3.3.2 In Class or Homework Exercise
1. A strong current is suddenly switched on in a wire, but no force acts on the
wire. Can you conclude that there is no magnetic field at the location of the
wire?
No. It is possible that there is a magnetic field but that it is parallel to the wire.
There is no force when a magnetic field and a wire carrying a current are
parallel.
2. A wire is carrying a current to the east in the earth's magnetic field. What is
the direction of the force on the wire?
The earth’s magnetic field is directed north. Using the third right hand rule
(fingers north, thumb east), the force must be up.
3. Find the direction of the force on the wire in each of the following magnetic
fields.
a.
Into the page.
b.
Out of the page.
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c.
To the left.
4. Electrons in a vertical wire are moving upward. The wire is placed in a
magnetic field directed from east to west. What is the direction of the force on
the wire?
Since the electrons are moving up, remember that the thumb of your right
hand must point down. The force is north.
5. A current carrying wire is pointing to the East. An external magnetic field is
directed vertically upward. What is the direction of the force on the wire?
South
6. If the force on the wire below is into the page, identify the poles of the
magnets.
Using the third right hand rule, we find that the magnetic field must be down
the page. Since magnetic field lines always travel from north to south, the top
pole must be a north pole.
7. A wire carrying a 30.0 A current has a length of 12 cm between the pole faces
of a magnet at an angle of 60.0o . The uniform magnetic field is approximately
0.90 T. What is the magnitude of the force on the wire?
I  30.0 A
l  12cm  0.12m
  60.0
B  0.90T
F ?
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F  IlB sin 
 (30.0)(0.12)(0.90) sin 60.0
 2.8 N
8. A copper wire 40.0 cm long carries a current 0f 6.0 A and weighs 0.35 N. A
certain magnetic field is strong enough to balance the force of gravity on the
wire. What is the strength of the magnetic field?
l  40.0cm
I  6.0 A
Fg  0.35 N
B?
Since no information is given regarding the angle, we must assume that the
angle is 90 .
Fg  FB
Fg  IlB sin 
0.35  (6.0)(0.400) B sin 90
B  0.15T
9. The magnetic field in a loudspeaker is 0.15 T. The wire consists of 250 turns
wound on a 2.5 cm diameter cylindrical form. The resistance of the wire is 8.0
 . Find the magnitude of the force on the wire when 15 V is placed across
the wire.
B  0.15T
N  250
d  2.5cm  0.025m
R  8.0
V  V
F ?
First we will find the total length of the wire:
C  2 r
 2 (0.0125)
 0.0785m
And since there are 250 turns, the total length is
l  250(0.0785)
 19.6m
Now we need the current:
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V
R
15

