aljabr (faa335) – Hw12 – Ross – (89251)
1
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printout
should
have
15
questions.
Multiplechoice questions may continue on
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before answering.
001
(part 1 of 3) 10.0 points
The magnetic field over a certain range is
given by
vector
B
=
B
x
ˆ
ı
+
B
y
ˆ
, where
B
x
= 9 T
and
B
y
= 3 T.
An electron moves into the
field with a velocity
vectorv
=
v
x
ˆ
ı
+
v
y
ˆ
+
v
z
ˆ
k
, where
v
x
= 8 m
/
s,
v
y
= 2 m
/
s and
v
z
= 2 m
/
s.
The charge on the electron is
−
1
.
602
×
10
−
19
C.
What is the ˆ
ı
component of the force ex
erted on the electron by the magnetic field?
Correct answer: 9
.
612
×
10
−
19
N.
Explanation:
Let :
v
x
= 8 m
/
s
,
v
y
= 2 m
/
s
,
v
z
= 2 m
/
s
,
q
=
−
1
.
602
×
10
−
19
C
,
B
x
= 9 T
,
and
B
y
= 3 T
.
The force exerted on the electron by the mag
netic field is given by
vector
F
=
qvectorv
×
vector
B
=
q
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
ˆ
ı
ˆ
ˆ
k
v
x
v
y
v
z
B
x
B
y
0
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
q
[
−
v
z
B
y
ˆ
ı
+
v
z
B
x
ˆ
+ (
v
x
B
y
−
v
y
B
x
)
ˆ
k
]
= (
−
1
.
602
×
10
−
19
C)
{−
(2 m
/
s) (3 T)ˆ
ı
+ (2 m
/
s) (9 T)ˆ
+ [(8 m
/
s) (3 T)
−
(2 m
/
s) (9 T)]
ˆ
k
}
= (9
.
612
×
10
−
19
N)ˆ
ı
+ (
−
2
.
8836
×
10
−
18
N)ˆ
+ (
−
9
.
612
×
10
−
19
N)
ˆ
k
002
(part 2 of 3) 10.0 points
What is the ˆ
component of the force?
Correct answer:
−
2
.
8836
×
10
−
18
N.
Explanation:
See the above part.
003
(part 3 of 3) 10.0 points
What is the
ˆ
k
component of the force?
Correct answer:
−
9
.
612
×
10
−
19
N.
Explanation:
See the above part.
004
10.0 points
A positively charged particle moving paral
lel to the
y
axis enters a magnetic field (point
ing into of the page), as shown in the figure
below.
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
y
z
v
x
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
ı
3.
hatwide
F
= +
ˆ
k
correct
4.
hatwide
F
= +ˆ
5.
vector
F
= 0 ; no deflection
6.
hatwide
F
=
−
ˆ
k
7.
hatwide
F
=
−
ˆ
Explanation:
Basic Concepts:
Magnetic Force on a
Charged Particle:
vector
F
=
qvectorv
×
vector
B
Righthand rule for crossproducts.
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aljabr (faa335) – Hw12 – Ross – (89251)
2
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
(
−
ˆ
ı
)
,
vectorv
=
v
(+ˆ
)
,
and
q >
0
,
therefore
,
vector
F
= +

q

vectorv
×
vector
B
= +

q

v B
[(+ˆ
)
×
(
−
ˆ
ı
)]
= +

q

v B
parenleftBig
+
ˆ
k
parenrightBig
hatwide
F
=
+
ˆ
k
.
This is the first of eight versions of the
problem.
005
10.0 points
A positively charged particle moving at 45
◦
angles to both the
z
axis and
x
axis enters a
magnetic field (pointing out of of the page),
as shown in the figure below.
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 Spring '10
 Gavrine
 Electron, Magnetic Force, Electric charge, Ion, Lorentz force

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