Electromagnetism

Class 10 Physics, unit 19, book pages 259–280

Electromagnetism studies how electricity and magnetism are connected. This unit shows that a current makes a magnetic field (around a wire and in a solenoid), that a magnetic field exerts a force on a current (the basis of the loudspeaker, relay and electric motor), and that a changing magnetic field induces an e.m.f. (Lenz's law, the a.c. generator and the transformer). It ends with the deflection of an electron beam and the cathode ray oscilloscope.

Learning outcomes (20)

Describe an experiment to demonstrate electromagnetic induction.
Use the fact that the magnitude of an induced e.m.f. is affected by (a) the rate of change of the magnetic field or the rate of cutting of magnetic field lines, and (b) the number of turns in a coil, to solve simple electromagnetic problems.
Use the fact that the effect of the current produced by an induced e.m.f. is to oppose the change producing it (Lenz's law).
Describe how a.c. generators work.
Sketch and interpret graphs of e.m.f. against time for simple a.c. generators.
Describe the pattern and direction of the magnetic field due to currents in straight wires and in solenoids.
State the effect on the magnetic field of changing the magnitude and direction of the current.
Describe how the magnetic effect of a current is used in relays and loudspeakers.
Describe an experiment to show that a force acts on a current-carrying conductor in a magnetic field.
State and use the relative directions of force, magnetic field and current.
Describe the magnetic field patterns between currents in parallel conductors and relate these to the forces on the conductors.
State that a current-carrying coil in a magnetic field may experience a turning effect and that the turning effect is increased by increasing (a) the number of turns on the coil, (b) the current, (c) the strength of the magnetic field.
Describe the operation of an electric motor, including the action of a split-ring commutator and brushes.
State that it is theorised that the Earth's magnetic field is generated by the rotation of the Earth and its molten iron core that contains charged particles in motion.
Explain the principle of operation of a simple iron-cored transformer.
Use the terms primary and secondary coils and step-up and step-down transformer.
Use the equation $V_{s} / V_{p} = N_{s} / N_{p}$ .
Justify the advantages of high-voltage transmission.
Describe the deflection of an electron beam by electric fields and magnetic fields.
Interpret waveforms on oscilloscopes.

19.1Magnetic field due to a current-carrying wireOften examined

Electromagnetism is the part of physics that studies how electricity and magnetism are connected. Ampere discovered that when a current passes through a conductor, it produces a magnetic field around it. You can see this for yourself with a simple activity.

Figure 19.1: Compass needles around a straight wire, seen from above. The needles show circular field lines. When the current is reversed, every needle turns round.

Conclusions from Activity 19.1

A current through a conducting wire creates a magnetic field around the wire.
The magnetic field lines are circles around the wire, and their direction depends on the direction of the current.

Direction of the magnetic field of a current-carrying wire

The direction of the magnetic field $B$ around a straight current-carrying wire is found with the right-hand rule, Figure 19.2(a).

Figure 19.2(a): Direction of the magnetic field around a current-carrying wire (right-hand rule).

Properties of magnetic field lines

Like electric field lines, magnetic field lines are imaginary lines and do not exist physically. Their properties are given below.

Magnetic field lines emerge from north poles and end on south poles.
Magnetic field lines form closed paths.
The tangent to a magnetic field line gives the direction of the magnetic field at any point.
Magnetic field lines never intersect or cross each other.
The density of magnetic field lines at a point shows the strength of the magnetic field at that point.

These properties help you understand and visualise magnetic fields. They also help you understand how magnets work in electric motors, generators and similar machines.

Forces on parallel current-carrying conductors

When the currents in two long parallel wires are in the same direction, the magnetic fields created along each wire are such that their opposite poles lie in between the two wires, Figure 19.2(b). Opposite poles attract each other. Thus two long parallel wires carrying current in the same direction attract each other.

What will happen if the currents are in opposite directions? Now the fields of the two wires point the same way in the space between the wires, so like poles face each other. The wires repel each other.

Figure 19.2(b): Top view of two long parallel wires. Dot: current towards you; cross: current away from you. Same direction: the wires attract. Opposite directions: they repel.

19.2Magnetic field due to a current-carrying solenoid

A solenoid is a long, tightly wound (spring-like) coil of wire, Figure 19.3.

Figure 19.3: A solenoid.

When current passes through a solenoid, a magnetic field is set up around it. The field from each loop adds to the fields of the other loops, so the total field is stronger. The magnetic field of a solenoid is strong and uniform along its axis and negligibly weak outside. If the magnitude of the current is increased, the magnetic field becomes stronger. If the direction of the current is reversed, the direction of the field (and so the poles) is reversed.

Figure 19.4: The field of (a) a solenoid and (b) a bar magnet. Outside, the lines leave the north end, curve round and enter the south end. The two patterns are the same.

When a current-carrying solenoid is brought close to a suspended bar magnet, one end of the solenoid repels the north pole of the magnet. So the current-carrying solenoid has a north and a south pole and behaves like a permanent bar magnet.

Figure 19.5: Right-hand rule for a solenoid. The thumb points along the axis to the north pole.

Worked example: poles of a solenoid

In a solenoid drawn on paper, the current in the front wire of each turn (the part facing you) flows upwards. (a) Which end of the solenoid is the north pole? (b) Draw the field inside and outside. (c) What happens to the poles and to the strength of the field if the battery is reversed and a larger current is used?

Take the right hand and curl the fingers the same way as the current in a turn. On the front wire the current goes up. Over the top of the turn it goes away from you, down the back and forward along the bottom. Curling the fingers round this loop, the thumb points to the left.
So the left end is the north pole and the right end is the south pole.
Outside the solenoid the field lines leave the north (left) end, curve round and enter the south (right) end. Inside, the field lines run along the axis from the south end to the north end, that is from right to left, and they are close, straight and parallel.
Reversing the battery reverses the current. The thumb then points the other way, so the left end becomes the south pole and the right end the north pole. A larger current makes the field stronger, but it does not change where the poles are.

Field and poles of the solenoid in the example. Outside the coil the lines run from N to S and inside from S to N. The accent arrow shows the current in the front wire of a turn.

Left end N, right end S. Reversing the current swaps the poles; a larger current only makes the field stronger.

19.3Earth's magnetic field

The Earth can be thought of as containing an enormous bar magnet in its core, Figure 19.6. The magnetic south pole of the Earth's magnet is at the geographic North Pole, and the magnetic north pole of the Earth's magnet is at the geographic South Pole. This is because opposite poles attract: the north end of a compass needle points towards the Earth's south magnetic pole, which lies near the geographic North Pole.

Figure 19.6 and 19.7: The Earth's magnetic field looks like that of a bar magnet tilted about 11° from the axis of rotation. The accent half of the magnet is its south pole, near the geographic North Pole.

The Earth's magnetic field appears like a magnetic dipole. It can be represented by the field of a bar magnet tilted at an angle of about $1 1^{\circ}$ with respect to the Earth's axis of rotation (Figure 19.7). The magnitude of the Earth's magnetic field at its surface ranges from $25$ to $65 μ T$ ( $0.25$ to $0.65 G$ ). It is several hundred times weaker than the field around a bar magnet. The magnetic field lines are parallel to the surface of the Earth at the equator, and normal (at right angles) to the surface at the magnetic poles.

Where does the Earth's magnetic field come from?

The Earth's magnetic field (also called the geomagnetic field) is generated by the rotation of the Earth and its molten iron core that contains charged particles in motion. The core is mainly made of molten magnetic material such as iron and nickel. Molten iron is a good conductor. Its movement, driven by convection currents and influenced by the Earth's rotation, creates electrical currents within the outer core. These currents produce a magnetic field, which maintains a stable and protective magnetic shield around the Earth. The field starts in the core, comes out into space and forms a magnetosphere.

19.4.1Loudspeakers

Many devices work on the magnetic effect of a current. Two of them are described here: the loudspeaker and the relay.

Loudspeakers

In a loudspeaker the magnetic effect of a current is used to change electrical energy into mechanical energy, which then produces sound. A loudspeaker has a permanent magnet and a coil of wire, called the voice coil, which is attached to the speaker cone (Figure 19.8).

Figure 19.8: A loudspeaker. The alternating current in the voice coil moves the coil, and the cone attached to it, back and forth.

When an electrical signal (corresponding to the sound to be produced) is passed through the voice coil, it creates a magnetic field around the coil. This field interacts with the field of the permanent magnet. Depending on the direction of the current, the voice coil is either attracted towards or repelled from the permanent magnet. This movement of the voice coil makes the attached speaker cone vibrate. The vibrations move the air particles around the speaker and create pressure waves that our ears perceive as sound.

From signal to sound

Alternating current that follows the shape of the voice or music flows in the voice coil.
The coil's magnetic field interacts with the field of the permanent magnet, so the coil is pushed in and out.
The paper or plastic cone attached to the coil vibrates with it.
The cone moves the air, producing sound waves.

19.4.2Relay

A relay is an electrical control device that lets an input circuit control a separate output circuit (Figure 19.9(a)). It uses the magnetic effect of a current in the input circuit to switch the output circuit. A mechanical relay consists of an electromagnet, a switch and some mechanical parts that change the state of the switch when the electromagnet is turned on.

Main parts of a relay

Coil and core. A coil wound around a magnetic core makes an electromagnet. It generates a magnetic field around it when an electric current passes through the coil.
Armature. The armature is a movable part of the relay. When the coil is energised, the magnetic field it produces attracts the armature and makes it move.
Return spring. The return spring is connected to the armature and gives a restoring force when the coil is de-energised. It brings the armature back to its original position when the current in the coil stops.
Moving contact. It is attached to the armature (through the return spring) and moves with it. When the armature is attracted by the energised coil, the moving contact changes its position, either making or breaking contact with the fixed contacts.

