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4 Electricity and Magnetism
4
Chapter 4 Electromagnetism
Electromagnetism
9
(a)
Practice 4.1 (p.166)
1
A
2
C
not.
B
4
D
5
C
S
N
S
N
S
S
N
(b)
Iron is a magnetic material and aluminium is
3
N
10
N
A
The field lines are in the same direction and
are uniformly spaced.
 (1) is correct.
B
P is a north pole and R is a south pole.
 (2) is incorrect.
11
The field lines between P and Q are more
closely packed together.
 (3) is correct.
6
(a)
S
(b)
The reading drops.
Practice 4.2 (p.183)
Iron is a magnetic material. The iron bar
1
and the magnet attract each other.
A
By right-hand grip rule, a magnetic field
7
pointing downwards is produced in the
N
electromagnet. It repels the bar magnet and
S
the reading of the scale increases.
2
A
3
D
The fields due to the two currents cancel out
each other.
8
(a)
It points towards north.
(b)
The bar magnet will rotate and its north
pole will point downwards as shown.
S
4
C
B = BP + BQ =
5
0 I
2πr

0  2I
2πr
=
3 0 I
2πr
B
The directions of the fields due to the four
wires are as shown below.
N
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1
4 Electricity and Magnetism
Chapter 4 Electromagnetism
BP
=
BS
= 4 × 10–5 T (out of page)
 I
BQ = 0
2 πr
4 π  10 7  4
=
2 π  0.03
45°
BR
BQ
The vertical components cancel each other.
The horizontal components are all equal to
0 I
cos 45 
2πr
 I
2 0 I
Resultant field = 0 cos 45   4 =
2πr
πr
6
4 π  10 7  4
2 π  0.02
= 2.67 × 10–5 T (into page)
10
(a)
X
(b)
C
The two identical solenoids can be regarded as
=
11
(a)
B remains unchanged.
7
(a)
N
S
N
4π  10 7  3 4π  10 7  3

2π  0.025
2π  0.075
= 3.2 × 10–5 T
turns per unit length and carries the same
current as the separate solenoids. By B = μ0nI,
Z
Magnitude of magnetic field at X
 I
 I
= 0 1 0 2
2πr1 2πr2
one when they are brought end-to-end. The
combined solenoid has the same number of
Y
P: to the left
Q: to the right
(b)
S
12
The magnitude of the field increases.
In the region between X and Y, the magnetic
field due to X and that due to Y both points
(b)
8
(a)
S
NN
S
into the page, i.e. no neutral point in this
region.
(top view)
In region above X, the field due to X is always
stronger than that due to Y, i.e. no neutral
point in this region.
 P must be below Y.
(b)
Let x be the distance between P and X.
0 I X 0 I Y

2πrX
2πrY
(top view)
5
2

x x  0.3
x  0.5 m (  50 cm)
13
9
2
BP =
0 I
2 πr
(a)
Magnitude of magnetic field
 NI
= 0
l
4π  10 7  10  5
=
= 4.19 × 10–3 T
0.015
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4 Electricity and Magnetism
(b)
Chapter 4 Electromagnetism
FS
Since the number of turns per unit length
FR
increases, the magnitude of the field
increases.
14
(a)
(b)
15
(a)
By B =
FQ
0 I
,
2 πr
2πrB 2π  0.02  5  10 6
I

 0.5 A
0
4π  10 7
Consider the resultant magnetic force per unit
If the current is doubled, the magnetic
Let I be the current and d be the side length of
field is also doubled.
the square. Taking the direction to the right as
 Magnetic field = 5 × 2 = 10 μT
positive, the horizontal component is
length.
The vertical component is zero.
FR – FS cos 45° – FQ cos45°
0 I 2
 I 2 1 0 I 2 1
 0

=
2π  2 d 2πd 2 2πd 2
Soft iron
(Or other reasonable answers)
(b)
X: south
=
Y: north
(c)
(i)
The current in the circuit decreases
6
For (2), A is doubled. For (1) and (3), there is
and hence the strength of the
no change in B, I, A or N.
electromagnet increases.
decreases, so the current increases
A
τ = BIAN sin θ
(ii) The current in the circuit increases
(iii) The total resistance of the circuit
7
(a)
Zero
(b)
F = BIl
= 1.2 × 4 × 0.2
and hence the strength of the
= 0.96 N (into page)
electromagnet increases.
