13.3 Solutions
625
13.66 U = −p · E = −p E cos θ = −(1.6 × 10 −29 )(1000) cos 30 ◦ = −1.38 ×
10 −26 J.
13.3.5 Faraday’s Law
13.67
dr
dt
α t
(by problem)
(1)
ξ = −
dφ
dt
=
d
dt
(BA)
(by Faraday’s law)
(2)
But
B ∝
1
r
and A = πr
2
(3)
∴
ξ ∝
dr
dt
(4)
or
ξ ∝ t
(5)
where we have used (1)
13.68 0.4 Oe = 0.4 × 80 A/m = 32 A/m
B = μ 0 H = 4π × 10
−7
× 32 = 4.02 × 10
−5 T
v = 720 km/h = 200 m/s
ξ = Blv = 4.02 × 10
−5
× 30 × 200 = 0.24 V
13.69 V = πr 2 B f = π(0.1) 2 × 2 × 10 −5 × 5 = 3.14 × 10 −6 V
13.70 R =
BAN
q
=
0.1 × 0.001 × 30
10 −5
= 300
13.71 ξ = −
φ
t
=
AB
t
=
0.2 × 0.25
10 −4
= 500 V
i =
ξ
R
=
500
20
= 25 A
13.72 The amplitude of the induced voltage
ξ 0 = ωB A = 2π fBA = 2π × 40 × 0.5 × (0.25)
2
= 7.85 V
Amplitude of the induced current
I 0 =
ξ 0
R
=
7.85
4
= 1.96 A
13.73 ξ = i R = 0.4 × 5 = 2 V