dzifanu (sd26397) – hw 9 – opyrchal – (121102)
1
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printout
should
have
15
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0points
An electron in a vacuum is first accelerated
by a voltage of 11000 V and then enters a
region in which there is a uniform magnetic
field of 0
.
697 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: 6
.
94626
×
10
−
12
N.
Explanation:
Let :
V
= 11000 V
,
B
= 0
.
697 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)(11000 V)
9
.
11
×
10
−
31
kg
= 6
.
22024
×
10
7
m
/
s
.
Then the force on it is
f
=
q v B
= (1
.
60218
×
10
−
19
C)
×
(6
.
22024
×
10
7
m
/
s) (0
.
697 T)
=
6
.
94626
×
10
−
12
N
.
002
10.0points
A negatively charged particle moving paral
lel to the
z
axis enters a magnetic field (point
ing toward the righthand side of the page),
as shown in the figure below.
z
x
v
y
vector
B
vector
B
−
q
Figure:
ˆ
ı
is in the
x
direction, ˆ
is
in the
y
direction, and
ˆ
k
is in the
z
direction.
What is the initial direction of deflection?
1.
hatwide
F
=
−
ˆ
ı
2.
hatwide
F
= +ˆ
3.
hatwide
F
=
−
ˆ
4.
vector
F
= 0 ; no deflection
correct
5.
hatwide
F
= +
ˆ
k
6.
hatwide
F
=
−
ˆ
k
7.
hatwide
F
= +ˆ
ı
Explanation:
Basic Concepts:
Magnetic Force on a
Charged Particle:
vector
F
=
qvectorv
×
vector
B
Righthand rule for crossproducts.
hatwide
F
≡
vector
F
bardbl
vector
F
bardbl
;
i.e.
, a unit vector in the
F
direc
tion.
Solution:
The force is
vector
F
=
qvectorv
×
vector
B
.
vector
B
=
B
parenleftBig
+
ˆ
k
parenrightBig
,
vectorv
=
v
parenleftBig
−
ˆ
k
parenrightBig
,
and
q <
0
,
therefore
,
vector
F
=
−
q

vectorv
×
vector
B
=
−
q

v B
bracketleftBigparenleftBig
−
ˆ
k
parenrightBig
×
parenleftBig
+
ˆ
k
parenrightBigbracketrightBig
=
−
q

v B
(0)
hatwide
F
=
0 no deflection
.
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dzifanu (sd26397) – hw 9 – opyrchal – (121102)
2
This is the seventh of eight versions of the
problem.
003(part1of2)10.0points
A proton moves perpendicularly to a uniform
magnetic field,
B
, with a speed of 5
.
4
×
10
7
m/s and experiences an acceleration of 5
.
1
×
10
13
m
/
s
2
in the positive
x
direction when its
velocity is in the positive
z
direction.
Find the magnitude of the field.
Correct answer: 0
.
00987535 T.
Explanation:
Let:
q
= 1
.
60
×
10
−
19
C
,
a
= 5
.
1
×
10
13
→
+
x ,
v
= 5
.
4
×
10
7
→
+
z ,
and
m
= 1
.
673
×
10
−
27
kg
.
F
net
=
F
mag
m a
=
q v B
B
=
m a
q v
=
(1
.
673
×
10
−
27
kg) (5
.
1
×
10
13
m
/
s
2
)
(1
.
6
×
10
−
19
C) (5
.
4
×
10
7
m
/
s)
=
0
.
00987535 T
.
004(part2of2)10.0points
What is its direction?
1.
None of these
2.
Positive
z
direction
3.
Positive
y
direction
4.
Negative
y
direction
correct
5.
Positive
x
direction
6.
Negative
x
direction
7.
Negative
z
direction
Explanation:
Righthand rule (positive charge):
Force
directed out of the palm of the hand, fingers
in the direction of the field, thumb in the
direction of the velocity.
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 Spring '09
 PRODAN
 Physics, Magnetic Force, Magnetic Field, Correct Answer, Electric charge

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