24. (a) We assume the flux is entirely due to the field generated by the long straight wire
(which is given by Eq. 29-17). We integrate according to Eq. 30-1, not worrying about
the possibility of an overall minus sign since we are asked to find the absolute value of
the flux.
µ 0ia § r + b2 ·
§ µ 0i ·
a
dr
=
(
)
ln ¨
.
b ¸
r − b / 2 ¨ 2πr ¸
2π
©
¹
©r−2¹
| Φ B |= ³
r +b / 2
When r = 1.5b , we have
| ΦB | =
(4π × 10 −7 T ⋅ m A)(4.7A)(0.022m)
ln(2.0) = 1.4 ×10−8 Wb.
2π
(b) Implementing Faraday’s law involves taking a derivative of the flux in part (a), and
recognizing that drdt = v . The magnitude of the induced emf divided by the loop resistance
then gives the induced current:
iloop =
ε
R
=−
µ 0ia d
§ r + b2 ·
µ 0iabv
=
ln ¨
b ¸
2πR dt © r − 2 ¹ 2πR[r 2 − (b / 2) 2 ]
(4π ×10 −7 T ⋅ m A)(4.7A)(0.022m)(0.0080m)(3.2 ×10−3 m/s)
=
2π (4.0 ×10−4 Ω)[2(0.0080m) 2 ]
= 1.0 ×10−5 A.