# Electromagnetic Induction

```2014 Sec 4 Physics Chapter 18
Electromagnetic Induction
Name: ____________________
(
) Class: 4 / ___
Levitating Barbecue! Electromagnetic Induction
18
18.1
18.2
18.3
Electromagnetic Induction
Principles of electromagnetic induction
The a.c. generator
The transformer
Specific Instructional Objectives
(a)
deduce from Faraday’s experiments on electromagnetic induction or other
appropriate experiments that a changing magnetic field can induce an e.m.f. in a
circuit
(b)
deduce from Faraday’s experiments on electromagnetic induction or other
appropriate experiments that the direction of the induced e.m.f. opposes the
change producing it
(c)
deduce from Faraday’s experiments on electromagnetic induction or other
appropriate experiments the factors affecting the magnitude of the induced e.m.f.
(d)
describe a simple form of a.c. generator (rotating coil or rotating magnet) and the
use of slip rings (where needed)
(e)
sketch a graph of voltage output against time for a simple a.c. generator
Chapter 18 : Electromagnetic Induction
1
(f)
describe the structure and principle of operation of a simple iron-cored
transformer as used for voltage transformations
(g)
recall and apply the equations V P / Vs = N P / Ns and V P I P = VsIs to new
situations or to solve related problems (for an ideal transformer)
(h)
describe the energy loss in cables and deduce the advantages of high voltage
transmission
(i)
recall and solve problems using Faraday's law of electromagnetic induction and
Lenz's law
(j)
explain simple applications of electromagnetic induction
• Textbook: “All About Physics ‘O’ Level” Chapter 22, page 432-453.
• http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html#c1
•
18.1
Principles of electromagnetic induction
According to Michael Faraday, a change in the magnetic environment will cause an
electromotive force (e.m.f.) to be &quot;induced&quot; in a conductor. The change could be
produced by moving a magnet towards or away from a conducting coil, moving a
conducting coil into or out of a magnetic field, rotating a coil relative to the magnet,
changing the magnetic field strength etc.
Faraday’s Law of Electromagnetic Induction states that the magnitude of the induced
e.m.f. in a conductor is directly proportional to the rate at which magnetic field lines
are cut by the conductor.
Lenz’s Law states that the polarity of the induced e.m.f. is such that it produces a
current whose magnetic field opposes the change which produces it.
An induced current is only present when there is a closed circuit
Terms commonly used:
Chapter 18 : Electromagnetic Induction
2
Terms
Electromotive
force (e.m.f.)
Induced e.m.f.
Induced current
•
Measured using a voltmeter
•
SI unit: volt (V)
•
E.g. e.m.f. of a battery = 1.5 V
An e.m.f. is “generated” in the coil. The coil behaves like a
“battery”.
• When a closed circuit is connected to the coil and an
induced e.m.f. is produced, an induced current flows.
• It can be detected by light bulb, a centre-zero
galvanometer, voltmeter, ammeter or cathode ray
oscilloscope (C.R.O.)
Factors affecting the magnitude of the induced e.m.f.
Factors
1
2
3
Change in speed of moving /
rotating coil (or magnet)
Strength of magnet
Number of turns in coil
4
Soft iron in the coil
To increase the induced e.m.f.
(or induced current)
Move / rotate coil or magnet faster
Use a stronger magnet
Increase number of turns in coil
Insert soft iron core in the coil
The right hand grip rule allows us to determine the direction of the induced current in a
coil and hence the polarity of the induced e.m.f.
Lenz’s Law is actually a law of conservation of energy. When an amount of energy is
used to push the magnet into the coil, the current induced at the the coil should produce
a magnetic field to repels the magnet, as this will weaken the resultant field and slows
down the magnet, thus reducing the induce current. The energy from the mechanical
work done against the opposing force experienced by the moving magnet is transformed
into electrical energy. [Consider otherwise: if the current induced at the the coil produces
a magnetic field to attract the magnet; this will strengthen the resultant field and cause
the magnet to accelerate and induce a larger current, thus resulting in an ever
increasing amount of energy; this is impossible!]
Chapter 18 : Electromagnetic Induction
3
Example 18.1.1
Fig. 18.1a below shows an insulated coil of copper wire (about 500 turns) that is connected
to a sensitive centre-zero galvanometer
galvanometer
N
S
coil of copper wire
Fig. 18.1a
(a)
When the N-pole of a bar magnet was plunged into the coil, the pointer of the
galvanometer was observed to deflect to one side. Explain this observation.