Figure 19.9: A relay. (b) Switch off: no current in the coil, the spring holds the armature away and the lamp circuit is open. (c) Switch on: the electromagnet pulls the armature down, the contacts touch and the lamp lights.

How a relay works

Take a lamp that is switched on using a switch and a relay.

Switch off. No current flows through the coil, so the armature stays in its normal position (Figure 19.9(b)). The lamp circuit remains open and the lamp does not glow.
Switch on. Current flows through the coil, so it becomes an electromagnet and the armature is attracted towards the coil (Figure 19.9(c)). The moving contact touches the fixed contact, the circuit of the lamp is closed and the lamp glows.

19.5Force on a current-carrying conductor placed in a magnetic fieldOften examined

When a current-carrying conductor is placed in a magnetic field, it experiences a force. You can observe this with a simple activity.

Figure 19.10 (idea of the activity, drawn end-on): the current in the rod is towards you and the field runs from N to S, so the force on the rod is upwards.

Why does the conductor feel a force? A current-carrying wire produces its own magnetic field. When such a wire is brought into an external magnetic field, the field of the wire interacts with the external field, and as a result the conductor experiences a force.

The force on the conductor is increased if:

the current in the wire is increased;
the strength of the magnetic field is increased;
the length of the wire inside the magnetic field is increased.

Fleming's left-hand rule

The direction of the force on a current-carrying wire in a magnetic field is always perpendicular to both the magnetic field and the current. It is found with Fleming's left-hand rule (Figure 19.11).

Figure 19.11: Fleming's left-hand rule. The three directions are at right angles to each other. The thumb (force) is the accent.

Worked example: direction of the force on a wire

A straight horizontal wire carries a current from left to right. It lies in a magnetic field that points into the page. (a) Use Fleming's left-hand rule to find the direction of the force. (b) What happens to the force if the current is reversed? (c) What happens if the current and the field are both reversed?

Match the fingers to the quantities. The forefinger points along the field, into the page. The middle finger points along the current, to the right.
Keep the thumb at right angles to both fingers. With the left hand and these two fingers, the thumb points upwards (up the page). So the force on the wire is upwards.
Reverse the current: the middle finger now points left, so the thumb points downwards. The force reverses.
Reverse both the current and the field: the force reverses twice, so it points upwards again. The force is unchanged.

The wire carries current to the right in a field into the page. The force (accent) is upwards.

(a) u pw a r d s; (b) d o w n w a r d s; (c) u pw a r d s a g ain .

19.6Current-carrying coil in a magnetic field

Consider a rectangular coil PQRS lying in the magnetic field of a permanent magnet, Figure 19.12. When the ends of the coil are connected to the terminals of a battery, a current starts to flow in it.

Apply Fleming's left-hand rule to each side of the coil. On the PQ side the force acts upwards, and on the RS side the force acts downwards, because the current in the two sides of the loop flows in opposite directions. The two forces are equal in size and opposite in direction, so they form a couple. The torque of this couple turns the loop.

Figure 19.12: End-on view of the coil in the field. The forces on the two sides form a couple and the coil turns clockwise.

Hence a current-carrying loop placed in a magnetic field experiences a couple, and the coil begins to rotate. This is the working principle of the electric motor.

19.7Electric motorOften examined

When a current-carrying coil is placed in a magnetic field it experiences a torque. This is the working principle of an electric motor. An electric motor is a device that converts electrical energy into mechanical energy (rotational kinetic energy). The motor in Figure 19.13 consists of:

a rectangular coil placed between the poles of a permanent magnet;
brushes, usually pieces of graphite, which allow current to flow into the coil;
a split-ring commutator (a ring split into two halves), which turns with the coil;
permanent magnets or an electromagnet to produce the magnetic field.

Figure 19.13: End view of a simple motor. (a) The brushes touch the two halves of the split ring and current flows. (b) The coil is vertical, the gaps are at the brushes, and the current is cut off for an instant.

Why a split-ring commutator is needed

A simple coil placed in a magnetic field **cannot rotate more than $9 0^{\circ}$ **. The forces push the PQ side up and the RS side down until the loop reaches the vertical position, where the plane of the loop is perpendicular to the magnetic field. The loop will not continue to turn because the forces are still up and down, so they are now balanced (the net force on the coil is zero and there is no turning effect).

To keep the coil turning, the direction of the current must be reversed at this moment. The coil is connected through the brushes and the split-ring commutator. The split ring is arranged so that each half of the commutator changes brushes just as the coil reaches the vertical position. Changing brushes reverses the current in the loop. As a result the direction of the force on each side of the coil is reversed and the coil continues to rotate. This process repeats at each half-turn, so the coil rotates continuously in the magnetic field.

One half-turn of a motor

The coil is horizontal. The left brush touches the left half of the ring, the right brush the right half. Current flows and the forces turn the coil.
The coil reaches the vertical position. Each brush now touches the gap between the halves. The coil keeps turning because of its motion.
The coil passes the vertical. Each half of the ring is now touching the other brush, so the current in the coil is reversed.
The forces on the two sides now push the coil to keep turning in the same sense. The cycle repeats every half-turn.

19.8Electromagnetic inductionOften examined

Electromagnetic induction is the production of an electromotive force (e.m.f.) across a conductor when it is exposed to a varying magnetic field. Because of this e.m.f. a current flows in the conductor. It can happen either when a conductor is placed in a moving magnetic field, or when a conductor is moving in a stationary magnetic field. Two activities show the effect.

Figure 19.14: A bar magnet and a coil joined to a galvanometer. The needle (accent) deflects only while the magnet is moving; the direction of the deflection depends on the direction of motion.

Now take an activity in which the magnet is at rest but the conductor is moving.

Figure 19.15: A wire (seen end-on) moved down or up through the field between the poles. The induced current reverses when the direction of motion reverses.

Factors affecting the induced e.m.f.

The magnitude of the induced e.m.f. is affected by:

(a) the rate of change of the magnetic field, or the rate of cutting of magnetic field lines (a faster movement gives a larger e.m.f.);
(b) the number of turns in a coil (more turns give a larger e.m.f.).

19.9Lenz's lawOften examined

In 1834 the physicist Heinrich Lenz stated that the direction of the induced current in a circuit is always such that it opposes the cause which produces it. This statement is called Lenz's law.

To see the law at work, bring the north pole of a magnet towards a coil, Figure 19.16(a). As the pole approaches, an e.m.f. is induced in the coil. The induced current must oppose this motion, so it flows in the direction that makes the end of the coil nearer the magnet a north pole, which repels the approaching north pole. By the right-hand grip rule for a coil, the induced current is anticlockwise (seen from the magnet's side).

Figure 19.16: Lenz's law. (a) A north pole moving towards the coil induces a north pole at the near end, which repels the magnet. (b) A north pole moving away induces a south pole at the near end, which attracts the magnet back.

Similarly, when the north pole is moved away from the coil, Figure 19.16(b), the induced current is clockwise. The left end of the coil now becomes a south pole, which attracts the receding magnet. So in both cases the induced current produces a magnetic field that opposes the motion of the bar magnet.

Worked example: apply Lenz's law to a south pole

The south pole of a bar magnet is moved quickly towards the left end of a coil. (a) What kind of pole must appear at the left end of the coil? (b) Is the induced current clockwise or anticlockwise as seen from the magnet's side? (c) What do you see on the galvanometer if the magnet is then held still near the coil?

The induced current opposes the cause, which is the approach of the magnet. So the coil must repel the approaching south pole.
Like poles repel, so the left end of the coil must become a south pole.
Use the right-hand grip rule for a solenoid backwards: the thumb points to the north pole of the coil. The south pole is at the left end, so the thumb points to the right, away from the magnet. Curling the fingers, the current is clockwise as seen from the magnet's side.
Once the magnet is held still there is no relative motion. The induced e.m.f. is zero, so the galvanometer returns to zero.

The approaching south pole induces a south pole at the near end of the coil (accent), so the two repel.

(a) so u t h p o l e; (b) c l oc k w i se; (c) z er o d e f l ec t i o n .

19.10Electric generatorOften examined

Electricity is one of the most important needs of everyday life, so generating it is our main concern. There are many ways to produce electricity, but the most common way is to use generators. A generator works on electromagnetic induction: an e.m.f. is induced in a coil when there is relative motion between the coil and a magnet.

An alternating current generator (Figure 19.17) consists of a coil (armature), two slip rings made of copper, two carbon brushes and a permanent magnet. When the coil is made to rotate in the magnetic field by some mechanical means, an e.m.f. is induced in it. When the coil is connected to an external circuit, the current flows through the circuit by way of the slip rings. As the armature rotates, the strength and the direction of the current change. Current can also be generated by rotating a magnet and keeping the coil stationary.

Figure 19.17: An a.c. generator. (a) The coil is turned in the field between the poles. (b) Each end of the coil is joined to its own slip ring, and carbon brushes pressed on the rings carry the current to the circuit.

Graph of e.m.f. against time for a simple a.c. generator

The strength and direction of the induced e.m.f. change as the loop turns, Figure 19.18. As the coil is rotated, the e.m.f. changes smoothly from zero to a maximum value and back to zero during each half-turn of the loop. In the next half-turn it does the same in the opposite direction. So the output is alternating: it is maximum when the coil is horizontal (its sides are moving across the field lines fastest) and zero when the coil is vertical (its sides are moving along the field lines).

The five coil positions marked on the graph. Vertical coil: e.m.f. zero. Horizontal coil: e.m.f. at its largest, in opposite directions in positions 2 and 4.

