Focus SPM KSSM 2021 Tingkatan 5 - Physics

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PHYSICS SPM

5Form

KSSM

Yew Kok Leh D ual L anguage
Chang See Leong • Abd Halim Bin Jama’in P rogramme

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PHYSICS SPM

Yew Kok Leh 5Form
Chang See Leong
Abd Halim Bin Jama’in KSSM

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4Chapter Electromagnetism

CHAPTER FOCUS
4.1 Force on a Current-carrying Conductor
in a Magnetic Field
4.2 Electromagnetic Induction
4.3 Transformer

If you are in the hospital, you will find most of the doors have been equipped with an
electromagnetic door lock system. With the push of a button, the door will be locked. So, what
is the mechanism behind the system?
It is common knowledge that the electricity supply in houses comes from a power station. How
to generate electricity? How can electricity be delivered to houses even if they are hundreds of
kilometres away from the power station?

129

  Physics Form 5  Chapter 4 Electromagnetism

4.1 Force on a Current-carrying (a) When the d.c. power supply is switched
Conductor in a Magnetic Field on
• the short copper wire will move.
1. When a current-carrying conductor is in a • the force from the interaction between
magnetic field, it will produce a force. the magnetic field of magnadur magnet
and the magnetic field formed by the
2. The force is due to the interaction between the current-carrying conductor causes the
magnetic field of the permanent magnet and short copper wire to move.
the magnetic field produced by the current-
carrying conductor. (b) When the direction of the current is
reversed
Magnadur magnet • the short copper wire will move in the
opposite direction.
D.c. power supply
(c) When the polarity of the magnadur magnet
Magnadur magnet Wooden block is reversed
Short copper wire • the short copper wire will move in the
opposite direction
Figure 4.1
(d) When the direction of the current flow
Figure 4.1 shows an activity demonstrating in the short copper wire is parallel to the
the force on a current-carrying conductor in a direction of the magnetic field
magnetic field. • the short copper wire does not move
• there is no interacting force produced.

Chapter Catapult Field
1. When current flows, it will produce its own magnetic field.

4 2. When the magnetic field produced from the current interacts with the magnetic field from the magnadur
magnet, a force is produced at the current-carrying conductor.

3. The pattern of the catapult field is an illustration to show the interaction of the magnetic field to produce
a force.

NS NS

• When two magnets of different • The pattern of the magnetic field for • Both magnetic fields interact with
poles are placed close to each the electric current in the conductor each other and produce a resultant
other, the pattern of the magnetic is in the form of rotation and its magnetic field known as catapult
field is in the form of a straight direction is according to the right- field.
line with its direction from north to hand grip rule.
south. • The region where both magnetic
fields have the same directions
produce stronger catapult field.
The region where both magnetic
fields have the opposite directions
produce weaker catapult field.

• The differences cause the current-
carrying conductor to move to the
weak region.

Figure 4.2 The pattern of the catapult field

130

Physics Form 5  Chapter 4 Electromagnetism  

Fleming’s left-hand rule
1. Fleming’s left-hand rule is used to determine the direction of the force.

2. (a) The thumb shows the direction of the force acting on the conductor.
(b) Index finger shows the direction of the magnetic field.
(c) Middle finger shows the direction of the electric current.

Magnadur Conductor Magnetic field Index finger (Field)
magnet Force Middle finger (Current)
S
U

Thumb (Force / Motion)

        Current   

Figure 4.3 Fleming’s left-hand rule

Factors Affecting the Magnitude of the Force Acting on a Current-carrying Conductor
in a Magnetic Field

ACTIVITY 4.1 Chapter

To study the factors affecting the magnitude of the force on a current-carrying conductor in a magnetic 4
field

Problem Statement:
What are the factors affecting the magnitude of the force on a current-carrying conductor in a magnetic
field?

A The magnitude of current

Inference:
The magnitude of force on a current-carrying conductor in a magnetic field is influenced by the
magnitude of current used.

Aims:
To investigate the relationship between the current and the magnitude of the force on a current-carrying
conductor in a magnetic field.

Hypothesis:
The greater the size of the current used, the greater the magnitude of the force on a current-carrying
conductor.

Material and Apparatus:
Wooden block, thick copper wire (s.w.g. 20) without insulator, thin copper wire (s.w.g. 26) without
insulator, magnets, connecting wire, rheostat, ammeter, power supply and half metre rule

131

  Physics Form 5  Chapter 4 Electromagnetism

Instructions:

Thick copper wire
Thin copper wire

Magnet Height of
swing, h

Rheostat

Figure 4.4

1. The apparatus and materials are setup as in Figure 4.4.

2. The current is set to 2.0 A by using rheostat.

3. When power supply is turned on, thin copper wire swung up. The height of the thin copper wire
swung is measured by using half metre rule.

4. Step 2 and step 3 is repeated with different magnitude of current (2.5 A, 3.0 A, 3.5 A and 4.0 A).

Chapter Result: Table 4.1
The result is recorded in Table 4.1.
The height of thin copper wire, l / cm
Current, I /A
2.0 1.1
2.5 1.5
3.0 2.0
3.5 2.3
2.6
4 4.0

Analysis:

The height of Graph of the height of thin copper wire against the current
thin copper
wire /cm

3.0

2.5

2

1.5

1

0.5

0 Current /A
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Conclusion:
The force on a current-carrying conductor increases linearly to the current supplied.

132

Physics Form 5  Chapter 4 Electromagnetism  

B The strength of the magnetic field

Inference:
The magnitude of force on a current-carrying conductor in a magnetic field is influenced by the strength
of the magnetic field.

Aim:
To investigate the relationship between the strength of the magnetic field and the magnitude of the
force on a current-carrying conductor.

Hypothesis:
The greater the strength of the magnetic field, the greater the magnitude of the force on a current-
carrying conductor.

Material and Apparatus:
Wooden block, thick copper wire (s.w.g. 20) without insulator, thin copper wire (s.w.g. 26) without
insulator, magnets, connecting wire, rheostat, ammeter, power supply and half metre rule

Instructions:

1. The apparatus and materials are setup as in Figure 4.4.
2. One pair of magnets is used and the current is fixed to 5.0 A.
3. When power supply is turned on, thin copper wire swung up. The height of the thin copper wire

swung is measured by using half metre rule.
4. Step 2 and step 3 is repeated with different number of magnet pairs (2, 3, 4 and 5).