(c)
B
2
D
= 4.5 N (upwards)
(d)
= 1.2 N (towards upper left and
the same as that of the current in X.
A
4
B
5
C
perpendicular to the wire)
(e)
F = BIl sin θ
= 2 × 5 × 0.2 × sin 30°
= 1 N (into page)
The forces acting on P by the other three wires
are as shown below.
F = BIl
= 0.5 × 12 × 0.2
The direction of the magnetic field due to Y is
3
F = BIl
= 1.5 × 15 × 0.2
Practice 4.3 (p.204)
1
2 2 πd
<0
and hence the strength of the
electromagnet decreases.
0 I 2
(f)
F = BIl sin θ
= 1.1 × 6 × sin (90  40°)
= 5.06 N (into page)
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3
4 Electricity and Magnetism
8
(a)
Chapter 4 Electromagnetism
The magnetic force applied on the wire
the coil and zero when the forces are
must be upwards to balance the weight
along the plane of the coil in this case.
of the rider. By Fleming’s left-hand rule,
(i)
the blue face is the south pole.
(b)
Take moment about the support.
BIl × rm = mg × rr
mgrr
B=
Ilrm
0.002  9.81  0.01
=
5  0.06  0.2
S
N
S
N
(ii)
= 3.27 × 10–3 T
The magnitude of the magnetic field is
3.27 × 10–3 T.
9
(a)
(c)
Any one of the following:
Use stronger magnets.
Increase the current.
(b)
Force acted on the wire
Increase the area of the coil within the
= BIl = 0.4 × 3 × 0.05 = 0.06 N
magnetic field
By Newton’s third law, a downward
Increase the number of turns of the coil.
force of 0.06 N is applied on the magnet
Use curved magnets.
by the wire, which would cause the
Use a soft-iron armature with several
balance reading to increase by
0.06 N
= 6.12 × 10–3 kg = 6.12 g.
9.81 m s 2
coils.
11
(a)
and hence no or only a small current
Mass of magnet = 120 – 6.12 = 114 g
(c)
(i)
pass through the coil.
The magnitude increases.
(ii) The direction reverses.
10
(a)
The coil is short-circuited by the ring
12
(b)
Replace the ring with a split-ring.
(c)
The turning effect increases.
(a)
The directions of the magnetic forces are
shown below.
S
N
FP
Q
(b)
The moment of a couple is proportional
to the perpendicular distance between
the forces, which is maximum when the
forces are perpendicular to the plane of
4
FR
For each metre,
0 I P I Q
FP =
2 π rP
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4 Electricity and Magnetism
=
Chapter 4 Electromagnetism
4π  10 7  2  2
2π  0.01
= 8 × 10–5 N
0 I R I Q
FR =
2 πrR
4π  10 7  2  2
=
2π  0.01
Since the force is at right angle to the moving
direction, no work is done and the particle
does not change its speed.
 (3) is incorrect.
3
Whether a particle carries positive or negative
charge, the direction of the electrostatic and
= 8 × 10–5 N
magnetic forces are opposite. The sign of the
Resultant force
charges cannot be determined.
= FP2  FR2
(b)
 (1) is not necessarily correct.
= 1.13 × 10–4 N (to the left)
0 I S I Q
(i) FS =
2 πrS
IS =
=
When a particle just enters the fields,
net force = QE – BQv = Q(E – Bv)
From the given information, it can only be
E
E
sure that v X 
and vY  . The
B
B
2 πrS FS
0 I Q
2 π  1  1.13  10 4
4 π  10  7  2
magnitudes of the charges cannot be
determined.
= 283 A
 (3) is correct and (2) is not necessarily
(ii) S may be on the left of Q with the
current flowing out of the page, or
B
correct.
4
(a)
it may be on the right of Q with the
13
(a)
(i)
Magnitude of force
= Bqv
current flowing into the page.
= 0.2 × 2 × 1.60 × 10–19 × 4 × 105
The motor may be overheated.
= 2.56 × 10–14 N
(ii) The motor requires a large voltage
to drive.
(iii) The motor is very large in size.
(Or other reasonable answers)
(b)
A.c. power source can be used.
(Or other reasonable answers)
Practice 4.4 (p.214)
1
D
2
B
There is magnetic force as long as the moving
(b)
Magnitude of force
= Bqv sin θ
= 0.2 × 1.60 × 10–19 × 4 × 105 × sin 30°
= 6.4 × 10–15 N
direction of the charge and the field are not
parallel.
