Homework 6
Ch20: Q 10, 18; P 3, 5, 9, 11, 15, 17
Questions:
10. Three particles, a, b, and c, enter a magnetic field as shown in Fig. 20–46. What can
you say about the charge on each particle?
Solution
Charge a experiences an initial upward force. By the right hand rule, charge a must be
positive.
Charge b experiences no force, and so charge b must be uncharged.
Charge c experiences an initial downward force. By the right hand rule, charge c must be
negative.
18.
A beam of electrons is directed toward a horizontal wire carrying a current from
left to right (Fig. 20–49). In what direction is the beam deflected?
Solution
The beam of electrons is deflected to the right. The magnetic field created by the current
in the wire, I, is directed into the page on the side of the wire where the electron beam is
located. The right hand rule tells us that the negative electrons will feel a force to the
right as they move toward the wire in a downward magnetic field.
Problems:
3.
(I) How much current is flowing in a wire 4.80 m long if the maximum force on it is
0.750 N when placed in a uniform 0.0800-T field?
Solution
Use Eq. 20-2.
Fmax  IlB  I 
Fmax
lB

0.750 N
 4.80 m  8.00 102 T 
 1.95 A
5.
(II) The force on a wire carrying 8.75 A is a maximum of 1.28 N when placed
between the pole faces of a magnet. If the pole faces are 55.5 cm in diameter, what
is the approximate strength of the magnetic field?
Solution
Use Eq. 20-2. The length of wire in the B-field is the same as the diameter of the pole
faces.
Fmax  IlB  B 
Fmax
Il

1.28 N
 8.75 A  0.555 m 
 0.264 T
9.
(I) Alpha particles of charge q  2e and mass m  6.6  10 27 kg are emitted from a
radioactive source at a speed of 1.6 107 m s . What magnetic field strength would be
required to bend them into a circular path of radius r  0.25 m?
Solution
The magnetic force is perpendicular to the velocity. In this scenario, the magnetic force
is causing centripetal motion, and so must have the form of a centripetal force.
Fmax  qvB  m
v2
r
 B
mv
qr
 6.6 10 kg 1.6 10 m s   1.3T

2 1.60  10 C   0.25 m 
27
7
19
11. (I) Find the direction of the force on a negative charge for each diagram shown in


Fig. 20–51, where v (green) is the velocity of the charge and B (blue) is the
direction of the magnetic field. (  means the vector points inward. means it
points outward, toward you.)
Solution
Since the charge is negative, the answer is the OPPOSITE of the result given from the
right hand rule applied to the velocity and magnetic field.
(a) left
(b) left
(c) upward
(d) inward into the paper
(e) no force
(f) downward
15.
(II) An electron experiences the greatest force as it travels 2.9 106 m s in a
magnetic field when it is moving northward. The force is upward and of magnitude
7.2 10 13 N. What are the magnitude and direction of the magnetic field?
Solution
The magnetic field can be found from Eq. 20-4, and the direction is found from the right
hand rule. Remember that the charge is negative.
Fmax  qvB  B 
Fmax
qv

7.2  1013 N
1.60 10
19

C 2.9 106 m s

 1.6 T
The direction would have to be East for the right hand rule, applied to the velocity and
the magnetic field, to give the proper direction of force.
17. (II) A doubly charged helium atom whose mass is 6.6 1027 kg is accelerated by a
voltage of 2100 V. (a) What will be its radius of curvature if it moves in a plane
perpendicular to a uniform 0.340-T field? (b) What is its period of revolution?
Solution
(a)
The velocity of the ion can be found using energy conservation. The electrical
potential energy of the ion becomes kinetic energy as it is accelerated. Then, since the
ion is moving perpendicular to the magnetic field, the magnetic force will be a maximum.
That force will cause the ion to move in a circular path.
2qV
Einitial  Efinal  qV  12 mv 2  v 
Fmax  qvB  m
r
mv
qB
m

v
m
2
r

2qV
1
m

qB
B
2mV
q



2 6.6  10 27 kg  2100 V 
1

19
2 1.60  10 C
0.340 T

 2.7  10 2 m
(b)
The period can be found from the speed and the radius. Use the expressions for
the radius and the speed from above.
v


2r
2r 2m
2 6.6  10 27 kg
T 


 3.8  10 7 s
19
T
v
qB
2 1.60  10 C 0.340T 

