MODULE: MOTORS AND GENERATORS
Chapter 6: The motor effect and DC electric motors (Questions, pages
112-115)
1. Like poles repel; opposite poles attract.
2. Diagram as in textbook, page 97, figure 6.2(a).
3. No. The north magnetic pole of the Earth attracts the north pole of magnets and is
therefore of the opposite polarity. It behaves like a south pole of a magnet and is not
located at the identical location on Earth as the north geographic pole.
4. <new diagram figure 6A to come>
5. <new diagram figure 6B(a) and (b) to come>
6.
(a) Repel. The magnetic fields due to each in the area between them are in the same
direction. Conductors with opposite direction currents repel, which decreases the
magnetic field intensity between them as they move apart.
(b) Attract. The magnetic fields due to each in the area between them are in the opposite
direction. Conductors with currents in the same direction attract, which decreases the
magnetic field intensity between them as they move together, cancelling their fields.
7.
(a)
<new figure 6C to come>
(b)
<new figure 6D to come>
8. <new figure 6E(a), (b) and (c) to come>
9. Current into the page; magnetic field down the page; this results in a force on current
to the left.
10.
(a) Up the page; (b) Down the page; (c) To the right; (d) Out of the page
11.
(a) F = BIl
= 4.5 x 10-3 x 3.5 x 0.43
= 6.8 x 10-3 N down the page
(B is perpendicular to I so no consideration of angle.)
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Physics 2: HSC Course, 2nd edition (Andriessen et al, 2003), Chapter 6
(b) Note that you need to apply the formula with sin θ carefully or use the component of
B perpendicular to I.
F = BIl
= 6.5 x 10-4 sin 63˚ x 2.5 x 0.105
= 1.5 x 10-4 N out of page
OR
= 6.5 x 10-4 cos 27˚ x 2.5 x 0.105
= 1.5 x 10-4 N out of page
12.
l = 0.02 m
I = 1.0 A
F = 0.25 N
(a)
F = BIl
0.25 = B x 1.0 x 0.02
B = 12.5 T
(b) Down the page
13.
I=2A
B = 0.60 T
lAB = 0.05 m
(a) FAB = BIl
= 0.60 x 2.0 x 0.05
= 6.0 x 10-2 N
(b) F is perpendicular to the wire, along radius, and also perpendicular to B, that is, in the
plane of the page towards C (by the right-hand rule), so the answer is (iii).
(c) The net force (sum of forces) is OD, so the answer is (v).
14.
l = 0.25 m
B = 4.0 x 10-2 T
I = 1.8 A
F = BIl
= 4.0 x 10-2 x 1.8 x 0.25
= 1.8 x 10-2 N
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Physics 2: HSC Course, 2nd edition (Andriessen et al, 2003), Chapter 6
15.
(a) d = 0.063 m
I1 = 3.4 A
I2 = 2.5 A
l = 0.45 m
k = 2.0 x 10-7 NA-2
F kI 1 I 2
=
l
d
F
2.0 × 10 −7 × 3.4 × 2.5
=
0.45
0.063
F = 1.2 x 10-5 N
(b) Opposite currents separate, so the force is repulsion.
16.
l =0.40 m
F kI 1 I 2
=
l
d
F
2 × 10 −7 × 5 × 5
(a) 0.4 =
0.15
F = 1.3 x 10-5 N
F
2 × 10 −7 × 4 × 2.5
=
(b) 0.4
0.25
F = 3.2 x 10-6 N
F
2 × 10 −7 × 1.5 × 2.0
=
(c) 0.4
0.42
F = 5.7 x 10-7 N
17.
The direction of the force on side A is down the page. The direction of the force on side
B is up the page.
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Physics 2: HSC Course, 2nd edition (Andriessen et al, 2003), Chapter 6
18.
(a) A = Brush; B = Split-ring commutator; C = External magnet; D = Armature coil.
(b)
A: The brush makes electrical contact with the split-ring commutator.
B: The split-ring commutator automatically changes the direction of the current through
the coil.
C: The magnet provides the external magnetic field.
D: The armature (iron core) coil rotates about an axis when it carries a current; increases
magnetic field through coil.
19.
n = 50
B = 0.025 T
I = 1.5 A
A = 0.18 x 0.15 m2
(a) Out of the page
(b) F = nBIl
= 50 x 0.025 x 1.5 x 0.18
= 0.34 N
(c) A = 2.7 x 10-2 m2
(d) τ = nBIA cos θ
= 50 x 0.025 x 1.5 x 0.027 cos 45º
= 3.6 x 10-2 N m
20.
(a)
n = 25
A = 0.050 x 0.030 m2
B = 0.45 T
I = 1.75 A
F = nBIl
= 25 x 0.45 x 1.75 x 0.05
= 0.98 N upwards (out of page)
(b) The direction of the force on side AB is downwards. The direction of the force on
side CD is upwards. Therefore the net force is zero. The forces on the sides provide a
torque on the coil which is free to rotate about the dotted line.
(c) The magnitude and direction of the force remain the same, but the torque decreases.
(d) Make a stronger magnetic field by using stronger magnets or moving the magnets
closer together; put more turns on the wire of the coil; increase the current flowing
through the coil by using a greater source of emf.
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Physics 2: HSC Course, 2nd edition (Andriessen et al, 2003), Chapter 6