Result: Table 4.2 Chapter
The result is recorded in Table 4.2.
The height of thin copper wire, l / cm
Number of magnet pair

1 1.5 4
2 1.7
3 2.1
4 2.3
5 2.5

Analysis:

Graph of the height of thin copper wire against the number of magnet pair

The height of
thin copper
wire / cm 2.5

2.0

1.5

1.0

0.5

0.0 The number
1.0 2.0 3.0 4.0 5.0 of magnet pair

Conclusion:

The force on a current-carrying conductor increases linearly to the strength of the magnetic field.

133

  Physics Form 5  Chapter 4 Electromagnetism

Factors Affecting the Magnitude of the Force Acting on a Current-carrying Conductor
in a Magnetic Field

The The larger the current, the larger the force produced.
magnitude of How to increase the current?
• Increase the magnitude of e.m.f.
current • Use thicker and shorter wire (lower resistance).
• Use coil instead of one straight wire.

The strength The stronger the magnetic field, the larger the force produced.
of the How to increase the strength of the magnetic field?
• Use stronger magnet.
magnetic • Place magnets closer each other.
field

The Turning Effect of a Current-carrying Coil in a Magnetic Field

Magnet

Axle

Wooden block Split pin
Rivet Base

Chapter

4 to battery Yoke
Coil

Figure 4.5 Simple motor

1. The turning effect can be seen when the coil and magnet are assembled into simple motor.
2. The wire is coiled around the wooden block and is placed between two magnets.
3. Axle is assembled to loosen with split pin so that the wooden block can be rotated.
4. The two ends of the wire connected to battery are not fixed together to the end of the coils.

F

bc U S
US

a
d

Current in Current out    F
Figure 4.6

5. When the power supply is turned on, current flows through the coil from a to b and then c to d as in
Figure 4.6, causing the catapult effect.

134

Physics Form 5  Chapter 4 Electromagnetism  

6. By using Fleming’s left-hand rule, the coil at a-b moves downward and the coil at c-d moves upward.
Therefore, the wooden block is turned anti-clockwise.

Working Principle of the Direct Current Motor

4. Magnetic fields due to
current and permanent
magnets combine to
form catapult fields

3. Current in the coil 5. Magnetic forces
produces a magnetic act on the sides of
field the coil beside the
poles of the magnets

2. Current flows into F B I 6. Magnetic forces
and out of the coil F produce turning
through the carbon I effect to rotate
brushes S the coil
N
Permanent magnet
Coil
Carbon brush
Permanent magnet
Carbon brush

Commutator B : Magnetic field
I : Current
1. DC power supply F : Magnetic Force

  is switched on

Figure 4.7 The direct current motor

1. The direct current motor is usually found at electrical devices which use low power consumption such as Chapter
toy car and mini fan.

2. The working principle of the direct current motor is explained in Figure 4.7 and Figure 4.8.

F 4
BC
UA Process 1
S • Current flows through the loop D, C, B and A.
• According to Fleming’s left-hand rule, the current flow in the coil has
D
F caused the force acts downward at D-C, while the force acts upward
at B-A.
• The acting force causes the coil to rotate clockwise.

B Process 2
• When the coil is in a vertical position, the current flow is interrupted
A
because the carbon brush does not touch the commutator.
U CS • However, the coil is still rotating clockwise due to the inertia effect.

D

135

  Physics Form 5  Chapter 4 Electromagnetism

UD F Process 3
CB • The coil continues to rotate until the current flows reversely through A,

S B, C and D.
• According to Fleming’s left-hand rule, the current flow in the coil has
A
F caused the force acts downward at A-B, while the force acts upward
at C-D.
• The acting force causes the coil continues to rotate clockwise.

C Process 4
• The coil rotates until it is in a vertical position.
D • At this time, the current flow is interrupted but the coil continues to

U BS rotate due to inertia until back to initial state.
• The coil will continue to rotate clockwise undergoing Process 1 to
A
Process 4 continuously.

Chapter Figure 4.8

Factors Affecting the Speed of Rotation of an Electric Motor
1. The rotation speed of a direct current (d.c.) motor depends on:

(a) The number of turns of wire coil

4 (b) The magnitude of current flow to the coil
(c) The strength of the magnetic field

The magnitude of current flow to the coil The strength of the magnetic field
The higher the magnitude of current flow to The higher the strength of the magnetic

the coil, the higher the speed of coil rotation. field, the higher the speed of coil rotation.

The number of turns of wire coil
The higher the number of turns of wire
coil, the higher the speed of coil rotation.

Figure 4.9 Components of an electric motor

136

Physics Form 5  Chapter 4 Electromagnetism  

Brushed Motor and Brushless Motor
1. There are two types of electric motor which are brushed and brushless motor.

U S To battery
U Magnetic field
Commutator
Axle Brush Coils

To battery

Armature

Brushed motor Brushless motor

Figure 4.10 Brushed motor and brushless motor

Need a pair Both use Does not
of brush to working principle need brush
commute with
commutator of d.c. motor and
commutator
Brushed Both use
motor coil and magnet Brushless
motor
as the main
component

The magnet is Turning effect The coil is Chapter
fixed but the from the fixed but the
coil rotates 4
interaction of magnet
catapult field rotates

Figure 4.11 The characteristic of brushed and brushless motor

Has no risk of
spark ion from

commutator
(ESD)

Higher torque Long lasting and
produced no maintenance

High efficiency of required
electric consumption
Less friction
and sound yield

Less
electromagnetic

interference

Figure 4.12 The advantages of using brushless motor

137

  Physics Form 5  Chapter 4 Electromagnetism

EXAMPLE 4.1 U

Figure 4.13 shows a short copper wire AB placed A
between two magnets.

U + +
– –
A
B

B S

S (b) If the current flows inversely, the direction of the
force can also be determined using Fleming’s
Figure 4.13 left-hand rule.

(a) Determine the direction of force applied to the U –
short copper wire AB. +
A
(b) What happens to the direction of force if
Chapter the direction of the current is changed in the B
opposite direction from the original direction?
S
Solutions
(a) The direction of force can be determined using

Fleming’s left-hand rule.

