1
~ ·B
~ = 0 for TE and TM waves.
Show E
For TE waves, Ez = 0, and Maxwell’s equations yields:
i
∂Bz
ω
2
2
(ω/c) − k
∂y
i
∂Bz
Ey = −
ω
2
2
(ω/c) − k
∂x
∂Bz
i
k
Bx =
(ω/c)2 − k 2 ∂x
i
∂Bz
By =
k
(ω/c)2 − k 2 ∂y
Ex =
Thus,
~ ·B
~ = Ex Bx + Ey By + Ez Bz
E
2
i
∂Bz ∂Bz
∂Bz ∂Bz
=
ωk
−
+ 0Bz
(ω/c)2 − k 2
∂y ∂x
∂y ∂x
=0
An analogous argument works for the TM case.
1
2
~ =
hSi
1
~ × B̃
~?
Re Ẽ
2µ0
Note that because the x and y components of Ẽ and B̃ are purely imaginary,
the only non-zero component of the Poynting vector is in the z direction.
So
mπx nπy 1
ωkπ 2 B02
n 2
~
hSi =
cos2
sin2
2
2µ0
b
a
b
(ω/c)2 − k 2
m 2
2 nπy
2 mπx
cos
+
sin
ẑ.
a
a
b
Z
2 2 2
2
~ · d~a = 1 ωkπ B0 ab n + m
hSi
2
8µ0
b
a
(ω/c)2 − k 2
In the last step I used
3
2
3