4.7 Magnetism and electromagnetism — revision question pack

10 specification points · notes, questions, answers and worked methods

Checked against AQA 8463 section 4.7. Review basis: the qualification registry sourced from the AQA GCSE Physics (8463) specification; registry verification recorded 17 July 2026.

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(physics only) means this content belongs to AQA GCSE Physics (8463), the separate-science qualification, but not AQA Combined Science: Trilogy (8464). It is not an exam tier: (HT only) separately marks Higher-tier content.

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4.7.1.1 · Poles of a magnet

Explanation

Magnetic forces are strongest at a magnet's north and south poles; like poles repel and unlike poles attract without touching.
A permanent magnet makes its own magnetic field, whereas an induced magnet becomes magnetic only while it is in another magnetic field.
For example, either pole of a permanent magnet attracts an unmagnetised iron nail because the nail becomes an induced magnet with the nearer end as the opposite pole.
A common error is to predict repulsion from an induced magnet: induced magnetism produces attraction, and most or all of it is lost quickly when the field is removed.

Worked example

A north pole is moved towards the north pole of a second permanent magnet. State the interaction and name the type of force involved.

1.Both approaching ends are north poles, so they are like poles and repel. The magnets exert this force across a gap, making it a non-contact force.

Answer: The poles repel. The interaction is a non-contact force.

Common mistakes

Don't predict repulsion from an induced magnet: induced magnetism produces attraction, and most or all of it is lost quickly when the field is removed.
Don't fall into the trap of saying a geographic north pole repels a magnet's north-seeking pole.

Exam tip

For magnetic poles, state both attraction or repulsion and the interacting pole names.

Tier 1 · Easy

State one difference between a permanent magnet and an induced magnet, including what happens when an external magnetic field is removed.
[2 marks]
Total for this question: 2
The north pole of a known magnet attracts end A of a suspended permanent magnet. Identify the pole at end A and predict what happens if the known north pole approaches the other end of the suspended magnet.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An unmagnetised iron pin is attracted to the south pole of a bar magnet. Explain why the pin is attracted and what happens after the magnet is taken far away.
[3 marks]
Total for this question: 3
Rods J and K are tested with a known magnet. K repels the known north pole at one end; J is attracted at either end and has almost no field once the known magnet is removed. State whether each rod is a permanent magnet or a material magnetised only by induction, and justify each answer from the test results.
[4 marks]
Total for this question: 4
Two facing ends of permanent bar magnets repel with forces $1.20\,\text{N}$ , $0.48\,\text{N}$ and $0.13\,\text{N}$ when their separation is $1.0\,\text{cm}$ , $2.0\,\text{cm}$ and $4.0\,\text{cm}$ respectively. State three conclusions supported by these observations.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Objects P and Q are each either an unmagnetised iron sample or a permanent magnet. P attracts the north pole of a known permanent magnet, while Q repels it. Decide which object must be a permanent magnet and explain why the evidence for P is inconclusive.
[4 marks]
Total for this question: 4
An unmagnetised iron washer is suspended below the south pole of a permanent magnet. A second unmagnetised iron washer hangs from the first, but both washers fall away soon after the magnet is removed. Explain the attractions in the chain and why the chain collapses.
[4 marks]
Total for this question: 4
A, B and C are permanent bar magnets. Their left ends are tested. A-left repels B-left, B-left attracts C-left, and a record says A-left repels C-left. The first two results were repeated and confirmed. State whether the three records are consistent. If A-left is north, identify B-left and C-left and correct any inconsistent record.
[5 marks]
Total for this question: 5
One pole of a permanent magnet is brought near separate samples of copper, iron, nickel and aluminium. Predict which samples are attracted and which experience no magnetic force, and explain the difference. State what could happen instead if the sample were an already-magnetised steel bar, and explain when this would occur.
[5 marks]
Total for this question: 5
Two unmagnetised iron pins touch opposite poles of the same permanent bar magnet. Pin P touches the north pole and pin Q touches the south pole. State the induced poles at both ends of each pin, predict the interaction when their free ends are brought together while they remain in contact with the magnet, and explain what happens after the magnet is removed.
[6 marks]
Total for this question: 6

4.7.1.2 · Magnetic fields

Explanation

A magnetic field is the region where a magnet exerts a force on another magnet or on iron, steel, cobalt or nickel.
Plot a field by placing a compass at successive positions, marking the direction of its north-seeking end, and joining the marks with smooth directed lines.
Outside a bar magnet, field lines point from north to south and are closest at the poles; compass alignment with Earth's field is evidence that Earth's core must be magnetic.
A common error is to draw crossing field lines or arrows from south to north outside the magnet; each point has one field direction, defined by the force on a north pole.

Magnetic field lines leave the north pole and enter the south pole outside a bar magnet.

Worked example

State the direction of the magnetic field immediately outside a bar magnet near its north pole.

1.Use the field-line convention outside a magnet: arrows run from the north pole to the south pole.

Answer: The field points away from the north pole, towards the south pole.

Common mistakes

Don't draw crossing field lines or arrows from south to north outside the magnet; each point has one field direction, defined by the force on a north pole.
Don't fall into the trap of drawing magnetic field lines that cross or point south to north outside a magnet.

Exam tip

Draw field arrows from north to south outside the magnet and make close spacing show greater strength.

Tier 1 · Easy

State where the magnetic field around a bar magnet is strongest and describe how a field-line diagram shows this.
[2 marks]
Total for this question: 2
The north-seeking end of a plotting compass points towards an unlabelled end of a bar magnet. Identify that end and state the field direction immediately outside it.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Describe how a student can map the direction of the magnetic field around a bar magnet using a small plotting compass.
[4 marks]
Total for this question: 4
A compass is placed at distances of $2.0\,\text{cm}$ , $4.0\,\text{cm}$ and $6.0\,\text{cm}$ from one pole of a bar magnet. Its deflections from north are $72^\circ$ , $21^\circ$ and $7^\circ$ . State the trend supported by the data and describe one way to make the comparison more reliable.
[3 marks]
Total for this question: 3
A nickel bead at position P moves towards a bar magnet without touching it. At a more distant position Q, no movement is detected. Explain what the observation at P shows about a magnetic field, and why the observation at Q does not prove that the field is exactly zero there.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Two field-pattern sketches are proposed for one bar magnet. Sketch A has widely spaced lines at the poles and crowded lines halfway between them; sketch B has crowded lines near the poles, no crossings, and arrows from north to south outside. Evaluate the sketches and relate a compass reading to your decision.
[4 marks]
Total for this question: 4
A student says, ‘The north-seeking end of a compass points towards geographic north, so that region of Earth must behave as a magnetic north pole.’ Evaluate the claim using magnetic-field direction and attraction between poles.
[4 marks]
Total for this question: 4
Describe an investigation to compare the field strengths of two bar magnets by measuring the greatest distance at which each attracts the same small iron ball. Include how the measurement is made fair and repeatable, and state how the results identify the stronger magnet.
[5 marks]
Total for this question: 5
At point X, $1.0\,\text{cm}$ to the left of an unlabelled bar magnet, the north-seeking end of a compass points left. At point Y, $3.0\,\text{cm}$ to the right, it also points left. The field lines are closer together at X than at Y. Identify both poles, state the field direction at each point, and compare the field strengths at X and Y.
[5 marks]
Total for this question: 5
Two bar magnets are placed with a north pole facing a south pole across a small gap. Describe the magnetic-field pattern and direction in the gap, state what a plotting compass placed at the midpoint would show, and explain what the line spacing indicates about the field strength there.
[5 marks]
Total for this question: 5

4.7.2.1 · Electromagnetism

Explanation

A current in a straight conductor creates concentric field lines; increasing current strengthens the field, increasing distance weakens it, and a nearby compass deflection demonstrates the effect.
Use the right-hand grip rule for direction: point the thumb in the conventional-current direction and the curled fingers show the field direction.
A solenoid's turns reinforce to make a strong, nearly uniform internal field and a bar-magnet-shaped external field; adding an iron core strengthens it and makes an electromagnet.
When interpreting an electromagnetic-device diagram, trace current to magnetic field and then to force or motion; a common error is to name the electromagnet without explaining that causal chain.

Worked example

Give two changes that would increase the magnetic field strength at a fixed point inside a current-carrying solenoid.

1.A larger current produces a stronger field from every turn. Easily magnetised iron concentrates and strengthens the solenoid's field.

Answer: Increase the current. Add an iron core.

Common mistakes

Don't make this mistake: When interpreting an electromagnetic-device diagram, trace current to magnetic field and then to force or motion; a common error is to name the electromagnet without explaining that causal chain.
Don't fall into the trap of reversing current direction without reversing the magnetic field direction.

Exam tip

Use a plotting compass to map field direction and identify where fields reinforce or oppose.

