lim (kl9356) – Homework 11 – Weathers – (22202)
1
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printout
should
have
12
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
A negatively charged particle moving paral
lel to the
x
axis enters a magnetic field (point
ing out of of the page), as shown in the figure
below.
x
y
v
z
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
k
2.
hatwide
F
= +ˆ
3.
hatwide
F
=
−
ˆ
k
4.
hatwide
F
= +ˆ
ı
5.
hatwide
F
=
−
ˆ
correct
6.
vector
F
= 0 ; no deflection
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
(
−
ˆ
ı
)
,
and
q <
0
,
therefore
,
vector
F
=
−
q

vectorv
×
vector
B
=
−
q

v B
bracketleftBig
(
−
ˆ
ı
)
×
parenleftBig
+
ˆ
k
parenrightBigbracketrightBig
=
−
q

v B
(
−
ˆ
)
hatwide
F
=
−
ˆ
.
This is the eighth of eight versions of the
problem.
002
10.0 points
A proton moving at 3
.
4
×
10
6
m
/
s through a
magnetic field of 6
.
6 T experiences a magnetic
force of magnitude 2
.
5
×
10
−
12
N.
The charge of proton is 1
.
60218
×
10
−
19
C
and the mass of proton is 1
.
67262
×
10
−
27
kg.
What is the angle between the proton’s
velocity and the field?
Correct answer: 44
.
0555
◦
.
Explanation:
Let :
E
= 2
.
5
×
10
−
12
N
,
B
= 6
.
6 T
,
v
= 3
.
4
×
10
6
m
/
s
,
q
p
= 1
.
60218
×
10
−
19
C
,
and
m
p
= 1
.
67262
×
10
−
27
kg
.
The
Lorentz
force
acting
on
a
moving
charged particle in a magnetic field is
F
=
q v B
sin
θ ,
where
q
is the charge,
v
is the speed,
B
is the
magnetic field. So thus the angle between the
velocity and the field is
θ
= arcsin
bracketleftbigg
F
q v B
bracketrightbigg
= arcsin
bracketleftbigg
(2
.
5
×
10
−
12
N)
(1
.
60218
×
10
−
19
C)
×
1
(3
.
4
×
10
6
m
/
s) (6
.
6 T)
bracketrightbigg
=
44
.
0555
◦
.
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lim (kl9356) – Homework 11 – Weathers – (22202)
2
003
(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
= 2 T
and
B
y
= 2 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
= 5 m
/
s,
v
y
= 4 m
/
s and
v
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 Spring '00
 Littler
 Magnetism, Force, Work, Magnetic Field, lim, Electric charge

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