Version PREVIEW – HW8 – feng – (58230)
1
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Holt SF 21Rev 31
001
10.0 points
A proton moves eastward in the plane of
Earth’s magnetic equator so that its distance
from the ground remains constant.
The acceleration of gravity is 9
.
81 m
/
s
2
and the charge on a proton is 1.60
×
10
−
19
.
What is the speed of the proton if Earth’s
magnetic Feld points north and has a magni
tude of 5
.
6
×
10
−
5
T?
Correct answer: 0
.
0018313 m
/
s.
Explanation:
Let :
m
= 1
.
673
×
10
−
27
kg
,
B
= 5
.
6
×
10
−
5
T
,
q
e
= 1
.
60
×
10
−
19
C
,
and
g
= 9
.
81 m
/
s
2
.
The magnetic and gravitational forces are
equal:
F
m
=
F
g
q v B
=
mg
v
=
mg
q B
=
(1
.
67262
×
10
−
27
kg) (9
.
81 m
/
s
2
)
(1
.
6
×
10
−
19
C) (5
.
6
×
10
−
5
T)
=
0
.
0018313 m
/
s
Circular Motion of a Particle
002
(part 1 of 3) 10.0 points
Consider the circular motion of a
positively
charged particle in the plane of this paper,
due to a constant magnetic Feld
V
B
which
points out of the paper. Neglect the e±ect
due to gravity.
B
B
What is the direction of the orbital motion
of the particle?
1.
clockwise
correct
2.
Insu²cient information
3.
counterclockwise
Explanation:
To maintain a circular orbit, the magnetic
force on the particle must be directed toward
the center of the circle. ³rom the right hand
rule we see that orbit must be in the
clock
wise
direction for the force to point radially
inwards.
003
(part 2 of 3) 10.0 points
What is the radius of the orbit?
1.
r
=
mv
q B
correct
2.
r
=
mv
2
q B
3.
r
=
q v
2
mB
4.
r
=
q m
v B
5.
r
=
q v
mB
6.
r
=
mB
q v
7.
r
=
v B
q m
8.
r
=
q B
mv
2
9.
r
=
q B
mv
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2
10.
r
=
v
2
B
q m
Explanation:
From Newton’s second law
F
=
q v B
=
m
v
2
r
r
=
mv
q B
.
004
(part 3 of 3) 10.0 points
Suppose the tangential speed of the charged
particle is increased to twice the original tan
gential speed. (
v
new
= 2
v
old
).
V
B
is ±xed.
What is the ratio of the new frequency to
the old frequency?
Note:
A unit of frequency
is a revolution per second.
1.
f
new
f
old
=
1
3
2.
f
new
f
old
=
1
4
3.
f
new
f
old
= 4
4.
f
new
f
old
=
1
2
5.
f
new
f
old
= 1
correct
6.
f
new
f
old
= 3
7.
f
new
f
old
= 2
8.
f
new
f
old
=
1
6
9.
f
new
f
old
= 5
10.
f
new
f
old
=
1
5
Explanation:
For uniform circular motion, the period
(
i.e.
, the time to complete one revolution)
is de±ned by
T
=
2
π r
v
,
and from Eq. 1 above,
v
r
=
q B
m
.
Frequency is
the reciprocal of the period (
i.e.
, the number
of revolutions per unit time), so
f
=
1
T
=
v
2
π r
=
1
2
π
q B
m
.
Hence, if
B
is kept ±xed, the frequency of the
motion of a charged particle is independent of
its tangential speed;
i.e.
,
f
new
f
old
= 1
.
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 Spring '08
 Turner
 Physics, Magnetic Field, Gravitational forces, τm, Rod Rolling

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