Magnetic Field and Magnetic Forces
2723
sin
sin60.0
B
BI
A
B
τμ
φ
==
°
()
gives
sin60.0
0.0400 m sin30.0
Bm
g
IAB
mg
ττ
=°
=
°
(
)
3
0.15 g/cm 2 8.00 cm
6.00 cm
4.2 g
4.2 10 kg
m
−
=+
=
=
×
(
)
32
0.800 m 0.0600 m
4.80 10 m
A
−
×
(
)
0.0400 m sin30.0
sin60.0
mg
B
IA
°
=
°
( )( )
4.2 10 kg 9.80 m/s
0.0400 m sin30.0
0.024 T
(8.2 A)(4.80 10 m )sin60.0
B
−
−
×°
EVALUATE:
As the loop swings up the torque due to
B
!
decreases to zero and the torque due to
mg
increases
from zero, so there must be an orientation of the loop where the net torque is zero.
27.76.
IDENTIFY:
The torque exerted by the magnetic field is
.
=
×
B
!
!
!
τ
μ
The torque required to hold the loop in place is
.
−
!
SET UP:
.
IA
=
!
is normal to the plane of the loop, with a direction given by the righthand rule that is
illustrated in Figure 27.32 in the textbook.
sin ,
IAB
=
where
is the angle between the normal to the loop and
the direction of
.
B
!
EXECUTE:
(a)
sin 60
(15.0 A)(0.060 m)(0.080 m)(0.48 T)sin60
0.030 N m
τ
IAB
=
°
=
⋅
, in the
&
−
j
direction.
To keep the loop in place, you must provide a torque in the
&
+
j
direction.
(b)
sin 30
(15.0 A)(0.60 m)(0.080 m)(0.48 T)sin30
0.017 N m,
τ
IAB
=
°
=
⋅
in the
&
+
j
direction. You must
provide a torque in the
&
−
j
direction to keep the loop in place.
EVALUATE:
(c)
If the loop was pivoted through its center, then there would be a torque on both sides of the loop
parallel to the rotation axis. However, the lever arm is only half as large, so the total torque in each case is identical
to the values found in parts (a)
and
(b).
27.77.
IDENTIFY:
Use Eq.(27.20) to calculate the force and then the torque on each small section of the rod and
integrate to find the total magnetic torque. At equilibrium the torques from the spring force and from the magnetic
force cancel. The spring force depends on the amount
x
the spring is stretched and then
2
1
2
Uk
x
=
gives the energy
stored in the spring.
(a) SET UP:
Divide the rod into infinitesimal sections of
length
dr
, as shown in Figure 27.77.
Figure 27.77
EXECUTE:
The magnetic force on this section is
I
dF
IBdr
=
and is perpendicular to the rod. The torque
d
due to
the force on this section is
.
I
dr
d
FI
B
r
d
r
The total torque is
2
1
2
0
0.0442 N m, clockwise.
l
dI
B
r
d
rI
l
B
=
⋅
∫∫
(b) SET UP:
I
F
produces a clockwise torque so the spring force must produce a counterclockwise torque. The
spring force must be to the left; the spring is stretched.
EXECUTE:
Find
x
, the amount the spring is stretched:
0,
=
∑
axis at hinge, counterclockwise torques positive
2
1
2
s
i
n
5
3
0
kx l
Il B
°−
=
(
)
( )
6.50 A 0.200 m 0.340 T
0.05765 m
2 sin53.0
2 4.80 N/m sin53.0
IlB
x
k
=
°°
23
1
2
7.98 10 J
x
−
×
EVALUATE:
The magnetic torque calculated in part (a) is the same torque calculated from a force diagram in
which the total magnetic force
I
l
B
=
acts at the center of the rod. We didn’t include a gravity torque since the
problem said the rod had negligible mass.
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Chapter 27
27.78.
IDENTIFY:
Apply
I
→
×
F=l B
!
!
to calculate the force on each side of the loop.
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 Spring '06
 Buchler
 Physics, Force

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