8.0
 1.9 A
I
F  IlB sin 
 (1.9)(19.6)(0.15) sin 90
 5.6 N
10. A straight 2.0 mm diameter copper wire can just “float” horizontally in air
because of the force of the earth's magnetic field B which is horizontal and of
magnitude 5.0 105 T. What current does the wire carry? The density of
copper is 8.9 103 kg / m3 .
d  2.0mm  0.0020m
B  5.0 105 T
  8.9 103 kg / m3
I ?
If the wire is floating, then the downward force of gravity must be balanced by
the upward magnetic force.
Fg  FB
mg  IlB sin 
Since   m / V , then m  V
Vg  IlB
 ( r 2 l ) g  IlB
8.9 103 ( )(0.0010) 2 (9.80)  I (5.0 10 5 )
I  5500 A
11. A proton having a speed of 5.0 106 m/s in a magnetic field feels a force of
8.0 1014 N toward the west when it moves vertically upward. When moving
horizontally in a northerly direction, it feels zero force. What is the magnitude
and direction of the magnetic field?
v  5.0 106 m / s
F  8.0 1014 N
q  1.60 1019 C
B?
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Using the third right hand rule, we get a direction of north for the magnetic
field; since it feels zero force when travelling horizontally in a northern
direction the original angle between the direction and the field must have been
90o.
F  qvB sin 
8.0 10
14
 (1.60 1019 )(5.0 106 ) B sin 90
B  0.10T
B  0.10T , north
12. An electron experiences the greatest force as it travels 2.1105 m/s in a
magnetic field when it is moving southward. The force is upward and of
magnitude 5.6 1013 N. What is the magnitude and direction of the magnetic
field?
Since it experiences the greatest force while travelling southward, this
direction must be at an angle of 90o with the magnetic field. Using the third
hand rule, we get a direction of west for the magnetic field.
q  1.60 1019 C
v  2.1105 m / s
F  5.6 1013 N
B?
F  qvB
5.6 10
13
 (1.60 1019 )(2.1105 ) B
B  17T
B  17T , west
13. A force of 5.78 1016 N acts on an unknown particle travelling at a 90.0o angle
through a magnetic field. If the velocity of the particle is 5.65 104 m/s and the
field is 0.032 T, how many elementary charges does the particle carry?
F  5.78 1016 N
  90.0
v  5.65  104 m / s
B  0.032T
q?
F  qvB sin 
5.78 1016  q(5.65 104 )(0.032) sin 90
q  3.2 1019 C
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3.2 1019 C
 2 elementary charges
1.60 1019 C / e
14. Describe the path (quantitatively and qualitatively) of a proton (
m  1.67 1027 kg ) that moves to the right in a 0.120 T magnetic field with a
speed of 9.25 106 m/s. The field points directly toward the observer.
Since the particle is moving in a uniform magnetic field, we know that the path
will be a circle. To describe this path quantitatively, we must find the radius of
the circle.
m  1.67  1027 kg
B  0.120T
v  9.25  106 m / s
r ?
Fc  mac
FB  mac
v2
r
mv
r
qB
qvB  m

(1.67 1027 )(9.25 106 )
(1.60 1019 )(0.120)
 0.805m
Since the field is pointing toward us and the proton is initially moving to the
right, the force on the proton is down. This will result in a clockwise circle.
15. A proton moves in a circular path perpendicular to a 1.10 T magnetic field.
The radius of its path is 4.5 cm. Calculate the energy of the proton.
B  1.10T
r  4.5cm  0.045m
m  1.67  1027 kg
q  1.60 1019 C
Ek  ?
To find the kinetic energy of the proton, we first need the speed.
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Fc  mac
FB  mac
v2
qvB  m
r
qBr
v
m
(1.60 1019 )(1.10)(0.045)

1.67  1027
 4.74  106 m / s
Ek  12 mv 2
 12 (1.67 1027 )(4.74 106 ) 2
 1.9 1014 J
16. A charged particle moves in a straight line through a particular region of
space. Could there be a nonzero magnetic field in this region? Why or why
not?
Yes, there could. The magnetic field could be parallel to the path of the
particle, in which case there would be no force on the particle.
17. Charged cosmic ray particles from outside the earth tend to strike the earth
more frequently at the poles than at lower latitudes. Explain.
At the poles, the field lines are nearly vertical so the charged cosmic ray
particles are roughly parallel to the field lines. They therefore experience no
force and continue to travel to the earth.
At lower latitudes, the field lines are more horizontal so they are perpendicular
to the incoming charged cosmic ray particles and the particles experience a
force which deflects them from hitting the earth.
18. An electron ( m  9.111031 kg) is accelerated from rest through a potential
difference of 2.00 104 V, which exists between the two parallel plates below.
The electron then passes through a small opening into a magnetic field of
uniform field strength 0.25 T.
a. What is the speed of the electron as it leaves the second plate?
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m  9.111031 kg
q  1.60  1019 C
V  2.00 104 V
v?
V
W  E
 Ekf  Eki
W
q
W
1.60 1019
W  3.20 1015 J
2.00 104 
 12 mv 2f  0
3.20 1015  12 (9.111031 )v 2f
v f  8.38 107 m / s
b. Describe the motion (radius and direction) of the electron.
Using the third right hand rule, the force on the electron is down so the
path will be clockwise.
m  9.11 1031 kg
q  1.60  1019 C
B  0.25T
v  8.38  107 m / s
r ?
Fc  mac
FB  mac
v2
r
mv
r
qB
qvB  m