Tier 1 · Easy

Describe the shape of the magnetic field around a straight current-carrying wire and how its strength changes with distance from the wire.
[2 marks]
Total for this question: 2
A compass beside a straight wire turns only while current flows in the wire and returns to its original direction when the switch is opened. State what this observation shows.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A vertical wire carries conventional current upwards. Describe the field-line shape and how a student determines its direction when viewed from above.
[3 marks]
Total for this question: 3
Four solenoids are tested with the same compass position. A has 100 turns, $0.50\,\text{A}$ and no core; B has 100 turns, $1.00\,\text{A}$ and no core; C has 200 turns, $1.00\,\text{A}$ and no core; D matches C but has an iron core. Their compass deflections are $12^\circ$ , $24^\circ$ , $41^\circ$ and $67^\circ$ . Use the paired comparisons to identify three changes that strengthen a solenoid's field.
[3 marks]
Total for this question: 3
A plotting compass is tested at three equally spaced positions inside a long current-carrying solenoid and once the same distance outside it. The deflections are $63^\circ$ , $61^\circ$ , $64^\circ$ and $9^\circ$ respectively. State two conclusions about the solenoid's field and explain what produces it.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Coil X has 80 closely spaced turns and no core. Coil Y has 80 identical turns carrying the same current but contains an iron core. Compare their fields, then explain why reversing the current through Y reverses its poles without removing its field-strength advantage.
[5 marks]
Total for this question: 5
An electromagnetic sorting crane must release steel cans immediately when switched off. A designer proposes a steel core carrying a small permanent magnet, supplied by a solenoid. Critique the design and give two changes that would increase lifting strength without preventing rapid release.
[6 marks]
Total for this question: 6
In an overcurrent switch, the circuit current passes through a solenoid beside an iron armature. When the current becomes too large, the armature moves and opens the circuit contacts. Explain the complete operating sequence and why the contacts can reset after the fault current is removed.
[5 marks]
Total for this question: 5
A vertical wire carries conventional current upwards. Viewed from above, compass north-seeking ends point west at a position north of the wire, north at a position east of it, west at a position south of it, and south at a position west of it. Use the right-hand grip rule to identify and correct the anomalous direction. Then state the effect of doubling the current and the effect of reversing it.
[6 marks]
Total for this question: 6
Two identical current-carrying solenoids, A and B, face end to end. A compass shows that the facing ends are both north poles. State their interaction and the field direction inside A if its north pole is on the right. Explain what happens to the pole and interaction when only B's current is reversed, then give one change that would strengthen B's field.
[5 marks]
Total for this question: 5

4.7.2.2 · Fleming's left-hand rule (HT only)

Explanation

The motor effect is the force produced when a current-carrying conductor lies in a magnetic field; the conductor and field-producing magnet exert forces on each other.
For Fleming's left-hand rule, hold the thumb, first finger and second finger mutually perpendicular: thumb is force, first finger is field from north to south, and second finger is conventional current.
For a conductor perpendicular to the field, calculate force with $F=BIl$ ; for example, $B=0.30\,\text{T}$ , $I=2.0\,\text{A}$ and $l=0.40\,\text{m}$ give $F=0.24\,\text{N}$ .
A common error is to use electron flow for the current finger or to apply $F=BIl$ unchanged when the conductor is not at right angles to the field.

Worked example

In Fleming's left-hand rule, state what the first finger and thumb represent.

1.Recall the ordered labels: first finger is field, second finger is conventional current, and thumb is force.

Answer: First finger: magnetic field direction. Thumb: force or motion direction.

Common mistakes

Don't use electron flow for the current finger or to apply $F=BIl$ unchanged when the conductor is not at right angles to the field.
Don't fall into the trap of using Fleming's right hand instead of the left hand for the motor effect.

Exam tip

Apply Fleming's left-hand rule with field, current and force mutually perpendicular.

Tier 1 · Easy

A current-carrying wire in a magnetic field experiences an upward force. The magnetic field is unchanged, but the current is reversed. State the new force direction and explain.
[2 marks]
Total for this question: 2
A horizontal wire lies in a magnetic field directed towards geographic north. The motor-effect force on the wire is vertically upwards. Use Fleming's left-hand rule to determine the conventional-current direction.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A $0.18\,\text{m}$ wire section is perpendicular to a $0.45\,\text{T}$ magnetic field and carries $3.2\,\text{A}$ . Calculate the force on the section.
[2 marks]
Total for this question: 2
Three perpendicular wire tests are recorded as $(B,I,l,F)$ : A $(0.20\,\text{T},1.5\,\text{A},0.40\,\text{m},0.12\,\text{N})$ ; B $(0.35\,\text{T},2.0\,\text{A},0.30\,\text{m},0.21\,\text{N})$ ; C $(0.50\,\text{T},1.2\,\text{A},0.25\,\text{m},0.10\,\text{N})$ . Use the Physics Equations Sheet to identify the inconsistent record and calculate its corrected force.
[3 marks]
Total for this question: 3
For a wire perpendicular to a $0.60\,\text{T}$ field, a graph of force against current is a straight line through the origin. The force increases from $0$ to $0.216\,\text{N}$ as the current increases from $0$ to $3.0\,\text{A}$ . Using the Physics Equations Sheet, calculate the graph gradient and determine the active wire length.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A wire of active length $0.35\,\text{m}$ experiences a $0.63\,\text{N}$ downward force while carrying $4.0\,\text{A}$ perpendicular to a uniform field. Determine the magnetic flux density. Then state the new force direction if only the current is reversed.
[4 marks]
Total for this question: 4
A wire has $24\,\text{cm}$ of its length perpendicular to a $0.35\,\text{T}$ field. The force is $0.168\,\text{N}$ . Convert the active length to metres and determine the current. The current is then doubled and the field direction is reversed. Determine the new force magnitude and compare its direction with the original force. Use the Physics Equations Sheet.
[5 marks]
Total for this question: 5
A horizontal wire is supported by an upward motor-effect force that exactly balances the weight of an $18\,\text{g}$ load. The wire's own weight is negligible. The active wire length is $0.12\,\text{m}$ and the current is $2.5\,\text{A}$ . Use $g=9.8\,\text{N/kg}$ and the Physics Equations Sheet to determine the magnetic flux density and state the required force direction.
[5 marks]
Total for this question: 5
A charge of $54\,\text{C}$ passes through a straight wire in $15\,\text{s}$ . A $0.25\,\text{m}$ section is perpendicular to a $0.48\,\text{T}$ magnetic field. Use the Physics Equations Sheet to calculate the current and the motor-effect force. State what happens to the force direction if only the magnetic field is reversed.
[5 marks]
Total for this question: 5
A $45\,\text{g}$ straight conductor is free to move on a horizontal surface. It carries $2.4\,\text{A}$ through a $0.18\,\text{m}$ active length perpendicular to a $0.75\,\text{T}$ field. Friction is negligible. Use the Physics Equations Sheet to calculate the motor-effect force and acceleration. State what happens to the acceleration direction if both the current and field are reversed.
[5 marks]
Total for this question: 5

4.7.2.3 · Electric motors (HT only)

Explanation

In a motor, opposite sides of a current-carrying coil experience forces in opposite directions because their currents run oppositely through the magnetic field.
These forces form a turning effect; use Fleming's left-hand rule separately on each active side to predict the rotation direction.
A split-ring commutator reverses the current every half-turn, so the forces swap sides and the turning effect continues in the same rotational direction.
A common error is to say the coil turns because unlike poles attract; the required explanation is the motor-effect force on current-carrying conductors in a magnetic field.

Worked example

State the energy transfer performed by an electric motor and name the effect that produces its turning force.

1.A motor uses current to produce motion, so its useful transfer is electrical to kinetic. The force on the current-carrying coil is the motor effect.

Answer: Electrical energy is transferred to kinetic energy. The motor effect produces the force.

Common mistakes

Don't say the coil turns because unlike poles attract; the required explanation is the motor-effect force on current-carrying conductors in a magnetic field.
Don't fall into the trap of saying the forces on a motor coil act in the same direction.

Exam tip

Explain motor rotation using opposite forces on the two current-carrying sides of the coil.

Tier 1 · Easy

In a simple electric motor the current in the coil is reversed every half-turn. Explain why this reversal is necessary.
[2 marks]
Total for this question: 2
A technician replaces a motor's split-ring commutator with two continuous slip rings and claims the coil will still be driven in one rotational direction. Identify the error.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain why a rectangular current-carrying coil between magnetic poles begins to rotate rather than simply moving sideways.
[4 marks]
Total for this question: 4
A motor coil is viewed before and after half a turn. Before the half-turn, its left side is forced up and its right side down. Predict the two force directions just after the half-turn if the commutator works correctly, and explain how this maintains rotation.
[4 marks]
Total for this question: 4
A motor initially rotates clockwise. An engineer reverses both the current in the coil and the direction of the magnetic field at the same time. Predict the rotation direction after the change, then state one change that would reverse it.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A simple motor is turning slowly. Explain how increasing the current and adding a stronger magnet affect its motion, and explain why its split-ring commutator must reverse the current after each half-turn.
[6 marks]
Total for this question: 6
Prototype P has a 60-turn coil carrying $0.80\,\text{A}$ between one pair of magnets. Prototype Q has a 120-turn coil carrying the same current in the same field. Prototype R matches Q but carries $1.20\,\text{A}$ . Rank their initial turning effects and explain the comparison. Then predict the symptom if R's split ring fails to swap connections after a half-turn.
[5 marks]
Total for this question: 5
The two active sides of a single rectangular loop of wire in an electric motor are perpendicular to a $0.50\,\text{T}$ field. Each active side is $0.18\,\text{m}$ long and carries $2.0\,\text{A}$ . The sides are $0.080\,\text{m}$ apart. Using the Physics Equations Sheet and $M=Fd$ , calculate the force on each side and the total moment of the two forces about the central axis, then explain why the loop turns.
[5 marks]
Total for this question: 5
A $12\,\text{V}$ d.c. motor draws $2.5\,\text{A}$ while lifting a $4.0\,\text{kg}$ load through $1.5\,\text{m}$ in $4.0\,\text{s}$ . Use $g=9.8\,\text{N/kg}$ and the Physics Equations Sheet to calculate the electrical energy supplied, useful energy transferred and efficiency. Explain why the motor's split-ring commutator is still needed during the lift.
[6 marks]
Total for this question: 6
A single rectangular loop in a d.c. motor must produce a moment of $0.020\,\text{N m}$ . The active sides are $0.10\,\text{m}$ apart, each has length $0.25\,\text{m}$ in a $0.50\,\text{T}$ field, and the sides are perpendicular to the field. Using $M=Fd$ and the Physics Equations Sheet, calculate the force on each active side and the required current, then state and explain the effect on the required current if the active sides were moved to $0.20\,\text{m}$ apart.
[6 marks]
Total for this question: 6

