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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
= 1
.
67262
×
10

27
kg
.
The Lorentz Force is the centripetal Force oF
the circular motion:
q v B
=
m
v
2
r
.
r
=
mv
q B
=
(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
A 3
.
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
= 0
.
0663511 T
,
and
R
= 1
.
96214 m
.
Kinetic energy is
K
=
1
2
mv
2
.
The magnetic feld supplies the centripetal
Force, so
F
c
=
F
mag
mv
2
R
=
q v B
mv
=
q RB
m
2
v
2
= (
q RB
)
2
m
p
1
2
mv
2
P
=
1
2
(
q RB
)
2
mK
=
1
2
(
q RB
)
2
m
=
(
q B R
)
2
2 K
=
(
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 Document garza (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
= 1
.
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
=
q R
2
B
R
=
r
L
q B
=
r
4
.
88
×
10

25
J
·
s
(1
.
602
×
10

19
C)(0
.
00113 T)
×
p
100 cm
1 m
P
=
5
.
19206 cm
.
005
(part 2 of 2) 10.0 points
±ind the speed of the electron.
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This note was uploaded on 07/30/2009 for the course CHEM 101 taught by Professor Williamson during the Spring '08 term at Texas A&M.
 Spring '08
 Williamson

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