E.m.f. against number of rotations

0.25

0.5

0.75

1

+ max

- max

number of rotations

induced e.m.f.

1

2

3

4

5

Figure 19.18: Graph of e.m.f. against the number of rotations for a simple a.c. generator. Points 1 to 5 are the coil positions in the figure above.

Worked example: a faster generator

The e.m.f. of a simple a.c. generator is sketched for one turn every second. The coil is now turned twice as fast by the same generator. Sketch the new e.m.f. on the same axes for the first second and say what changes. A student says the shape of the graph changes. Is that right?

The e.m.f. depends on the rate at which the sides of the coil cut the magnetic field lines. Turning twice as fast doubles this rate, so the largest e.m.f. is twice as big.
One turn now takes half the time. In the first second the coil makes two turns, so the graph shows two complete cycles in the time the first graph showed one.
The shape is still a smooth wave that is zero when the coil is vertical, largest when it is horizontal and reverses each half-turn. Only the height and the spacing change, so the student is wrong.

Slow and fast rotation

0.25

0.5

0.75

+ 1

- 1

+ 2

- 2

time (s)

e.m.f.

slow

fast

The same generator at one turn per second (accent) and at two turns per second (ink). The e.m.f. is in units of the largest e.m.f. of the slow turn.

Peak value doubles and the cycle repeats twice as often; the shape stays the same.

19.11TransformerOften examined

Many devices in our homes, such as game systems, printers, mobile chargers and stereos, use transformers. A transformer is an electrical device used to increase or decrease the value of an alternating voltage.

A transformer has three main parts: a primary coil, a secondary coil and a laminated iron core. The primary and secondary coils are wound on two different sides (limbs) of the iron core. The primary coil is connected to the input electrical supply, which has to be a.c. for the transformer to work properly.

Figure 19.19: A transformer. The primary coil (

N_{p}

turns, voltage

V_{p}

, current

I_{p}

) and the secondary coil (

N_{s}

turns, voltage

V_{s}

, current

I_{s}

) are wound on a laminated iron core.

An alternating current in the primary coil creates a changing magnetic field, which is concentrated in the iron core. When the primary coil is connected to an a.c. source of voltage $V_{p}$ , the changing field is carried through the core to the secondary coil. In the secondary coil the changing field induces an alternating e.m.f. called the secondary voltage $V_{s}$ . The secondary voltage is proportional to the primary voltage. It also depends on the ratio of the number of turns $N_{s}$ on the secondary coil to the number of turns $N_{p}$ on the primary coil:

\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}

19.1

the basic transformer equation

Example 19.1: turns of a step-down transformer

A step-down transformer converts

240 V

into

12 V

. If there are

2000

turns on the primary coil, find the number of turns on the secondary coil.

Given

$V_{p} = 240 V$ primary voltage
$V_{s} = 12 V$ secondary voltage
$N_{p} = 2000$ turns on the primary

To find

turns on the secondary, $N_{s}$

Write the basic transformer equation.
$\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$
Make $N_{s}$ the subject.
$N_{s} = \frac{V _{s}}{V _{p}} \times N_{p}$
Put in the values.
$N_{s} = \frac{12 V}{240 V} \times 2000$
Calculate.
$N_{s} = 100$

The secondary coil needs only $100$ turns compared with $2000$ turns in the primary.

N_{s} = 100

Types of transformer

Step-up transformer. The number of turns in the secondary coil is greater than in the primary ( $N_{s} > N_{p}$ ), so the voltage across the secondary coil is greater than across the primary ( $V_{s} > V_{p}$ ). Its function is to increase the voltage.
Step-down transformer. The number of turns in the secondary coil is less than in the primary ( $N_{s} < N_{p}$ ), so $V_{s} < V_{p}$ . Its function is to decrease the incoming voltage.

Figure 19.20: Types of transformer. Many turns on the secondary coil give a step-up transformer; few turns give a step-down transformer.

Secondary voltage against number of secondary turns

200

400

600

800

1000

100

120

140

N_{s}

V_{s} (V)

N_{s} = 100, V_{s} = 12 V

N_{s} = 1000, V_{s} = 120 V

V_{s} \propto N_{s}

For a primary of

N_{p} = 100

turns on

12 V

, the secondary voltage is directly proportional to the number of secondary turns:

V_{s} = 0.12 N_{s}

19.11.2Ideal transformerOften examined

In an ideal transformer the electric power delivered to the secondary circuit is equal to the power supplied to the primary circuit. An ideal transformer dissipates no power itself. For such a transformer, power in the primary = power in the secondary, or

I_sV_s=I_pV_p

19.2

power in the secondary = power in the primary

Combining this with Equation 19.1 shows how the current changes. From I_sV_s=I_pV_p it follows that \dfrac{I_s}{I_p}=\dfrac{V_p}{V_s}, and from Equation 19.1, \dfrac{V_p}{V_s}=\dfrac{N_p}{N_s}. So in an ideal transformer

\dfrac{I_s}{I_p}=\dfrac{V_p}{V_s}=\dfrac{N_p}{N_s}

Worked example: voltage and current in an ideal step-down transformer

An ideal transformer has 1200 turns on the primary and 60 turns on the secondary. The primary is connected to a 240\,\mathrm{V} a.c. supply and the secondary delivers a current of 2.0\,\mathrm{A}. Find (a) the secondary voltage, (b) the primary current, (c) the power in the primary and in the secondary. A student says: "the voltage goes down by 20 times, so the current also goes down by 20 times." Is that right?

Given

N_p = 1200
N_s = 60
V_p = 240\,\mathrm{V}
I_s = 2.0\,\mathrm{A}

To find

secondary voltage V_s
primary current I_p
power in primary and secondary

(a) Use the turns equation.
\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}\Rightarrow V_s=V_p\times\dfrac{N_s}{N_p}
Substitute.
V_s=240\,\mathrm{V}\times\dfrac{60}{1200}=12\,\mathrm{V}
(b) For an ideal transformer the power is the same on both sides.
I_pV_p=I_sV_s
Make I_p the subject and substitute.
I_p=\dfrac{I_sV_s}{V_p}=\dfrac{2.0\,\mathrm{A}\times12\,\mathrm{V}}{240\,\mathrm{V}}=0.10\,\mathrm{A}
(c) Find the power on each side.
P_s=I_sV_s=2.0\times12=24\,\mathrm{W},\qquad P_p=I_pV_p=0.10\times240=24\,\mathrm{W}
The student is wrong. The voltage falls 20 times but the current rises 20 times, from 0.10\,\mathrm{A} to 2.0\,\mathrm{A}, so that the power stays 24\,\mathrm{W}.

The ideal step-down transformer of the example: 24\,\mathrm{W} goes in and 24\,\mathrm{W} comes out.

V_s=12\,\mathrm{V};\ I_p=0.10\,\mathrm{A};\ P_p=P_s=24\,\mathrm{W}

19.11.3High electric power transmission

Electric power is usually generated at places far away from where it is used. It is transmitted over long distances at high voltage to keep down the loss of energy as heat during transmission. The heat produced in a transmission cable of resistance R carrying a current I for a time t is I^2Rt. So, by reducing the current in the cable, the heat loss can also be reduced.

\text{heat lost in the cable}=I^{2}Rt

heat dissipated in the transmission cable

A transformer makes this possible: the power station steps the voltage up before transmission, and step-down transformers lower it again near the consumers (see Equation 19.2: raising the voltage lowers the current for the same power). High-voltage transmission also allows more efficient transmission over longer distances. It reduces infrastructure costs, because it reduces the need for intermediate substations and needs fewer lines and towers, which brings economic benefits.

Worked example: why transmit at high voltage?

A power station sends 11\,\mathrm{kW} of power along cables of total resistance 2\,\Omega. Compare the power wasted as heat in the cables if the power is sent (a) at 220\,\mathrm{V} and (b) at 11\,000\,\mathrm{V}. Use P=IV for the current. By what factor does the loss fall?

Given

P = 11\,\mathrm{kW}=11\,000\,\mathrm{W} power sent
R = 2\,\Omega cable resistance

To find

current and power lost at 220\,\mathrm{V}
current and power lost at 11\,000\,\mathrm{V}

(a) Find the current from P=IV.
I=\dfrac{P}{V}=\dfrac{11\,000}{220}=50\,\mathrm{A}
Heat lost in the cable each second is I^2R.
I^2R=(50)^2\times2=5000\,\mathrm{W}
So 5000\,\mathrm{W} of the 11\,000\,\mathrm{W} is wasted, about 45\%.
(b) Current at the higher voltage.
I=\dfrac{11\,000}{11\,000}=1\,\mathrm{A}
Heat lost each second.
I^2R=(1)^2\times2=2\,\mathrm{W}
The voltage was raised 50 times, so the current fell 50 times and the loss fell 50^2=2500 times.
\dfrac{5000}{2}=2500

Power lost in the cable against current

Power lost =I^2R for R=2\,\Omega. At 220\,\mathrm{V} the line carries 50\,\mathrm{A}; at 11\,000\,\mathrm{V} it carries 1\,\mathrm{A}.

Loss: 5000 W at 220 V, 2 W at 11 000 V (2500 times smaller).

19.12Deflection of an electron beam

When a charged particle passes through an electric or a magnetic field, it is deflected, that is, its path is bent away from the straight line.

Deflection of electrons by an electric field

When an electron beam passes through an electric field, the electrons are deflected towards the positive plate, Figure 19.21(a). The reason is that electrons are attracted by positive charges and repelled by negative charges, because of the electric force. The amount of deflection of the electrons from their original direction is **proportional to the strength of the applied electric field E**.