4.7.2.4 · Loudspeakers (physics only) (HT only)

Explanation

A moving-coil loudspeaker uses the motor effect to convert variations in electrical current into pressure variations in a sound wave.
The alternating current in the voice coil repeatedly reverses, so the motor-effect force reverses and the coil moves backwards and forwards in the permanent magnet's field.
The coil is attached to a cone: larger current variations produce larger cone displacements and a louder sound, while faster variations produce a higher-frequency sound.
A common error is to describe electromagnetic induction in a loudspeaker; induction is used by a microphone, whereas a loudspeaker requires force on a supplied current.

Worked example

Name the effect used by a moving-coil loudspeaker and state what the cone transfers energy to.

1.Current in a magnetic field gives a force by the motor effect. The moving cone makes pressure variations in the air, transferring energy as sound.

Answer: The motor effect. The cone transfers energy to the surrounding air as sound waves.

Common mistakes

Don't describe electromagnetic induction in a loudspeaker; induction is used by a microphone, whereas a loudspeaker requires force on a supplied current.
Don't fall into the trap of saying the cone moves without the current changing direction.

Exam tip

Link alternating current to a reversing force and therefore vibration of the loudspeaker cone.

Tier 1 · Easy

For the same moving-coil loudspeaker, state how increasing the amplitude and increasing the frequency of the alternating current affect the sound.
[2 marks]
Total for this question: 2
A student says a moving-coil loudspeaker creates sound by the generator effect. Correct the statement by naming the effect used and identifying which quantity must be supplied to the coil.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain how an alternating electrical signal makes the cone of a moving-coil loudspeaker vibrate.
[4 marks]
Total for this question: 4
Signal X completes 5 cycles in $0.020\,\text{s}$ and has a peak current of $0.32\,\text{A}$ . Signal Y completes 8 cycles in the same time and has a peak current of $0.20\,\text{A}$ . Both drive the same ideal loudspeaker. Calculate both frequencies and compare the sounds produced.
[4 marks]
Total for this question: 4
A moving-coil loudspeaker is connected to a steady direct current. Its cone moves once and then remains displaced. Explain why this does not produce a continuous tone and state the required change to the supply.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Signal A has twice the frequency of signal B but a smaller current amplitude. Compare the sounds produced when each signal drives the same ideal loudspeaker, and justify both comparisons using the coil's motion.
[4 marks]
Total for this question: 4
Two otherwise identical loudspeakers receive $320\,\text{Hz}$ signals. Coil P has $B=0.40\,\text{T}$ , peak current $0.80\,\text{A}$ and active wire length $0.20\,\text{m}$ . Coil Q has $B=0.50\,\text{T}$ , peak current $1.6\,\text{A}$ and active length $0.20\,\text{m}$ . Use the Physics Equations Sheet to compare their peak forces, then predict their relative pitch and loudness.
[5 marks]
Total for this question: 5
An ideal loudspeaker coil is perpendicular to a $0.50\,\text{T}$ field and has active wire length $0.16\,\text{m}$ . Its current is $+0.75\,\text{A}$ for $1.0\,\text{ms}$ , then $-0.75\,\text{A}$ for $1.0\,\text{ms}$ , and this pattern repeats. Use the Physics Equations Sheet to determine the force magnitude, force-reversal interval and cone-vibration frequency, then explain the motion.
[5 marks]
Total for this question: 5
A moving-coil loudspeaker requires a peak motor-effect force of $0.060\,\text{N}$ . Its $0.25\,\text{m}$ active wire is perpendicular to a $0.40\,\text{T}$ field. The input completes nine cycles in $0.030\,\text{s}$ . Use the Physics Equations Sheet to calculate the peak current and signal frequency, then explain how the alternating current produces the sound.
[6 marks]
Total for this question: 6

4.7.3.1 · Induced potential (physics only) (HT only)

Explanation

The generator effect induces a potential difference when a conductor moves relative to a magnetic field or when the magnetic field around it changes; a complete circuit then carries an induced current.
Increase the induced potential difference by increasing relative speed, field strength or coil turns; reversing either the relative motion or the magnetic-field direction reverses the induced potential difference and current.
The induced current creates its own magnetic field opposing the change that produced it, so pushing a magnet into a coil produces a magnetic effect that resists the push.
A common error is to say a stationary magnet permanently induces current in a stationary coil: induction requires relative motion or a changing magnetic field.

Worked example

A bar magnet is held motionless inside a coil connected to a sensitive voltmeter. State the reading after the magnet has stopped moving and explain it.

1.The generator effect needs a changing magnetic field around the conductor. Once the magnet is stationary, that change stops, so no potential difference is induced.

Answer: The reading is zero. There is no relative motion or change in magnetic field through the coil.

Common mistakes

Don't say a stationary magnet permanently induces current in a stationary coil: induction requires relative motion or a changing magnetic field.
Don't fall into the trap of claiming a stationary magnet inside a coil induces a potential difference.

Exam tip

For induction, state the change in magnetic field through the conductor and how to increase it.

Tier 1 · Easy

State two changes that increase the induced potential difference when a bar magnet is moved into a coil.
[2 marks]
Total for this question: 2
Moving a magnet's north pole into a coil gives a positive voltmeter pulse. Predict the pulse when the magnet's south pole is moved into the coil in the same way, and explain the change.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A magnet is pushed into a 200-turn coil and then withdrawn at the same speed. Describe the two voltmeter pulses and state two changes that would increase their magnitudes.
[4 marks]
Total for this question: 4
A magnet is moved through the same coil at $0.20\,\text{m/s}$ , giving peak voltages $0.82\,\text{V}$ , $0.80\,\text{V}$ and $0.81\,\text{V}$ . At $0.40\,\text{m/s}$ the readings are $1.61\,\text{V}$ , $1.59\,\text{V}$ and $2.40\,\text{V}$ . Identify the anomalous reading, calculate a mean for each speed without it, and state the supported relationship.
[4 marks]
Total for this question: 4
A magnet is moved through a coil connected only to a high-resistance voltmeter. The meter shows a potential-difference pulse, but no appreciable current flows. Explain why a pulse can exist without a current and state what change would allow an induced current.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A student feels a resisting force while pushing the north pole of a magnet into a short-circuited coil. Explain the resistance, predict what happens to the induced current direction when the magnet is pulled out, and state why pulling it out faster requires more work per second.
[6 marks]
Total for this question: 6
Four coil tests give $(\text{turns},\text{magnet},\text{speed},\text{peak voltage})$ : A $(200,\text{standard},0.15\,\text{m/s},0.60\,\text{V})$ ; B $(400,\text{standard},0.15\,\text{m/s},1.20\,\text{V})$ ; C $(400,\text{strong},0.15\,\text{m/s},1.80\,\text{V})$ ; D $(400,\text{strong},0.30\,\text{m/s},3.60\,\text{V})$ . Use controlled comparisons to explain the pattern, then predict the reading if the magnet pauses inside coil D.
[6 marks]
Total for this question: 6
Coil P is connected to a direct-current supply and switch. Nearby coil Q is connected to a centre-zero voltmeter, with neither coil moving. Q shows a brief pulse when the switch is closed, zero while the current in P is steady, and an opposite pulse when the switch is opened. Explain all three readings and state how adding turns to Q would affect the pulses.
[5 marks]
Total for this question: 5
A magnet's north pole passes through a coil. It enters at $0.30\,\text{m/s}$ and produces a $+0.90\,\text{V}$ pulse, then leaves at $0.50\,\text{m/s}$ and produces a $-1.5\,\text{V}$ pulse. Explain the polarity and magnitude changes, then predict the effect of using a stronger magnet at the same two speeds.
[5 marks]
Total for this question: 5
A generator-effect demonstration produces an approximately constant induced potential difference of $2.4\,\text{V}$ across an $8.0\,\Omega$ resistor for $0.50\,\text{s}$ while a magnet is pulled from a coil. Use the Physics Equations Sheet to calculate the current, electrical power and energy transferred. Explain why the person pulling the magnet must transfer energy mechanically.
[6 marks]
Total for this question: 6

4.7.3.2 · Uses of the generator effect (physics only) (HT only)

Explanation

An alternator rotates a coil in a magnetic field and generates alternating potential difference, so its output reverses every half-turn.
A dynamo uses a split-ring commutator to exchange coil connections every half-turn, producing a direct potential difference of one polarity.
For steady rotation, the alternator graph crosses zero when the coil sides move parallel to the field and reaches maximum magnitude when they cut field lines at the greatest rate.
A common error is to draw a constant dc line for a dynamo: its output keeps one polarity but normally varies in magnitude as the coil rotates.

