Page 320 - Electromagnetics

˜
2
2
c
Using k = ω ˜µ˜ we find that the two Hankel function terms cancel. Solving for E z0 we
have
˜
E z0 =− jω ˜µI ˜
and therefore
ω ˜µ
˜ ˜ (2) ˜ ˜
E z (ρ, ω) =− I(ω)H 0 (kρ) =− jω ˜µI(ω)G(x, y|0, 0; ω). (4.345)
4
˜
Here G is called the two-dimensional Green’s function and is given by
1
˜ (2) 2 2
G(x, y|x , y ; ω) = H 0 k (x − x ) + (y − y ) . (4.346)
4 j
Green’s functions are examined in greater detail in Chapter 5
It is also possible to determine the field amplitude byevaluating
"
˜
lim H · dl.
a→0
C
This produces an identical result and is a bit simpler since it can be argued that the
˜
surface integral of E z vanishes as a → 0 without having to perform the calculation
directly[83, 8].
˜
For a magnetic line source I m (ω) aligned along the z-axis we proceed as above, but
note that the source onlyproduces TE fields. By(4.333) and (4.334) we have
˜
j (2) k H 0z (2)
˜
˜
˜
H z (ρ, ω) =− H z0 (ω)H 0 (kρ), E φ =− H 1 (kρ).
4 ω˜ c 4
˜
We can find H z0 by applying Faraday’s law
"
˜ ˜ ˜
E · dl =− J m · dS − jω B · dS
C S S
about a circle of radius a in the z = 0 plane. We have
˜
a
k H z0 (2) j (2)
˜
˜
− H 1 (ka)2πa =−I m − jω ˜µ − H z0 2π lim H 0 (kρ)ρ dρ.
ω˜ c 4 4 δ→0 δ
Proceeding as above we find that
˜ c ˜
H z0 = jω˜ I m
hence
ω˜ c (2)
˜ ˜ c ˜ ˜
H z (ρ, ω) =− I m (ω)H 0 (kρ) =− jω˜ I m (ω)G(x, y|0, 0; ω). (4.347)
4
Note that we could have solved for the magnetic field of a magnetic line current by
using the field of an electric line current and the principle of duality. Letting the magnetic
current be equal to −η times the electric current and using (4.198), we find that
˜
c
1 I m (ω) 1 ω ˜µ ω˜
˜ ˜ (2) ˜ (2)
H z0 = − − − I(ω)H 0 (kρ) =−I m (ω) H 0 (kρ) (4.348)
η ˜ I(ω) η 4 4
as in (4.347).

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