P31_024.PDF

24. (a) We assume the ﬂux is entirely due to the ﬁeld generated by the long straight wire (which is given

by Eq. 30-19). We integrate according to Eq. 31-3, not worrying about the possibility of an overall

minus sign since we are asked to ﬁnd the absolute value of the ﬂux.

Φ B |

= r + b/ 2

r − b/ 2

µ 0 i

2 πr

( adr )= µ 0 ia

2 π

ln r + 2

dt = v . The magnitude of the induced emf divided by the loop resistance then gives the induced

current:

i loop =

dt ln r + 2

−

µ 0 ia

2 πR

µ 0 iabv

2 πR ( r 2

( b/ 2) 2 ) .

−

(b) Implementing Faraday’s law involves taking a derivative of the ﬂux in part (a), and recognizing that