cos θ), (5.183)
n
n
kr
n=1
˜ "
∞
j E 0
˜ ˆ 1
E φ = sin φ a n J (kr)P (cos θ), (5.184)
n
n
kr sin θ
n=1
˜ "
∞
E 0
˜ ˆ 1
H θ =− sin φ a n J n (kr)P (cos θ), (5.185)
n
krη sin θ
n=1
˜ "
∞
˜ E 0 ˆ 1
H φ = sin θ cos φ a n J n (kr)P (cos θ). (5.186)
n
krη
n=1
Here
d d

ˆ ˆ [xj n (x)] = xj (x) + j n (x)
J (x) =
n J n (x) = n
dx dx
and
j −n (2n + 1)
a n = . (5.187)
n(n + 1)
Similarly, we have the TE fields from (5.176)–(5.179):
˜
∞
j E 0 sin φ "
˜ −n ˆ 1
H r =− j (2n + 1)J n (kr)P (cos θ), (5.188)
n
η(kr) 2
n=1
˜ E 0 "
∞
˜
1
ˆ
H θ = j sin θ sin φ a n J (kr)P (cos θ), (5.189)
n
n
ηkr
n=1
˜ "
∞
E 0
˜
1
ˆ
H φ =− j cos φ a n J (kr)P (cos θ), (5.190)
n
n
ηkr sin θ
n=1
© 2001 by CRC Press LLC