Page 325 - Electromagnetics

∞ jn 1
˜ #
H ρ (r,ω) =− B n (ω)J n (kρ) sin nφ,
Z TM k ρ
n=0
∞ j
˜ #
H φ (r,ω) =− B n (ω)J (kρ) cos nφ,
n
Z TM
n=0
and the field outside the cylinder (ρ > a) from
∞
# (2)
− jk 0 ρ cos φ
˜
˜
E z (r,ω) = E 0 (ω)e + D n (ω)H (k 0 ρ) cos nφ,
n
n=0
˜
∞
E 0 (ω) − jk 0 ρ cos φ # jn 1 (2)
˜
H ρ (r,ω) =− sin φ e − D n (ω)H n (k 0 ρ) sin nφ,
η 0 η 0 k 0 ρ
n=0
˜
∞
E 0 (ω) − jk 0 ρ cos φ # j
˜
H φ (r,ω) =− cos φ e − D n (ω)H (2) (k 0 ρ) cos nφ.
n
η 0 η 0
n=0
We can easilyspecialize these results to the case of a perfectlyconducting cylinder by
allowing ˜σ →∞. Then

η 0 µ 0 ˜ c
= →∞
Z TM ˜ µ 0
and
J m (k 0 a)
−m
˜
B n → 0, D n →−E 0 m j .
(2)
H m (k 0 a)
In this case it is convenient to combine the formulas for the impressed and scattered fields
when forming the total fields. Since the impressed field is z-independent and obeys the
homogeneous Helmholtz equation, we mayrepresent it in terms of nonuniform cylindrical
waves:
∞
#
− jk 0 ρ cos φ
˜
˜ i
E = E 0 e = [E n sin nφ + F n cos nφ] J n (k 0 ρ),
z
n=0
where we have chosen the Bessel function J n (k 0 ρ) since the field is finite at the origin
and periodic in φ. Applying orthogonality we see immediately that E n = 0 and that
2π π − jk 0 ρ cos φ −m
˜
˜
F m J m (k 0 ρ) = E 0 cos mφe dφ = E 0 2π j J m (k 0 ρ).
m −π
˜
Thus, F n = E 0 n j −n and
∞
#
˜ i ˜ −n
E = E 0 n j J n (k 0 ρ) cos nφ.
z
n=0
Adding this impressed field to the scattered field we have the total field outside the
cylinder,
∞ −n
# n j
˜ ˜ (2) (2)
E z = E 0 (2) J n (k 0 ρ)H n (k 0 a) − J n (k 0 a)H n (k 0 ρ) cos nφ,
H n (k 0 a)
n=0
while the field within the cylinder vanishes. Then, by (4.350),
∞
j # n j −n
˜ ˜ (2) (2)
H φ =− E 0 (2) J (k 0 ρ)H n (k 0 a) − J n (k 0 a)H n (k 0 ρ) cos nφ.
n
η 0 H n (k 0 a)
n=0
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