song (shs546) – Practice HW 11 Solutions – weathers – (22202)
1
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printout
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have
14
questions.
Multiplechoice questions may continue on
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before answering.
001
0.0points
A positively charged particle passes through a
laboratory traveling in an easterly direction.
There are both electric and magnetic field
in the room and their effects on the charged
particle cancel.
If the electric field points upward, what
must be the direction of the magnetic field?
1.
east
2.
west
3.
south
correct
4.
north
5.
upward
6.
downward
Explanation:
The force on the object due to the magnetic
field must point down in order to cancel the
force due to the electric field. Using the right
hand rule, the magnetic field must point south
in order for the magnetic force to be directed
downward.
002
0.0points
An electron in a vacuum is first accelerated
by a voltage of 81700 V and then enters a
region in which there is a uniform magnetic
field of 0
.
444 T at right angles to the direction
of the electron’s motion.
The mass of the electron is 9
.
11
×
10
−
31
kg
and its charge is 1
.
60218
×
10
−
19
C.
What is the magnitude of the force on the
electron due to the magnetic field?
Correct answer: 1
.
20591
×
10
−
11
N.
Explanation:
Let :
V
= 81700 V
,
B
= 0
.
444 T
,
m
= 9
.
11
×
10
−
31
kg
,
q
e
= 1
.
60218
×
10
−
19
C
.
The kinetic energy
K
gained after acceler
ation is
K
=
1
2
m v
2
=
q
e
V
, so the velocity
is
v
=
radicalbigg
2
q
e
V
m
=
radicalBigg
2 (1
.
60218
×
10
−
19
C)(81700 V)
9
.
11
×
10
−
31
kg
= 1
.
6952
×
10
8
m
/
s
.
Then the force on it is
f
=
q v B
= (1
.
60218
×
10
−
19
C)
×
(1
.
6952
×
10
8
m
/
s) (0
.
444 T)
=
1
.
20591
×
10
−
11
N
.
003(part1of2)0.0points
An electron is projected into a uniform mag
netic field given by
vector
B
=
B
z
ˆ
k
+
B
x
ˆ
ı
, where
B
z
= 4
.
9 T and
B
x
= 1
.
7 T.
The magnitude of the charge on an electron
is 1
.
60218
×
10
−
19
C
.
z
x
y
v
= 3
.
9
×
10
5
m
/
s
electron
4
.
9 T
1
.
7 T
B
Find the direction of the magnetic force
when the velocity of the electron is
v
ˆ
ı
, where
v
= 3
.
9
×
10
5
m
/
s.
1.
hatwide
F
= ˆ
correct
2.
hatwide
F
=
−
ˆ
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song (shs546) – Practice HW 11 Solutions – weathers – (22202)
2
3.
hatwide
F
= ˆ
ı
4.
hatwide
F
=
−
ˆ
ı
5.
hatwide
F
=
−
ˆ
k
6.
hatwide
F
=
ˆ
k
7.
hatwide
F
=
1
√
2
parenleftBig
ˆ
k
+ˆ
ı
parenrightBig
8.
hatwide
F
=
1
√
2
parenleftBig
ˆ
ı
−
ˆ
k
parenrightBig
9.
hatwide
F
=
1
√
2
parenleftBig
ˆ
k
−
ˆ
ı
parenrightBig
Explanation:
Let :
q
= 1
.
60218
×
10
−
19
C
,
B
z
= 4
.
9 T
,
and
B
x
= 1
.
7 T
.
Basic Concepts:
Magnetic force on a mov
ing charge is given by
vector
F
=
qvectorv
×
vector
B .
Solution:
vector
B
= (4
.
9 T)
ˆ
k
+ (1
.
7 T) ˆ
ı
v
= (3
.
9
×
10
5
m
/
s) ˆ
ı
for the electron.
Find:
The vector expression for the force on
the electron. This solves both part 1 and part
2.
We will go through two methods of doing
the problem.
The first is more mathematically oriented
and the second uses more of a reasoning argu
ment.
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 Spring '00
 Littler
 Charge, Magnetic Force, Magnetic Field, Magnetic moment, Fmag

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