31. (a) Letting
x
be the distance from the right end of the rails to the rod, we Fnd an expression for the
magnetic ﬂux through the area enclosed by the rod and rails. By Eq. 3019, the Feld is
B
=
µ
0
i/
2
πr
,
where
r
is the distance from the long straight wire. We consider an inFnitesimal horizontal strip of
length
x
and width
dr
, parallel to the wire and a distance
r
from it; it has area
A
=
xdr
and the
ﬂux
d
Φ
B
=(
µ
0
ix/
2
)
dr
. By Eq. 313, the total ﬂux through the area enclosed by the rod and
rails is
Φ
B
=
µ
0
ix
2
π
Z
a
+
L
a
dr
r
=
µ
0
ix
2
π
ln
µ
a
+
L
a
¶
.
According to ±araday’s law the emf induced in the loop is
E
=
d
Φ
B
dt
=
µ
0
i
2
π
dx
dt
ln
µ
a
+
L
a
¶
=
µ
0
iv
2
π
ln
µ
a
+
L
a
¶
=
(4
π
×
10
−
7
T
·
m
/
A)(100A)(5
.
00 m
/
s)
2
π
ln
µ
1
.
00 cm + 10
.
0cm
1
.
00 cm
¶
=2
.
40
×
10
−
4
V
.
(b) By Ohm’s law, the induced current is
i
`
=
E
/R
=(2
.
40
×
10
−
4
V)
/
(0
.
400 Ω) = 6
.
00
×
10
−
4
A. Since
the ﬂux is increasing the magnetic Feld produced by the induced current must be into the page in
the region enclosed by the rod and rails. This means the current is clockwise.
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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