BROCK UNIVERSITY
Test 3: February 2015
Course: PHYS 1P22/1P92
Examination date: 28 February 2015
Time of Examination: 13:00–13:50
Number of pages: 6 + formula sheet
Number of students: 100
Instructor: S. D’Agostino
No aids are permitted except for a non-programmable, non-graphics calculator.
Solve all problems in the space provided.
Total number of marks: 24
SOLUTIONS
1. [4 point] A wire has a length of 7.00 cm and is used to make a circular coil of one
turn. There is a current of 4.30 A in the wire. In the presence of a 2.50-T magnetic
field, what is the maximum torque that this coil can experience?
Solution: First we should figure out how large an area A is enclosed by a wire of
length L = 7.00 cm bent into a circle.
L = 2πr
=⇒
L2 = 4π 2 r2
Noting that A = πr2 , we can determine a relation between A and L:
πr2
A
=
L2
4π 2 r2
1
A
=
2
L
4π
L2
A=
4π
The torque exerted by the magnetic field on the coil is
τ = IAB sin φ
Thus, the maximum torque is obtained when the coil is oriented relative to the direction
of the magnetic field in such a way that the sine function attains its maximum value,
which is 1. Thus, the maximum torque exerted by the magnetic field on the coil is
τ = IAB
2
L
τ =I
B
4π
IL2 B
τ=
4π
(4.30)(7.00 × 10−2 )2 (2.50)
τ=
4π
−3
τ = 4.19 × 10 N · m
2. [4 point] A flat circular coil with 105 turns, a radius of 4.00 × 10−2 m, and a resistance
of 0.480 Ω is exposed to an external magnetic field that is directed perpendicular to
the plane of the coil. The magnitude of the external magnetic field is changing at a
rate of 0.783 T/s, thereby inducing a current in the coil. Find the magnitude of the
magnetic field at the centre of the coil that is produced by the induced current.
Solution: Let B1 represent the magnitude of the external magnetic field and let B2
represent the magnitude of the magnetic field produced by the current in the coil.
First determine the magnitude of the induced emf in the coil using Faraday’s law of
electromagnetic induction; then determine the magnitude of the induced current in the
coil using Ohm’s law; finally use the formula for the magnetic field at the centre of a
coil of radius r.
The induced emf is
∆B
E = NA
∆t
The induced current is therefore
E
R N A ∆B
I=
R
∆t
I=
Thus, the magnetic field at the centre of the coil produced by the induced current is
N µ0 I
2r
N µ0 N A ∆B
=
·
2r
R
∆t
2
N µ0 A ∆B
=
2rR
∆t
2
2
N µ0 (πr ) ∆B
=
2rR
∆t
2
N µ0 (πr) ∆B
=
2R
∆t
2
(105) (4π × 10−7 )π(4.00 × 10−2 )
=
(0.783)
2(0.480)
= 1.42 × 10−3 T
B2 =
B2
B2
B2
B2
B2
B2
3. [4 point] A charged particle enters a uniform magnetic field and follows the circular
path shown in the figure.
(a) Is the particle’s charge positive or negative?
(b) The particle’s speed is 140 m/s, the magnitude of the magnetic field is 0.48 T,
and the radius of the path is 0.960 m. Determine the mass of the particle, given
that its charge has a magnitude of 8.2 × 10−4 C.
Solution: (a) The particle’s charge is negative, by the right-hand rule.
(b) Use Newton’s second law of motion. For the force, use the expression for the force
that a magnetic field exerts on a moving charged particle, and for the acceleration use
the expression for centripetal acceleration (because the particle moves along part of a
circle).
F = ma
v2
r
qvBr
m=
v2
qBr
m=
v
(8.2 × 10−4 )(0.48)(0.96)
m=
140
−6
m = 2.7 × 10 kg
qvB = m ·
4. [4 point] A generating station is producing 1.2 × 106 W of power that is to be sent to a
small town located 7.0 km away. Each of the two wires that comprise the transmission
line has a resistance per kilometre of 5.0 × 10−2 Ω/km.
(a) Find the power dissipated as heat in the wires if the power is transmitted at 1200
V.
(b) A 100:1 step-up transformer is used to raise the voltage before the power is transmitted. How much power is now dissipated as heat in the wires?
Solution: First determine the total resistance of the two wires:
5.0 × 10−2 Ω/km × 14 km = 0.70 Ω
(a) The current in the wires is
P
V
1.2 × 106
I=
1200
I = 1.0 × 103 A
I=
Thus, the power dissipated in the wires is
∆P = I 2 R
∆P = (1.0 × 103 )2 (0.70)
∆P = 7.0 × 105 W
Although the question didn’t ask us to calculate this quantity as a percentage of the
transmitted power, you can’t help but notice that over half of the generated power is
lost in heating the wires.
(b) The step-up transformer ratio of 100:1 means that power is transmitted at a voltage
of
1200 × 100 = 120, 000 V
Thus, the current in the wires is now less than in Part (a):
P
V
1.2 × 106
I=
120, 000
I = 10 A
I=
Thus, the power dissipated in the wires is
∆P = I 2 R
∆P = (10)2 (0.70)
∆P = 70 W
This is a much better scenario, as the percentage of generated power that is dissipated
as heat is minuscule.
Answer each question briefly and clearly, in at most a few sentences. Your explanation
may include formulas or diagrams, if you wish. Remember, brevity and clarity are
courtesy.
5. [2 points] A charged particle is moving in a region of space where there may or may
not be a magnetic field. If the net force exerted by the magnetic field on the charged
particle is zero, can you conclude that the magnitude of the magnetic field is zero?
Explain.
Solution: No. It could be that the charged particle is moving parallel to the magnetic
field, in which case the force exerted by the magnetic field on the charged particle is
zero even though the magnitude of the magnetic field may not be zero.
6. [2 points] How should two long, straight, current-carrying wires be oriented in space
so that neither exerts a magnetic force on the other? Explain.
Solution: Place the wires perpendicular to each other. In this way, the magnetic field
of each wire is parallel to the current in the other wire, which means that the magnetic
force exerted by the magnetic field of each wire on the current in the other wire is zero.
(At least this is true at the points on the wires that are closest to each other. At other
points the explanation is a little more complex; draw a diagram and you’ll be able to
deduce that the net force exerted by each wire on the other is zero, although each wire
does exert a net torque on the other.)
7. [2 point] A small cylindrical bar magnet is placed inside the top of a vertical copper
tube and released so that its North end is down and its South end is up. The magnet
falls much more slowly than a similar non-magnetic cylinder. Explain why.
Solution: This was demonstrated in lecture. Remember that the moving magnet
causes a changing flux in imaginary loops in the copper tube that are coaxial with
the tube’s axis. Thus, eddy currents are produced in the copper tube; by Lenz’s law,
the magnetic fields produced by the eddy currents oppose the change in flux, which
means they oppose the motion of the magnet, thus slowing it down. The non-magnetic
cylinder produces no eddy currents, and therefore falls freely.
8. [2 point] Briefly explain how an electrical generator works.
Solution: Rotate a loop of wire in an external magnetic field. Because the magnetic
flux through the loop continually changes, electric current is induced in the loop, by
Faraday’s law of electromagnetic induction.