homework 07 – SMITH, TAYLOR – Due: Mar 2 2008, 4:00 am
1
Question 1, chap 30, sect 1.
part 1 of 1
10 points
Suppose a new particle is discovered, and
it is found that a beam of these particles
passes undeflected through “crossed” electric
and magnetic fields, where
E
= 302 V
/
m and
B
= 0
.
00288 T. If the electric field is turned
off, the particles move in the magnetic field in
circular paths of radius
r
= 1
.
33 cm.
Determine
q
m
for the particles from these
data.
Correct answer: 2
.
7376
×
10
9
C
/
kg (tolerance
±
1 %).
Explanation:
Let :
E
= 302 V
/
m
,
q
e
=
q
p
= 1
.
60218
×
10
−
19
C
,
m
e
= 9
.
10939
×
10
−
31
kg
,
m
p
= 1
.
67262
×
10
−
27
kg
,
B
= 0
.
00288 T
,
and
r
= 0
.
0133 m
.
Since the particle passes undeflected, the
electric force on it is equal to the magnetic
force.
When the electric field is turned off,
only the magnetic force is exerted on it as the
centripetal force. So we get two equations
q E
=
B q v
B q v
=
m v
2
r
.
So, the charge to mass ratio for this new
particle is
q
m
=
E
r B
2
=
302 V
/
m
(0
.
0133 m) (0
.
00288 T)
2
=
2
.
7376
×
10
9
C
/
kg
.
For comparison, the electron chargetomass
ratio is
q
e
m
e
=
1
.
60218
×
10
−
19
C
9
.
10939
×
10
−
31
kg
= 1
.
75882
×
10
11
C
/
kg
,
and the proton chargetomass ratio is
q
p
m
p
=
1
.
60218
×
10
−
19
C
1
.
67262
×
10
−
27
kg
= 9
.
57883
×
10
7
C
/
kg
.
Question 2, chap 29, sect 1.
part 1 of 2
10 points
An electron is projected into a uniform
magnetic field given by
vector
B
=
B
x
ˆ
ı
+
B
y
ˆ
,
where
B
x
= 4
.
2 T and
B
y
= 1
.
8 T.
The magnitude of the charge on an electron
is 1
.
60218
×
10
−
19
C
.
x
y
z
v
= 370000 m
/
s
electron
4
.
2 T
1
.
8 T
B
Find the direction of the magnetic force
when the velocity of the electron is
v
ˆ
, where
v
= 370000 m
/
s.
1.
hatwide
F
= ˆ
2.
hatwide
F
=
1
√
2
(ˆ
−
ˆ
ı
)
3.
hatwide
F
=
1
√
2
(ˆ
ı
−
ˆ
)
4.
hatwide
F
=
−
ˆ
k
5.
hatwide
F
=
ˆ
k
correct
6.
hatwide
F
= ˆ
ı
7.
hatwide
F
=
−
ˆ
8.
hatwide
F
=
−
ˆ
ı
9.
hatwide
F
=
1
√
2
(ˆ
ı
+ ˆ
)
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homework 07 – SMITH, TAYLOR – Due: Mar 2 2008, 4:00 am
2
Explanation:
Let :
q
= 1
.
60218
×
10
−
19
C
,
B
x
= 4
.
2 T
,
and
B
y
= 1
.
8 T
.
Basic Concepts:
Magnetic force on a mov
ing charge is given by
vector
F
=
qvectorv
×
vector
B .
Solution:
vector
B
= (4
.
2 T) ˆ
ı
+ (1
.
8 T) ˆ
v
= (370000 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
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
.
2 T)
×
(370000 m
/
s)
ˆ
k
= (2
.
48978
×
10
−
13
N)
ˆ
k ,
and the direction is +
ˆ
k .
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 Spring '08
 Turner
 Physics, Charge, Mass, Work, Magnetic Field, Ampere

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