PHYSICS 120 : ELECTRICITY AND MAGNETISM
TUTORIAL QUESTIONS
MAGNETIC FIELDS AND FORCES
Question 96
A charge of 12.0 µC, travelling with a speed of 9.00 × 106 ms−1 in a direction perpendicular
to a magnetic field, experiences a magnetic force of 8.70 × 10−3 N. Calculate the magnitude
of the magnetic field.
The magnitudes of the magnetic field B and the magnetic force F are related by
F = |q|vB sin θ
F
∴B =
|q|v sin θ
8.70 × 10−3
(12.0 × 10−6 ) × (9.00 × 106 ) × sin 90°
= 8.06 × 10−5 T
=
Question 97
A particle, moving with a velocity of 8.00 × 104 ms−1 at an angle of 30° with respect to a
magnetic field of 5.60 × 10−5 T, experiences a force of 2.00 × 10−4 N. Calculate the magnitude
of the particle’s charge.
F = |q|vB sin θ
F
∴ |q| =
vB sin θ
2.00 × 10−4
(8.00 × 104 ) × (5.60 × 10−5 ) × sin 30°
= 8.93 × 10−5 C
=
page 1 of 7
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 98
Particles 1, 2 and 3 follow the paths shown below as they pass through a magnetic field.
What can one conclude about each particle?
Since particle 2’s trajectory is unchanged, it must be a neutral particle. The paths of particles
1 and 3 is affected by the magnetic field, and by the right-hand rule it can be seen that 1
must be a negatively charged particle, while 3 is positively charged.
Question 99
Three particles have identical charges and masses. They enter a uniform magnetic field and
follow the paths shown below. Which particle is moving the fastest and which is moving the
slowest? Justify your answers.
The force is given by
F = |q|vB sin θ
and since q, B and sin θ are the same for each particle, then F ∝ v. Also, the resultant force
in circular motion is
v2
F =
r
so that v ∝ r, since the particles have the same mass. Particle 2’s trajectory has the smallest
r and so it has the smallest v. Particle 1 has the largest v.
Question 100
A wire 1.0 m long carries a current of 10 A and makes an angle of 30° with a uniform magnetic field with B = 1.5 T. Calculate the magnitude and direction of the force on the wire.
F = BIl sin θ = (1.5) × (10) × (1) × sin 30° = 7.5 N
F is perpendicular to B and the wire.
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 101
(a) Calculate the force per unit length on a wire carrying a current of 0.50 A when perpendicular to a 4.0 T magnetic field.
(b) What if the angle between the wire and the field is 45°?
(a)
F
= BI sin θ = (4.0) × (0.50) × sin 90° = 2.0 N m−1
l
(b) When the angle is 45°
√
2.0
F
= BI sin θ = (4.0) × (0.50) × sin 45° = √ = 2 N m−1
l
2
Question 102
The force on a wire carrying 20 A is a maximum of 3.6 N when placed between the pole faces
of a magnet. If the wire is 15 cm long, what is the approximate magnitude of B?
The maximum force implies that the angle is 90°. The magnitude of B is
B=
3.6
F
=
= 1.2 T
Il
20 × 0.15
Question 103
How far from a long, straight wire carrying 10 A will the magnetic field be 1.0 × 10−2 T?
B=
µ0 I
2π r
∴r=
µ0 I
10
= 2.0 × 10−7 −2 = 2.0 × 10−4 m
2π B
10
Question 104
Determine the magnetic field midway between two long straight wires 10 cm apart if one
carries 10 A and the other 8.0 A and these currents are
(a) in the same direction, and
(b) in opposite directions.
(a) The magnetic field is additive. Midway between the wires, the fields are either in the
same direction or in the opposite direction. Let both wires carry current out of the
plane of the paper.
page 3 of 7
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
The fields are in opposite directions. Hence
B = BA − BB =
µ0
2.0 × 10−7
(IA − IB ) =
(10 − 8.0) = 8.0 × 10−6 T
2π
5.0 × 10−2
Since BA was taken as positive, the resultant magnetic field points in the same direction
BA .
(b) For currents in opposite directions, the field midway points in the same direction for
boh wires.
B = BA + BB =
µ0
(IA + IB ) = 7.2 × 10−5 T
2π
Question 106
Calculate the magnitude and direction of the force between two parallel wires 85.0 m long
and 30.0 cm apart, each carrying 60.0 A in the same direction.
