Page 329 - Electromagnetics

where n is Neumann’s number (A.132). The magnetic fields can also be found by
substituting the coefficients into (4.369) and (4.370).
The fields produced byan impressed plane wave maynow be obtained byletting the
line source recede to infinity. For large ρ 0 we use the asymptotic form (E.62) and find
that
$ %
∞ 2 j
˜ # ˜ η 0 ν n − jk 0 ρ 0
E z (ρ, φ, ω) =− I πk 0 n J ν n (k 0 ρ) j e sin ν n φ sin ν n φ 0 , ρ < ρ 0 .
2ψ
n=0 πk 0 ρ 0
(4.374)
−1/2
Since the field of a line source falls off as ρ , the amplitude of the impressed field
0
approaches zero as ρ 0 →∞. We must compensate for the reduction in the impressed
field byscaling the amplitude of the current source. To obtain the proper scale factor,
we note that the electric field produced at a point ρ bya line source located at ρ 0 may
be found from (4.345):

˜
e
E z =−I ˜ k 0 η 0 H 0 (2) (k 0 |ρ − ρ 0 |) ≈−I ˜ k 0 η 0 2 j e − jk 0 ρ 0 jkρ cos(φ−φ 0 ) , k 0 ρ 0 1.
4 4 πk 0 ρ 0
But if we write this as
˜
˜
E z ≈ E 0 e jk·ρ
˜
then the field looks exactlylike that produced bya plane wave with amplitude E 0 trav-
˜
˜
eling along the wave vector k =−k 0 ˆ x cos φ 0 − k 0 ˆ y sin φ 0 . Solving for I in terms of E 0 and
substituting it back into (4.374), we get the total electric field scattered from a wedge
with an impressed TM plane-wave field:
∞
2π #
˜ ˜ ν n
E z (ρ, φ, ω) = E 0 n j J ν n (k 0 ρ) sin ν n φ sin ν n φ 0 .
ψ
n=0
Here we interpret the angle φ 0 as the incidence angle of the plane wave.
To determine the field produced byan impressed TE plane-wave field, we use a mag-
˜
netic line source I m located at ρ 0 ,φ 0 and proceed as above. Byanalogywith (4.367) we
write

&
[A ν sin νφ + B ν cos νφ] J ν (k 0 ρ), ρ < ρ 0 ,
˜ ν
[C δ sin δφ + D δ cos δφ] H (k 0 ρ), ρ > ρ 0 .
H z (ρ, φ, ω) = & (2)
δ δ
By(4.351) the tangential electric field is

& Z TE 1
− [A ν cos νφ − B ν sin νφ] j ν J ν (k 0 ρ), ρ < ρ 0 ,
˜ ν k ρ
E ρ (ρ, φ, ω) = & Z TE 1 (2)
− [C δ cos δφ − D δ sin δφ] j δH (k 0 ρ), ρ > ρ 0 .
δ k ρ δ
Application of the boundaryconditions on the tangential electric field at φ = 0,ψ results
˜
in A ν = C δ = 0 and ν = δ = ν n = nπ/ψ, and thus H z becomes

&
∞ (k 0 ρ), ρ < ρ 0 ,
˜ n=0 B n cos ν n φJ ν n (4.375)
∞ D n cos ν n φH (2) (k 0 ρ), ρ > ρ 0 .
H z (ρ, φ, ω) = &
n=0 ν n
Application of the boundaryconditions on tangential electric and magnetic fields across
the magnetic line source then leads directlyto

& (2)
∞ ˜ η 0
I
− n=0 m πk 0 n J ν n (k 0 ρ)H (k 0 ρ 0 ) cos ν n φ cos ν n φ 0 ,ρ < ρ 0
˜ 2ψ ν n
H z (ρ, φ, ω) =
∞ ˜ η 0 (2)
&
I
− n=0 m 2ψ πk 0 n H ν n (k 0 ρ)J ν n (k 0 ρ 0 ) cos ν n φ cos ν n φ 0 ,ρ > ρ 0 .
(4.376)
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