(9.111031 )(8.38 107 )
(1.60 1019 )(0.25)
 1.9 103 m
19. A doubly charged helium atom whose mass is 6.7 1027 kg is accelerated by
a voltage of 2800 V.
a. What is its period of revolution if it then encounters a 0.240 T uniform
magnetic field?
q  3.20 1019 C
m  6.7  1027 kg
V  2800V
B  0.240T
T ?
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V
W  E
 Ekf  Eki
W
q
W
3.20 1019
W  9.0 1016 J
2800 
 12 mv 2f  0
9.0 1016  12 (6.7 1027 )v 2f
v f  5.2 105 m / s
Fc  mac
FB  mac
2 r
T
2 (4.5 102 )
5.2 105 
T
v
v2
qvB  m
r
mv
r
qB

27
(6.7 10 )(5.2 10 )
(3.20 1019 )(0.240)
5
T  5.4 107 s
 4.5  102 m
b. Show that it does the speed of the particle actually has no effect on the
period of revolution.
Since we know that
mv
from part a), we can rearrange this so that
r
qB
r m

v qB
We also know that
v
2 r
T
or
T
2 r
v
Combining these equations gives
T  2
m
qB
which does not depend on the speed of the particle. Substituting our
values in we get
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T  2
 2
m
qB
6.71027
(3.20 1019 )(0.240)
 5.5 107 s
which is the same answer that was obtained in a) if we ignore rounding
errors.
20. A beam of singly charged ions move in a region of space where there is a
uniform electric field, E=1000 N/C, and a uniform magnetic field, B=0.02 T.
The electric and magnetic fields are at right angles to each other and both are
perpendicular to the ion beam so that the electric and magnetic forces on an
ion oppose each other. If an ion is to pass through these fields without being
deflected, what must be the speed of the ion?
E  1000 N / C
B  0.02T
q  1.60 1019 C
v?
If the ion is not to be deflected, the electric force and the magnetic force must
have the same magnitude.
Fe  FB
qE  qvB
1000  v(0.02)
v  50000m / s
21. Protons move in a circle of radius 8.10 cm in a 0.385 T magnetic field. What
value of electric field could make their path straight?
m  1.67 1027 kg
q  1.60 1019 C
r  8.10cm  0.0810m
B  0.385T
E ?
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Fc  mac
FB  mac
v2
qvB  m
r
qBr
v
m
(1.60 1019 )(0.385)(0.0810)

1.67 1027
 2.99 106 m / s
To make their path straight, the force of the electric field must balance the
force from the magnetic field so that there is no net force.
Fe  FB
qE  qvB
E  2.99 106 (0.385)
E  1.15 106 N / C
22. A particle with a charge of 2.0 1018 C is accelerated by 400. V. It then enters
a magnetic field (B=0.40 T) and follows a path with a radius of 0.080 m.
Calculate the mass of the particle.
Looking at the electric field,
q  2.0 1018 C
V  400.V
V
W
q
W
2.0 1018
W  8.0 1016 J
400. 
W  E
 Ekf  Eki
 12 mv 2f  0
8.0 1016  12 mv 2f
mv 2  1.6 1015
Looking at the magnetic field information,
q  2.0 1018 C
B  0.40T
r  0.080m
m?
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Fc  mac
FB  mac
v2
qvB  m
r
qBr
v
m
(2.0 1018 )(0.40)(0.080)

m
20
6.4 10

m
Substituting this into the equation from the electric field part of the problem
gives
mv 2  1.6 1015
2
 6.4 1020 
15
m
  1.6 10
m


m
4.096 1039
 1.6 1015
2
m
m  2.6 1024 kg
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