Worked example

State which device generates ac, an alternator or a dynamo, and identify the output produced by the other device.

1.Match the devices to their outputs: continuous slip rings retain the alternator's reversal, while a split-ring commutator makes the dynamo output unidirectional.

Answer: An alternator generates ac. A dynamo generates dc.

Common mistakes

Don't draw a constant dc line for a dynamo: its output keeps one polarity but normally varies in magnitude as the coil rotates.
Don't fall into the trap of confusing an alternator's slip rings with a motor's split-ring commutator.

Exam tip

Compare generators by naming the rotating part and whether the output is alternating or direct.

Tier 1 · Easy

Explain why the potential difference produced by a rotating-coil alternator reverses every half-turn.
[2 marks]
Total for this question: 2
An oscilloscope trace from a hand generator consists of repeating positive pulses that return to zero but never cross below zero. Decide whether the device uses slip rings or a split-ring commutator, and name the output type.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An alternator coil completes five rotations each second. Determine the frequency and period of its output potential difference, then state how the graph changes if the rotation rate doubles.
[4 marks]
Total for this question: 4
An alternator trace has readings $(t,V)$ of $(0\,\text{ms},0\,\text{V})$ , $(5\,\text{ms},+6\,\text{V})$ , $(10\,\text{ms},0\,\text{V})$ , $(15\,\text{ms},-6\,\text{V})$ and $(20\,\text{ms},0\,\text{V})$ , after which the pattern repeats. Determine the period, frequency and peak potential difference. Use the Physics Equations Sheet.
[4 marks]
Total for this question: 4
At position A in a rotating-coil alternator, the active sides move parallel to the magnetic field lines. A quarter-turn later at B, they cut the field lines at the greatest rate. Compare the instantaneous potential differences at A and B and explain the difference.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Two otherwise identical rotating-coil generators use the same field and speed. Generator P uses slip rings; generator Q uses a split-ring commutator. Compare their potential-difference-against-time graphs and explain the role of Q's commutator.
[5 marks]
Total for this question: 5
Generator M has 300 coil turns, rotates slowly in a weak field and uses slip rings. Generator N has 600 turns, rotates faster in a stronger field and uses a split-ring commutator. Compare the expected amplitudes and polarities of their output traces. Also compare the frequency of M's ac trace with the pulse repetition rate of N's varying-dc trace, explaining each design link.
[6 marks]
Total for this question: 6
A cyclist transfers $24\,\text{J}$ of mechanical energy to a bicycle generator in $4.0\,\text{s}$ . The generator supplies electrical power of $4.5\,\text{W}$ . Using the Physics Equations Sheet, calculate the mechanical input power, the efficiency and the power dissipated, then explain why generating the current makes pedalling harder.
[5 marks]
Total for this question: 5
Alternator G completes 25 rotations in $0.50\,\text{s}$ . Calculate the frequency and period of its output. It is then driven at 40 rotations in $0.50\,\text{s}$ in the same magnetic field. Calculate the new frequency and period, then describe and explain the two changes to the potential-difference graph.
[6 marks]
Total for this question: 6
A hand-driven d.c. generator supplies a mean current of $0.40\,\text{A}$ and mean electrical power of $2.4\,\text{W}$ for $30\,\text{s}$ . Use the Physics Equations Sheet to calculate the charge delivered and electrical energy transferred. Explain why its split-ring output is varying d.c.
[4 marks]
Total for this question: 4

4.7.3.3 · Microphones (physics only) (HT only)

Explanation

A moving-coil microphone uses the generator effect to convert pressure variations in sound into variations in current in an electrical circuit.
Sound pressure makes a diaphragm vibrate; its attached coil moves relative to a permanent magnet's field, inducing a varying potential difference.
The induced signal follows the diaphragm: greater sound amplitude gives greater coil speed and signal amplitude, while sound frequency sets the signal frequency.
A common error is to supply a driving current to the microphone coil; the sound-driven motion induces the signal, whereas supplied current drives a loudspeaker.

Worked example

State the input and output of a moving-coil microphone.

1.A microphone is a transducer from sound to electrical signal, so identify air-pressure variation as the input and a varying electrical signal as the output.

Answer: Input: pressure variations in a sound wave. Output: variations in electrical potential difference or current.

Common mistakes

Don't supply a driving current to the microphone coil; the sound-driven motion induces the signal, whereas supplied current drives a loudspeaker.
Don't fall into the trap of describing a microphone as converting electrical energy into sound.

Exam tip

A microphone explanation should follow sound vibration to coil motion to induced electrical signal.

Tier 1 · Easy

A louder sound of the same pitch enters a moving-coil microphone. Compare the new electrical signal with the original signal.
[2 marks]
Total for this question: 2
A student connects a battery to a moving-coil microphone so that its coil can create the incoming electrical signal. Explain why the battery is unnecessary and name the effect that creates the signal.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain the sequence by which a singer's sound produces an electrical signal in a moving-coil microphone.
[4 marks]
Total for this question: 4
Microphone trace A has peak voltage $0.08\,\text{V}$ and period $2.0\,\text{ms}$ . Trace B has peak voltage $0.12\,\text{V}$ and period $4.0\,\text{ms}$ . Calculate both frequencies and compare the sounds that produced them. Use the Physics Equations Sheet.
[4 marks]
Total for this question: 4
Describe an investigation using a signal generator, loudspeaker, moving-coil microphone and oscilloscope to test whether microphone output frequency follows sound frequency. Include one control variable and how the traces are analysed. Use the Physics Equations Sheet.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Tones P and Q have the same pitch, but P is louder. Tones R and S have the same loudness, but R has the higher pitch. All four are recorded by one ideal moving-coil microphone. Compare the relevant electrical signals and explain the generator-effect link.
[5 marks]
Total for this question: 5
The same steady tone is recorded in three tests. A 200-turn coil with a standard magnet gives $24\,\text{mV}$ peak. A 400-turn coil with the same magnet gives $47\,\text{mV}$ . The 400-turn coil with a stronger magnet gives $81\,\text{mV}$ . Evaluate how the two design changes affect microphone sensitivity, and explain why all three signals should retain the same frequency.
[5 marks]
Total for this question: 5
Devices M and N each contain a diaphragm, coil and permanent magnet. Pushing M's diaphragm produces a voltage pulse at its terminals. Applying an alternating current to N makes its diaphragm vibrate. Identify each device and explain the different electromagnetic effect and energy-transfer direction in each.
[6 marks]
Total for this question: 6
A moving-coil microphone records the same note from a sound source first at $0.5\,\text{m}$ and then at $2.0\,\text{m}$ . Describe and explain the change to the oscilloscope trace at the greater distance, then give one change to the microphone that could restore the original trace amplitude.
[5 marks]
Total for this question: 5
Three moving-coil microphone faults are tested with the same sound. In M the diaphragm vibrates but the coil is detached from it. In N the coil moves but the external circuit is open. In P the coil moves normally but the permanent magnet is much weaker. Explain the electrical observation expected for each microphone.
[5 marks]
Total for this question: 5

4.7.3.4 · Transformers (physics only) (HT only)

Explanation

A transformer has primary and secondary coils wound on an easily magnetised iron core; alternating current in the primary produces a changing core field that induces a potential difference in the secondary.
Use $\dfrac{V_{\mathrm{p}}}{V_{\mathrm{s}}}=\dfrac{n_{\mathrm{p}}}{n_{\mathrm{s}}}$ : more secondary turns than primary turns gives a step-up transformer, and fewer gives a step-down transformer.
For an ideal transformer, input and output powers are equal, so $V_{\mathrm{p}}I_{\mathrm{p}}=V_{\mathrm{s}}I_{\mathrm{s}}$ ; raising transmission potential difference reduces current for the same power and therefore reduces heating losses.
A common error is to connect a transformer to steady dc: without a changing primary current there is no continuously changing core field and therefore no continuous secondary potential difference.

Worked example

A transformer has 300 turns on its primary coil and 900 turns on its secondary coil. State whether it is step-up or step-down and give the potential-difference factor.

1.The turn ratio is $n_{\mathrm{s}}/n_{\mathrm{p}}=900/300=3$ . The secondary has more turns, so $V_{\mathrm{s}}=3V_{\mathrm{p}}$ and the transformer is step-up.

Answer: It is step-up. The secondary potential difference is three times the primary potential difference.

Common mistakes

Don't connect a transformer to steady dc: without a changing primary current there is no continuously changing core field and therefore no continuous secondary potential difference.
Don't fall into the trap of using the turns ratio upside down when calculating secondary potential difference.

Exam tip

Write $V_p/V_s=N_p/N_s$ and use conservation of power for an ideal transformer.