 (1) is incorrect.
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5
4 Electricity and Magnetism
(c)
Chapter 4 Electromagnetism
The magnetic force is zero since the
V
Bd
3000
=
0.015  0.1
direction of motion is along the field.
= 2 × 106 m s–1
The magnetic force is zero since the
=
charge is not moving.
(d)
5
(a)
The magnetic force provides the
centripetal force for circular motion.
mv 2
By Bqv =
,
r
mv
r=
Bq
31
=
9.11  10  8.0  10
2  10 3  1.60  10 19
6
The speed of the particle is 2 × 106 m s–1.
(c)
Change the strength of the electric or
magnetic field.
Revision exercise 4
Concept traps (p.218)
1
F
= 0.022775 m
The magnetic field at the centre of a long
 0.0228 m
solenoid depends on the number of turns per
N
unit length
.
l
The radius of the path is 0.0228 m.
(b)
2
T
3
F
v and B are not required to be perpendicular.
Multiple-choice questions (p.218)
(c)
Distance travelled = π × 0.022775
= 0.0715 m
4
D
5
B
As charge flows from one place to another, a
Time of staying in the field
distance
=
speed
0.0715
=
8.0  10 6
current is formed and this creates a magnetic
field.
 (1) and (2) are correct.
Light travels outwards from the lightning bolt
–9
6
(a)
= 8.94 × 10 s
in straight lines.
No, whether the particle carries positive
 (3) is incorrect.
or negative charge, the electrostatic and
(b)
D
magnetic forces it experiences are
The electromagnet applies an attractive force
opposite. Both cases are possible.
on the bar magnet. To increase the scale
The magnetic force and the electrostatic
reading, the force should be reduced.
force balance each other.
Bqv = qE
E
v=
B
6
6
7
A
Copper is a non-magnetic material and does
not affect the magnetic field strength.
 (1) is incorrect.
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4 Electricity and Magnetism
8
Chapter 4 Electromagnetism
By B = μ0nI, the field strength can be raised by
14
(HKDSE 2013 Paper 1A Q26)
increasing the number of turns per unit length.
15
(HKDSE 2013 Paper 1A Q28)
 (2) is correct and (3) is incorrect.
16
(HKDSE 2014 Paper 1A Q26)
D
In (1) and (2), the current through coil Y
Conventional questions (p.220)
increases, while in (3) magnetic field strength
17
(a)
through coil Y increases.
9
D
10
A
The magnetic field applies a force on the
(i)
electron perpendicular to its motion, so it
accelerates but has no change in kinetic
(b)
energy.
 (1) is incorrect and (2) is correct.
Since the magnetic force is always
(Current out of the page)
1A
(ii) (Correct field lines)
 I
Magnetic field = 0
2 πr
4π  10 7  2
=
2π  0.05
1A
= 8 × 10–6 T
perpendicular to the magnetic field, the
1A
magnetic field will never change and thus its
Q: Into page
1A
direction of motion will not become
R: No field
1A
perpendicular to the field.
S: Out of page
 I
By B = 0 ,
2 πr
1A
 (3) is incorrect.
11
(a)
(b)
D
Magnitude of B-field at Q
4π  10 7  0.5
2
=
2π  0.04
The force between the wires is attractive.
= 5 × 10–6 T
19
12
B
(a)
(i)
1M
1A
Electromagnet A and resistor R1 are
connected to the battery in parallel.
If the charge is positive, both the electrostatic
1A
and magnetic forces are in the upward
Therefore, the voltage across the
direction. If the charge is negative, both forces
electromagnet never changes and it
are downwards.
keeps attracting the door.
 (2) is correct.
13
1A
P: No field
component of the electron’s velocity along the
18
1M
1A
(ii) R3 should be placed in series with
The magnitudes of the forces and the velocity
the battery.
cannot be deduced from the given
When the springy metal contacts
information.
are closed, R1 is connected in
 (1) and (3) are not necessarily correct.
parallel with the electromagnet.
(HKDSE 2012 Paper 1A Q29)
Their equivalent resistance
New Senior Secondary Physics at Work (Second Edition)
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1A
7
4 Electricity and Magnetism
Chapter 4 Electromagnetism
decreases and the electromagnet
A higher voltage results in a larger
has a smaller share of voltage. 1A
current and hence stronger magnetic
The current passing through the
field.