4 4.2 Electromagnetic Induction

Checkpoint! 4.1 1. Electromagnetic induction is the phenomenon
of inducing an electromotive force (e.m.f)
Figure 4.14 shows the structure of a direct current across the wire by cutting a magnetic field.
motor.
2. Figure 4.15 shows the activity to study
U B C electromagnetic induction in a straight wire.
D
A S Galvanometer
+–

Figure 4.14 U
S Straight wire
(a) By using arrows, show the direction of force at
wire AB and CD in the figure. Magnet

(b) What happen to the moving direction of AB and
CD if the magnet pole is inverted?

Figure 4.15

138

Physics Form 5  Chapter 4 Electromagnetism  

Table 4.3 Electromagnetic induction in a straight wire

(a) Copper wire is moved quickly upwards. (b) Copper wire is moved quickly downwards.

U U
S S

• The pointer of the galvanometer is deflected to • The pointer of the galvanometer is deflected to
one side. the opposite site as compared to Activity (a).

• There is a current flow in the wire. • This shows that there is a current flow in the
opposite direction.
(c) Copper wire is moved horizontally rapidly in parallel
with the direction of the magnetic field. (d) Copper wire is moved horizontally rapidly
perpendicular to the direction of the magnetic field.

U U
S S

• The pointer does not deflect. • The pointer does not deflect. Chapter
• This shows that no current flows in the wire. • This shows that no current flows in the wire.
4

3. Figure 4.16 shows the activity to study electromagnetic induction in a solenoid.

Galvanometer

Pemanent
bar magnet

Solenoid with
600 turns

Figure 4.16

Table 4.4 Electromagnetic induction in a solenoid

Activity Explanation

(a) Bar magnet is pushed – The pointer of the galvanometer is deflected to one side.
quickly into the solenoid. – This shows that there is a current flow in the wire.

(b) Bar magnet is pulled quickly – The pointer deflected to the opposite side as compared to Activity (a).

out from the solenoid. – This shows that there is a current flow in the opposite direction.

139

  Physics Form 5  Chapter 4 Electromagnetism

(c) Bar magnet is stationary – No deflection is observed.
inside a solenoid. – This shows that no current flows in the wire.

(d) The solenoid is moved – The pointer of the galvanometer is deflected to one side.
quickly towards bar – There is a current flow in the wire.
magnet.
– The pointer is deflected to the opposite side as compared to Activity (d).
(e) The solenoid is moved – This shows that there is a current flow in the opposite direction.
quickly away from bar
magnet.

Chapter Factors Affecting the Magnitude of the Induced Electromotive Force (e.m.f.)
There are three factors that affect the magnitude of the induced e.m.f.

1. The strength of magnetic field

2. The speed of relative motion

3. The number of turns of the coil

ACTIVITY 4.2

To study the factors affecting the magnitude of the induced e.m.f.

Identify the problem:
What are the factors affecting the magnitude of the induced e.m.f?

Inference:
The magnitude of the induced e.m.f is influenced by the strength of magnetic field, the speed of relative
motion and the number of turns of the coil.

Aim:

4 1. To investigate the relationship between the magnitude of induced e.m.f. and the strength of the
magnetic field.
2. To investigate the relationship between the magnitude of induced e.m.f. and the speed of relative
motion.
3. To investigate the relationship between the magnitude of induced e.m.f and the number of turns
of the coil.

Apparatus and Material:
Solenoids with 600 turns and 1200 turns, bar magnets, galvanometer and wire connector

Instruction:

Pemanent
bar magnet

Solenoid
with 600 turns

Figure 4.17

1. The apparatus is setup as in Figure 4.17.

140

Physics Form 5  Chapter 4 Electromagnetism  

2. A bar magnet is moved into a solenoid with 600 turns.
3. The maximum deflection of galvanometer reading is recorded.
4. Step 2 and 3 is repeated but 2 bar magnets are used.
5. Step 2 and 3 is repeated but the bar magnet is moved with higher speed.
6. Step 2 and 3 is repeated but the solenoid is changed to 1200 turns.

Result:
The galvanometer reading is recorded in Table 4.5.

Table 4.5

Number of Speed of Number Maximum reading of galvanometer / V
magnets magnet of turns of
solenoid Reading 1 Reading 2 Average

1 Low 600 0.2 0.3 0.25
0.4 0.5 0.45
2 Low 600 0.3 0.5 0.40

1 Higher 600

1 Low 1200 0.3 0.4 0.35

Analysis: Chapter

1. When the number of bar magnets is increased, the strength of the magnetic field is increased.
Therefore, the magnitude of the induced e.m.f also is increased.

2. When the speed of bar magnet is increased, the cutting of magnetic flux is increased. Therefore,
the magnitude of the induced e.m.f also is increased.

3. When the number of turns of coil is increased, the number of wires to cut magnetic flux is increased.
Therefore, the magnitude of the induced e.m.f also is increased.

Conclusion 4
1. The higher the magnetic field strength, the higher the induced e.m.f. is produced.
2. The higher the speed of relative motion, the higher the induced e.m.f. is produced.

3. The higher the number of turns of the coil, the higher the induced e.m.f. is produced.

Faraday’s Law
1. Based on Activity 4.2, the induced e.m.f is increased when:

(a) the speed of relative motion is increased.
(b the number of turns of the coil is increased.
(c) the strength of the magnetic field is increased.
2. All three factors cause the rate of cutting the magnetic flux to increase, therefore the magnitude of the
induced e.m.f also increases.

Faraday’s law states that the magnitude of the induced e.m.f is directly proportional to the rate of
cutting of magnetic flux.


Direction of Induced Current in a Straight Wire and Solenoid
1. For a straight wire which cutting through a magnetic field, the direction of the induced current is

determined by using Fleming’s right-hand rule.
2. For solenoid which cutting magnetic field, the polarity of the ends of solenoid is determined by using

Lenz’s law.

141

  Physics Form 5  Chapter 4 Electromagnetism Solenoid
Straight wire
SU
Magnetic Direction of motion of wire
field lines

S U

Induced Magnet
current
Magnet Figure 4.19

Figure 4.18

Fleming’s right-hand rule Lenz’s law

Len’s law states that the induced current always
flows in a direction that opposes the change of
magnetic flux that causes it.