Tier 1 · Easy

Explain why connecting a transformer primary to a steady direct current does not produce a continuous potential difference across the secondary.
[2 marks]
Total for this question: 2
The primary and secondary coils of a transformer have no electrical connection. Explain how an alternating current in the primary can still produce a potential difference across the secondary.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An ideal transformer has $1200$ primary turns and $80$ secondary turns. Its primary potential difference is $230\,\text{V}$ . Calculate the secondary potential difference.
[3 marks]
Total for this question: 3
A table claims three ideal transformer results $(n_p,n_s,V_p,V_s)$ : P $(800,40,240\,\text{V},12\,\text{V})$ ; Q $(600,150,240\,\text{V},60\,\text{V})$ ; R $(1000,50,230\,\text{V},13.8\,\text{V})$ . Use the Physics Equations Sheet to identify the inconsistent row and calculate its corrected secondary potential difference.
[3 marks]
Total for this question: 3
An ideal transformer has a $240\,\text{V}$ primary and a $12\,\text{V}$ secondary. The secondary current is normally $5.0\,\text{A}$ but rises to $10.0\,\text{A}$ during a fault. A $0.40\,\text{A}$ fuse protects the primary. Calculate the primary current in each case and state whether the fuse melts in each case. Use the Physics Equations Sheet.
[4 marks]
Total for this question: 4

Tier 3 · Hard

An ideal step-up transformer supplies $18.0\,\text{kW}$ at $12.0\,\text{kV}$ from a $240\,\text{V}$ primary. Calculate the secondary current, the primary current and the secondary-to-primary turn ratio. Explain one transmission advantage of the high output potential difference.
[6 marks]
Total for this question: 6
A transformer delivers $24\,\text{V}$ and $6.0\,\text{A}$ to a load at $80\%$ efficiency. Its primary is connected to $240\,\text{V}$ and has 1500 turns. The Physics Equations Sheet gives the transformer equations. Assume the coil potential differences are in the same ratio as the turns. Calculate the output power, input power, primary current, secondary turns and power dissipated by the transformer.
[6 marks]
Total for this question: 6
A step-up transformer must transmit $2.0\,\text{MW}$ through cables of total resistance $4.0\,\Omega$ while keeping cable heating at or below $40\,\text{kW}$ . Its primary potential difference is $500\,\text{V}$ . Using the Physics Equations Sheet, calculate the maximum cable current, the minimum transmission potential difference and the minimum secondary-to-primary turn ratio. Assume the transformer is ideal.
[6 marks]
Total for this question: 6
An ideal transformer steps down $11\,\text{kV}$ to $230\,\text{V}$ . The primary has $11\,000$ turns and the secondary supplies $44\,\text{A}$ . Use the Physics Equations Sheet to calculate the secondary turn count, output power and primary current, then explain why it is a step-down transformer.
[5 marks]
Total for this question: 5
A power of $600\,\text{kW}$ is transmitted through cables of total resistance $5.0\,\Omega$ . Use the Physics Equations Sheet to calculate the transmission current and cable-heating power at $15\,\text{kV}$ and at $3.0\,\text{kV}$ . Determine the factor by which the heating increases at the lower potential difference and explain this factor.
[5 marks]
Total for this question: 5

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.7.1.1 · Poles of a magnet

Tier 1 · Easy

Mark scheme for 4.7.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	A permanent magnet produces its own magnetic field. An induced magnet is magnetic only while it is in an external field and loses most or all of its magnetism when that field is removed.	Compare whether the magnetism is retained. A permanent magnet remains magnetised, whereas induced magnetism depends on the external field and is usually lost quickly when the field is removed.	2
Total Question 1			2
02.1	End A is a south pole because unlike poles attract. The other end is north, so it repels the known north pole.	Use the first interaction to identify A as the pole unlike north. A permanent magnet has the opposite pole at its other end, so that end is north and gives repulsion.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The magnetic field induces magnetism in the iron pin. The end nearest the south pole becomes a north pole, so there is attraction. The pin quickly loses most or all of its magnetism when the field is removed.	Identify the pin as an induced magnet. Its near end becomes the opposite pole to the magnet's south pole, producing attraction. Iron does not retain most of this induced magnetism once the external field is removed.	3
Total Question 1			3
02.1	K is a permanent magnet. Its repulsion of a known pole shows that it already has a fixed like pole; induction can produce attraction but not repulsion in this test. J behaves as an induced magnet. It is attracted at either end only while the external field magnetises it and loses almost all of that magnetism when the known magnet is removed.	Use repulsion with a known pole as the decisive test for a permanent magnet, then remove the known magnet and test for retained magnetism. K gives the permanent-magnet result; J's temporary attraction identifies induced magnetism.	4
Total Question 2			4
03.1	The facing ends are like poles because they repel. The magnets exert a force across a gap, so the magnetic force is non-contact. The force becomes weaker as the separation increases.	Use the interaction type to classify the pole relationship, then use the non-zero separation and the ordered force measurements as separate pieces of evidence about contact and distance.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.7.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Q must be a permanent magnet because repulsion can occur only between like magnetic poles. P could be an induced magnetic material because induced magnetism causes attraction to either pole. P could instead present the south pole of a permanent magnet, so attraction alone is not conclusive.	Use repulsion as the decisive test: a like pole is required, so Q already has a permanent pole. Attraction does not distinguish a permanent south pole from an induced magnetic material, because either can be attracted in the stated observation.	4
Total Question 1			4
02.1	The permanent magnet's field induces magnetism in the first iron washer. Its upper face becomes a north pole and is attracted to the magnet's south pole. The magnetised first washer induces an opposite pole at the nearer face of the second washer, so they attract. When the external field is removed, the iron washers lose most or all of their induced magnetism, so the attractions disappear and they fall.	Follow the induction from the permanent magnet through the two washers. Each nearer face takes the opposite pole to the object above it; removing the original field removes the temporary magnetism that maintained both attractions.	4
Total Question 2			4
03.1	A-left and B-left must be like poles because they repel. If A-left is north, B-left is also north. B-left and C-left are unlike because they attract, so C-left is south. A-left north and C-left south must attract. The A-left and C-left repulsion record is therefore inconsistent and should record attraction.	Translate repulsion into same polarity and attraction into opposite polarity. Propagate the known north pole through the two relationships, then test the third observation against the resulting pole labels.	5
Total Question 3			5
04.1	The iron and nickel samples are attracted. Iron and nickel are magnetic materials; induced magnetism makes the nearer end an unlike pole, causing attraction. Copper and aluminium are not magnetic materials, so they experience no magnetic force in this test. An already-magnetised steel bar could instead be repelled. Repulsion occurs if the pole of the steel bar facing the permanent magnet is a like pole.	First separate the samples into magnetic and non-magnetic materials. An unmagnetised magnetic sample is attracted by induction, whereas an already-magnetised steel bar has a fixed pole that can produce the pole test unique to magnets: repulsion between like poles.	5
Total Question 4			5
05.1	The end of P touching the magnet's north pole becomes south. P's free end becomes north. The end of Q touching the magnet's south pole becomes north. Q's free end becomes south. The free ends attract because they are induced unlike poles. After the permanent magnet is removed, the iron pins lose most or all of their induced magnetism, so the magnetic attraction is lost.	Work away from each permanent pole: the touching end becomes unlike that pole and the far end takes the opposite induced polarity. The two free ends are therefore unlike until the inducing field is removed.	6
Total Question 5			6

4.7.1.2 · Magnetic fields

Tier 1 · Easy

Mark scheme for 4.7.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The field is strongest near the poles. The field lines are closest together where the field is strongest.	Use field-line spacing to represent strength. The lines crowd most closely near the north and south poles, matching the regions of greatest magnetic force.	2
Total Question 1			2
02.1	The unlabelled end is the south pole. Immediately outside it, the field points towards the magnet.	The north-seeking compass end follows the local field direction. Outside a bar magnet, field lines enter the south pole, so a compass pointing towards an end identifies that end as south.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Place the compass beside the magnet and mark the direction indicated by the north-seeking end. Move the compass so its tail is at the previous mark and repeat. Join the marks with a smooth line and add an arrow from north to south. Repeat from several starting positions to build the pattern.	The north-seeking end shows the field direction at one point. Repeated adjacent readings trace one field line; repeating the process from other starting points maps the whole pattern.	4
Total Question 1			4
02.1	The compass deflection decreases as distance from the pole increases. This supports the conclusion that the magnet's field becomes weaker with distance. Repeat the reading at each distance and calculate a mean, keeping the compass and magnet orientations fixed.	Use compass deflection as the indicator of the magnet's influence relative to Earth's field. Compare the ordered readings, then improve reliability using repeat measurements and a controlled geometry.	3
Total Question 2			3
03.1	P is in a magnetic field because the magnet exerts a force on the magnetic material there. The force acts across a gap, so no contact is required. At Q the field may be too weak to produce a detectable movement rather than being exactly zero.	Use the operational definition of a field: it is a region where a magnetic object experiences a force. Treat a null measurement as a detection limit, not automatic proof of no field.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.7.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Sketch B is consistent with the field being strongest at the poles. Crowded field lines represent a stronger field and field lines must not cross. The arrows outside the magnet should run from north to south. A compass north-seeking end would align tangentially with those arrows at its position.	Test each sketch against three rules: strength is greatest where lines are closest, one point cannot have two directions, and the external direction is north to south. A compass then provides the local direction shown by the tangent to a valid line.	4
Total Question 1			4
02.1	The north-seeking end of the compass is itself a magnetic north pole. It is attracted towards the geographic north region, so that region behaves like a magnetic south pole. The compass aligns with Earth's magnetic field at its position. Therefore the observation is evidence that Earth's core is magnetic, but the student's pole label is reversed.	Treat the compass as a small magnet. Because unlike poles attract and the compass's north-seeking end points geographically north, the magnetic polarity responsible there must be south-like rather than north-like.	4
Total Question 2			4
03.1	Place each magnet in the same orientation and move the same iron ball towards the same pole along a measured scale. Record the greatest separation at which the ball first moves towards the magnet. Use the same ball, surface, pole position and approach method for both magnets. Repeat the measurement for each magnet and calculate a mean distance. The magnet with the greater mean attraction distance has the stronger field under the controlled conditions.	Choose one response threshold and change only the magnet. A common test object and geometry make the distances comparable; repeats reduce random variation before the mean distances are ranked.	5
Total Question 3			5
04.1	The left end of the magnet is north because the field at X points away from it. The right end is south because the field at Y points towards it. At X the magnetic field is directed to the left. At Y the magnetic field is also directed to the left, into the south pole. The field is stronger at X because the field lines are closer together there.	Treat the compass north-seeking end as a local field-direction arrow. Outside a bar magnet the field leaves north and enters south; line density then provides the separate strength comparison.	5
Total Question 4			5
05.1	Field lines run from the north pole across the gap to the south pole. The field lines are approximately straight and parallel within the small gap. The magnetic field is directed from the north pole towards the south pole. The compass aligns with the field, with its north-seeking end pointing towards the south pole. The field is strong in the gap because the field lines are close together.	Join the facing unlike poles using the external north-to-south field convention. In the narrow central region the lines are nearly uniform, so a compass follows their direction and their close spacing represents a strong field.	5
Total Question 5			5

