P31_024.PDF
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Chapter 31 - 31.24
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
r
b
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
R
=
−
µ
0
ia
2
πR
d
=
µ
0
iabv
2
πR
(
r
2
(
b/
2)
2
)
.
r
−
b
2
−
|
−
(b) Implementing Faraday’s law involves taking a derivative of the flux in part (a), and recognizing that
dr
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