Page 328 - Electromagnetics

thus

(k 0 ρ 0 ) = C m H (2) (k 0 ρ 0 ). (4.371)
A m J ν m
ν m
˜
˜
˜
The boundarycondition ˆ n 12 × (H 1 − H 2 ) = J s requires the surface current at ρ = ρ 0 .We
can write the line current in terms of a surface current densityusing the δ-function:
δ(φ − φ 0 )
˜
J s = ˆ zI ˜ .
ρ 0
This is easilyverified as the correct expression since the integral of this densityalong
˜
the circular arc at ρ = ρ 0 returns the correct value I for the total current. Thus the
boundarycondition requires
δ(φ − φ 0 )
˜
˜
+
H φ (ρ ,φ,ω) − H φ (ρ ,φ,ω) = I ˜ .
−
0 0
ρ 0
By(4.370) we have
∞ ∞
# j # j δ(φ − φ 0 )

− C n sin ν n φH (2) (k 0 ρ 0 ) + A n sin ν n φJ (k 0 ρ 0 ) = I ˜
ν n ν n
η 0 η 0 ρ 0
n=0 n=0
and orthogonalityyields
ψ j ψ j sin ν m φ 0

− C m H (2) (k 0 ρ 0 ) + A m J (k 0 ρ 0 ) = I ˜ . (4.372)
ν m ν m
2 η 0 2 η 0 ρ 0
The coefficients A m and C m thus obeythe matrix equation

(2)
(k 0 ρ 0 ) −H (k 0 ρ 0 ) 0
J ν m A m
ν m =

J (k 0 ρ 0 ) −H (2) (k 0 ρ 0 ) C m − j2I ˜ η 0 sin ν m φ 0
ν m ν m ψ ρ 0
and are
j2I ˜ η 0 sin ν m φ 0 H (2) (k 0 ρ 0 )
ψ ρ 0 ν m
A m = ,
(2) (2)
(k 0 ρ 0 )
H ν m (k 0 ρ 0 )J ν m (k 0 ρ 0 ) − J (k 0 ρ 0 )H ν m
ν m
j2I ˜ η 0 sin ν m φ 0 J ν m (k 0 ρ 0 )
ψ ρ 0
C m = .
(2) (2)
(k 0 ρ 0 )
H ν m (k 0 ρ 0 )J ν m (k 0 ρ 0 ) − J (k 0 ρ 0 )H ν m
ν m
Using the Wronskian relation (E.93), we replace the denominators in these expressions
by 2/( jπk 0 ρ 0 ):
A m =−I ˜ η 0 πk 0 sin ν m φ 0 H (2) (k 0 ρ 0 ),
ψ ν m
C m =−I ˜ η 0 πk 0 sin ν m φ 0 J ν m (k 0 ρ 0 ).
ψ
Hence (4.368) gives

&
∞ ˜ η 0 (2)
− I πk 0 n J ν n (k 0 ρ)H (k 0 ρ 0 ) sin ν n φ sin ν n φ 0 ,ρ < ρ 0 ,
˜ n=0 2ψ ν n
E z (ρ,φ,ω) = & (4.373)
∞ ˜ η 0 (2)
− I πk 0 n H (k 0 ρ 0 ) sin ν n φ sin ν n φ 0 ,ρ > ρ 0 ,
n=0 2ψ ν n (k 0 ρ)J ν n
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