4.7.2.1 · Electromagnetism

Tier 1 · Easy

Mark scheme for 4.7.2.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The field lines are concentric circles centred on the wire. The magnetic field becomes weaker as the distance from the wire increases.	Represent the field in a plane perpendicular to the wire. Each field line circles the conductor, and increasing separation from the conductor reduces the field strength.	2
Total Question 1			2
02.1	A current in the wire produces a magnetic field around the wire. Opening the switch stops the current, so that magnetic field disappears and the compass returns to its original alignment.	Link the one changed condition, current on or off, to the compass response. Deflection only with current is evidence for the magnetic effect of a current.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.2.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The field lines are concentric circles centred on the wire. Point the right thumb upwards, in the current direction. Viewed from above, the curled fingers show an anticlockwise field.	Use the right-hand grip rule with the thumb along the upward current. From above, the fingers curl anticlockwise around the wire, matching the circular field lines.	3
Total Question 1			3
02.1	A to B shows that increasing current strengthens the field. B to C shows that increasing the number of turns strengthens the field. C to D shows that adding an iron core strengthens the field.	Compare only pairs in which one stated design feature changes. The larger compass deflection in each pair identifies the current, turn count and iron core as field-strengthening changes.	3
Total Question 2			3
03.1	The similar internal deflections show that the field inside is nearly uniform over the tested region. The much smaller external deflection shows that the field is stronger inside than at the tested outside position. The fields produced by the current in the many coil turns reinforce inside the solenoid.	Compare the spread of the three internal readings before comparing them with the outside reading. Link the strong, nearly uniform internal result to reinforcement by the fields of the turns.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.7.2.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Both coils have a strong, nearly uniform field inside and a bar-magnet-shaped external field. Y has the stronger field because its iron core becomes magnetised. Reversing current reverses the direction of the field made by every turn. The north and south poles therefore swap. The iron core is still easily magnetised, so it continues to strengthen the reversed field.	First compare like-for-like turn count and current: only the iron core differs, so Y is stronger. Current direction sets field direction by the grip rule, whereas the core controls the extra strength; reversing one does not remove the effect of the other.	5
Total Question 1			5
02.1	A permanent magnet could retain attraction after the current is switched off, so it conflicts with rapid release. Steel retains its magnetism or is not easily demagnetised, so the cans would not release; replace the steel core with iron. Remove the permanent magnet so that the core is magnetised by the solenoid only while current flows. The iron core loses most or all of its induced magnetism when the current stops, allowing the cans to be released. Increase the current to strengthen the electromagnet while it is on. Increase the number of solenoid turns as another way to strengthen the field.	Separate on-state strength from off-state retention. Remove the permanent magnet, replace the steel core with easily demagnetised iron, then strengthen the switched field through current and coil-turn changes.	6
Total Question 2			6
03.1	Current in the solenoid produces a magnetic field. A larger current produces a stronger field. The iron armature becomes magnetised by induction and is attracted towards the solenoid. Its movement opens the contacts and interrupts the circuit. When the fault current is removed, the field and most or all of the induced magnetism are lost, allowing the armature and contacts to reset.	Trace the device from electrical input to magnetic field, induced magnetism, force and mechanical switching. The temporary magnetism of iron explains the return after current stops.	5
Total Question 3			5
04.1	Viewed from above, the upward current is directed out of the page. The right-hand grip rule gives an anticlockwise magnetic field. The reading south of the wire is anomalous. At that position the north-seeking end should point east, not west. Doubling the current strengthens the field but does not change its direction. Reversing the current reverses the field direction, making it clockwise when viewed from above.	Map the anticlockwise tangent at each compass position before comparing the observations. Current magnitude controls strength, whereas current direction controls the sense of the circular field.	6
Total Question 4			6
05.1	The facing solenoids repel because their facing ends are like poles. Inside A the field is directed towards its north pole, so it points to the right. Reversing B's current reverses B's magnetic field and swaps its poles. B's facing end becomes south, so A and B then attract. Increasing the current in B would strengthen its field.	Use the bar-magnet field pattern for each solenoid, then change only B's current direction. Reversing current changes B's polarity, while increasing its current magnitude strengthens the field without changing its poles.	5
Total Question 5			5

4.7.2.2 · Fleming's left-hand rule (HT only)

Tier 1 · Easy

Mark scheme for 4.7.2.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The force is now downward. Reversing the current while keeping the field fixed reverses the motor-effect force.	Use Fleming's left-hand rule with the same first-finger field direction and the opposite second-finger current direction. The thumb then points in the opposite force direction.	2
Total Question 1			2
02.1	The conventional current is directed towards the east. With the first finger north and the thumb upwards, the second finger points east.	Set the first finger along the northward field and the thumb along the upward force. The mutually perpendicular second finger then gives an eastward conventional current.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.2.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$0.26\,\text{N}$	Use $F=BIl=0.45\times3.2\times0.18=0.2592\,\text{N}$ . To two significant figures, the force is $0.26\,\text{N}$ .	2
Total Question 1			2
02.1	Record C is inconsistent. Its corrected force is $0.15\,\text{N}$ .	Apply $F=BIl$ . A gives $0.20\times1.5\times0.40=0.12\,\text{N}$ and B gives $0.35\times2.0\times0.30=0.21\,\text{N}$ . C should give $0.50\times1.2\times0.25=0.150\,\text{N}$ , not $0.10\,\text{N}$ .	3
Total Question 2			3
03.1	The graph gradient is $0.072\,\text{N/A}$ . For $F=BIl$ , the gradient of force against current is $Bl$ . The active length is $0.12\,\text{m}$ .	Calculate the gradient as $0.216/3.0=0.072\,\text{N/A}$ . Since $F/I=Bl$ , rearrange to $l=0.072/0.60=0.12\,\text{m}$ .	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.7.2.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$0.45\,\text{T}$ The force becomes upward.	Rearrange $F=BIl$ to $B=F/(Il)=0.63/(4.0\times0.35)=0.45\,\text{T}$ . Fleming's rule shows that reversing one of current or field reverses the force, so the downward force becomes upward.	4
Total Question 1			4
02.1	$24\,\text{cm}=0.24\,\text{m}$ . The original current is $2.0\,\text{A}$ . The new force magnitude is $0.336\,\text{N}$ , or $0.34\,\text{N}$ to two significant figures. The new force is opposite to the original force.	Convert using $24/100=0.24\,\text{m}$ . Rearrange $F=BIl$ to $I=F/(Bl)=0.168/(0.35\times0.24)=2.0\,\text{A}$ . Doubling current doubles the unrounded force to $0.336\,\text{N}$ , which is $0.34\,\text{N}$ to two significant figures. Reversing only the field direction reverses the force direction.	5
Total Question 2			5
03.1	The load mass is $0.018\,\text{kg}$ . Its weight is $0.1764\,\text{N}$ . At balance, the motor-effect force has magnitude $0.1764\,\text{N}$ . The magnetic flux density is $0.588\,\text{T}$ , or $0.59\,\text{T}$ to two significant figures. The motor-effect force must act vertically upwards.	Convert the mass using $18/1000=0.018\,\text{kg}$ , then $W=mg=0.018\times9.8=0.1764\,\text{N}$ . At equilibrium $F=W$ . Rearrange $F=BIl$ to $B=0.1764/(2.5\times0.12)=0.588\,\text{T}$ , giving $0.59\,\text{T}$ to two significant figures.	5
Total Question 3			5
04.1	$I=Q/t=54/15=3.6\,\text{A}$ . $F=BIl$ because the wire is perpendicular to the field. $F=0.48\times3.6\times0.25=0.432\,\text{N}$ . The force is $0.43\,\text{N}$ to two significant figures. Reversing only the field reverses the force direction without changing its magnitude.	Use the charge-flow equation to obtain the current before substituting into the sheet equation $F=BIl$ . Fleming's left-hand rule then shows the effect of reversing only the field.	5
Total Question 4			5
05.1	$F=BIl=0.75\times2.4\times0.18=0.324\,\text{N}$ . The conductor's mass is $45\,\text{g}=0.045\,\text{kg}$ . Using $F=ma$ , $a=F/m=0.324/0.045=7.2\,\text{m/s}^2$ . Reversing the current reverses the motor-effect force once, and reversing the field reverses it a second time. The two reversals cancel, so the acceleration direction and magnitude are unchanged.	Calculate the magnetic force before using it as the resultant force in $F=ma$ . Treat the current and field reversals separately: each reverses the force, so doing both preserves its direction.	5
Total Question 5			5

