13.3 Solutions
617
(b) B = μ 0 ni =
μ 0 Ni
l
U B = u B Al =
B 2
2μ 0
· Al =
(μ 0 Ni) 2 Al
2μ 0 l 2
=
μ 0 N 2 i 2 πr 2
2l
=
(4π × 10 −7 ) (100) 2 (5) 2 π (0.01) 2
2 × 0.1
= 4.93 × 10
−4 J
(c) φ = B A =
μ 0 Ni
l
· πr
2
=
4π × 10 −7 × 100 × 5 × π × (0.1) 2
0.1
= 1.972 × 10
−4 Wb
ξ =
−φ
t
=
−1.972 × 10 −4
5
= −3.94 × 10
−5 V
13.46 (a) B =
μ 0 i
2πr
=
4π × 10 −7 × 3
2π × 50 × 10 −3 = 1.2 × 10 −7 T
(b) In the vector form the Biot and Savart law can be written as
dB =
μ 0 i
4π
dl × r
r 3
(i) dB =
(4π × 10 −7 )(2)(2 ˆ
k × 3 ˆ
i)
4π
3 3
= 4.44 × 10 −8 ˆ
j
(∵ ˆ
k × ˆ
i = ˆ
j)
Thus dB = 4.44 × 10 −8 T along positive y-axis
(ii) dB =
(4π × 10 −7 )(2)(2 ˆ
k × (−6 ˆ
i))
4π
6 3
= −1.11 × 10 −8 ˆ
j
Thus dB = 1.11 × 10 −8 T along negative y-axis
(iii) dB = 0 (∵ ˆ
k × ˆ
k = 0)
(iv) dB =
(4π × 10 −7 )(2)(2 ˆ
k × 3 ˆ
j)
4π
3 3
= −4.44 × 10
−8 ˆ
i
dB = 4.44 × 10
−8 T along negative x-axis.
13.47 H = n 0 i =
Ni
2πr
=
100 × 2
2π × 0.1
= 318.47 A/m for both vacuum and material.
B = K μ 0 H = 1 × 4π × 10
−7
× 318.47 = 4 × 10
−4 T (vacuum)
B = 500 × 4π × 10
−7
× 318.47 = 0.2 T (material)
M =
B − μ 0 H
μ 0
= 0
(vacuum)
M =
B
μ 0
− H =
0.2
4π × 10 −7 − 318 = 1.59 × 10
5
(material)
13.48 Use the formula for B(z) on the axis of a circular coil of radius r carrying
current i: