14.3 Solutions
669
Consider a volume element dV for the cylindrical shell of radii r and
r + dr and of length l. The energy contained in the volume element is
dU = u dV =
μ 0 i 2
8π 2 r 2 (2πrl dr ) =
μ 0 i 2 l
4π
dr
r
(6)
U =
dU =
μ 0 i 2 l
4π
b
a
dr
r
=
μ 0 i 2 l
4π
ln(b/a)
(7)
But U =
1
2
Li
2
(8)
Comparing (7) and (8)
L =
μ 0 l
2π
ln(b/a)
∴ Inductance per unit length of the cable is
L
l
=
μ 0
2π
ln(b/a)
14.42 (a) By prob. (14.41) the potential difference between the conductors is
V =
q
2πε 0 l
ln(b/a) =
λ
2πε 0
ln(b/a)
(1)
where λ = q/l is the charge density.
Now E =
λ
2πε 0 r
=
ξ
r ln(b/a)
(2)
where we have put V = ξ .
(b) B =
μ 0 i
2πr
=
μ 0 ξ
2π r R
(a < r < b)
(3)
(c) The Poynting vector
S =
1
μ 0
E × B =
1
μ 0
EB =
ξ 2
2πr 2 R ln(b/a)
where we have used (2) and (3).
14.43 u B =
1
2
B · H
(1)
B = μH
(2)