The force on a current carrying wire in a magnetic field is F = BIl sin θ. Here, magnetic
field is due to the other wire, and the angle is 90°. Hence
µ0 I 1
60.0 × 60.0 × 85.0
F =
I2 l = (2.0 × 10−7 ) ×
= 0.204 N
2π r
30.0 × 10−2
The wires carry current in the same direction so the force is attractive.
Question 107
A vertical, straight wire carrying 5.0 A exerts an attractive force per unit length 6.0 ×
10−4 N m−1 on a second, parallel wire 8.0 cm away. What current (magnitude and direction)
flows in the second wire?
The force is attractive therefore the second wire carries current in the same direction as
the first.
F
l
∴ I2
µ0 I 1 I 2
2π r
F r 2π
=
l I 1 µ0
=
= (6.0 × 10−4 ) ×
= 48 A
8.0 × 10−2
(2.0 × 10−7 ) × 5.0
page 4 of 7
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 109
A solenoid 120 cm long and 3.0 cm in mean diameter has five layers of windings with 840
turns on each. The current in the solenoid is 5.0 A. Calculate (i) the magnetic field at the
centre of the solenoid and (ii) the magnetic flux per turn for a cross-section of the solenoid
at its centre.
(i) The total number of turns is
N = 840 × 5 = 4200
The magnetic field is
B = µ0 nI = 4π × 10−7 ×
4200
× 5.0 = 2.2 × 10−2 T
120 × 10−2
(ii) The total flux inside the solenoid is
φ = BAN cos θ
so the flux per turn is
φ
= BA cos θ = B(πr2 ) cos θ = (2.2×10−2 )× π
N
3.0
2
2 !
cos 0° = 1.6×10−5 Wb/turn
Question 112
A closely-wound rectangular coil of 50 turns has dimensions of 12.0 cm × 25.0 cm. The plane
of the coil is rotated in time t = 0.100 s from a position where it makes an angle of 60.0° with
a uniform magnetic field of strength 2.00 T to a position perpendicular to the field. What is
the average emf induced in the coil?
Let the average emf hEi be
∆φ
dφ
hEi = −
=−
dt
∆t
The component of the field perpendicular to the area A goes from A cos 30° to A cos 0° in
0.10 s. Thus the average emf is given by
∆φ
∆t
BAN (cos φF − cosI )
= −
∆t
2.00 × (12.0 × 10−2 ) × (25.0 × 10−2 ) × (50.0) × (cos 0° − cos 30°)
= −
0.100
= −4.02 V
hEi = −
page 5 of 7
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 113
A rectangular coil having 10 turns with dimensions of 20.0 cm × 30.0 cm rotates with an
angular velocity of 600 rpm in a uniform magnetic field of strength 0.100 T. The axis of rotation is perpendicular to the field. Find the maximum emf produced.
With a diagram similar to Question 112, here B, A and N are constants. The magnitude of
the induced emf is
dφ
dt
d
(BAN cos θ)
=
dt
d
= BAN (cos θ)
dt
d cos θ dθ
= BAN
dθ dt
|E| =
Now
d cos θ
= − sin θ
dθ
and can have a maximum value of 1. Also
dθ
dt
is the angular frequency ω, where
ω = 2πf = 2π(600) rad/min = 20π rad/s
The maximum emf is then
EMAX = BAN ω = (0.100) × (20.0 × 10−2 ) × (30.0 × 10−2 ) × (10) × (20π) = 3.77 V
Question 115
A closely-wound coil has an area of 4.0 cm2 , 160 turns and a resisntance of 50 Ω. It is connected to a charge-measuring instrument whose resistance is 30 Ω. When the coil is rotated
quickly from a position parallel to a uniform magnetic field to one perpendicular to the field,
the instrument indicates a charge of 4.0 × 10−5 C/ What is the magnitude of the field?
Ignoring the reactance of the coil, the average current hIi is
hEi =
∆q
4.0 × 10−5
=
A
∆t
∆t
The average potential difference for the coil and charge-measuring instrument is
hV i = hIi R =
3.2 × 10−3
4.0 × 10−5
× (50 + 30) =
∆t
∆t
page 6 of 7
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
The magnitude of the average emf is
|hEi| =
=
∴
3.2 × 10−3
=
∆t
B =
=
∆φ ∆t BAN
(cos 0° − cos 90°)
∆t
BAN
(cos 0° − cos 90°)
∆t
3.2 × 10−3
4.0 × 10−4 × 160
5.0 × 10−2 T
page 7 of 7