4.7.2.3 · Electric motors (HT only)

Tier 1 · Easy

Mark scheme for 4.7.2.3 Tier 1 · Easy
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01.1	After half a turn the two sides of the coil have swapped positions, so without reversal the forces would turn the coil back the other way. Reversing the current reverses those forces, so the turning effect continues in the same rotational direction.	After half a turn the active sides of the coil have exchanged positions. Reversing the current also reverses their forces, so the pair of forces continues to turn the coil the same way.	2
Total Question 1			2
02.1	Continuous slip rings do not reverse the current every half-turn. The motor needs a split-ring commutator to reverse the coil current so the turning effect continues in the same rotational direction.	Track the coil through half a turn. Its sides exchange positions, so the current connections must swap at that point; continuous rings cannot perform that switching action.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.2.3 Tier 2 · Standard
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01.1	The two active sides carry currents in opposite directions. Each side experiences a motor-effect force in the magnetic field. The two forces act in opposite directions. Because they act on different sides of the axis, they form a turning effect on the coil.	Apply Fleming's left-hand rule to each side. Reversing current while keeping the field fixed reverses the force, so the separated forces form a couple and rotate the coil.	4
Total Question 1			4
02.1	The physical side now on the left is again forced upwards. The physical side now on the right is again forced downwards. The split-ring commutator reverses the current as the coil sides exchange positions. The forces therefore continue to produce a turning effect in the original rotational direction.	After half a turn, each conductor occupies the opposite side. The commutator reverses current and hence each conductor's motor-effect force, recreating the same up-left and down-right force pattern in space.	4
Total Question 2			4
03.1	The motor continues to rotate clockwise because reversing both current and field reverses the motor-effect force twice. Reversing only the current would reverse the rotation. Alternatively, reversing only the magnetic field would reverse the rotation.	Apply Fleming's left-hand rule one change at a time. Reversing either current or field reverses each force, but reversing both restores the original force directions; one reversal alone changes the turning direction.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.7.2.3 Tier 3 · Hard
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01.1	Increasing current increases the force on each active side of the coil. A stronger magnet gives a larger magnetic flux density and therefore a larger force. The larger forces produce a larger turning effect, tending to increase the rotation rate. After half a turn, each side of the coil has exchanged positions. The commutator reverses current so the force on each side also reverses. The turning effect therefore remains in the same rotational direction rather than reversing every half-turn.	Use $F=BIl$ qualitatively: raising $I$ or $B$ raises the forces and hence the turning effect. Track one side through half a rotation; because its position swaps, its current must also swap to keep its force driving the same sense of rotation.	6
Total Question 1			6
02.1	P has the smallest initial turning effect, Q is larger and R is largest. Q has more active wire sections in the field than P, so the total turning effect is greater. R also has the greater current, increasing the motor-effect force on each active section. If the split ring fails, the current does not reverse at the required time. The turning effect then reverses after the half-turn, so the coil tends to stop or rock rather than continue rotating in one direction.	Compare one design change at a time: P to Q isolates turn count, and Q to R isolates current. For the fault, follow a conductor into its exchanged position without current reversal; its force then acts in the wrong rotational sense.	5
Total Question 2			5
03.1	The force on each active side is $0.18\,\text{N}$ . The currents on the two sides run in opposite directions, so their motor-effect forces act in opposite directions. Each force acts $0.040\,\text{m}$ from the central axis. The total moment is $2\times0.18\times0.040=0.0144\,\text{N m}$ . The separated opposite forces form a turning effect on the loop.	Use $F=BIl=0.50\times2.0\times0.18=0.18\,\text{N}$ on each side. Each moment arm is half the $0.080\,\text{m}$ separation, so the two moments add to $2F\times0.040=0.0144\,\text{N m}$ .	5
Total Question 3			5
04.1	The electrical input power is $P=VI=12\times2.5=30\,\text{W}$ . The electrical energy supplied is $E=Pt=30\times4.0=120\,\text{J}$ . The useful gravitational energy transfer is $mgh=4.0\times9.8\times1.5=58.8\,\text{J}$ . The efficiency is $58.8/120=0.49$ , or $49\%$ . After each half-turn the active sides of the coil exchange positions. The split-ring commutator reverses the coil current so the motor-effect turning direction continues rather than reversing.	Compare the electrical input energy with the load's gain in gravitational potential energy. The energy calculation does not remove the commutation requirement: current must still reverse after each half-turn for continuous rotation.	6
Total Question 4			6
05.1	For the named single loop, $M=Fd$ with $d=0.10\,\text{m}$ . $F=M/d=0.020/0.10=0.20\,\text{N}$ on each active side. Using the sheet equation $F=BIl$ , $I=F/(Bl)$ . $I=0.20/(0.50\times0.25)=1.6\,\text{A}$ . Doubling the separation to $0.20\,\text{m}$ halves the required force to $F=0.020/0.20=0.10\,\text{N}$ . With $B$ and $l$ unchanged, the required current also halves to $I=0.10/(0.50\times0.25)=0.80\,\text{A}$ .	Work backwards from the required moment on the single loop using $M=Fd$ , then use $F=BIl$ from the sheet to find current. At fixed moment, doubling the force separation halves the force and therefore halves the current when field strength and active length are unchanged.	6
Total Question 5			6

4.7.2.4 · Loudspeakers (physics only) (HT only)

Tier 1 · Easy

Mark scheme for 4.7.2.4 Tier 1 · Easy
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01.1	A larger current amplitude gives larger cone vibrations and a louder sound. A higher current frequency gives faster cone vibrations and a higher-frequency sound.	The alternating current controls the reversing motor-effect force. Its amplitude controls the cone's displacement and loudness, while its frequency controls how rapidly the cone vibrates.	2
Total Question 1			2
02.1	A loudspeaker uses the motor effect. A varying or alternating current must be supplied to the coil.	Distinguish a driven output device from an inducing input device. Current supplied in a magnetic field produces the force that moves the loudspeaker coil.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.2.4 Tier 2 · Standard
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01.1	The signal current flows through a coil in a permanent magnetic field. The motor effect exerts a force on the coil. As the alternating current reverses, the force direction reverses. The attached cone therefore moves backwards and forwards, producing pressure variations in the air.	Follow the causal chain from alternating current to alternating motor-effect force, then to coil and cone vibration, and finally to air-pressure variations.	4
Total Question 1			4
02.1	Signal X has frequency $250\,\text{Hz}$ and signal Y has frequency $400\,\text{Hz}$ . Y produces the higher-pitched sound because it has the higher frequency. X produces the louder sound because its larger peak current produces a larger force and cone amplitude.	Use cycles divided by time: $f_X=5/0.020=250\,\text{Hz}$ and $f_Y=8/0.020=400\,\text{Hz}$ . Frequency sets pitch, while current amplitude controls motor-effect force and hence sound amplitude.	4
Total Question 2			4
03.1	A steady current produces a motor-effect force in one constant direction. After the initial movement, the cone does not keep moving backwards and forwards, so it does not make repeating pressure variations. The coil needs an alternating or varying current so that the force reverses and the cone vibrates.	Distinguish a one-off displacement from vibration. Continuous sound needs repeated pressure changes, which require the supplied current and motor-effect force to vary with time.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.7.2.4 Tier 3 · Hard
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01.1	Signal A produces the higher-pitched sound because the cone vibrates at a higher frequency. Signal A produces the quieter sound because its smaller current gives a smaller motor-effect force and cone amplitude.	The cone follows the frequency of the alternating force, so doubling signal frequency raises pitch. Force size depends on current, so the smaller current amplitude produces smaller pressure variations and hence lower loudness.	4
Total Question 1			4
02.1	P has peak force $0.064\,\text{N}$ . Q has peak force $0.16\,\text{N}$ . Q should be louder because it has the larger peak force and cone amplitude. Their pitches are the same because both signals have frequency $320\,\text{Hz}$ .	Use $F=BIl$ . For P, $F=0.40\times0.80\times0.20=0.064\,\text{N}$ . For Q, $F=0.50\times1.6\times0.20=0.160\,\text{N}$ . Force amplitude controls loudness, while the shared signal frequency fixes the same pitch.	5
Total Question 2			5
03.1	The force magnitude is $0.060\,\text{N}$ . The force reverses every $1.0\,\text{ms}$ . One full current cycle takes $2.0\,\text{ms}=0.0020\,\text{s}$ . The cone-vibration frequency is $500\,\text{Hz}$ . Reversing the current reverses the motor-effect force, driving the attached cone backwards and forwards.	Use $F=BIl=0.50\times0.75\times0.16=0.060\,\text{N}$ . A full cycle contains one positive and one negative interval, so $T=2.0\,\text{ms}=0.0020\,\text{s}$ and $f=1/T=500\,\text{Hz}$ .	5
Total Question 3			5
04.1	Using $F=BIl$ , $I=F/(Bl)$ . $I=0.060/(0.40\times0.25)=0.60\,\text{A}$ . $f=9/0.030=300\,\text{Hz}$ . The alternating current repeatedly reverses direction. The motor-effect force on the coil therefore reverses, moving the attached cone backwards and forwards. The cone produces pressure variations at $300\,\text{Hz}$ in the surrounding air.	Use the required force to obtain the current amplitude, and cycles per unit time to obtain frequency. Current reversal links the electrical signal to the alternating force and cone vibration.	6
Total Question 4			6

