P31_031.PDF

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Chapter 31 - 31.31
31. (a) Letting x be the distance from the right end of the rails to the rod, we find an expression for the
magnetic flux through the area enclosed by the rod and rails. By Eq. 30-19, the field is B = µ 0 i/ 2 πr ,
where r is the distance from the long straight wire. We consider an infinitesimal 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 πr ) dr . By Eq. 31-3, the total flux through the area enclosed by the rod and
rails is
Φ B = µ 0 ix
2 π
a + L
dr
r = µ 0 ix
ln a + L
a
.
2 π
a
According to Faraday’s law the emf induced in the loop is
E
=
d Φ B
dt
= µ 0 i
2 π
dx
dt ln a + L
= µ 0 iv
2 π
ln a + L
a
a
ln 1 . 00cm+10 . 0cm
1 . 00cm
=
(4 π
×
10 − 7 T
·
m / A)(100A)(5 . 00m / s)
2 π
=2 . 40
×
10 − 4 V .
10 − 4 A. Since
the flux is increasing the magnetic field 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.
(c) Thermal energy is being generated at the rate P = i 2 R =(6 . 00
E
/R =(2 . 40
×
10 − 4 V) / (0 . 400Ω) = 6 . 00
×
10 − 7 W.
(d) Since the rod moves with constant velocity, the net force on it is zero. The force of the external
agent must have the same magnitude as the magnetic force and must be in the opposite direction.
The magnitude of the magnetic force on an infinitesimal segment of the rod, with length dr at a
distance r from the long straight wire, is dF B = i Bdr =( µ 0 i i/ 2 πr ) dr . We integrate to find the
magnitude of the total magnetic force on the rod:
×
10 − 4 A) 2 (0 . 400Ω) = 1 . 44
×
µ 0 i i
2 π
a + L
dr
r = µ 0 i i
ln a + L
a
F B =
2 π
a
ln 1 . 00cm+10 . 0cm
1 . 00cm
=
(4 π
×
10 − 7 T
·
m / A)(6 . 00
×
10 − 4 A)(100A)
2 π
=2 . 87
×
10 − 8 N .
Since the field is out of the page and the current in the rod is upward in the diagram, the force
associated with the magnetic field is toward the right. The external agent must therefore apply a
force of 2 . 87
×
×
10 − 8 N)(5 . 00m / s) =
10 − 7 W. This is the same as the rate at which thermal energy is generated in the rod. All
the energy supplied by the agent is converted to thermal energy.
×
(b) By Ohm’s law, the induced current is i =
10 − 8 N, to the left.
(e) By Eq. 7-48, the external agent does work at the rate P = Fv =(2 . 87
1 . 44
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