The magnet was plunged into the coil, there is a momentary increase in
the magnetic lines of force linked to the coil. By Faraday’s Law and
Lenz’s Law, an emf is induced in the coil which drives a current to
oppose the change, hence the galvanometer deflected to one side.
(b)
When the bar magnet was held stationary inside the coil, the pointer of the
galvanometer did not deflect. Explain this observation.
When the magnet was held stationary inside the coil, there is no change
in magnetic lines of force linked to the coil and hence no e.m.f. and no
current is induced.
(c)
When the N-pole of a bar magnet was withdrawn out of the coil, the pointer of the
galvanometer was observed to deflect momentarily to the opposite side to that in
(a) above. Explain this observation.
When the N-pole was withdrawn from the coil, there is momentary
decrease in the magnetic lines of force linked to the coil. By Faraday’s law
and Lenz’s Law, an e.m.f. is induced in the coil which drives a current in
the coil to oppose the change producing it. As the change (motion of
magnet) is in the opposite direction to (a) above, the induced e.m.f. is in
the opposite direction, the galvanometer deflects in the opposite direction.
(d)
Using the right hand grip rule, determine the direction of the induced current in
Fig. 18.1a.
Chapter 18 : Electromagnetic Induction
4
Example 18.1.2
Fig. 18.1b below shows an anemometer which is a device for measuring wind speed.
When it is positioned on the rooftop, the spindle will rotate when the wind blows and the
pointer of the galvanometer will deflect to give a reading.
rubber cups
spindle
bearing
galvanometer
magnet
Coil of wire
Fig. 18.1b
(a)
Explain how the rotating spindle causes the galvanometer pointer to be deflected.
The rotating spindle caused the magnet attached to it to rotate. This
caused the cutting of magnetic lines of force with the coil. By Faraday’s
law, an e.m.f. is induced in the coil which is registered in the voltmeter.
(b)
When the wind speed increases, what difference (if any) would you expect in the
magnitude of deflection of the pointer on the galvanoeter? Explain
When the spindle turns with a greater speed, the rate of cutting of
magnetic lines of force by the rotating magnet increases. This induces a
greater e.m.f. and hence a larger reading is shown on the voltmeter.
(c)
State and explain two ways in which the apparatus could be modified so as to
obtain a bigger reading on the same galvanometer.
• use a stronger magnet/ have more turns in the coil of wire to produce a
greater rate of cutting of magnetic lines of force.
• Use bigger cups to have greater force acting on them and hence greater
turning moment. So the spindle will turn at a greater speed.
Induced Current in a Straight Conductor
The relative directions of induced e.m.f. in a
straight conductor can be determined using
Fleming’s right hand rule
Chapter 18 : Electromagnetic Induction
5
18.2 The a.c. generator
An a.c. generator is a device that uses the principle of electromagnetic induction to
transform mechanical energy into electrical energy.
Example 18.2
Fig. 18.2 below shows an alternating current (a.c.) generator.
Fig. 18.2
(a) State the functions of the following components:
slip rings: to allow firm contact between the coil and carbon brush
and at the same time maintain freedom of rotation
(w/o twisitng the wires of the coil).
carbon brush: to maintain good electrical contact between the external
circuit and the rotating slip rings so as to transfer the
alternating e.m.f. generated to the external circuit
(b)
Explain how an e.m.f. is induced in the coil as it rotates.
When the coil is rotating from the horizontal to the vertical position, it cuts
through the magnetic field lines of permanent magnetic poles. The rate of
cutting of magnetic field lines decreases from a maximum (when the coil is
horizontal) to a minimum (when the coil is vertical) and then increases in the
opposite direction. By Faraday's Law, a varying e.m.f is induced in the coil.
(c)
At the instance shown in the above diagram, indicate clearly:
(i)
the direction of current flowing in the coil,
(ii)
the direction of currnt flowing through the bulb in the external circuit,
(iii)
the polarity of the carbon brushes.
Chapter 18 : Electromagnetic Induction
6
(d)
Sketch the graph which shows the variation of the induced e.m.f. of the a.c.
generator when the coil ABCD is rotated through one complete turn.