Deflection of electrons by a magnetic field

When an electron beam passes through a magnetic field, the electrons are also deflected, Figure 19.21(b). The amount of deflection from the original direction depends on three things:

the velocity of the electrons;
the strength of the applied magnetic field;
the angle at which the electrons enter the magnetic field.

The direction of the deflection is found with Fleming's left-hand rule. Remember that the current is taken in the direction of conventional current, which is opposite to the motion of the electrons.

Figure 19.21: Deflection of an electron beam. (a) In an electric field the beam bends towards the positive plate. (b) In a magnetic field (here directed out of the page, shown by dots) the beam bends in a direction given by Fleming's left-hand rule.

19.13Cathode ray oscilloscope (CRO)

The cathode ray oscilloscope (CRO) is an electrical instrument that converts electric signals into visual signals, for displaying and analysing waveforms and other electronic phenomena.

The block diagram of a CRO is in Figure 19.22(a). The electron beam from the electron gun is focused onto a fluorescent screen by the focusing electrode. This gives a visible spot at the centre of the screen where the beam hits. The spot is visible because the screen is coated with a thin layer of phosphor, which emits light when struck by electrons. The horizontal and vertical plates (the deflection system), placed between the focusing electrode and the screen, deflect the beam according to the input signal.

Figure 19.22(a): Block diagram of a cathode ray oscilloscope. The electron gun makes the beam, the focusing electrode narrows it, the Y and X plates deflect it, and it lights a spot on the fluorescent screen.

The deflector has two mechanisms, one to change the vertical direction and one to change the horizontal direction of the beam. This allows the electron beam to sweep over the entire screen.

Working principle

The working of a CRO is based on the movement of an electron beam caused by electrostatic force. When the beam strikes the phosphor screen, it makes a bright spot. The CRO acts on the electron beam with electric fields from two perpendicular directions (vertical and horizontal). The spot on the screen moves as a result of these two perpendicular electrostatic forces and traces out the required waveform of the input signal.

Waveforms on an oscilloscope

A typical waveform of a signal from an a.c. source is displayed on the oscilloscope as in Figure 19.23. It shows time on the horizontal axis and the potential difference on the vertical axis. The waveform on an oscilloscope can be controlled by applying an electric field of suitable magnitude.

Waveform of an a.c. signal on the screen

Figure 19.23: Waveform on the screen. The horizontal axis is time and the vertical axis is the potential difference (p.d.).

Formula sheet

Transformer equation

\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}

V_s, V_p: secondary and primary voltages; N_s, N_p: turns on the secondary and primary coils (19.1)

Units: \mathrm{V} for voltages; turns are numbers

Ideal transformer: power in = power out

I_sV_s=I_pV_p

I_s, I_p: secondary and primary currents; V_s, V_p: secondary and primary voltages (19.2)

Units: \mathrm{A} for currents, \mathrm{V} for voltages

Current ratio of an ideal transformer

\dfrac{I_s}{I_p}=\dfrac{V_p}{V_s}=\dfrac{N_p}{N_s}

follows from Equations 19.1 and 19.2

ratio, no units

Heat lost in a transmission cable

H=I^{2}Rt

I: current in the cable; R: resistance of the cable; t: time

Unit: \mathrm{J}

Key terms

Electromagnetism: The part of physics that studies how electricity and magnetism are connected.
Magnetic field lines: Imaginary lines that show a magnetic field: they emerge from north poles, end on south poles, form closed paths, never cross, and are closer together where the field is stronger.
Right-hand rule (straight wire): Grasp the wire with the right hand, thumb along the current: the fingers curl in the direction of the magnetic field.
Solenoid: A long, tightly wound (spring-like) coil of wire. Its field is strong and uniform along the axis and weak outside, like that of a bar magnet.
Right-hand rule (solenoid): Grip the coil with the right hand, fingers curling along the conventional current: the thumb points to the north pole of the coil.
Geomagnetic field: The Earth's magnetic field, generated by the rotation of the Earth and its molten iron core that contains charged particles in motion.
Magnetosphere: The region of space shaped by the Earth's magnetic field, which acts as a protective magnetic shield around the Earth.
Loudspeaker: A device with a permanent magnet and a voice coil joined to a cone; it changes electrical energy into sound.
Relay: An electrical control device in which a small input current, through an electromagnet, switches a separate output circuit.
Armature: The movable part of a relay; also the coil of many loops mounted on a shaft in a motor or generator.
Fleming's left-hand rule: Thumb, forefinger and middle finger of the left hand at right angles: forefinger along the field, middle finger along the current, thumb along the force.
Electric motor: A device that converts electrical energy into mechanical energy by using the couple on a current-carrying coil in a magnetic field.
Split-ring commutator: A ring split into two halves that turns with the coil of a motor and reverses the current in the coil every half-turn.
Brushes: Pieces of graphite or carbon that touch the commutator or slip rings and let current flow into or out of the coil.
Electromagnetic induction: The production of an e.m.f. across a conductor when it is exposed to a varying magnetic field.
Lenz's law: The direction of the induced current in a circuit is always such that it opposes the cause which produces it.
A.c. generator: A device that converts mechanical energy into electrical energy by turning a coil in a magnetic field.
Slip rings: Two rings joined to the ends of the coil of an a.c. generator; the brushes press on them to carry the current out.
Transformer: An electrical device used to increase or decrease the value of an alternating voltage.
Primary coil: The coil of a transformer connected to the a.c. input supply.
Secondary coil: The coil of a transformer in which the output voltage is induced.
Step-up transformer: A transformer with more turns on the secondary than on the primary; it increases the voltage.
Step-down transformer: A transformer with fewer turns on the secondary than on the primary; it decreases the voltage.
Ideal transformer: A transformer that dissipates no power itself, so the power in the secondary equals the power in the primary.
Cathode ray oscilloscope (CRO): An instrument that converts electric signals into visual signals for displaying and analysing waveforms.

Multiple choice questionsp. 277–278

The presence of a magnetic field can be detected by a

Asmall mass
Bstationary positive charge
Cstationary negative charge
Dmagnetic compassAnswer

Why

A magnetic compass needle is a small magnet. It turns to a definite direction in a magnetic field, so it shows that the field is there (19.1.2).

A small mass (A) feels only gravity. A stationary charge (B or C) produces an electric field, not a magnetic field, so it cannot detect a magnetic field.

A DC motor converts:

Amechanical energy into electrical energy
Bmechanical energy into chemical energy
Celectrical energy into mechanical energyAnswer
Delectrical energy into chemical energy

Why

A DC motor takes electrical energy from the supply and gives rotational mechanical energy. A current-carrying coil in a magnetic field experiences a couple and turns (19.7).

Option A describes a generator, which does the reverse (19.10). Options B and D mention chemical energy, which a motor does not change.

Note. The book prints 'ADC motor'. The question is read as 'A DC motor'.

Which of the following device works on the principle of electromagnetic induction?

Amagnetic compass
Bmotor
CtransformerAnswer
Doscilloscope

Why

A transformer works because the changing magnetic flux in its iron core induces an alternating emf in the secondary coil (19.11). This is electromagnetic induction.

A magnetic compass only turns in a field and induces nothing (A). A motor works by the force on a current-carrying coil (B). An oscilloscope deflects an electron beam with electric fields (D).

Voltage in the secondary coil of transformer does not depend upon

Afrequency of source
Bprimary voltage
Cpower lossesAnswer
Dturns ratio

Why

The basic transformer equation is \dfrac{V_s}{V_p}=\dfrac{N_s}{N_p} (19.1). The secondary voltage therefore depends on the primary voltage (B) and on the turns ratio (D).

The frequency of the source (A) sets how quickly the flux in the core changes, and the induced emf depends on that rate, so frequency does affect the secondary voltage. Power losses (C) reduce the power that reaches the load, but they are not one of the quantities in the transformer equation, so the secondary voltage does not depend on them.

Which of the following quantities remain constant in an ideal step-up transformer?

Acurrent
Bvoltage
CpowerAnswer
Dheat

Why

An ideal transformer dissipates no power, so the power in the secondary equals the power in the primary: V_sI_s=V_pI_p (19.2). A step-up transformer raises the voltage, so the current falls by the same ratio.

Current (A) and voltage (B) both change in a step-up transformer. Heat (D) is not the quantity the transformer relation keeps fixed; the constant quantity the book gives is power.

A step-up transformer has a transformation ratio of 3:2. What is the voltage in secondary, if voltage in primary is 30\,\mathrm{V}?

A15\,\mathrm{V}
B45\,\mathrm{V}Answer
C90\,\mathrm{V}
D300\,\mathrm{V}

Why

The ratio 3:2 is read as N_s:N_p=3:2, which is the usual meaning for a step-up transformer. Then:

V_s=V_p\,\dfrac{N_s}{N_p}=30\,\mathrm{V}\times\dfrac{3}{2}=45\,\mathrm{V}

Option A (15 V) divides by 2 and ignores the 3. Option C (90 V) multiplies by 3 and ignores the 2. Option D has no basis in the ratio.

Toward which magnetic pole of Earth is the north pole of a compass needle attracted?

AThe north pole of a compass needle is attracted to the north magnetic pole of Earth, which is located near the geographic North Pole of Earth.
BThe north pole of a compass needle is attracted to the south magnetic pole of Earth, which is located near the geographic North Pole of Earth.Answer
CThe north pole of a compass needle is attracted to the north magnetic pole of Earth, which is located near the geographic South Pole of Earth.
DThe north pole of a compass needle is attracted to the south magnetic pole of Earth, which is located near the geographic South Pole of Earth.

