Magnetic Field and Magnetic Forces
2713
27.44.
IDENTIFY:
sin
IAB
τ
φ
=
, where
is the angle between
B
!
and the normal to the loop.
SET UP:
The coil as viewed along the axis of rotation is shown in Figure 27.44a for its original position and in
Figure 27.44b after it has rotated 30.0
°
.
EXECUTE: (a)
The forces on each side of the coil are shown in Figure 27.44a.
12
0
+
=
FF
!
!
and
34
0
+=
!!
. The
net force on the coil is zero.
0
=
°
and sin
0
=
, so
0
=
. The forces on the coil produce no torque.
(b)
The net force is still zero.
30.0
=
°
and the net torque is
(1)(1.40 A)(0.220 m)(0.350 m)(1.50 T)sin30.0
0.0808 N m
==
⋅
°
. The net torque is clockwise in Figure 27.44b
and is directed so as to increase the angle
.
EVALUATE:
For any current loop in a uniform magnetic field the net force on the loop is zero. The torque on the
loop depends on the orientation of the plane of the loop relative to the magnetic field direction.
Figure 27.44
27.45.
IDENTIFY:
The magnetic field exerts a torque on the currentcarrying coil, which causes it to turn. We can use
the rotational form of Newton&s second law to find the angular acceleration of the coil.
SET UP:
The magnetic torque is given by
=
×
B
!
!
!
τμ
, and the rotational form of Newton&s second law is
I
α
=
∑
. The magnetic field is parallel to the plane of the loop.
EXECUTE: (a)
The coil rotates about axis
A
2
because the only torque is along top and bottom sides of the coil.
(b)
To find the moment of inertia of the coil, treat the two 1.00m segments as pointmasses (since all the points in
them are 0.250 m from the rotation axis) and the two 0.500m segments as thin uniform bars rotated about their
centers. Since the coil is uniform, the mass of each segment is proportional to its fraction of the total perimeter of
the coil. Each 1.00m segment is 1/3 of the total perimeter, so its mass is (1/3)(210 g) = 70 g = 0.070 kg. The mass
of each 0.500m segment is half this amount, or 0.035 kg. The result is
22
2
1
12
2(0.070 kg)(0.250 m)
2 (0.035 kg)(0.500 m)
0.0102 kg m
I
=+
=
⋅
The torque is
=
sin90
(2.00 A)(0.500 m)(1.00 m)(3.00 T) = 3.00 N m
IAB
=×
°
=
⋅
B
!
Using the above values, the rotational form of Newton&s second law gives
2
290 rad/s
I
EVALUATE:
This angular acceleration will not continue because the torque changes as the coil turns.
27.46.
IDENTIFY:
B
!
and
cos
UB
μ
=−
, where
NIB
=
.
sin
B
μφ
=
.
SET UP:
is the angle between
B
!
and the normal to the plane of the loop.
EXECUTE: (a)
±± ±
90 .
sin(90 )
, direction
.
cos
0.
τ
NIAB
NIAB
U
B
φμ
=° =
°
=
× −
=
−
=
kj
=i
(b)
0.
sin(0)
0, no direction.
cos
.
τ
NIAB
U
B
NIAB
=
=
−
=
−
(c)
±±±
90 .
sin(90 )
, direction
.
cos
0.
τ
NIAB
NIAB
U
B
°
=
−×
=
−
=
=
i
(d)
180 :
sin(180 )
0, no direction,
cos(180 )
.
τ
NIAB
U
B
NIAB
=°
=
°
=
=
−
°
=
EVALUATE:
When
is maximum,
0
U
=
. When
U
is maximum,
0
=
.
27.47.
IDENTIFY
and
SET UP:
The potential energy is given by Eq.(27.27):
.
U
=
⋅
B
!
!
The scalar product depends on
the angle between
and .
B
!
!
EXECUTE:
For
and
parallel,
0 and
cos
.
B
B
⋅ =
=
BB
μμ
For
and
antiparallel,
B
!
!
180 and
cos
.
B
B
⋅
=
=
−
!
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 Buchler
 Physics, Force, Magnetic Field, m/s, Magnetic Forces, Magnetic Field and Magnetic Forces

Click to edit the document details