Emf / V
rotation
0
&frac14;
B
&frac34;
1
C
A
(e)
&frac12;
D
Sketch the graph which shows the variation of the induced e.m.f. of the a.c.
generator over time for the same complete turn which took 4.0 s.
Emf / V
g(i)
2.0
0
time / s
4.0
g(ii)
B
A
(f)
(g)
C
D
Sketch and label on the same axes in (f), the graphs which show the variation
of the induced e.m.f. of the a.c. generator when
(i)
the number of turns in the coil ABCD is halved, and
(ii)
the frequency of rotation of the coil ABCD is halved.
(iii)
the polarity of the carbon brushes.
State the changes to be made to convert the a.c. generator to a d.c motor.
i. When Replace the slip rings with a split-ring commutator
ii. Put a d.c. power supply in the external circuit.
Chapter 18 : Electromagnetic Induction
7
18.3.1 Mutual Induction
A changing current in a coil generates a varying magnetic field which can induce an e.m.f. in
another coil nearby. This effect is known as mutual induction.
Example 18.3a
Fig. 18.3a below shows two coils of copper wire wound around a soft-iron rod.
Q
P
soft-iron rod
S
Fig. 18.3a
Both coils can slide easily on the rod. Coil P is connected in series to a battery, a rheostat
and a switch S. Coil Q is connected to a sensitive centre-zero galvanometer.
(a)
As the switch S is closed, a momentary deflection is seen on the galvanometer
and the coils slide apart a little. Explain briefly:
(i)
why a momentary deflection is seen on the galvanometer, and
When the switch S is closed, the current flowing in coil P induces
law, this increasing change in magnetic flux induces an e.m.f in
coil Q that opposes the change. Hence, a momentary current
flows in Q which cause a momentary deflection in the
galvanometer.
(ii)
why the coils slide apart a little.
The current in coil P induces a magnetic field in coil P with a S
pole on its right. To oppose the change in magnetism, the induced
current in coil Q flows in a direction to set up a S pole on its left
end, by Lenz's law. Since both coils have S-poles facing each
other, they repel a bit.
Chapter 18 : Electromagnetic Induction
8
(b)
State and explain what you would observe as S is opened.
As the switch S is opened, there is decreasing magnetic field linking
coil Q. The induced current set up in coil Q to oppose the change, is in
the opposite direction to that in (a) and N-pole is induced on the left of
coil Q. Since opposite poles are facing each other, the two coils slide
towards each other a little.
(c)
State what would be the change(s) to the observations in part (a) above if the
soft-iron rod was replaced by a wooden rod
The deflection will be smaller as there is weaker magnetic flux linkage
between the two coils. Unlike wood, soft iron will intensify the magnetic
effect through the coil.
18.3.2 The transformer
A transformer is a device that changes a high alternating voltage (at low current) to a
low alternating voltage (at high current) and vice versa.
It is used for:
• Electrical power transmission from power stations to households and factories,
and
• Regulating voltages for proper operation of electrical appliances, e.g. the
television and CD player
Example 18.3b
Fig. 18.3b below shows a simple transformer.
Fig. 18.3b
The coil on the left is connected to an a.c. power source. At the instant shown, the current
is at its maximum positive value.
Chapter 18 : Electromagnetic Induction
9
(a)
(b)
On the diagram above, draw the magnetic field pattern setup by the current at this
instant.
The current in the left-hand coil goes through one complete cycle. Sketch graphs
to show the variation of
(i) current in the left-hand coil against time.
current
time
(ii) magnetic field strength set up by the left-hand coil against time.
magnetic
field
strength
time
(c)
Describe the effect of the current in the left-hand coil on the right-hand coil.
An e.m.f. is induced in the right hand coil to oppose this change of
magnetic flux. Hence the emf and the current induced varies similarly
to the magnetic field variation in the iron core shown above.
Example 18.3c
A transformer has 3300 turns in its primary coil and is used to operate a 12 V 24 W lamp
from the 220 V a.c. mains.
(a)
Draw a labelled circuit diagram showing the above connection.
220 V
a.c.
mains
(b)
3300
turns
12 V
24 W
Assuming there is no power loss in the transformer, find
(i) the number of turns in the secondary coil,
Vs / Vp = Ns / Np
Ns = 180
Chapter 18 : Electromagnetic Induction
10
(ii) the current in the secondary coil,
I = P/V
= 24/12 = 2.0 A
(iii) the current in the primary coil.