Why

Unlike poles attract. The north pole of the needle is therefore attracted to a south pole, and the south magnetic pole of Earth lies near the geographic North Pole (19.3, Figure 19.6). Option B says this.

Option A names a north magnetic pole, which would repel the needle. Options C and D are wrong about position: the magnetic south pole is near the geographic North Pole, not the South Pole.

The step-up transformer:

Aincreases the input current
Bincreases the output voltageAnswer
Chas more turns in the primary
Dhas less turns in the secondary coil

Why

A step-up transformer has more turns in the secondary than in the primary, N_s>N_p, so the output voltage is larger than the input voltage (B).

Option A is wrong: for the same power, a larger output voltage means a smaller input current. Options C and D reverse the turns: a step-up transformer has fewer turns in the primary and more turns in the secondary.

A step up transformer is used before electricity is transmitted by overhead cables. Select the statement that explains why this is done.

AIt increases the voltage to increase the speed at which the electricity travels.
BIt increases the voltage to reduce energy loss in the cables.Answer
CIt increases the current to increase the speed at which the electricity travels.
DIt increases the current to reduce energy loss in the cables.

Why

The heat lost in a cable is I^{2}R. For a given power P=VI, a higher voltage means a smaller current, so the loss falls sharply (19.11.3).

Options A and C are wrong because the voltage and the current do not change the speed of electricity. Option D is wrong because a larger current increases the loss, I^{2}R.

Which of the following does not change in an ordinary transformer?

Apower
BfrequencyAnswer
Ccurrent
Dvoltage

Why

A transformer changes the voltage and the current (options C and D). The induced emf follows the alternating supply, so the output has the same frequency as the input, and the frequency does not change (B).

Power (A) is almost unchanged in a good transformer, but some power is lost in the core and coils, so the power in the secondary is a little less. Frequency is never changed.

Constructed response questionsp. 278–279

In which direction the current must be flowing through the conductor to create the magnetic field shown in the diagram? Up or Down?

Solution

The field lines are circles around a vertical conductor. Use the right-hand rule for a straight wire (19.1.1): grip the wire with the right hand, thumb in the direction of the current, and the fingers curl in the direction of the magnetic field.

Working

Read the direction of the field from the arrows. On the far side of each circle the arrows point to the left, and on the near side they point to the right. Seen from above, the field circulates anticlockwise.
Put the thumb pointing up. With the thumb up, the fingers curl anticlockwise when seen from above, which matches the field.
The thumb therefore points upwards, so the current must flow up through the conductor.

Seen from above, the field circles the wire anticlockwise, so the current flows upwards.

\text{Up}

In the following diagram, will there be a North pole or a South pole at the left side of the solenoid? Also draw its magnetic field having arrows to show its direction.

Solution

The right-hand rule for a solenoid (19.4, Figure 19.5): grip the coil with the right hand, with the fingers curling in the direction of the conventional current. The thumb then points to the north pole of the coil.

Working

In the diagram the current leaves the coil through the lead at the left end, so inside the coil it flows from right to left.
On the front of the coil the current therefore moves upwards. Curl the fingers that way and the thumb points to the left, along the axis.
The thumb points to the left, so the left end is the North pole.

So there is a North pole at the left side. Inside the solenoid the field runs from the South pole to the North pole, so it points to the left. Outside, the field lines leave the North pole at the left, curve round and enter the South pole at the right.

Field of a solenoid: the field inside points from the South pole to the North pole, and the field outside leaves the North pole.

\text{North pole}

Note. The book does not print an answer. The answer assumes the current leaves the coil through the left lead.

A current carrying conductor is placed in the magnetic field, as shown.

Solution

(a) In which direction will the conductor move? Up, down, left or right?

The battery symbol has alternating long and short lines, and the long line is the positive terminal. The positive terminal is on the right of the battery, so current leaves it to the right and goes round the circuit. Along the conductor, which is the right-hand edge of the loop, the current flows from the near corner to the far corner, away from the viewer.

The magnet lies flat on the board with its North pole facing the conductor. The field at the conductor therefore points away from the magnet, to the right and across the conductor.

Direction of the force

Fleming's left-hand rule: point the forefinger of the left hand in the direction of the field, which is to the right.
Point the middle finger in the direction of the current, which is away from the viewer along the conductor.
The thumb then points downwards, into the board. The force is perpendicular to both the conductor and the field, so the conductor is pushed down through the board.

\vec F = I\,\vec L\times\vec B

Top view: the current I along the conductor, the field B from the North pole, and the force F into the board.

\text{Down, into the board}

(b) If the magnetic field was reversed then in which direction will the conductor move?

Reversing the field reverses the force, because the force is \vec F = I\,\vec L\times\vec B and changes sign when \vec B changes sign. The current is unchanged, so the conductor is pushed the opposite way, up out of the board (Activity 19.2).

\text{Up}

Reversing both the field and the current changes the sign of the force twice, so the force keeps its original direction. The conductor moves down, as in part (a).

\text{Down}

Short response questionsp. 279

How does a generator differ from a motor, and what is their relationship to each other?

Solution

A generator converts mechanical energy into electrical energy. A coil is turned in a magnetic field, and an emf is induced in it because the coil and the magnet move relative to each other (19.10).

A motor does the reverse. Electrical energy is supplied to a current-carrying coil in a magnetic field; the coil experiences a couple and turns, giving rotational mechanical energy (19.7).

They are related because they use the same parts, a coil and a magnetic field, and the same principle. A motor is a generator run in reverse: a generator is driven by mechanical energy and gives electrical energy, while a motor is supplied with electrical energy and gives mechanical energy.

Why is a transformer ineffective with DC voltage?

Solution

A transformer works only when the magnetic flux in the iron core changes. An alternating current in the primary produces a changing magnetic field, which induces an alternating emf in the secondary (19.11).

A direct current is steady, so its magnetic field is steady and the flux linked with the secondary coil does not change. Electromagnetic induction needs a changing field (19.8), so no emf is induced and the transformer gives no output.

Explain the relationship between the direction of a current-carrying conductor and the magnetic field it produces.

Solution

A current-carrying conductor produces a magnetic field around itself. The field lines are circles centred on the wire, lying in planes at right angles to it (Activity 19.1, Figure 19.2).

Right-hand rule for a straight wire

Grip the wire with the right hand, with the thumb pointing in the direction of the conventional current.
The fingers curl in the direction of the magnetic field around the wire (19.1.1).

Reversing the current reverses the field, and the compass needle turns through 180 degrees (Activity 19.1). The field is stronger near the wire and weaker farther away.

Describe the interaction between a current-carrying conductor and a magnetic field, and how to determine the direction of the force exerted.

Solution

A current-carrying conductor placed in a magnetic field experiences a force. The magnetic field of the current interacts with the external field, and the conductor is pushed (19.5, Activity 19.2). The force is increased if the current is increased, the field is strengthened, or more of the wire is placed in the field.

Fleming's left-hand rule

Stretch the thumb, forefinger and middle finger of the left hand so that each is at right angles to the other two.
Point the forefinger in the direction of the magnetic field.
Point the middle finger in the direction of the current.
The thumb gives the direction of the force on the conductor.

The direction of the force is reversed by reversing the current or the direction of the magnetic field, but not both together.

Why is high voltage used for long-distance transmission of electrical power?

Solution

Power is sent along cables that have resistance R, and the power lost as heat in the cables is I^{2}R (19.11.3). The current needed to send a power P at voltage V follows from P=VI.

Working

Write the current in terms of the power and the voltage.
I=\dfrac{P}{V}
Substitute this current into the heat lost in the cable.
P_{\text{loss}}=I^{2}R=\left(\dfrac{P}{V}\right)^{2}R=\dfrac{P^{2}R}{V^{2}}
For the same power and cable, the loss falls as the square of the voltage rises. Doubling the voltage cuts the loss to one quarter.

High voltage therefore reduces the energy wasted as heat. It also needs fewer lines and towers and fewer intermediate substations, which lowers the infrastructure cost (19.11.3).

Design an experiment to generate an electric current using a coil of wire and a bar magnet.

Solution

Aim: to generate an electric current in a coil by moving a bar magnet.

Apparatus: a coil of insulated copper wire, a galvanometer with its zero in the centre of the scale, connecting wires and a bar magnet (Activity 19.3, Figure 19.14).

Procedure

Connect the two ends of the coil to the galvanometer. Keep the magnet still: the galvanometer shows no deflection, so there is no current.
Push the North pole of the magnet into the coil. The galvanometer deflects in one direction.
Hold the magnet still inside the coil. The deflection returns to zero.
Pull the magnet out of the coil. The galvanometer deflects in the opposite direction.
Repeat with a faster movement, and with a coil that has more turns, to see how the deflection changes.

Conclusion: the galvanometer shows a current only while the magnet moves relative to the coil. The current is produced by the changing magnetic field through the coil, which is electromagnetic induction (19.8).

What factors influence the strength of an induced electromotive force (EMF)?

Solution

The magnitude of the induced emf is affected by two factors (19.8, B):

The rate of change of the magnetic field, or the rate at which the coil cuts the field lines. A faster movement of the magnet or the coil gives a larger emf.
The number of turns in the coil. More turns give a larger emf.

A stronger magnet changes the field through the coil by a larger amount in the same time, so it also gives a larger emf. The direction of the induced emf is given by Lenz's law, not by these factors (19.9).

Compare and contrast the two primary types of transformers.

Solution

The two types are the step-up and the step-down transformer (19.11.1). Both have a primary and a secondary coil wound on a laminated iron core, both need an alternating supply, and both obey \dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}.

