Physics 112
Homework 6 (Ch20)
Question:
1. Three particles, a, b, and c, enter a magnetic field as shown in figure below. 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.
Problems:
2. 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
Fmax  IlB  I 
Fmax
lB

0.750 N
 4.80 m  8.00 102 T 
 1.95 A
3. 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
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
Physics 112
Homework 6 (Ch20)
4. 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
5. Find the direction of the force on a negative charge for each diagram shown in figure


below, 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
Physics 112
Homework 6 (Ch20)
6. 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.
7. 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 


Physics 112
Homework 6 (Ch20)

8. An electron enters a uniform magnetic field B  0.23 T at a 45° angle to B.
Determine the radius r and pitch p (distance between loops) of the electron’s helical
path assuming its speed is 3.0 106 m s . See figure below.
Solution
The centripetal force is caused by the magnetic field, and is given by Eq. 20-3.
F  qvB sin   qv B  m
r
mv
v2
r

 9.1110 kg  3.0 10 m s  sin 45

1.60 10 C   0.23T 
31
6
19
qB
o
 5.251 105 m  5.3  105 m
The component of the velocity that is parallel to the magnetic field is unchanged, and so
the pitch is that velocity component times the period of the circular motion.
T
2 r
v
2

mv
qB
v

2 m
qB
2  9.11  1031 kg 
 2 m 
6
o

3.0

10
m
s
cos
45
 3.3  10 4 m



19
1.60 10 C   0.23T 
 qB 
p  v T  v cos 45o 
Note!
p  v||T  2r
v||
v
 2r tan 45  2r
9. A circular coil 16.0 cm in diameter and containing nine loops lies flat on the ground.
The Earth’s magnetic field at this location has magnitude 5.50 10 5 T and points into
the Earth at an angle of 56.0° below a line pointing due north. If a 7.20-A clockwise
current passes through the coil, (a) determine the torque on the coil, and (b) which
edge of the coil rises up: north, east, south, or west?
Solution
(a) The torque is given by Eq. 20-10. The angle is the angle between the B-field and the
perpendicular to the coil face.
2
  NIAB sin   9  7.20 A    0.080 m    5.50  105 T  sin 34o  4.01  10 5 m N
(b) If the coil is free to turn, it will rotate toward the orientation so that the angle is 0
(torque is 0). In this case, that means the north edge of the coil will rise, so that a
perpendicular to its face will be parallel with the Earth’s magnetic field.
Physics 112
Homework 6 (Ch20)
10. A rectangular loop of wire lies in the same plane as a straight wire, as shown in figure
below. There is a current of 2.5 A in both wires. Determine the magnitude and
direction of the net force on the loop.
Solution
The magnetic field at the loop due to the long wire is into the page, and can be calculated
by equation
F
 0 I1 I 2
l
2 d
The force on the segment of the loop closest to the wire is towards the wire, since the
currents are in the same direction. The force on the segment of the loop farthest from the
wire is away from the wire, since the currents are in the opposite direction.
Because the magnetic field varies with distance, it is difficult to calculate the total force
on either the left or right segments of the loop. Using the right hand rule, the force on
each small piece of the left segment of wire is to the left, and the force on each small
piece of the right segment of wire is to the right. If left and right small pieces are chosen
that are equidistant from the long wire, the net force on those two small pieces is zero.
Thus the total force on the left and right segments of wire is zero, and so only the parallel
segments need to be considered in the calculation.
Fnet  Fnear  Ffar 

 1
 0 I1 I 2
 II

1 
lnear  0 1 2 lfar  0 I1 I 2l 


2 d near
2 d far
2
 d near d far 
4  107 T m A
2


 2.6  10 6 N, towards wire

 0.030 m 0.080 m 
 2.5 A   0.100 m  
2
1

1
Physics 112
Homework 6 (Ch20)
11. Two long wires are oriented so that they are perpendicular to each other. At their
closest, they are 20.0 cm apart (see figure below). What is the magnitude of the
magnetic field at a point midway between them if the top one carries a current of 20.0
A and the bottom one carries 5.0 A?
Solution
The magnetic fields created by the individual currents will be at right angles to each
other. The field due to the top wire will be to the right, and the field due to the bottom
wire will be out of the page. Since they are at right angles, the net field is the hypotenuse
of the two individual fields.
2
 0 I top   0 I bottom 
0
4  107 T m A
2
2
Bnet  

I

I

 

top
bottom
2  0.100 m 
 2 rtop   2 rbottom  2 r
 4.12  105 T
2
 20.0 A 2   5.0 A 2