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Physics 2213
HW #7 – Solutions
Spring 2010
F
B
30
o
60
o
B
down
up
N
S
#1 [Magnetic Force on a Cable]
(a)
The force on the cable due to the Earth’s magnetic field
is
FI
LB
=
±
±±
.
The direction will be due South and 30
o
below the horizontal (see diagram on right).
The
magnitude of this force is
F
B
=ILBsin90
o
(90
o
is the angle between
B
±
and
L
±
.)
=(120A)(30m)(0.5x10
4
T)(1)=0.18N
(b)
The weight of the cable is given by
F
g
=mg=
±
Vg=
±
(
²
r
2
L)g=(8.9x10
3
kg/m
3
)(
²
)(0.0025m)
2
³
(30m)(9.8N/kg)
=51.4N
The magnetic force is much smaller than the cable’s weight, so it can be ignored when designing
the cables.
27.19. [Fusion Reactor]
In part (a), apply conservation of energy to the motion of the two nuclei. In part (b) apply
2
/.
qvB mv R
=
(a)
Let point 1 be when the two nuclei are far apart and let point 2 be when they are at their closest
separation.
Apply conservation of energy to the motion of the two nuclei.
11 2 2
KUKU
+=+
.
12
0,
UK
==
so
KU
=
.
Now
U
2
=ke
2
/r
, while
2
1
1
2
2
Km
v
=
±
(because there are
two deuterium nuclei, each of mass
m
, moving at speed
v
) so
22
mv
ke r
=
.
19
6
27
15
(1.602 10
C)
8.4 10 m s
(3.34 10
kg)(1.0 10
m)
kk
ve
mr
±
²
=
²
²²
(3% of the speed of light!)
(b)
Recall that a charged particle moving perpendicular to a uniform magnetic field will travel in a
circle with
2
=
Then
27
6
19
(3.34 10
kg)(8.4 10 m/s)
0.14 T
(1.602 10
C)(1.25 m)
mv
B
qr
±
±
=
²
.
27.75. [Swinging Loop]
The setup is shown to the right.
For the loop to be in
equilibrium the net torque on it must be zero. The total torque about the pivot
point A is the sum of the torque from gravity and the torque from the
magnetic force.
Use
B
±
μ
=
²
±
to calculate the torque due to the magnetic field
and use
rF
=
²
±
for the torque due to the gravity force.
Note that, if the
angle with the
yz
plane were zero, the gravitational torque would be zero, and
the magnetic torque would dominate.
If the angle were 90
o
, the magnetic torque would vanish,
and gravity would dominate.
Therefore, there must be an equilibrium angle in between, and this
angle will increase with
B
.
First, we compute the gravitational torque:
sin
(0.400 m)sin30.0
mg
mgr
mg
±²
°
The torque is clockwise;
mg
±
is
directed into the paper.
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For the loop to be in equilibrium the torque due to
B
±
must be counterclockwise (opposite to
mg
±
±
)
and it must be that
.
Bm
g
±±
=
.
B
=
±
B
±
²
μ
For this torque to be
counterclockwise (
B
±
directed out of
the paper),
B
±
must be in the
direction.
y
+
sin
sin 60.0
B
BI
A
B
==
°
()
gives
sin60.0
0.0400 m sin30.0
g
IAB
mg
=°
=
°
(
)
3
0.15 g/cm 2 8.00 cm
6.00 cm
4.2 g
4.2 10 kg
m
±
=+
=
=
²
(
)
32
0.080 m
0.0600 m
4.80 10 m
A
±
²
(
)
0.0400 m
sin30.0
sin60.0
mg
B
IA
°
=
°
(
)
9.80 m/s
0.0400 m sin30.0
0.024 T
(8.2 A)(4.80 10 m )sin60.0
B
±
±
²
°
²
°
27.28. [Crossed E & B Fields]
(a)
For no deflection the magnetic and electric forces must be equal in magnitude and opposite in
direction.
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 Spring '08
 PERELSTEIN
 Force

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