Step-up and step-down transformers compared (ideal transformer).
Feature	Step-up transformer	Step-down transformer
Turns	N_s>N_p, more turns on the secondary	N_s<N_p, fewer turns on the secondary
Output voltage	V_s>V_p, increases the incoming voltage	V_s<V_p, decreases the incoming voltage
Output current (ideal)	I_s<I_p, the current is smaller	I_s>I_p, the current is larger
Example in the unit	Raises the voltage before transmission (19.11.3)	Installed in streets to reduce the voltage to 220 V (Information box, 19.11.1)

The output currents follow from the equal power in both coils, V_pI_p=V_sI_s (19.2).

Identify three devices that operate on the principle of electromagnetic induction.

Solution

The unit describes two devices that work on electromagnetic induction. The first is the AC generator, which converts mechanical energy into electrical energy by turning a coil in a magnetic field (19.10). The second is the transformer, which changes an alternating voltage through the changing magnetic field in its iron core (19.11).

The unit does not describe a third device that works on this principle, so only these two can be named from the unit.

Note. The book names only two such devices in this unit: the AC generator and the transformer. The Information box on page 271 says that generators and transformers work on electromagnetic induction. No third device is given in the unit.

Long response questionsp. 279

Design an experiment to qualitatively demonstrate the existence of a magnetic field around a current-carrying conductor.

Solution

Aim

To show that a current-carrying conductor has a magnetic field around it, that the field lines are circles centred on the conductor, and that reversing the current reverses the field.

Apparatus

A straight copper wire, passed vertically through a hole at the centre of a stiff piece of cardboard.
A battery, a switch and connecting wires.
Several small compass needles, placed on the cardboard around the wire.

Procedure

Pass the wire vertically through the cardboard and connect its ends to the battery through the switch.
Place a compass needle at several points around the wire on the cardboard. With the switch open, each needle points to the Earth's field only.
Close the switch. Each needle turns to lie along a circle centred on the wire.
Move a needle closer to the wire and then farther away. The needle turns more strongly near the wire.
Reverse the battery terminals. Each needle turns through 180 degrees and points the other way.

Top view of the cardboard: the compass needles lie along circles around the wire, and the current comes out towards the reader.

Conclusion

A current in the wire creates a magnetic field around it. The field lines are circular, the field is stronger near the wire, and its direction depends on the direction of the current (Activity 19.1). The direction is found with the right-hand rule (19.1.1).

Describe an experimental setup to induce an EMF in a circuit using a changing magnetic field. How can you measure the induced EMF?

Solution

Setup

Take a coil C of many turns of insulated wire and connect its two ends to a galvanometer G, which shows a deflection on either side of its zero (Activity 19.3, Figure 19.14). Use a bar magnet as the source of the changing field.

A bar magnet moved towards a coil connected to a galvanometer. The induced current deflects the galvanometer needle.

Procedure

Hold the magnet still near the coil. The galvanometer shows no deflection, because there is no induced emf.
Move the North pole towards the coil. The galvanometer deflects in one direction.
Move the magnet away from the coil. The galvanometer deflects in the opposite direction.
Move the magnet faster. The deflection is greater, because the rate of change of the field is greater.

Measuring the induced emf

The induced emf drives a current round the closed circuit. The galvanometer shows this current: its deflection is larger when the induced emf is larger, and the direction of the deflection shows the direction of the induced emf. To read the emf in volts, the galvanometer is replaced by a voltmeter connected across the ends of the coil.

The size of the emf depends on the rate of change of the magnetic field and on the number of turns in the coil (19.8). A large deflection is therefore obtained with a fast movement of a strong magnet and a coil with many turns.

Explain Lenz's law in your own words, and provide a real-world example to illustrate its application.

Solution

Statement in words

Lenz's law says that the induced current in a circuit always flows in the direction that opposes the change which produces it (19.9). The induced current creates its own magnetic field, and this field pushes against the motion that causes it.

Example with a bar magnet

Move the North pole of a bar magnet towards a coil (Figure 19.16 a).
The increasing flux in the coil induces a current. The current must make the end of the coil facing the magnet a North pole, so that it repels the approaching North pole.
By the right-hand grip rule for a solenoid, this induced current flows anticlockwise as seen from the magnet's end.
If the magnet is withdrawn, the induced current reverses and the coil end facing the magnet becomes a South pole, which attracts the magnet back.

Real-world example

Lenz's law applies in the AC generator (19.10). As the coil turns, the induced current opposes the motion that produces it. The turning therefore needs mechanical work, and the more current is drawn from the generator, the harder it is to turn the coil. The generator converts the mechanical work into electrical energy.

Define an alternating current (AC) generator. Draw a labeled diagram and explain its components and operation. How does it convert mechanical energy into electrical energy?

Solution

Definition

An AC generator is a device that converts mechanical energy into electrical energy. Its output emf and current alternate: they change size and reverse direction periodically (19.10).

Components

Coil (armature): a coil of wire that is rotated in the magnetic field.
Permanent magnet: produces the magnetic field, with its North and South poles on either side of the coil.
Two slip rings: copper rings that turn with the coil.
Two carbon brushes: fixed contacts pressing on the slip rings, which carry the current to the external circuit.

Schematic of an AC generator. The coil turns between the poles; the slip rings and brushes carry the induced current to the external circuit.

Operation

Steps

The coil is turned by a mechanical means. The mechanical work done on it is the energy source.
As the coil turns, the magnetic flux through it changes, so an emf is induced in the coil (19.8).
The slip rings keep the connection to the coil while it turns, and the brushes carry the induced current to the external circuit.
The strength and the direction of the current change as the coil turns. The emf reverses every half-turn.

Emf against time

The emf is zero when the coil is vertical and a maximum when the coil is horizontal (19.10.1). The graph shows the emf for one complete turn of the coil, starting from the vertical position.

Emf of a simple AC generator

The emf varies sinusoidally with time. It is zero when the coil is vertical and maximum when it is horizontal.

Energy conversion

The mechanical energy supplied to turn the coil is converted into electrical energy. The induced current opposes the turning (Lenz's law), so the mechanical work is what drives the current.

5FBISE SSC-II 2026 Annual I, Set D, Q.6

What is an AC motor? Draw a labeled diagram and explain its components and operation. How does it convert electrical energy into mechanical energy?

Solution

Definition

An electric motor converts electrical energy into mechanical energy, that is, rotational kinetic energy (19.7). An AC motor is a motor that runs from an alternating supply.

Components

Coil (armature): a rectangular coil of many turns, free to rotate on an axle.
Permanent magnets: produce the magnetic field between their North and South poles.
Split-ring commutator: a ring split into two halves, which changes the connections to the coil as it turns.
Brushes: pieces of graphite that carry the current from the battery to the commutator.

Labelled diagram of an electric motor. The coil sits between the poles, and the split ring and brushes supply the current.

Operation

Steps

A current flows through the coil. Each side of the coil experiences a force, given by Fleming's left-hand rule.
The two forces on opposite sides are equal and opposite, so they form a couple that turns the coil (19.6).
A single coil stops when its plane is vertical, because the forces are then balanced.
At that point the split ring changes the brushes, which reverses the current in the coil. The forces reverse, and the coil keeps turning.

Energy conversion

Electrical energy supplied to the coil is converted into mechanical energy: the couple does work as the coil turns (19.7).

6FBISE SSC-II 2026 Annual I, Set B, Q.6

Define a transformer. Explain its principle of operation, components, and applications. How does it change the AC voltage?

Solution

Definition

A transformer is an electrical device used to increase or decrease the value of an alternating voltage (19.11).

Principle of operation

An alternating voltage applied to the primary coil produces an alternating current and a changing magnetic field. The iron core carries this field to the secondary coil, where it induces an alternating emf (19.11). A steady DC supply would not work, because its field does not change.

Components

Primary coil: connected to the AC supply, with N_p turns.
Secondary coil: gives the output voltage, with N_s turns.
Laminated iron core: carries the magnetic field from the primary coil to the secondary coil.

A transformer: the iron core carries the changing magnetic field from the primary coil to the secondary coil.

How the voltage changes

\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}

19.1

If N_s>N_p the output voltage is greater than the input voltage, which is a step-up transformer. If N_s<N_p the output voltage is smaller, which is a step-down transformer. As an example from the book, a step-down transformer with 2000 primary turns that converts 240 V into 12 V has N_s=\dfrac{12}{240}\times2000=100 turns (Example 19.1).

For an ideal transformer the power is the same in both coils, V_pI_p=V_sI_s (19.2). A step-up increases the voltage and reduces the current in the same ratio.

Applications

Step-up transformers increase the voltage for high-voltage transmission of electrical power, which reduces the energy lost in the cables (19.11.3).
Step-down transformers in the streets reduce the voltage to 220 V for homes (Information box, 19.11.1).
Many devices in homes, such as game systems, printers, mobile chargers and stereos, use transformers (19.11).

Numerical response questionsp. 280

A step-up transformer has a turn ratio of 1:100. An alternating supply of 20\,\mathrm{V} is connected across the primary coil. What is the secondary voltage?

Solution

Given

V_p = 20\,\mathrm{V} primary voltage
N_s/N_p = 100 turn ratio of a step-up transformer

To find

V_s (secondary voltage)

For an ideal transformer the voltages are in the same ratio as the turns.
\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}
Make V_s the subject.
V_s=V_p\,\dfrac{N_s}{N_p}
Substitute the values.
V_s=20\,\mathrm{V}\times100
Evaluate.
V_s=2000\,\mathrm{V}

V_s=2000\,\mathrm{V}

Book answer. 2000 volts

A step-down transformer has a turns ratio of 100:1. An ac voltage of amplitude 170\,\mathrm{V} is applied to the primary. What is the voltage across the secondary?