Ip Vp = Is Vs
Ip = 0.11 A
Example 18.3d
In Fig. 18.3d below, a lamp is connected to the secondary coil of the transformer by long
leads (or which have a total resistance of 2.5 Ω.The resistance of the leads is represented
by a single resistor of 2.5 Ω.
The input power of the transformer is 40 W, and the transformer is 100% efficient.
2.5 Ω
input voltage
240 V
lamp
primary coil
1600 turns
Fig. 18.3d
secondary coil
80 turns
Determine
(a) the turns ratio of the transformer.
80:1600 = 1:20 = 0.050
(b)
the voltage across the secondary coil of the transformer.
VS / VP = 0.050
(c)
the current in the secondary coil of the transformer.
Ip Vp = Is Vs
(d)
Vs = 240 x 0.050 = 12 V
40 W = Is x 12; Is = 3.33 ≈ 3.3 A
the voltage across the 2.5 Ω resistor.
V = IR = 3.33 x 2.5 = 8.325 ≈ 8.3 V (2 sf)
(e)
the electrical power dissipated in the long leads.
leads: Pleads = I2R = (3.33)2 (2.5) = 27.7 ≈ 28 W (2 sf
(f)
the electrical power dissipated in the lamp.
Ps = Plamp + Pleads, P leads = Ps - Plamp = Is Vs - Plamp
= (3.33 x 12) - 27.7 ≈ 12 W (2 sf)
Chapter 18 : Electromagnetic Induction
11
18.3.3 Power Transmission
Example 18.3e
In the Fig. 18.3e shown below, electrical energy is conveyed from a 10 V supply to a
filament lamp via two ideal transformers. PR and QS are two long cables, each of
resistance 5.0 Ω. The current drawn from the supply is 2.0 A.
R
P
.
10 V
supply
100
100
turns
turns
1000 turns
1000 turns
Q
100
turns
S
Fig. 18.3e
(a)
Determine
(i) the potential difference across PQ,
the potential across PQ = 10 x 10 = 100 V
(ii) the current in cable PR,
2.0 A x 10 V = Is x 100 V
Is = 0.2 A
(iii) the potential difference across cable PR,
VPR = 0.2 A x 5.0 Ω = 1 V
(iv) the potential difference between terminals R and S.
V across RS = 100 V - 1 x 2 V = 98 V
(b)
Determine
(i) the power from the supply,
P = IV = 2.0 A x 10 V = 20 W
(ii) the total power dissipated by the cables.
power dissipated = heat generated
= 0.22 x 5.0 x 2 Ω = 0.4 W
Chapter 18 : Electromagnetic Induction
12
Example 18.3f
The diagram below represents, in simplified outline, an electrical supply system from a
power station generator to a house.
power station
generator
transmission lines of
resistance 150 Ω
X
Y
transformer X supplies 50 kV
to transmission lines
transformer Y at
substation supplies
240 V to houses
Fig. 18.3f
(a)
Calculate the current passing through the element of a 3.0 kW electric kettle when
it is used in a house in which the supply voltage is 240 V.
P = IV I =
(b)
300 A
240 V
= 12.5 A
The distribution cable connecting transformer Y to the house has a resistance of
0.5 Ω. Calculate the rate at which heat is produced in this cable when the kettle is
in use.
Heat = I2 R = 12.52 x 0.5 Ω
= 78.125 = 78 W(2 sf)
(c)
The transmission lines have a resistance of 150 Ω and carry a current of 0.062 A.
Determine the rate at which heat is produced in the transmission lines.
Heat = 0.0622 x 150 Ω
= 0.5766 W = 0.58 W (2 sf)
(c)
Comment on the fact that the result for part (b) is much greater than that for part
(c).
As the current flows in the distribution cable is about 200 times more
than that flows in the transmission lines so the heat generated in (b) is
much more than in (c) as heat = I2R.
Chapter 18 : Electromagnetic Induction
13
Example 18.3g
The alternating voltage chosen for the transmission of electrical power over a large
distance is many times greater than the voltage of the domestic supply
(a)
State two reasons why electrical power is transmitted at high voltage.