(1) Magnet is moved
towards solenoid

S UU S

(2) Induced current produces a North
pole to repel the magnet coming
towards the solenoid

Chapter (1) Magnet is moved
away from solenoid

U SU S

4 (2) Induced current produces a
South pole to attract the
magnet moving away from it

Direct Current Generator and Alternating Current Generator
1. The main application of electromagnetic induction is to generate electricity.

2. There are two types of generators:
• Direct current (d.c.) generator
• Alternating current (a.c.) generator

V S Coil is rotated V S Coil is rotated
BU BU
C

0 t AD 0 tA D

Commutator Coil Slip rings Coil
+ Carbon brush Carbon
– brushes
R
   R
Figure 4.20

142

Physics Form 5  Chapter 4 Electromagnetism  

Direct Current (d.c.) Generator
1. A direct current generator consists of coil mounted on an axle between the poles of a magnet.
2. There are commutators at both ends of the coil, while two carbon brushes are connected to output such

as electric device.
3. The brushes are contacted to commutators loosely.
4. The coil structure on the axis is rotated so that there is a magnetic flux cutting, and thus produces an

induced current and induced e.m.f.

Mechanism of D.C. Generator

Process 1 Process 2

US US
A A
B
B

• When the coil is in a vertical position, the sides of • When the coil is in a horizontal position, the sides Chapter
the coil will move in parallel with the magnetic flux. of the coil will move perpendicular to the magnetic
flux. 4
• Therefore, there is no cutting of magnetic flux.
• In addition, the carbon brushes do not touch the • Therefore, there is a cutting of magnetic flux.
• The carbon brushes touch the commutator. Induced
commutator. Side A of commutator is above and
side B of commutator is below. current is produced.
• Based on Fleming’s right-hand rule, the current
Process 4
flows from side A to side B.

Process 3

US US
B B
A
A

• When the coil is in a horizontal position, the sides • When the coil is in a vertical position, the sides of
of the coil will move perpendicular to the magnetic the coil will move in parallel with the magnetic flux.
flux.
• Therefore, there is no cutting of magnetic flux.
• Therefore, there is a cutting of magnetic flux. • In addition, the carbon brushes do not touch the
• The carbon brushes touch the commutator. Induced
commutator. Side B of commutator is above and
current is produced. The position of the commutators side A of commutator is below.
is reversed.
• Based on Fleming’s right-hand rule, the current
flows from side B to side A.

Figure 4.21

143

  Physics Form 5  Chapter 4 Electromagnetism Alternating Current (a.c.) Generator
1. An alternating current generator consists of a
5. If the output of d.c. generator is connected to
C.R.O, a V against t graph is plotted as shown coil mounted on an axle between the poles of
in Figure 4.22. a magnet.
2. Both ends of the coil are fixed with two
V different slip rings, while two carbon brushes
are connected to output such as electric device.
0 t 3 The brushes are contacted to slip rings loosely.
Figure 4.22 4. The coil structure on the axis is rotated so
that there is a magnetic flux cutting, and thus
Mechanism of A.C. Generator produces an induced current and induced e.m.f.

Process 1 Process 2

U S U S
B A – B+

A

Chapter • When the coil is in a vertical position, the sides of • When the coil is in a horizontal position, the sides
the coil will move in parallel with the magnetic flux. of the coil will move perpendicular to the magnetic
flux.
4 • Therefore, there is no cutting of magnetic flux.
• The carbon brushes always touch the slip ring. • Therefore, there is a cutting of magnetic flux.
However, no induced current is produced because • The carbon brushes always touch the slip ring.
there is no current flows out to the wire.
Induced current is produced.
• Based on Fleming’s right-hand rule, the current

flows from brush B to brush A.

Process 4 Process 3

U S U S
A + B– B

A

• When the coil is in a horizontal position, the sides • When the coil is in a vertical position, the sides of
of the coil will move perpendicular to the magnetic the coil will move in parallel with the magnetic flux.
flux.
• Therefore, there is no cutting of magnetic flux.
• Therefore, there is a cutting of magnetic flux. • The carbon brushes always touch the slip ring.
• The carbon brushes always touch the slip ring.
However, no induced current is produced because
Induced current is produced. there is no current flows out to the wire.
• Based on Fleming’s right-hand rule, the current

flows from brush A to brush B.

Figure 4.23

144

Physics Form 5  Chapter 4 Electromagnetism  

5. If the output of a.c. generator is connected to C.R.O, 4.3 Transformer Soft iron
a V against t graph is plotted as shown in Figure core
4.24. Primary
coil output
V a.c.
input
0t a.c.

Soft iron core Secondary  
coil

Figure 4.24

EXAMPLE 4.2

Figure 4.25 shows + Input
a conductor wire
is connected to a 10 0 10 20 30 35 Output
galvanometer and
– 20 G
30

35

U
S

located in between Primary coil Secondary coil

two magnets. Figure 4.25 Figure 4.27 The structure and symbol of the transformer
(a) Explain what
1. A transformer is made up of two coils wound
happen to the galvanometer if the wire is on a soft-iron core.

moved upward rapidly. 2. The primary coil is connected to the a.c.
power supply while the secondary coil is
(b) Predict the effect of galvanometer if the wire connected to the output terminals. Chapter

is moved downward rapidly.

Solutions

(a) If the wire moved upward, induced current Working Principle of a Simple Transformer 4
is produced. The galvanometer pointer is
deflected to one side. A.c. power Soft iron core
supply
(b) If the wire moved downward, induced current
is produced in opposite direction. Thus, the Primary Secondary
galvanometer pointer is deflected to the circuit circuit
opposite side.
Primary Output
coil terminal

Checkpoint! 4.2 Secondary
coil
Figure 4.26 shows a bar magnet and a solenoid which
connected to a galvanometer. Figure 4.28

– 1. When power supply is switched on, the current
in primary coil increases, the growth of the
35 30 20 10 0 10 + magnetic flux causes the magnetic field lines
20 to cut the secondary coil. An e.m.f. is induced
30 in the secondary coil.
35

Permanent
bar magnet

Solenoid with 2. When the current in the primary circuit
600 turns decreases, the magnetic flux collapses and
the field lines again cut the secondary coil.
Figure 4.26 An e.m.f. acting in the opposite direction is
induced in the secondary coil.
(a) Explain what happen to the galvanometer if the
bar magnet is pushed quickly into the solenoid. 3. The alternating current in the primary coil
produces a changing magnetic flux that induces
(b) Predict what happen to the galvanometer pointer an alternating e.m.f. of the same frequency in
if the bar magnet is stay stationary inside the the secondary coil.
solenoid.