1A
(a)
From Y to X
1A
(b)
By F = BIl sinθ,
F
I=
Bl sin 
0.01
=
0.4  0.05  sin 30 
1M
electromagnet decreases, and so the
21
strength of the electromagnet
decreases.
(b)
1A
Make the coils in A more closely spaced.
1A
By B =
 0 NI
l
, the magnetic field
strength increases with
Or
=1A
(c)
N
.
l
1A
1A
1A
increases the magnetic field.
22
(a)
Magnetic force FB
= BQv
= 1.8 × 10 × 1.60 × 10
1A
–19
× 1.22 × 107
= 3.51 × 10–15 N
1A
The direction is downwards.
1A
1A
Iron core
1A
0.4 m wire with diameter 2 mm
1A
(b)
Downwards
9–V cell
1A
(c)
The electrostatic force must balance the
magnetic force.
(b)
QE = FB
F
E= B
Q
wire coil
core
=
(Wire wound on core and connected to
battery correctly)
1A
(Correct labels)
1A
Iron is the most readily magnetized
material among the three choices.
1A
1M
3.51  10 15
1.60  10 19
= 2.20 × 104 N C–1
1A
4
–1
The field strength is 2.20 × 10 N C .
(d)
The 0.4 m wire with diameter 2 mm has
(i)
The electron deflects due to the
magnetic force.
1A
(ii) It performs uniform circular motion
the lowest resistance among the three,
in the clockwise direction.
which results in the largest current and
The magnetic force provides the
hence the strongest magnetic field. 1A
centripetal force. Therefore,
mv 2
BQv =
r
mv
r=
BQ
(thickness of the wire does not affect the
number of turns per unit length as the
wire can be wound in multiple layers)
8
1M
–3
(Or other reasonable answers)
(c)
1A
and the direction remains unchanged.
This increases the current through A and
(a)
The magnitude increases
Replace R2 with a resistor of lower
resistance.
20
1A
1A
1M
New Senior Secondary Physics at Work (Second Edition)
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4 Electricity and Magnetism
=
Chapter 4 Electromagnetism
9.11 10 31 1.22 10 7
1.8 10 3 1.60 10 19
25
(a)
magnetic field is set up.
= 0.0386 m
1A
(a)
the coil and the frame are attracted
The poles of the electromagnet which
towards the soft-iron block.
have the greatest attractive force are not
The scale reading thus increases.
facing the materials to be sorted.
1A
(b)
closer to each other.
of the following):
Use a thicker wire to reduce the
1A
Turning the electromagnet by 90 so that
resistance of the coil.
one end of the coil faces the materials to
(Or other reasonable answers)
(c)
Replace the soft-iron block by a
permanent magnet.
By doing so, the coil is attracted towards
materials to be sorted.
the magnet when current flow in one
Steel retains part of its magnetic
direction
1A
1A
property after the external magnetic field
and is repelled by the magnet when
is removed.
current flow in the other direction.
1A
The rider should be placed on XY.
1A
1A
26
(a)
core by an iron core.
1A
By Fleming’s left-hand rule, the wire XY
Increase the current.
1A
experiences a magnetic force pointing
upwards.
length in the coil.
1A
(a)
The rod rolls towards A.
1A
(b)
(i)
(b)
= mg
1M
= 0.002 × 9.81
1M
= 0.04 × 5 × 0.035
= 0.007 N (towards A)
= 0.0196 N
0.035
sin30
1M
1A
(c)
By F = BIl,
1M
F
0.0196
d=l=
=
= 0.0327 m 1A
BI 0.05  12
(d)
(i)
1A
(ii) Magnetic force
= BIl = 0.04  5 
1A
Magnetic force = weight of rider
Magnetic force
= BIl
The rider should be put on WZ 1A
because the direction of the
= 0.014 N
magnetic force is reversed.
The direction is at 60° to AB. 1A
(c)
1A
electromagnet with both ends facing the
Increase the number of turns per unit
24
1A
Using a U-shape core in the
This can be solved by replacing the steel
(c)
1A
Compress the coil so that the turns are
The design can be improved by (anyone
be sorted.
(b)
1A
The coil becomes an electromagnet and
The radius of the path is 0.0386 m.
