13.3 Solutions
621
an external field of magnetic induction B 1 . A length l of the segment
AB will experience a sideway magnetic force equal to il × B whose
magnitude is
F 2 = I 2 l B 1 =
μ 0 l I 1 I 2
2π d
(2)
The vector rule of signs shows that F 2 lies in the plane of the coil and
points to the right. A similar reasoning shows that the force on the segment CD will be
F 3 =
μ 0 l I 1 I 2
2π (d + l)
(3)
to the left in the plane of the coil. There is no force on the segments BC
and AD as they are perpendicular on PQ:
F net =
μ 0 l I 1 I 2
2π
1
d
−
1
d + l
(4)
to the right
F net =
(4π × 10 −7 )(0.05)(1)(4)(10)
2π
1
0.1
−
1
0.1 + 0.05
= 1.335 × 10
−5 N
where we have multiplied (4) by 10 for the number of coils.
13.59 (a) The flux φ B enclosed by the loop, Fig. 13.30, is
φ B = Blx
(1)
where lx is the area of that part of the loop in which B is not zero. By
Faraday’s law
ξ = −
dφ B
dt
= −
d
dt
(Blx) = −Bl
dx
dt
= Bl v
(2)
Fig. 13.30
621
an external field of magnetic induction B 1 . A length l of the segment
AB will experience a sideway magnetic force equal to il × B whose
magnitude is
F 2 = I 2 l B 1 =
μ 0 l I 1 I 2
2π d
(2)
The vector rule of signs shows that F 2 lies in the plane of the coil and
points to the right. A similar reasoning shows that the force on the segment CD will be
F 3 =
μ 0 l I 1 I 2
2π (d + l)
(3)
to the left in the plane of the coil. There is no force on the segments BC
and AD as they are perpendicular on PQ:
F net =
μ 0 l I 1 I 2
2π
1
d
−
1
d + l
(4)
to the right
F net =
(4π × 10 −7 )(0.05)(1)(4)(10)
2π
1
0.1
−
1
0.1 + 0.05
= 1.335 × 10
−5 N
where we have multiplied (4) by 10 for the number of coils.
13.59 (a) The flux φ B enclosed by the loop, Fig. 13.30, is
φ B = Blx
(1)
where lx is the area of that part of the loop in which B is not zero. By
Faraday’s law
ξ = −
dφ B
dt
= −
d
dt
(Blx) = −Bl
dx
dt
= Bl v
(2)
Fig. 13.30