145

  Physics Form 5  Chapter 4 Electromagnetism

The relationship between number of turns and voltage: Where,
Np = Number of turns in primary coil
Np = Vp Ns = Number of turns in secondary coil
Ns Vs Vp= Input voltage or primary voltage
Vs= Output voltage or secondary voltage

Type of Transformer Step-down Transformer

Step-up Transformer Soft iron core

Soft iron core

Input Output Input Output

Primary Secondary Primary Secondary
coil coil coil coil

(a) The number of turns of the secondary coil is greater (a) The number of turns of the secondary coil is less than

than in the primary coil. in the primary coil.

(b) The induced output voltage in the secondary coil is (b) The induced output voltage in the secondary coil is

Chapter larger than the input voltage. smaller than the input voltage.

(c) The induced output current in the secondary coil is (c) The induced output current in the secondary coil is

smaller than the input current. larger than the input current.

4 (d) Increase the input voltage. (d) Decrease the input voltage.

Ideal Transformer 4. In reality, it is almost impossible to make an
ideal transformer.
1. An ideal transformer transfers all input power
from the primary circuit to the secondary 5. Therefore, efficiency of the transformer is
circuit (output). measured by using the following formula:

2. There are no power loss during the process of Efficiency, η = Output power × 100%
transforming the voltage. Input power

3. The output power is equal to the input power.

Therefore,

Output power = Input power 6. Substitute P = VI, so the formula becomes:
That is:
η= VsIs × 100%
VsIs = VpIp VpIp



146

Physics Form 5  Chapter 4 Electromagnetism  

Energy Loss in a Transformer Table 4.6

Causes of energy loss How to minimise energy loss?

1. Resistant • Use thicker wire.
Transformer coil consists of a long thin wire. High • Use other less resistance materials such as copper.

resistance causes the energy loss in heat form when
current flows.

2. Eddy current • Use a laminated iron core to increase surface area,
Magnetic field from a.c. current always changes thus reduce eddy current.

and induces eddy currents in the iron core. The
eddy currents heat up the iron core. Energy lost to
surrounding in heat form.

Chapter

3. Hysteresis • Use soft iron core because less energy is used to 4
Energy loss due to magnetisation and demagnetisation magnetise soft iron.

of iron core. When magnetic field from a.c. current
passes through the iron core, some of the input energy
lost when it changes to kinetic energy of the iron core
molecules.

4. Leakage of magnetic flux • Wind the secondary coil on the primary coil so that all
Secondary coil does not cut the magnetic flux with the the magnetic flux produced by the primary current will
pass through the secondary coil.
optimum capacity.

147

  Physics Form 5  Chapter 4 Electromagnetism Explanation

Uses of Transformers in Daily Life

Application

Electrical appliances

• Decrease the voltage so that the electrical appliance is
safe to use.

Electrical energy transmission system • Reduce the power loss caused by large amount of
current and resistance during transmission.
Chapter
• Power loss can be calculated by using the following
4 formula:
National grid network P = I2R

275 kW or 500 kV Where
P = Power loss
11 kV 132 kV I = Current
R = Resistance
Power station Town Grid • Based on Ohm’s law, increase in the voltage will reduce
Light industry
Residential the current.
area Heavy industry • So, a transformer can increase the voltage to reduce

11 kV the power loss.

Substation 33 kV • A national grid network connects many power stations
415 kV and all different types of consumers such as industries
and domestic users.

• If a power station is shutdown due to breakdown
or maintenance, other power station can cover the
electrical supply.

• Easy to manage the capacity usage of electric.
• Step-up and / or step-down transformers are used

depend on the usage.

240 V

Substation Main substation

148

Physics Form 5  Chapter 4 Electromagnetism  

EXAMPLE 4.3 EXAMPLE 4.4

Ip = 0.01 A IS = ? Calculate the efficiency of the following transformer

in Figure 4.29. IS = 0.3 A
Ip = 15 A

240 V 5V

VP = 240 V VS = 8000 V

Figure 4.28 Figure 4.29

An ideal transformer decreases the voltage from Solutions
240 V to 5 V. Calculate the current in the secondary Step 1: List all the information given.
voltage if the primer current is 0.01 A.

Solutions Vp = 240 V
Step 1: List all the information given. Vs = 8000 V
Vp = 240 V Ip = 15 A
Vs = 5 V Is = 0.3 A
Ip = 0.01 A
Step 2: Identify and write the formula that will be
Step 2: Identify and write the formula that will
be used. used. VsIs
VpIp
Vs Is = Vp Ip Efficiency, η = × 100% Chapter

Step 3: Substitute the values in the formula and
solve it.
Step 3: Substitute the values in the formula and
5 × Is = 240 × 0.01 solve it. 4
240 × 0.01
Is = 5 (8000) × (0.3)
η = (240) × (15) ×
Is = 0.48 A Efficiency, 100%

= 66.6%

Checkpoint! 4.3

Figure 4.30 shows a transformer is stepping up the voltage from 240 V to 6 000 V. Find the efficiency of the
transformer.

