14.3 Solutions
679
increase of magnetic energy. The last term corresponds to the power converted into Joule’s heat. Energy conservation demands that these three terms
be balanced by flow of energy into the volume and this is accounted for by
the term on the left.
The quantity S = E × H, representing the power density, is known as the
Poynting vector. The minus sign in (9) means that ds is the outward normal
and that (E × H) · ds represents power flowing outwards rather than inwards
for energy balance.
14.68 In one dimension the given equation reduces to
∂ 2 E
∂ x 2 − μ 0 ε 0 ε r
∂ 2 E
∂t 2 − μ 0 σ N
∂E
∂t
= 0
(1)
Let the travelling wave be given by
E = E 0 e
i(kx−ωt)
(2)
∂ 2 E
∂ x 2 = −k
2 E,
∂ E
∂t
= −iωE and
∂ 2 E
∂t 2 = −ω
2 E
(3)
Inserting (3) in (1) and re-arranging and cancelling E
k
2
= μ 0 ε 0 ε r ω
2
+ iμ 0 σ N ω
(4)
14.69 F = x
2 z
3 ˆ
i
∇ × F =
ˆ
i
ˆ
j ˆ
k
∂
∂ x
∂
∂ y
∂
∂z
x 2 z 3 0 0
= 3x
2 z
2 ˆ
j
∇ × 3x
2 z
2 ˆ
j =
ˆ
i
ˆ
j
ˆ
k
∂
∂ x
∂
∂ y
∂
∂z
0 3x 2 z 2 0
= −6x
2 z ˆ
i + 6xz
2 ˆ
k
∴ ∇ × (∇ × F) = −6x
2 z ˆ
i + 6xz
2 ˆ
k
(1)
− ∇
2 F = −
∂ 2
∂ x 2 +
∂ 2
∂ y 2 +
∂ 2
∂z 2
x
2 z
3 ˆ
i = −(6x
2 z + 2z
3
) ˆ
i
(2)
∇(∇ · F) = −
ˆ
i
∂
∂ x
+ ˆ
j
∂
∂ y
+ ˆ
k
∂
∂z
ˆ
i
∂
∂ x
+ ˆ
j
∂
∂ y
+ ˆ
k
∂
∂z
· (x
2 z
3 ˆ
i)
=
ˆ
i
∂
∂ x
+ ˆ
j
∂
∂ y
+ ˆ
k
∂
∂z
(2xz
3
) = 2z
3 ˆ
i + 6xz
2 ˆ
k
(3)