kavo (ak22862) – HW09 – Janow – (12121)
1
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001
10.0 points
Find the magnitude of the magnetic force on
a proton moving with velocity 4
.
38 Mm
/
s in
the positive
x
direction in a magnetic field of
1
.
8 T in the positive
z
direction.
Correct answer: 1
.
26316 pN.
Explanation:
Let :
q
= 1
.
60218
×
10
−
19
C
,
vectorv
= (4
.
38 Mm
/
s) ˆ
ı
= (4
.
38
×
10
6
m
/
s) ˆ
ı ,
and
vector
B
= (1
.
8 T)
ˆ
k .
vector
F
=
qvectorv
×
vector
B
= (1
.
60218
×
10
−
19
C)
×
bracketleftbig
(4
.
38
×
10
6
m
/
s) ˆ
ı
bracketrightbig
×
bracketleftBig
(1
.
8 T)
ˆ
k
bracketrightBig
×
10
12
pN
1 N
=
−
(1
.
26316 pN) ˆ
,
so the magnitude is
1
.
26316 pN
.
002
(part 1 of 2) 10.0 points
An electron is projected into a uniform mag
netic field given by
vector
B
=
B
x
ˆ
ı
+
B
y
ˆ
, where
B
x
= 4
.
7 T and
B
y
= 1
.
9 T.
The magnitude of the charge on an electron
is 1
.
60218
×
10
−
19
C
.
x
y
z
v
= 3
.
9
×
10
5
m
/
s
electron
4
.
7 T
1
.
9 T
B
Find the direction of the magnetic force
when the velocity of the electron is
v
ˆ
, where
v
= 3
.
9
×
10
5
m
/
s.
1.
hatwide
F
=
1
√
2
(ˆ
ı
+ ˆ
)
2.
hatwide
F
=
1
√
2
(ˆ
ı
−
ˆ
)
3.
hatwide
F
=
−
ˆ
4.
hatwide
F
=
−
ˆ
ı
5.
hatwide
F
=
1
√
2
(ˆ
−
ˆ
ı
)
6.
hatwide
F
=
−
ˆ
k
7.
hatwide
F
=
ˆ
k
correct
8.
hatwide
F
= ˆ
ı
9.
hatwide
F
= ˆ
Explanation:
Let :
q
= 1
.
60218
×
10
−
19
C
,
B
x
= 4
.
7 T
,
and
B
y
= 1
.
9 T
.
Basic Concepts:
Magnetic force on a mov
ing charge is given by
vector
F
=
qvectorv
×
vector
B .
Solution:
vector
B
= (4
.
7 T) ˆ
ı
+ (1
.
9 T) ˆ
v
= (3
.
9
×
10
5
m
/
s) ˆ
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.
Method 1:
The force acting on a charge
q
with velocity
vectorv
in the presence of an external
magnetic field
vector
B
is given by
vector
F
=
qvectorv
×
vector
B
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kavo (ak22862) – HW09 – Janow – (12121)
2
Taking the cross product of
vectorv
with
vector
B
we
obtain
vector
F
=
qvectorv
×
vector
B
=
q
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
ˆ
ı
ˆ
ˆ
k
0
v
0
B
x
B
y
0
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
q
braceleftBig
[(
B
y
)(0)
−
(
B
x
)(
v
)]
ˆ
k
−
[(0)(0)
−
(
B
x
)(0)] ˆ
+ [(
v
)(0)
−
(
B
x
)(0)] ˆ
ı
bracerightBig
=
−
q B
x
v
ˆ
k
=
−
(
−
1
.
60218
×
10
−
19
C)(4
.
7 T)
×
(3
.
9
×
10
5
m
/
s)
ˆ
k
= (2
.
93679
×
10
−
13
N)
ˆ
k ,
and the direction is +
ˆ
k .
Method 2:
The other method is to real
ize that the only component of the magnetic
field which affects the electron is the compo
nent perpendicular to its velocity. Therefore,
F
=
q

vectorv
×
vector
B

=
q v B
⊥
with the direction
given by the right hand rule to be in the neg
ative
ˆ
k
direction; but recalling to reverse the
direction because the electron has a negative
instead of positive charge.
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 Fall '08
 Opyrchal
 Physics, Force, Magnetic Field, Fgrav

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