Solution

Given

V_p = 170\,\mathrm{V} amplitude of the primary voltage
N_p/N_s = 100 turns ratio of a step-down transformer

To find

V_s (amplitude of the secondary voltage)

The voltages are in the same ratio as the turns, so V_s=V_p\,\dfrac{N_s}{N_p}.
Here \dfrac{N_s}{N_p}=\dfrac{1}{100}.
V_s=170\,\mathrm{V}\times\dfrac{1}{100}
Evaluate.
V_s=1.7\,\mathrm{V}

V_s=1.7\,\mathrm{V}\ \text{(amplitude)}

Book answer. 1.7 volts

Note. 170 V is the amplitude, so 1.7 V is the amplitude on the secondary. The root-mean-square value of the output is \dfrac{1.7}{\sqrt{2}}\approx1.2\,\mathrm{V}. The book gives the amplitude.

Calculate the input voltage if it draws 0.5\,\mathrm{A}. Also calculate turn ratio of primary to secondary coil if a transformer is rated for an output of 100\,\mathrm{W} and 8.3\,\mathrm{A}. Also identify the type of transformer.

Solution

Given

P_s = 100\,\mathrm{W} output power
I_s = 8.3\,\mathrm{A} output current
I_p = 0.5\,\mathrm{A} input current

To find

V_p (input voltage)
N_p:N_s (turn ratio)
type of transformer

(i) Calculate the input voltage if it draws 0.5\,\mathrm{A}.

The transformer is ideal, so the input power equals the output power.
P_p=P_s=100\,\mathrm{W}
The input power is P_p=V_pI_p, so the input voltage is:
V_p=\dfrac{P_p}{I_p}=\dfrac{100\,\mathrm{W}}{0.5\,\mathrm{A}}
Evaluate.
V_p=200\,\mathrm{V}

V_p=200\,\mathrm{V}

(ii) Calculate turn ratio of primary to secondary coil if a transformer is rated for an output of 100\,\mathrm{W} and 8.3\,\mathrm{A}.

The output voltage follows from the output power and the output current.
V_s=\dfrac{P_s}{I_s}=\dfrac{100\,\mathrm{W}}{8.3\,\mathrm{A}}=12.05\,\mathrm{V}
The turn ratio is the same as the voltage ratio, \dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}.
\dfrac{N_p}{N_s}=\dfrac{200\,\mathrm{V}}{12.05\,\mathrm{V}}=16.6
So the turn ratio is 16.6 to 1 (primary to secondary).

N_p:N_s=16.6:1 \quad(\text{equivalently } N_s:N_p=1:16.6)

(iii) Identify the type of transformer.

The output voltage, 12\,\mathrm{V}, is less than the input voltage, 200\,\mathrm{V}. So N_s<N_p and the transformer is a step-down transformer (19.11.1).

\text{Step-down transformer}

Book answer. 200 volt, 1:16.6, step down

Note. The book prints the turn ratio as 1:16.6, which is secondary to primary. The ratio of primary to secondary turns is 16.6:1, as worked above.

Calculate the output power and output current of the following step-down transformer. The value of input current is 0.050\,\mathrm{A} (assuming that the transformer is 100% efficient).

Step-down transformer: 230 V mains on the primary, 8000 turns, 6 V output.

Solution

Given

V_p = 230\,\mathrm{V} mains voltage on the primary
I_p = 0.050\,\mathrm{A} input current
V_s = 6\,\mathrm{V} output voltage
\eta = 100\% efficiency

To find

P_s (output power)
I_s (output current)

With 100% efficiency all the input power reaches the output, so P_s=P_p=V_pI_p.
P_s=230\,\mathrm{V}\times0.050\,\mathrm{A}=11.5\,\mathrm{W}
The output current follows from the output power and the output voltage.
I_s=\dfrac{P_s}{V_s}=\dfrac{11.5\,\mathrm{W}}{6\,\mathrm{V}}=1.92\,\mathrm{A}
Round to two significant figures.
I_s\approx1.9\,\mathrm{A}

P_s=11.5\,\mathrm{W}\qquad I_s\approx1.9\,\mathrm{A}

Book answer. 11.5 W, 1.9 A

Note. The number of turns is not needed, because the output voltage is given directly.

A transformer in an adapter steps up a 115\,\mathrm{V} to the 230\,\mathrm{V} needed for a hair dryer. A current of 5\,\mathrm{A} flows through the hair dryer. Assuming that the transformer is 100% efficient, calculate the current drawn from the mains supply.

Solution

Given

V_p = 115\,\mathrm{V} mains voltage
V_s = 230\,\mathrm{V} voltage needed for the hair dryer
I_s = 5\,\mathrm{A} current through the hair dryer

To find

I_p (current drawn from the mains)

The power delivered to the hair dryer is:
P_s=V_sI_s=230\,\mathrm{V}\times5\,\mathrm{A}=1150\,\mathrm{W}
With 100% efficiency, the mains supplies the same power.
P_p=P_s=1150\,\mathrm{W}
The mains current is:
I_p=\dfrac{P_p}{V_p}=\dfrac{1150\,\mathrm{W}}{115\,\mathrm{V}}=10\,\mathrm{A}

I_p=10\,\mathrm{A}

Book answer. 10 A

FBISE 2026 questions

Questions from the FBISE SSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

FBISE SSC-II 2026 Annual I, Set B, MCQ 61 mark

If the direction of current in a conductor is reversed, the direction of force on it will:

AReverseAnswer
BRemain the same
CBecome 45^{\circ}
DIncrease

Why

The force on a current-carrying conductor in a magnetic field is perpendicular to both the field and the current (Fleming's left-hand rule). Reversing the current turns the middle finger the other way, so the thumb, which shows the force, also turns the other way. The force reverses (A) but its size does not change, so options B, C and D are wrong. Option B would be true only if both the current and the field were reversed together.

FBISE SSC-II 2026 Annual I, Set B, MCQ 111 mark

Which instrument converts electrical energy into mechanical energy?

ACapacitor
BElectric motorAnswer
CDC Generator
DTransformer

Why

An electric motor takes electrical energy from the supply and gives rotational mechanical energy, because a current-carrying coil in a magnetic field experiences a couple and turns. The DC generator (C) is the tempting wrong option: it does the reverse and converts mechanical energy into electrical energy. A capacitor stores charge and a transformer only changes an alternating voltage, so neither gives mechanical energy.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (v)1+2 = 3 marks

State Lenz's law. How it determines the direction of induced current? Explain briefly.

Solution

Lenz's law: the direction of the induced current in a circuit is always such that it opposes the cause which produces it. (1 mark)

How it fixes the direction (2 marks)

The induced current produces its own magnetic field. Lenz's law says this field must oppose the change (the movement of the magnet) that causes the current, so the direction of the current is the one whose field does this.

Bar magnet and coil

Move the north pole of a magnet towards a coil. The induced current flows so that the end of the coil facing the magnet becomes a north pole, which repels the approaching magnet. By the right-hand grip rule the induced current is anticlockwise as seen from the magnet's end.
Move the north pole away from the coil. The current reverses, the near end of the coil becomes a south pole, and it attracts the receding magnet back. The current is now clockwise.

In each case the rule gives the pole formed at the near end of the coil, and the right-hand grip rule then gives the direction of the current round the coil.

FBISE SSC-II 2026 Annual I, Set B, Q.61+2+2 = 5 marks

a. On which principle do the transformers work? b. Explain the working of a simple iron-core transformer. Also give its formula. c. Describe the construction of a simple iron-core transformer with a neat diagram.

Solution

(a) Principle (1 mark)

A transformer works on the principle of electromagnetic induction: a changing magnetic field in the iron core, produced by an alternating current in the primary coil, induces an alternating e.m.f. in the secondary coil.

(b) Working and formula (2 marks)

An a.c. voltage V_p is applied to the primary coil. The alternating current in it produces a changing magnetic field.
The laminated iron core concentrates this field and carries it to the secondary coil.
The changing field in the secondary coil induces an alternating e.m.f. V_s across it. The secondary voltage depends on the ratio of the numbers of turns.

\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}

19.1

If N_s>N_p the transformer is step-up (V_s>V_p); if N_s<N_p it is step-down (V_s<V_p). A steady d.c. supply cannot be used, because its field does not change and no e.m.f. is induced.

(c) Construction (2 marks)

A primary coil of N_p turns, connected to the a.c. supply.
A secondary coil of N_s turns, which gives the output voltage.
A laminated iron core; the two coils are wound on two different limbs of this core.

Diagram of a simple iron-core transformer: the laminated core carries the changing magnetic field from the primary coil (N_p turns) to the secondary coil (N_s turns).

FBISE SSC-II 2026 Annual I, Set D, MCQ 61 mark

Reversing the direction of the magnetic field while keeping the current constant will cause the force to:

AReverse directionAnswer
BRemain unchanged
CIncrease
DDecrease

Why

The direction of the force depends on the directions of both the field and the current. Reversing only the field turns the forefinger of Fleming's left hand the other way, so the thumb (force) reverses (A). The size of the force stays the same, so C and D are wrong. The force would remain unchanged (B) only if the current were reversed as well.

FBISE SSC-II 2026 Annual I, Set D, MCQ 111 mark

Which component of an electric motor reverses the direction of current?

ABrushes
BArmature
CField magnet
DSplit-ring commutatorAnswer

Why

The split-ring commutator turns with the coil. Each half of the ring changes brushes as the coil passes the vertical position, so the current in the coil is reversed every half-turn and the torque keeps the same sense. The brushes (A) are the tempting wrong option: they only carry current to the ring and stay fixed. The armature (B) is the rotating coil and the field magnet (C) only supplies the magnetic field.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (v)1+2 = 3 marks

State Lenz's law. Why does the induced current oppose the change producing it?

