14.3 Solutions
667
14.39 E = E 0 cos(kx − ωt)
(1)
∂ E y
∂ x
= −k E 0 sin(kx − ωt)
(2)
B = B 0 cos(kx − ωt)
(3)
∂ B Z
∂t
= ωB 0 sin(kx − ωt)
(4)
But
∂ E y
∂ x
= −
∂ E z
∂t
(5)
Combining (2), (4) and (5), we get
E 0 =
ω
k
B 0 = cB 0
14.40 Maxwell’s equations for a non-ferromagnetic homogeneous isotropic
medium can be written as
∇ · E =
ρ
ε
(1)
∇ · B = 0
( 2 )
∇ × E = −
∂B
∂t
(3)
∇ × B = μσ E + με
∂E
∂t
(4)
Taking the curl of (4)
∇ × (∇ × B) = μσ (∇ × E) + με
∂
∂t
(∇ × E)
(5)
where the time and space derivatives are interchanged as E is assumed to be
a well-behaved function. Expression (3) can be substituted in (5) to obtain
∇ × (∇ × B) = −μσ
∂B
∂t
− με
∂ 2 B
∂t 2
(6)
Using the vector identity
∇ × (∇ × B) = ∇ (∇ · B) − ∇
2 B
(7)
By virtue of (2), ∇ · B = 0 and (6) becomes
∇
2 B = με
∂ 2 B
∂t 2 + μσ
∂B
∂t
(8)