6.25 A 0.20 A

240 V 6000 V

Figure 4.30

149

Chapter
  Physics Form 5  Chapter 4 Electromagnetism
CONCEPT MAP
4
ELECTROMAGNETISM
150
Force on a current-carrying Electromagnetic Transformer
conductor in a magnetic field Induction
Working
The effect of current- The effect of current- Factor affecting Current principle
carrying conductor in a carrying coil in a the induced generator
magnetic field Ideal
magnetic field electromotive force transformer

Catapult Factors affecting Direct Factors affecting Speed or Direct Energy
current the speed of magnet current loss
motor rotation of a
Field the magnitude of motor Uses of
transformer
the force Current Number of
turns of the
The direction of Alternating
coil current
current Current

The direction of Strength of Magnetic Magnetic
magnetic field magnetic field field field

The direction of Fleming’s left- Direction of
force hand rule induced current

Right-hand grip Number of Straight wire Solenoid
rule turns of the

coil

Fleming’s right-hand rule Lenz’s law

Physics Form 5  Chapter 4 Electromagnetism  

SPM Practice 4

Objective Questions

1. Which of the following is the 4. Figure 3 shows the set up of D U
correct rule to determine apparatus to study the effect S
the direction of force on a of force on a current-carrying
current-carrying conductor in conductor in a magnetic field. 6. Figure 4 shows the inside of
a magnetic field? It shows that the conductor an ammeter structure. HOTS
A Fleming’s right-hand rule moves to the left at a small
B Fleming’s left-hand rule angle. HOTS 123 4
C Right hand grip rule 05
D Left hand grip rule Current, I
Point
2. Figure 1 shows Fleming’s of needle
left-hand rule.
Conductor U S Magnet
S Coil
S
Figure 1 Figure 4 Chapter
Magnet
What is represented by S? Which of the following actions 4
A Current U is able to increase the
B Force sensitivity of the ammeter?
C Magnetic field Figure 3 A increase the distance
between magnet and coil
3. Figure 2 shows a conductor What do you need to do so B reduce the number of
located between two that the wire moves to the turns of coil
magnadur magnet. HOTS opposite direction with larger C increase the number of
angle of deflection? turns of coil
A Change the polarity of the D change the magnet
magnet and increase the polarity
magnitude of current.
B Maintain the polarity of 7. Figure 5 shows a direct
the magnet and reduce current motor. HOTS
the magnitude of current.
Conductor Wire C Change the direction of Conductor
Magnadur magnet current and reduce the coil
U magnetic field strength.
D Maintain the direction of U S
current and increase the
magnetic strength. Magnet

Direction of S 5. Which of the following shows
current the correct pattern of catapult
flow X Z field?
Y W A

US

Figure 2 B Figure 5
S
Which direction of the U What happen to the coil when
the switch is turned on?
conductor will move? A Remains stationary
A W C Y B Rotates clockwise
B X D Z C C Rotates anti-clockwise
U D Rotates and stops with
S vertical position

151

  Physics Form 5  Chapter 4 Electromagnetism

8. What is the advantage of 10. Figure 7 shows a magnet 12. A transformer is used to step-
using brushless motor? moved into the coil and an down the potential difference
induced current is produced. from 240 V to 5 V. If the
A Motor is able to rotate efficiency of the transformer
faster. HOTS is 75% and the current flows
at primary coil is 120 mA,
B There is no contact 0 Galvanometer what is the output current?
between the brush and 10 10 A 4.32 A
B 4.42 A
the commutator Coil C 5.76 A
D 5.82 A
C There is an increase in U
current flow. 13. Figure 9 shows a symbol of a
Magnet transformer.
D There is less friction.

9. Figure 6 shows a deflection
of a galvanometer pointer
when a magnet is moved
towards solenoid.

Magnet

Solenoid U S

Figure 7

Galvanometer reading will Primary Secondary
increase when coil coil

Chapter Figure 6 A the relative movement is Figure 9
reduced.
Which law or rule describes The function of this
the direction of the B the number of turns of transformer is to decrease
deflection? coil is reduced. A the output energy.
A Fleming’s right-hand rule B the output power.
B Right hand grip rule C the strength of magnetic C the output current.
C Lenz’s law field is reduced. D the output potential
D Faraday’s law difference.
D the resistance of the wire
is reduced. 14. The following is a
conversation between Ahmad
4 11. Figure 8 shows a generator is connected to a cathode ray and David. HOTS
oscilloscope (C.R.O.).

US Ahmad: David, my cell
phone’s charger
Figure 8 has broken. Can I
charge my phone
Which of the following will show on the monitor if the coil rotates? by connecting to the
socket with a copper
HOTS C wire?

A David: No, It is wrong! Your
phone will damage
and can cause fire.

B D Why a cell phone cannot be
charged by using a wrong
charger?
A cell phone’s charger has
a step-up transformer to
reduce the voltage output.

152

Physics Form 5  Chapter 4 Electromagnetism  
15. Figure 10 shows a transmission power system. HOTS
B cell phone’s charger has
a step-down transformer Power station P Q
to reduce the voltage
output. Figure 10

C cell phone’s charger has Which of the following methods is suitable to reduce the power loss
a step-up transformer during the transmission?
to increase the voltage
output. A Use high resistance cable.
B Use tungsten as a transmission cable.
D cell phone’s charger has C Increase the voltage of transmission.
a step-down transformer D Increase the current flow along the cable.
to increase the voltage
output.

Subjective Questions

Section A
1. Figure 1.1 shows a conductor is located between two strong magnets.

Soft spring U Chapter
S Conductor

4

Figure 1.1 [2 marks]
(a) Explain what happen to the conductor when the switch is turned on. [1 mark]
(b) Based on your answer in 1(a), state whether the soft spring is extended or contracted.
(c) Identify two ways to increase the extension of the soft spring. [2 marks]

(d) Figure 1.2 shows an activity to study the force on a current-carrying conductor in a magnetic field. The
conductor moves to the left when current flows through it.

Current, I

Conductor [2 marks]
U 153

Magnet

S

Figure 1.2
State two ways to make the conductor to move to the opposite side.

  Physics Form 5  Chapter 4 Electromagnetism

2. (a) Figure 2 shows a conductor located in between two magnets. When the current flows through the
conductor, it will be displaced to a direction.

U

Conductor
B
A CS

D

Figure 2

(i) What is the suitable rule used to identify the direction of the conductor displaced? [1 mark]
(ii) Choose A, B, C or D for the direction of the conductor movement. [1 mark]
(iii) What happen to the conductor if the dry cells are inverted? [1 mark]
(iv) Give two ways to increase the distance of the conductor displaced. [2 marks]
(b) Table 1 shows three electric motors design, A, B and C, used to build a toy car.