Solution

Lenz's law: the direction of the induced current in a circuit is always such that it opposes the cause which produces it. (1 mark)

Why it must oppose the change (2 marks)

Suppose the induced current helped the motion instead. A magnet pushed towards a coil would then be pulled in faster, the current would grow, and the magnet would go on speeding up with no one doing work. That would create energy from nothing. Energy cannot be created, so the induced current must oppose the change; the work done against this opposition is the energy that appears as the electrical energy of the induced current.

For example, a north pole moving towards a coil makes the near end of the coil a north pole, which repels the magnet. The person moving the magnet must do work against this repulsion, and this work becomes the electrical energy of the current.

FBISE SSC-II 2026 Annual I, Set D, Q.53+2 = 5 marks

a. How mutual induction is involved in the working of a transformer? Explain. Also draw diagram. b. Describe the energy transformation that occurs in a transformer.

Solution

(a) Mutual induction in a transformer (3 marks)

Mutual induction means that a changing current in one coil induces an e.m.f. in a neighbouring coil, because the changing magnetic field of the first coil passes through the second. This is exactly how the two coils of a transformer are linked.

An alternating voltage V_p is applied to the primary coil, so an alternating current flows in it and makes a changing magnetic field.
The laminated iron core carries this changing field through the secondary coil. The two coils are not connected electrically; they are linked only by the field.
The changing field in the secondary coil induces an alternating e.m.f. V_s. The voltages are related by the turns:

\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}

Diagram of a transformer: the alternating current in the primary coil sets up a changing field in the iron core, which induces an e.m.f. in the secondary coil.

(b) Energy transformation (2 marks)

Electrical energy supplied to the primary coil is changed into the energy of the magnetic field in the iron core, and this field then gives electrical energy in the secondary coil. So the energy goes electrical, then magnetic, then electrical again, at a different voltage.

In an ideal transformer no energy is lost, so the power in the primary equals the power in the secondary:

I_pV_p=I_sV_s

A step-up transformer raises the voltage and lowers the current in the same ratio; a step-down transformer does the opposite. The total energy is not changed.

FBISE SSC-II 2026 Annual I, Set D, Q.64+1 = 5 marks

Describe the operation of an electric motor, including the action of split-ring, commutator and brushes. Also draw a neat sketch.

Solution

Principle

An electric motor converts electrical energy into mechanical energy. A current-carrying coil placed in a magnetic field experiences a couple, so the coil rotates.

Parts

Coil (armature): a rectangular coil of many turns on an axle, between the poles.
Magnets: permanent magnets (or an electromagnet) giving the magnetic field.
Split-ring commutator: a ring split into two halves, which turns with the coil.
Brushes: pieces of graphite that touch the two halves of the ring and carry current from the battery into the coil.

Labelled sketch of an electric motor: the coil lies between the poles of the magnet, and the split ring and brushes supply the current.

Operation

Current flows through the coil. By Fleming's left-hand rule the force is up on one side of the coil and down on the other, so the equal and opposite forces form a couple that turns the coil.
When the coil reaches the vertical position, the forces are balanced and a simple coil would stop; a coil without a commutator cannot turn more than 90^{\circ}. Here each brush touches the gap between the two halves of the ring, and the coil keeps turning because of its motion.
As the coil passes the vertical, each half of the split ring slides under the other brush. The split ring and brushes therefore reverse the current in the coil.
The forces on the two sides now reverse, so they still turn the coil the same way. This happens at every half-turn, so the coil rotates continuously.

FBISE SSC-II 2026 Annual I, Set M, MCQ 61 mark

The force on a current-carrying conductor in a magnetic field is maximum when the conductor is placed:

AAlong the current direction
BParallel to the field
CAt any angle
DPerpendicular to the fieldAnswer

Why

The force is largest when the conductor cuts the field lines at right angles, that is, when it is perpendicular to the field (D). When the conductor lies parallel to the field (B) it cuts no field lines and the force is zero. Option A has no meaning, since the conductor itself carries the current, and the force is not the same at every angle (C).

FBISE SSC-II 2026 Annual I, Set M, MCQ 111 mark

If the split-ring commutator is removed, the motor will:

AProduce electrical energy
BRotate faster
CStop after half rotationAnswer
DContinue rotating smoothly

Why

Without the commutator the current in the coil cannot be reversed. The coil turns only until it reaches the vertical position, where the forces on its two sides are balanced, so it stops at the end of its first half-turn swing (a simple coil cannot turn more than 90^{\circ} on its own). Option D is the tempting wrong option: smooth rotation needs the current to be reversed every half-turn, which is the commutator's job. The motor still uses electrical energy, so A is wrong.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (v)1+2 = 3 marks

State Lenz's law. Explain with an example how Lenz's law obeys the law of conservation of energy.

Solution

Lenz's law: the direction of the induced current in a circuit is always such that it opposes the cause which produces it. (1 mark)

Conservation of energy with an example (2 marks)

Take a north pole of a magnet moved towards a coil. The induced current makes the near end of the coil a north pole, which repels the magnet, so the magnet must be pushed against a force. The work done in pushing it is the energy supplied, and it appears as the electrical energy of the induced current in the coil. When the magnet is pulled away, the near end becomes a south pole which attracts the magnet back, and again work must be done to pull it away.

If the law were opposite, the induced current would attract the approaching magnet and push away a receding one. The magnet would speed up by itself and the current would keep growing, so energy would be gained for nothing. Because the induced current opposes the motion, energy is only converted from the work done into electrical energy, none is created or lost, and the law of conservation of energy is obeyed.

FBISE SSC-II 2026 Annual I, Set M, Q.63+2 = 5 marks

a. Explain the step-up and step-down transformers. b. A step-down transformer converts 240 V into 12V. If there are 2000 turns on the primary coil, then find the number of turns on the secondary coil.

Solution

(a) Step-up and step-down transformers (3 marks)

A transformer is a device that increases or decreases an alternating voltage. It has a primary coil, a secondary coil and a laminated iron core. The changing field made by the a.c. in the primary induces an e.m.f. in the secondary, and

\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}

	Step-up transformer	Step-down transformer
Turns	N_s>N_p	N_s<N_p
Voltage	V_s>V_p	V_s<V_p
Function	increases the voltage	decreases the voltage
Current (ideal)	decreases (I_s<I_p)	increases (I_s>I_p)
Use	raising the voltage for transmission over long distances	street transformers lowering the voltage to 220\,\mathrm{V} for homes

In an ideal transformer I_pV_p=I_sV_s, so when the voltage is raised the current falls, and when the voltage is lowered the current rises.

Step-up and step-down transformers differ only in the numbers of turns on the two coils.

(b) Numerical (2 marks)

Given

V_p = 240\,\mathrm{V} primary voltage
V_s = 12\,\mathrm{V} secondary voltage
N_p = 2000 turns on the primary

To find

turns on the secondary, N_s

Write the transformer equation (given on the paper as V_s/V_p=N_s/N_p).
\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}
Make N_s the subject.
N_s=\dfrac{V_s}{V_p}\times N_p
Put in the values.
N_s=\dfrac{12\,\mathrm{V}}{240\,\mathrm{V}}\times2000
Calculate. The answer is smaller than N_p, as it must be for a step-down transformer.
N_s=100

N_s=100

Exam focus

Describe the construction and working of a simple iron-core transformer with a diagram; give V_s/V_p=N_s/N_p and explain step-up and step-down types. (long)
Transformer long questions appeared in all three 2026 sets: Q.6 in Set B (principle, working, construction), Q.5 in Set D (mutual induction, energy transformation) and Q.6 in Set M (step-up and step-down, with a turns numerical); also comprehensive question 6 of the book.
Go to answer
Describe the operation of an electric motor, including the action of the split-ring commutator and brushes, with a sketch. (long)
Set D Q.6 (5 marks) asked it directly; the motor and commutator also gave MCQ 11 in all three sets; same as comprehensive question 5.
Go to answer
State Lenz's law and explain how it fixes the direction of the induced current and agrees with conservation of energy. (short)
Q.2 (v) (3 marks, 1+2) in every 2026 set: Set B (direction of induced current), Set D (why the current opposes the change) and Set M (conservation of energy with an example); also comprehensive question 3.
Go to answer
Number of turns or voltage of a transformer from V_s/V_p=N_s/N_p. (numerical)
Set M Q.6 (b) is Example 19.1 with the same numbers (240 V to 12 V, 2000 turns); the book's numerical problems 1 to 5 use the same equation and I_pV_p=I_sV_s.
Go to answer
Effect of reversing the current or the field on the force, and when the force is greatest (Fleming's left-hand rule). (mcq)
MCQ 6 in all three 2026 sets tested this (reverse current in Set B, reverse field in Set D, maximum force in Set M).
Go to answer
Which part of the motor reverses the current, and what happens without the split-ring commutator? (mcq)
MCQ 11 in all three 2026 sets (energy conversion in Set B, commutator in Sets D and M); book exercise MCQ 2 is the same energy-conversion idea.
Go to answer
Why is a transformer ineffective with DC, and why is high voltage used for transmission? (short)
Not asked as a separate item in 2026, but the answers sit inside the transformer long questions; they are book short questions 2 and 5 and an SLO (advantages of high-voltage transmission).
Go to answer
Compare a generator with a motor and describe how an AC generator works. (long)
No 2026 question, but it is book short question 1 and comprehensive question 4, and an SLO on a.c. generators; a likely next question.
Go to answer