Page 324 - Electromagnetics

where n is Neumann’s number (A.132) and where we have used (E.83) and (E.39) to
evaluate the integral.
˜
We must also have continuityof the tangential magnetic field H φ at ρ = a.Thus
∞
# j

− [A n sin nφ + B n cos nφ] J (ka) =
n
Z TM
n=0
∞ ˜
# j E 0 − jk 0 a cos φ
− [C n sin nφ + D n cos nφ] H (2) (k 0 a) − cos φ e
n
η 0 η 0
n=0
must hold for all −π ≤ φ ≤ π. Byorthogonality
˜
j j E 0 π − jk 0 a cos φ

π A m J (ka) − π C m H (2) (k 0 a) = sin mφ cos φe dφ = 0 (4.359)
m m
Z TM η 0 η 0 −π
and
˜
j j E 0 π − jk 0 a cos φ
2π B m J (ka) − 2π D m H (2) cos mφ cos φe dφ.

m m (k 0 a) = m
Z TM η 0 η 0 −π
The integral maybe computed as
π d π

cos mφ cos φe − jk 0 a cos φ dφ = j cos mφe − jk 0 a cos φ dφ = j2π j −m J (k 0 a)
m
−π d(k 0 a) −π
and thus
˜
1 1 E 0
(2) −m
B m J (ka) − D m H (k 0 a) = m j J (k 0 a). (4.360)
m m m
Z TM η 0 η 0
We now have four equations for the coefficients A n , B n , C n , D n . We maywrite (4.357)
and (4.359) as

(2)
J m (ka) −H m (k 0 a) A m
J (ka) −H m (k 0 a) C m
η 0 (2) = 0, (4.361)
m
Z TM
and (4.358) and (4.360) as
˜

(2)

−m
J m (ka) −H m (k 0 a) B m E 0 m j J m (k 0 a)
J (ka) −H m (k 0 a) D m E 0 m j J (k 0 a)
η 0 (2) = ˜ −m . (4.362)
m
Z TM m
Matrix equations (4.361) and (4.362) cannot hold simultaneouslyunless A m = C m = 0.
Then the solution to (4.362) is
$ (2) %
H m (k 0 a)J (k 0 a) − J m (k 0 a)H m (2) (k 0 a)

m
−m
˜
B m = E 0 m j , (4.363)
η 0 (2) (2)
J (ka)H m (k 0 a) − H m (k 0 a)J m (ka)
Z TM m
$ %

η 0 J (ka)J m (k 0 a) − J (k 0 a)J m (ka)
˜
D m =−E 0 m j −m Z TM m m . (4.364)
η 0 (2) (2)
J (ka)H m (k 0 a) − H m (k 0 a)J m (ka)
Z TM m
With these coefficients we can calculate the field inside the cylinder (ρ ≤ a) from
∞
#
˜
E z (r,ω) = B n (ω)J n (kρ) cos nφ,
n=0
© 2001 by CRC Press LLC