Table 1

Electric motor A

Fan blade

Number of
turns of coil = 50 turns
S

U

Chapter Semi circular-shaped Battery
permanent magnet
4 Electric motor B S Number of
Fan blade turns of coil = 10 turns
Battery
U

Rectangular-shaped
permanent magnet

Electric motor C

Fan blade

U S Number of
turns of coil = 50 turns

Battery

Rectangular-shaped
permanent magnet

154

Physics Form 5  Chapter 4 Electromagnetism  

Based on Table 1, state the suitable characteristics of an electric motor to be used to rotate the toy car

wheel with a greater speed. Give reason for the suitability of the following characteristics: [2 marks]
(i) Number of turns of coil [2 marks]
(ii) Number of batteries used [2 marks]
(iii) Shape of permanent magnet

(iv) Determine the most suitable design of an electric motor that can be used to rotate the toy car wheel
[1 mark]
with greater speed.

3. Figure 3.1 shows the pattern of the magnetic field for magnets and current in a conductor.

US

Figure 3.1

(a) What is the catapult field? [1 mark]

(b) In Figure 3.1, label L for the region with the weakest catapult field and K for the region with the strongest
[2 marks]
catapult field.

(c) Explain why the weak catapult field region exists? [3 marks]

(d) Figure 3.2 shows the interaction of the magnetic field patterns.

US Chapter

4

Figure 3.2

(i) Name the rule used to determine the direction of force on a current-carrying conductor in a magnetic
[1 mark]
field.

(ii) Label with an arrow in Figure 3.2 to show the movement direction of the conductor wire. [1 mark]

4. Figure 4 shows an electric motor. [1 mark] CB
(a) Name the type of the motor shown in Figure 4.

(b) Underline the correct answer.

(i) Based on the motor in Figure 4, when the switch is turned on, U S
current flows through (A - B - C - D /D - C - B - A). D A

(ii) Therefore, AB conductor will move (upward / downward) while
CD conductor will move (upward / downward).

(iii) The motor will rotate (clockwise / anti-clockwise). [4 marks]


(c) Give two advantages of using brushless motor. Figure 4
[2 marks]

155

  Physics Form 5  Chapter 4 Electromagnetism

5. Figure 5 shows a bar magnet in a state of free fall into a solenoid. It is found that the galvanometer pointer
was deflected when the bar magnet passes through the solenoid.
S
Magnet
U

Wire
Coil

Galvanometer

Figure 5

(a) What is the physical quantity measured by galvanometer? [1 mark]
(b) Explain why the galvanometer pointer was deflected. [2 marks]
(c) State one way to increase the deflection of galvanometer pointer. Explain your answer. [2 marks]

(d) What happen to the galvanometer pointer if the bar magnet is stay inside the solenoid? Explain your
[2 marks]
answer.

6. Figure 6 shows a solenoid positioned in between two magnets X and Y.

X Y
S
U P QU S

Chapter

4R

Figure 6 [2 marks]
(a) Name the polarity at P and Q if magnet X and magnet Y is moved into the solenoid. [2 marks]
(b) Name the polarity at P and Q if magnet X and magnet Y is moved out from the solenoid.
[1 mark]
(c) Name the law used to identify the polarity of solenoid.

Section B

7. Figure 7 shows the transmission of electricity from the power station to a school using transmission cables
and transformers.

Hight voltage transmission cables

X School
Y

Power station

Figure 7

(a) What is the meaning of national grid network? [1 mark]

156

Physics Form 5  Chapter 4 Electromagnetism  

(b) Based on Figure 7, identify the type of transformer used at X and Y. [2 marks]
(c) Give three advantages of using national grid network. [3 marks]

(d) The output power and the output voltage at the transmission cables are 10 MW and 144 kV respectively.

The total resistance of the transmission cable is 300 Ω. [2 marks]
(i) Calculate the current flows through the cable [2 marks]
(ii) Calculate the power loss

(e) In a school laboratory, a step-down transformer is used to light up a bulb of 6 V. S, T, U and V are four

transformers that can be used to step down the voltage from 240 V to 6 V.

The following table shows the characteristics of those transformers.

Transformer Core material Core structure Shape of the core Np : Ns
1 : 40
S Iron Solid K-shaped

Secondary
coil

T Soft iron Solid Primary 20 : 1
coil
Secondary Chapter
L-shaped coil

Primary
coil

U Soft iron Laminated K-shaped 40 : 1 4

Primary
coil

Secondary
coil

V Iron Laminated L-shaped 1 : 20

Primary Secondary
coil coil

Study the specifications of all the four transformers based on the following aspects: [10 marks]
(i) Core material
(ii) Core structure
(ii) Shape of the core
(iiv) Ratio Np : Ns
Then choose the best transformer and give the reason for the answer.



157

  Physics Form 5  Chapter 4 Electromagnetism

Section C
8. Figure 8.1 and Figure 8.2 show a bar magnet dropped into two different coils from the same height.

Coils U Galvanometer Coils U Galvanometer
S S

Figure 8.1 Figure 8.2

(a) What is the meaning of induced current? [1 mark]

(b) Using Figure 8.1 and Figure 8.2,

(i) Compare the strength of magnetic field of the bar magnet, the number of turns of the coil and the
[3 marks]
deflection of the galvanometer pointer.

(ii) State the relationship between the number of turns of the coil with the deflection of the galvanometer

pointer and the rate of cutting of magnetic flux. Then deduce the relationship between the rate of
[3 marks]
cutting of magnetic flux with the magnitude of induced current.

(c) Figure 8.3 shows a simple alternating current generator.

Chapter

US

4

Slip rings

Carbon
brush

Figure 8.3

Explain the working principle of the alternating current generator. [3 marks]

(d) Figure 8.4 shows a direct current motor used to rotate a mini fan blade.

