Brock University
Physics 1P22/1P92
Winter 2014
Dr. D’Agostino
Solutions for Tutorial 8: Magnetic Fields and Forces (Chapter 24)
1. One long solenoid is placed inside another long solenoid with twice the diameter but
the same length. Each solenoid carries the same current but in opposite directions, as
shown in the figure. If the two solenoids have the same number of turns, what is the
direction of the magnetic field in the centre of the inside solenoid? Explain. [2 points]
Solution: The net magnetic field inside the inner solenoid is approximately zero. The
field inside either solenoid does not depend on its radius, but only on the number of
turns per unit length, which is the same for each solenoid. Thus, the magnitude of the
magnetic field inside each solenoid (if they were far away from each other) is the same.
When the two solenoids are arranged as in the figure, the net field inside the inner
solenoid is the sum of the fields from each solenoid (superposition principle). Because
the current moves in the opposite sense in each solenoid, the magnetic fields due to
each solenoid (in the region within the inner solenoid) are equal in magnitude but
oppositely directed. Thus, the net magnetic field inside the inner solenoid is zero.
2. Two positive charges are moving in a uniform magnetic field with directions as shown
in the figure. The magnetic force on each charge is also shown. Determine the direction
of the magnetic field. [2 points]
Solution: Use the right-hand rule and you will find that magnetic field is in the
positive x-direction. (Place the appropriate fingers in the direction of the velocity and
force and the remaining finger will point in the direction of the magnetic field.)
3. A proton moves with speed 1.0 × 107 m/s in the direction shown. A 0.50 T magnetic
field points in the positive x-direction. Determine the force exerted on the proton by
the magnetic field. [2 points]
Solution: The magnitude of the force can be calculated as follows:
F = qv⊥ B
F = q (v sin 45◦ ) B
1
7
C 1.0 × 10 m/s × √ (0.50 T)
2
N
−19
F = 1.6 × 10
F = 5.7 × 10−13
The force exerted by the magnetic field on the proton acts in the positive y-direction,
as you can verify using the right-hand rule.
4. An antiproton is moving in a region of space that has a uniform electric field and a
uniform magnetic field, as shown in the figure. Determine the magnitude and direction
of the antiproton’s acceleration. [4 points]
Solution: The force exerted by the electric field on the antiproton acts in the upward
direction, which we’ll take to be the positive direction. The force exerted by the
magnetic field on the antiproton is in the downward direction, as you can verify using
the right-hand rule. Thus, the net force acting on the antiproton is
F = |q|E − |q|vB
F = |q| (E − vB)
F = 1.6 × 10−19 C (1000 V/m − [500 m/s] [2.5 T])
F = −4 × 10−17 N
The negative sign indicates that the force is directed downwards (i.e., in the negative
direction).
The magnitude of the acceleration of the antiproton can be determined using Newton’s
second law of motion:
F
m
4 × 10−17 N
a=
1.67 × 10−27 kg
a = 2.4 × 1010 m/s
a=
The acceleration is directed downwards, in the same direction as the force. Note how
enormous the acceleration is; although the force is tiny, it acts on a particle that has
a truly minuscule mass.