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garza (olg97) – Homework No. 6 – jimenez – (130201)
1
This printout should have 19 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
(part 1 oF 2) 10.0 points
The magnetic feld oF the Earth at a certain
location is directed vertically downward and
has a magnitude oF 7
.
1
×
10

5
T. A proton
is moving horizontally toward the west in this
feld with a speed oF 2
.
4
×
10
6
m
/
s.
The charge on the proton is 1
.
60218
×
10

19
C and its mass is 1
.
67262
×
10

27
kg.
What is the magnitude oF the magnetic
Force the feld exerts on this charge?
Correct answer: 2
.
73011
×
10

17
N.
Explanation:
Let :
q
=1
.
60218
×
10

19
C
,
v
=2
.
4
×
10
6
m
/
s
,
and
B
=7
.
1
×
10

5
T
.
The Lorentz Force is
F
=
q v B
= (1
.
60218
×
10

19
C) (2
.
4
×
10
6
m
/
s)
×
(7
.
1
×
10

5
T)
=
2
.
73011
×
10

17
N
.
002
(part 2 oF 2) 10.0 points
What is the radius oF the circular arc Followed
by this proton?
Correct answer: 352
.
891 m.
Explanation:
Let :
m
.
67262
×
10

27
kg
.
The Lorentz Force is the centripetal Force oF
the circular motion:
q v B
=
m
v
2
r
.
r
=
mv
qB
=
(1
.
67262
×
10

27
kg) (2
.
4
×
10
6
m
/
s)
(1
.
60218
×
10

19
C) (7
.
1
×
10

5
T)
=
352
.
891 m
.
003
10.0 points
A3
.
09013
μ
C charged particle with a kinetic
energy oF 0
.
0407885 J is placed in a uniForm
magnetic feld oF magnitude 0
.
0663511 T.
IF the particle moves in a circular path oF
radius 1
.
96214 m, fnd its mass.
Correct answer: 1
.
984
×
10

12
kg.
Explanation:
Let :
q
=3
.
09013
μ
C = 3
.
09013
×
10

6
C
,
K
=0
.
0407885 J
,
B
.
0663511 T
,
and
R
.
96214 m
.
Kinetic energy is
K
=
1
2
2
.
The magnetic feld supplies the centripetal
Force, so
F
c
=
F
mag
2
R
=
q v B
=
q RB
m
2
v
2
=(
)
2
m
±
1
2
2
²
=
1
2
(
)
2
mK
=
1
2
(
)
2
m
=
(
q B R
)
2
2K
=
(
3
.
09013
×
10

6
C
)
2
(0
.
0663511 T)
2
(1
.
96214 m)
2
2 (0
.
0407885 J)
=
1
.
984
×
10

12
kg
.
004
(part 1 oF 2) 10.0 points
An electron moves in a circular path perpen
dicular to a constant magnetic feld oF magni
tude 1
.
13 mT.
The charge on an electron is
q
e
.
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View Full Documentgarza (olg97) – Homework No. 6 – jimenez – (130201)
2
If the angular momentum of the elec
tron about the center of the circle is
4
.
88
×
10

25
J
·
s, Fnd the radius of the circu
lar path.
Correct answer: 5
.
19206 cm.
Explanation:
Let :
q
=1
.
602
×
10

19
C
,
L
=4
.
88
×
10

25
J
·
s
,
and
B
.
13 mT = 0
.
00113 T
.
The magnetic force on a moving charge is
q v B
=
mv
2
R
,
q RB
=
mv ,
and the angular momentum is
L
=
mv R
=
qR
2
B
R
=
±
L
qB
=
±
4
.
88
×
10

25
J
·
s
(1
.
602
×
10

19
C) (0
.
00113 T)
×
²
100 cm
1m
³
=
5
.
19206 cm
.
005
(part 2 of 2) 10.0 points
±ind the speed of the electron.
Correct answer: 1
.
03183
×
10
7
m
/
s.
Explanation:
Let :
m
=9
.
109
×
10

31
kg
.
L
=
v
=
L
mR
=
4
.
88
×
10

25
J
·
s
(9
.
109
×
10

31
kg) (5
.
19206 cm)
×
²
100 cm
³
=
1
.
03183
×
10
7
m
/
s
.
006
10.0 points
An electron is accelerated by a 2
.
8 kV poten
tial di²erence.
The charge on an electron is 1
.
60218
×
10

19
C and its mass is 9
.
10939
×
10

31
kg.
How strong a magnetic Feld must be expe
rienced by the electron if its path is a circle of
radius 5
.
3 cm?
Correct answer: 0
.
00336672 T.
Explanation:
Let :
r
=5
.
3 cm = 0
.
053 m
,
V
=2
.
8 kV = 2800 V
,
m
.
10939
×
10

31
kg
,
and
q
.
60218
×
10

19
C
.
±rom Newton’s second law,
F
=
q v B
=
2
r
v
=
q B r
m
.
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