Figure 8.4

Suggest and explain how to improve the efficiency of the direct current motor based on the following

aspect:

(i) The number of turns of coil used

(ii) The strength of magnet

(iii) The shape of magnet [10 marks]
(iv) The use of brush or brushless motor

158

JAWAPAN

1Chapter Q3 (a) Magnitude of the resistance Checkpoint 1.3

= 20 N Q1 (a) The forces are in equilibrium if
(b) Resultant force, F = 30 – 20 = 10 N the resultant force produced is
Force and Motion II F = ma → 10 = 4 × a zero in all directions.

a = 2.5 m s–2 (b) (i) Triangle of forces
FR B
Checkpoint 1.1 Q4 (a) Resultant force, F = 0 N;
Tension, T = weight of the fish O
Q1 (a) 17 N, in the same direction as
= 0.8 × 10
the direction of the acting force
=8N
(b) 7 N, in the same direction as the (b) Resultant force, F = 0 N;

(c) dFir=ect5io2n+o1f2122=N13 N Tension, T = weight of the fish = 8 N
(c) Resultant force, F = ma = 0.8
12 × 0.5 = 0.4 N; T – 8 = 0.4, 120 N
q = tan–1 5
therefore T = 8.4 N 100 N

= 67.4° with the horizontal Q5 (a) F = ma → 2 × 10 = (6 + 2) × a

force of 5 N a = 2.5 m s–2 A

Q2 (a) Scale 1 cm : 2 N (b) T = m × a = 6 × 2.5 = 15 N Resultant force,
FR = 1002 + 1202 = 156.2 N
B (ii) Free body diagram

4.6 cm Checkpoint 1.2

3 cm Q1 (a) Resolution of a force means
dividing a single force into two
component forces perpendicular 120 N
to each other.
O 27° 45° FR
2 cm A (b) (i) Vertical component upward 39.8°
= 12 cos 55°
Resultant force OB = 4.6 cm × 2 N
Horizontal component
= 9.2 N = 12 sin 55°
(ii) Vertical component
Direction of the resultant force = 100 N
27° with the force of 4 N. downward = 25 sin 45°
Horizontal component
(b) Scale 1 cm : 2 N = 25 cos 45° Direction of resultant force, q
(iii) Component parallel to the
B = tan–1 100 = 39.8°
inclined plane = 30 cos 60° 120
Component perpendicular
Therefore, a force of 156.2 N
to the inclined plane =
4.4 cm 3 cm 30 sin 60° needed to act in the

Q2 (a) Net force = 0 N as the car is at direction 129.8° (39.8°
rest
+ 90°) with the force of
(b) Frictional force = component
O 37° 120° weight of the car down the slope 100 N for the box to be in
A
5 cm = 12 000 sin 20° = 4 104 N equilibrium.
(c) The supporting force
Resultant force OB = 4.4 cm × 2 N = 12 000 cos 20° = 11 276 N Q2 Magnitude of the force
= 400 cos 30° + 400 cos 30°
= 8.8 N Q3 (a) Free body diagram:
= 692.8 N

Direction of the resultant force = Q3 (a) (i)
37° with the force of 10 N

(c) Scale 1 cm : 2 N 40° T
150 N 1
O A
3 cm Tension, T
10°
88° 120°

5.2 cm 6 cm Air resistance, R (ii) T2
Weight, W
30°
60° P

(b) Vertical component of T = T cos 10° 70°

B Horizontal component of Q = 100 N
T = T sin 10°
Resultant force OB = 5.2 cm × 2 N R
(c) T cos 10° = 1000 × 10

= 10.4 N T = 10 000 = 10154 N 50°
cos 10° 40°
Direction of the resultant force =
88° with the force of 6 N

247

  Physics Form 5  Answers

(iii) Energy needed, E = 1 kx2 (b) F = ma
2 T sin 10° = 1 kg × a
a = 10.2 sin 10°
F 100° F = 21 × 200 N m–1
C A = 1.77 m s–2

50° 30° × 0.052 m2 (c) (i)
FB = 10 kN
= 0.25 J

(b) (i) Scale 1 cm : 20 N Q3 (a) Spring constant,
k = gradient of graph

O = 8 N
16 cm

40° = 0.5 N cm–1 (ii) Direction of motion
a = – 1.77 m s-1
7.5 cm 9.7 cm (b) Potential elastic energy of spring 10°
W T1
= Area under the graph

= 1 × 8 × 0.16 = 0.64 J
2

T2 50° Q4 (a) (i) Hooke’s law
A 6.1 cm B
T1 = 9.7 cm × 20 N = 194 N (ii) Mass of load of 1 kg is
T2 = 6.1 cm × 20 N = 122 N
equivalent to a force of 10 N

Substitute into formula, 1 kg
F = kx → 10 N = k × 2 cm
Due to deceleration, the
(ii) Scale 1 cm : 20 N Therefore, spring constant, direction of action is in the
k = 5 N cm–1 atau 500 N m–1 opposite direction.

(b) – Increase the stiffness 3. (a) Weight due to force of gravity
act on the spring.
60° P 4.1 cm of spring by replacing
(b) Vector has magnitude and
the copper spring with a direction.

stronger steel spring. (c) (i) I = lO + x

Q 70° – Replace spring with spring Panjang spring/cm
5 cm R 4.6 cm
made of thicker wire. 140
50°
SPM Practice 1 120
Objective Questions
100
1. A 2. D 3. A 4. C 5. B 80
P = 4.1 cm × 20 N = 82 N 6. C 7. A 8. D 9. A 10. C I = 76 cm
11. A 12. B 13. B 14. A 15. A 60
R = 4.6 cm × 20 N = 92 N 16. B 17. C 18. D 19. C 20. C 40 I0
(iii) Scale 1 cm : 20 kN 20
Subjective Questions 9N
0
F Section A 2 4 6 8 10 12 14
A 1. (a) daya/N
3.9 cm (ii) l = lO + x
F TT 76 = 40 + x
C x = 76 – 40 = 36 cm
(d) Curve upwards after 10 N (refer
2.6 cm the above graph)

50° 30° Section B
4. (a) Scale 1 cm : 500 N
FB 5 cm 45° 45°
FA = 3.9 cm × 20 kN = 78 kN 5 cm
FC = 2.6 cm × 20 kN = 72 kN 10 N

(b) T2 + T2 = 102
Checkpoint 1.4 2T2 =
T = 15000 = 7.07 N
Q1 (a) Spring extension, x = 18 cm –
2. (a) Direction of motion
15 cm = 3 cm = 0.03 m
Magnitude of force, F = kx =

100 N m–1 × 0.03 m = 3 N

(b) Length of compression, T
F 10°
x = k 2N F1
= 100 N m–1 1 kg 7 cm

= 0.02 m R
10.5cm
= 2 cm 10 N 7 cm

Therefore, length of spring Resolving vertically
l = 15 cm – 2 cm = 13 cm T cos 10° = 10 N

Q2 Length of compression, T = 10 36°
x = 20 cm – 15 cm cos 10° 60°
= 5 cm F2
= 0.05 m = 10.2 N 5 cm

248

PHYSICS SPM CC035541 FOCUS SPM

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