• When electrical power is transmitted at high voltage, the current in
the transmission cables would be small. This would minimize the
power loss in the cables.
• Smaller currents would allow relatively thinner cables to be used for
power transmission to save cost.
(b)
.State two advantages of using alternating voltages.
• A .c. voltages can easily be stepped up or down by transformers
cheaply and efficiently with very little loss of power. (D.c. voltages
can be changed, but it is difficult and expensive.)
• A.c. voltages produced by a.c. generators are cheaper to produce
than direct voltages that are converted from chemicals.
Physics around you
The picture below shows a number of rectangular loops cut into the asphalt surface
on the road commonly seen at traffic junctions. Discuss why such markings are
Summary
Chapter 18 : Electromagnetic Induction
14
Examples of EM induction
Equipment / Description
Change
1 Magnet + straight wire
Magnet or wire
moves
2 Magnet + coil/solenoid
When magnet
moves into/out
of coil/solenoid
quickly
Effect
Cutting of magnetic
the wire
⇒ e.m.f. induced
across wire
to the coil changes
⇒ e.m.f. induced in
the coil, hence
induced current flows
(or coil moves
towards/away
from magnet)
3 Two coils
1 coil to produce magnetic field e.g. transformer
When the
switch is
opened or
closed, current
in 1st coil
changes, so
magnetic field in
this coil
changes
to the 2nd coil
changes
⇒ e.m.f. induced in
the 2nd coil, hence
induced current flows
4 Magnet + rectangular coil (armature)
e.g. a.c. generator
When coil
rotates
to the coil changes
⇒ e.m.f. induced in
the coil, hence
induced current flows
in the external circuit
5 Magnet + coil
e.g. bicycle dynamo
Coil
Chapter 18 : Electromagnetic Induction
When bicycle
moves, it
causes the
magnet to spin
Permanent magnet or rotate
Driving wheel
to the coil changes
⇒ e.m.f. induced in
the coil, hence
induced current flows
to light up the bicycle
lamp
15
Electromagnetism vs Electromagnetic Induction
Topics
Key
question
17, 18
Electromagnetism &amp;
Motor effect
Is there motion of a conductor
or a force on a conductor?
19
Electromagnetic Induction
“generator” (or dynamo)
Is there induced e.m.f. produced or
induced current flowing?
e.g. a wire moves, a coil rotates
Key physics
principles or
laws
Key rules of
thumb
and usage
e.g. a galvanometer deflects, a light bulb
lights up, a voltmeter gives a reading
1. A magnetic field is set up when 3. Faraday’s law of electromagnetic
a current flows in a conductot
induction
or when a charge moves
4. Lenz’s
law of electromagnetic
2. Force is experienced by a
induction
current carrying conductor / a
charge when a magnetic field
is present
Right hand grip rule (or corkscrew rule)
1. Given current in straight wire  find magnetic field around wire
(thumb)
(other fingers)
OR
2. Given current in coil/solenoid  find polarity at each end of coil/solenoid
(other fingers)
(thumb  North)
• OR aNticlockwise current = North pole
3. Fleming’s left hand rule
(motor rule)
Given
• current in a conductor
&amp;
• magnetic field
4. Fleming’s right hand rule
(dynamo rule)
Given
• motion of a conductor (force on it)
&amp;
• magnetic field
Produce  force or motion
Produce  induced emf and
induced current (if there is a closed
circuit)
Motion (Force)
Magnetic field
Current
Applications
1.
2.
3.
4.
5.
d.c. motor
electric bell / chime
speakers
electrical meters
particle accelerators
Chapter 18 : Electromagnetic Induction
Magnetic field
Motion (Force)
Current
1.
2.
3.
4.
5.
a.c. generator
transformer
microphone
induction cookers
mechanically powered flashlight
16
d.c. motor vs a.c. generator
d.c. motor
a.c. generator
B
N
P
A
X
C
S
D
Q
Y
Similarities
Similar basic structure
•
•
•
•
permanent magnet
rectangular coil of wire
external circuit
carbon brushes
Differences
•
electrical energy is used to cause
the rotation of the coil to produce
mechanical energy
•
mechanical energy is used to
physically rotate the coil to
produce electrical energy
Current
•
•
d.c. (direct current) flows
from battery in external circuit to coil
•
•
a.c. (alternating current) flows
from coil to external circuit
Sliding
contact
•
One split-ring commutator
•
Two slip-rings
Others
aspects
•
Fleming’s left hand rule (motor rule)
is applicable
•
Fleming’s right hand rule
(dynamo rule) is applicable
Energy
change
Chapter 18 : Electromagnetic Induction
17
Exercises
1
2
As shown in the diagram, a bar magnet is moving towards a wire loop. The
electron flow induced in the loop
A.
