Page 327 - Electromagnetics

(scattered) field. This scattered field, also z-invariant, can be found bysolving a bound-
aryvalue problem. We do this byseparating space into the two regions ρ< ρ 0 and
ρ> ρ 0 , 0 <φ <ψ. Each of these is source-free, so we can represent the total field using
nonuniform cylindrical waves of the type (4.353). The line source is brought into the
problem byapplying the boundarycondition on the tangential magnetic field across the
cylindrical surface ρ = ρ 0 .
Since the impressed electric field has onlya z-component, so do the scattered and total
˜
electric fields. We wish to represent the total field E z in terms of nonuniform cylindrical
waves of the type (4.353). Since the field is not periodic in φ, the separation constant
k φ need not be an integer; instead, its value is determined bythe positions of the wedge
boundaries. For the region ρ< ρ 0 we represent the radial dependence of the field using
the functions J ν since the field must be finite at the origin. For ρ> ρ 0 we use the
(2)
outward-propagating wave functions H δ .Thus

&
[A ν sin νφ + B ν cos νφ] J ν (k 0 ρ), ρ < ρ 0 ,
˜ ν (4.367)
[C δ sin δφ + D δ cos δφ] H (k 0 ρ), ρ > ρ 0 .
E z (ρ, φ, ω) = & (2)
δ δ
The coefficients A ν , B ν , C δ , D δ and separation constants ν, δ maybe found byapplying
the boundaryconditions on the fields at the surface of the wedge and across the surface
˜
ρ = ρ 0 . On the wedge face at φ = 0 we must have E z = 0, hence B ν = D δ = 0.On
˜
the wedge face at φ = ψ we must also have E z = 0, requiring sin νψ = sin δψ = 0 and
therefore
ν = δ = ν n = nπ/ψ, n = 1, 2,....
So

∞ (k 0 ρ), ρ < ρ 0 ,
&
˜ n=0 A n sin ν n φJ ν n (4.368)
E z = & (2)
∞ C n sin ν n φH (k 0 ρ), ρ > ρ 0 .
n=0 ν n
The magnetic field can be found from (4.349)–(4.350):

∞ ν n (k 0 ρ), ρ < ρ 0 ,
& j
˜ n=0 A n η 0 k 0 ρ cos ν n φJ ν n
∞ ν n cos ν n φH (2) (k 0 ρ), ρ > ρ 0 ,
H ρ = & j (4.369)
n=0 C n η 0 k 0 ρ ν n

∞ sin ν n φJ (k 0 ρ), ρ < ρ 0 ,

& j
˜ − n=0 A n η 0 ν n
H φ = (4.370)
∞ j (2)
&
− C n sin ν n φH (k 0 ρ), ρ > ρ 0 .
n=0 η 0 ν n
The coefficients A n and C n are found byapplying the boundaryconditions at ρ = ρ 0 .
Bycontinuityof the tangential electric field
∞ ∞
# # (2)
(k 0 ρ 0 ) = C n sin ν n φH (k 0 ρ 0 ).
A n sin ν n φJ ν n
ν n
n=0 n=0
We now applyorthogonalityover the interval [0,ψ]. Multiplying by sin ν m φ and inte-
grating we have
∞ ψ ∞ ψ
# # (2)
(k 0 ρ 0 ) sin ν n φ sin ν m φ dφ = C n H (k 0 ρ 0 ) sin ν n φ sin ν m φ dφ.
A n J ν n
ν n
n=0 0 n=0 0
Setting u = φπ/ψ we have
ψ π
ψ ψ
sin ν n φ sin ν m φ dφ = sin nu sin mu du = δ mn ,
0 π 0 2
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