23
When a current flows through the coil, a
1A
(ii) The rider should be put between Y
For a higher resistance, the current will
and the support. / A lighter rider
be smaller
should be used.
1A
and hence the magnetic force will be
smaller.
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
1A
Since the magnetic field is not
1A
perpendicular to the current, by
9
4 Electricity and Magnetism
27
Chapter 4 Electromagnetism
F  BIl sin , the magnetic force
Moment of coil
decreases.
= Fl
(a)
Same as I1
(b)
(i)
By B =
1A
0 I
,
2 πr
 I
B1 = 0 1 (r constant)
2πr
0 I1
r=
2πB1
1A
= 3.6 × 10 × 0.04
1M
= 1.44 × 10–4 N m
the magnetic field,
(ii) Into page
(c)
1A
moment of coil
= 1.44 × 10–4 N m
4 π  10  3.5
2 π  1.4  10  4
= 0.005 m
1A
(ii) Since the coil is always parallel to
1M
7
=
1M
–3
(d)
Any two of the following:
1A
1A + 1A
Use stronger magnets.
1A
Increase the current.
1A
Increase the area of the coil within the
At P, the magnetic field produced by I1
magnetic field.
has the same magnitude as that produced
Increase the number of turns of the coil.
by I2.
 0 I 1 0 I 2
=
1M
2 πr1 2πr2
I1 r1
r
0.005
1
= =
=
=
I 2 r2 0.03  r 0.03  0.005 5
Use a soft-iron armature with several
coils.
29
(a)
The magnetic forces act along the plane
of the coil and have no turning effect.
1A
1A
(d)
I2 = 5I1 = 5  2 = 10 A
Magnetic force per unit length
II
= 0 1 2
2πr
4π  10 7  2  10
=
2π  0.03
28
(a)
The coil should not be vertical before the
1M
switch is closed.
(b)
1M
of the current through the coil every half
It vibrates a few times
1A
direction.
(c)
and comes to rest lying along the vertical.
(c)
(i)
10
that the coils stay longer in parallel with
the magnetic field.
1A
(Or other reasonable answers)
(d)
1M
1A
A practical motor use curved magnets so
1A
Force on one side of coil
= BIl
1A
so that the coil always turns in the same
1A
Use a split-ring.
The commutator reverses the direction
turn
= 1.33 × 10–4 N
(b)
1A
1A
Replace the permanent magnets with
electromagnets that are connected to the
= 0.02 × 3 × 0.06
a.c. power supply.
= 3.6 × 10–3 N
When the direction of the current
The magnetic forces acting on the
through the coil reverses, the direction of
two side of the coil form a couple.
the magnetic field also reverses.
1A
1A
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4 Electricity and Magnetism
30
(a)
Chapter 4 Electromagnetism
As a result, the direction of the magnetic
Physics in article (p.225)
force will not reverse due to the
36
alternating current.
1A
No,
1A
(a)
Opposite
1A
(b)
No,
1A
the currents in the live and the neutral
the magnetic force is always
wires always flow in opposite directions.
perpendicular to the direction of motion
1A
of the particle.
1A
(b)
No change
1A
ground fault interrupter can cut off the
(c)
(i)
1A
power immediately while an earth wire
Uniform circular motion
(c)
(ii) The magnetic force provides the
centripetal force.
mv 2
 BQv
r
BQr
v
m
2 πr
T=
v
2 πm
=
BQ
=
(d)
1M
If there is a leakage of current, the
may not be able to do so.
1A
No.
1A
As long as the current in the live and the
neutral wires remains the same, the
interrupter does not response. Hence, it
1M
cannot detect a large current due to a
short circuit.
1A
2π  6.64  10 27
1.6  2  1.60  10 19
= 8.15 × 10–8 s
1A
–8
The period is 8.15 × 10 s.
31
(HKCEE 2007 Paper 1 Q11)
32
(HKALE 2007 Paper 1 Q3)
33
(HKDSE 2013 Paper 1B Q4)
Experiment questions (p.225)
34
Bring them close to each other pair-by-pair.
The two bars which are neither attracted nor
repelled by any of the other are the copper
bars.
1A
Two of them may attract each other, but repel
each other if one of them is reversed. They are
the bar magnets.
1A
The remaining two bars are the iron bars. 1A
(Or other reasonable answers)
35
(HKCEE 2011 Paper 1 Q6)
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
11