B.
C.
moves in an anti-clockwise direction.
moves in a clockwise direction.
attracts the magnet.
D.
speeds up the magnet.
When a magnet moves past an object, it will produce eddy currents in the object if
the object is
A.
B.
C.
D.
3
a solid.
an insulator.
a conductor.
As shown in the diagram, a conducting wire XY moves between the magnets.
Which of the following motions of XY will make the galvanometer deflect the most?
A.
B.
C.
D.
It moves sideways along the magnetic field quickly.
It moves sideways along the magnetic field slowly.
It moves perpendicular to the magnetic field quickly.
It moves perpendicular to the magnetic field slowly.
Chapter 18 : Electromagnetic Induction
18
4
Rotating a wire loop between the poles of a magnet can generate electricity. Which
of the following positions would induce the greatest current in the loop?
A.
B.
C.
D.
5
In a transformer, the secondary loop has twice as many turns as the primary. If the
alternating current in the primary loop is 2.0 A, what is the current in the secondary
loop?
A.
B.
C.
D.
6
7
8
The plane of the loop is parallel to the magnetic field.
The plane of the loop is perpendicular to the magnetic field.
The plane of the loop makes an angle of 45&deg; with the magnetic field.
The induced current is the same in all positions.
1.0 A
2.0 A
4.0 A
It cannot be determined.
Which of the following statements about generators and motors are correct?
A. The working principle of motors is based on electromagnetic induction.
B. Motors convert electrical energy to mechanical energy.
C.
The working principle of generators is based on the magnetic effect of a
current.
D.
Generators convert electrical energy to mechanical energy.
The voltage across the input terminals of a transformer is 110 V. The primary has
50 loops and the secondary has 120 loops. The output voltage of the transformer is
A.
B.
46 V.
55 V.
C.
D.
180 V.
264 V.
In which of the following situations is a voltage induced in a conductor?
A. The conductor moves through the air.
B. The conductor is connected to a battery.
C. The conductor is connected to a motor.
D. The conductor is moved in a magnetic field.
Chapter 18 : Electromagnetic Induction
19
9
Faraday’s law of electromagnetic induction describes how an electric field can be
reduced at a point in space by
A.
B.
C.
D.
10
11
an electric charge.
a constant magnetic field.
a changing magnetic field.
Which of the following statements about d.c. and a.c generators is / are correct?
(1)
An a.c. generator can be converted to a d.c. generator by replacing the
commutator with two slip rings.
(2)
Both of them convert mechanical energy to electrical energy.
(3)
The methods to increase the induced voltage in these two types of generators
are in totally opposite ways.
A.
B.
C.
D.
(1) only
(2) only
(1) and (2) only
(2) and (3) only
Figure (a) and Figure (b) show two set-ups using the same horseshoe core.
Figure (a)
Figure (b)
(a) What are the applications of the set-ups in Figure (a) and Figure (b)
respectively?
(b) What is the pole of the end P in each case?
(c) In Figure (b), explain why there is a reading in the voltmeter.
(d) State one method in each figure to increase the voltage across the coil on Q.
Chapter 18 : Electromagnetic Induction
20
11
(a)
In Figure (a), it is an electromagnet.
In Figure (b), it is a transformer.
(b)
In Figure (a), P is the north pole. It is decided by the right
hand grip rule.
In Figure (b), P is alternately north pole and south pole as the
current flowing through it is a.c.
(c)
As P is connected to the a.c., the alternating current causes a
changing magnetic field which is linked to Q through the
horseshoe core. By electromagnetic induction, there is a
current induced in the coil wound on Q and thus a voltage
induced across the coil.
(d)
For the set-up in Figure (a), the rheostat is adjusted so that
more current can flow through the coil.
For the set-up in Figure (b), increase the number of turns of
coil on Q or decrease the number of turns of coil on P.
Chapter 18 : Electromagnetic Induction
21
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