4.7.3.1 · Induced potential (physics only) (HT only)

Tier 1 · Easy

Mark scheme for 4.7.3.1 Tier 1 · Easy
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01.1	Move the magnet faster, giving a greater relative speed. Use a stronger magnet, or a coil with more turns.	Increase how quickly or how strongly the magnetic field through the conductor changes. Greater relative speed, field strength or turn count increases the induced potential difference.	2
Total Question 1			2
02.1	The voltmeter gives a negative pulse of the same magnitude if the speed and magnet strength are unchanged. Reversing the magnetic-field direction reverses the induced potential difference.	Only the pole, and hence field direction, is reversed. The generator effect therefore reverses the polarity while unchanged speed and field strength keep the magnitude the same.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.3.1 Tier 2 · Standard
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01.1	Insertion and withdrawal produce pulses in opposite directions. At equal speeds their magnitudes are equal, for the same motion range. Moving the magnet faster increases the magnitude. Using a stronger magnet or more coil turns also increases the magnitude.	Reversing the relative motion reverses the induced potential difference. Its size depends on how quickly the magnetic field changes and on field strength and turn count.	4
Total Question 1			4
02.1	$2.40\,\text{V}$ is anomalous. The mean at $0.20\,\text{m/s}$ is $0.81\,\text{V}$ . The mean at $0.40\,\text{m/s}$ is $1.60\,\text{V}$ to two decimal places. Doubling the relative speed approximately doubles the induced potential difference in these results.	The first mean is $(0.82+0.80+0.81)/3=0.81\,\text{V}$ . Excluding $2.40\,\text{V}$ , the second is $(1.61+1.59)/2=1.60\,\text{V}$ . Their ratio is $1.60/0.81=1.98$ to two decimal places, so the faster value is about twice the slower one.	4
Total Question 2			4
03.1	The changing magnetic field through the coil induces a potential difference across it. An appreciable current does not flow because the circuit is effectively open or has very high resistance. Connecting the coil in a complete low-resistance circuit would allow the induced potential difference to drive a current.	Separate induction of potential difference from the condition for current. Relative motion produces the voltage; a complete conducting path is additionally required for appreciable current.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.7.3.1 Tier 3 · Hard
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01.1	The changing magnetic field induces a current in the complete coil circuit. That current produces its own magnetic field. The induced field opposes the magnet's motion or the change producing it. Pulling the magnet out reverses the change, so the induced current reverses. Faster motion produces a larger induced potential difference and current. The stronger opposing magnetic effect means more mechanical energy is transferred each second.	Apply the opposition principle to the change, not merely to the magnet's field. Reversing motion reverses the change and current. Increasing speed raises the induction rate, so the opposing force and mechanical power input rise.	6
Total Question 1			6
02.1	A to B shows that more coil turns increase the induced potential difference. B to C shows that a stronger magnetic field increases the induced potential difference. C to D shows that greater relative speed increases the induced potential difference. Each comparison changes only one stated factor. The reading becomes zero when the magnet pauses. A stationary magnet gives no continuing change of magnetic field through the coil, so no potential difference is induced.	Use A-B, B-C and C-D as one-variable comparisons. Each larger reading follows a faster or larger change of magnetic field through the conductor; pausing removes that change even though the magnet remains inside the coil.	6
Total Question 2			6
03.1	Closing the switch makes the current and magnetic field around P increase, so the changing field induces a pulse in Q. Once the current in P is steady, its magnetic field is unchanging and Q reads zero. Opening the switch makes P's field decrease. The opposite change in field induces a pulse of opposite polarity in Q. Adding more turns to Q would increase the magnitude of the induced pulses.	Track change rather than field presence. The field grows at switch-on, stays constant, then collapses at switch-off; the generator effect acts only during the changes and reverses polarity when the change reverses.	5
Total Question 3			5
04.1	Entering and leaving change the magnetic field through the coil in opposite ways. The induced potential difference therefore reverses polarity, giving pulses on opposite sides of zero. The magnet leaves faster than it enters, so the magnetic field changes more quickly during the exit. This explains why the exit pulse has the greater magnitude, $1.5\,\text{V}$ rather than $0.90\,\text{V}$ . A stronger magnet would produce larger-magnitude entry and exit pulses at the same speeds, without changing which pulse is positive or negative.	Separate direction from magnitude. Reversing the field change reverses pulse polarity, while a greater relative speed or stronger field increases the rate of change and hence the induced-potential magnitude.	5
Total Question 4			5
05.1	$I=V/R=2.4/8.0=0.30\,\text{A}$ . $P=VI=2.4\times0.30=0.72\,\text{W}$ . $E=Pt=0.72\times0.50=0.36\,\text{J}$ . The induced current produces its own magnetic field. That field opposes the change or motion producing the current. The person must work against this resisting effect, supplying the mechanical energy transferred to the circuit.	Follow the energy chain after calculating $I$ , $P$ and $E$ from sheet equations. The opposing induced magnetic effect accounts for the required mechanical input rather than creating electrical energy without a source.	6
Total Question 5			6

4.7.3.2 · Uses of the generator effect (physics only) (HT only)

Tier 1 · Easy

Mark scheme for 4.7.3.2 Tier 1 · Easy
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01.1	After each half-turn, each side of the coil moves through the magnetic field in the opposite direction. Reversing the relative motion reverses the induced potential difference.	Track one side of the coil. Its motion across the field reverses after half a rotation, so the generator effect induces the opposite polarity and the output alternates.	2
Total Question 1			2
02.1	The device uses a split-ring commutator. It produces a varying direct potential difference, or dc, because the polarity does not reverse.	Use the trace polarity as the evidence. A split ring swaps the external connections every half-turn, keeping all pulses on one side of zero.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.3.2 Tier 2 · Standard
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01.1	Frequency $=5\,\text{Hz}$ . Period $=0.20\,\text{s}$ . Doubling rotation rate doubles the frequency and halves the period.	One complete rotation produces one complete ac cycle, so $f=5\,\text{Hz}$ . Then $T=1/f=1/5=0.20\,\text{s}$ . At twice the rotation rate, $f=10\,\text{Hz}$ and $T=0.10\,\text{s}$ .	4
Total Question 1			4
02.1	The period is $20\,\text{ms}=0.020\,\text{s}$ . The frequency is $50\,\text{Hz}$ . The peak potential difference is $6\,\text{V}$ .	One complete positive-and-negative cycle takes $20\,\text{ms}$ . Convert using $20/1000=0.020\,\text{s}$ , then $f=1/T=1/0.020=50\,\text{Hz}$ . The greatest magnitude in the table is $6\,\text{V}$ .	4
Total Question 2			4
03.1	The instantaneous potential difference at A is zero. Moving parallel to the field lines gives no cutting of field lines at that instant. The magnitude of the potential difference at B is maximum. At B the conductors cut the field lines at the greatest rate.	Judge the rate at which the conductor cuts field lines at each orientation. Zero cutting gives a zero crossing; the greatest cutting rate gives a peak magnitude.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.7.3.2 Tier 3 · Hard
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01.1	P has an alternating graph with equal positive and negative half-cycles. Q has pulses on one side of zero, so its polarity does not reverse. Both outputs vary from zero to a maximum magnitude as the coil rotates. Q's split ring swaps the external connections every half-turn. That swap reverses the connection exactly when the coil's induced potential reverses, keeping the external polarity unchanged.	Begin with the same alternating induced potential in each rotating coil. Slip rings pass that reversal to P. Q's split ring swaps connections at each half-turn, rectifying the external output into varying dc pulses.	5
Total Question 1			5
02.1	N should have the larger output amplitude because it has more turns, a stronger field and a greater rotation speed. N should have the higher pulse repetition rate because it rotates faster and its split ring produces a pulse every half-turn. M produces alternating potential difference with positive and negative half-cycles. Its slip rings pass the changing coil polarity to the external circuit. N produces varying dc pulses that remain on one side of zero. Its split-ring commutator swaps the external connections every half-turn.	Separate trace size, repetition rate and polarity. Turns, field strength and rotation speed affect amplitude; rotation rate sets M's ac frequency and, with the split ring, N's pulse repetition rate; the ring arrangement decides whether the coil's reversal appears externally as ac or is rectified to varying dc.	6
Total Question 2			6
03.1	The mechanical input power is $6.0\,\text{W}$ . The efficiency is $0.75$ , or $75\%$ . The power dissipated is $1.5\,\text{W}$ . The induced current produces its own magnetic field that opposes the change or motion producing it. The cyclist must therefore provide a force and transfer mechanical energy to keep the generator rotating.	Use $P_{\text{in}}=E/t=24/4.0=6.0\,\text{W}$ . Then efficiency $=P_{\text{out}}/P_{\text{in}}=4.5/6.0=0.75=75\%$ , and dissipated power $=6.0-4.5=1.5\,\text{W}$ . Link the extra effort to the opposing effect of the induced current.	5
Total Question 3			5
04.1	The initial frequency is $f=25/0.50=50\,\text{Hz}$ . The initial period is $T=1/f=1/50=0.020\,\text{s}$ . The new frequency is $f=40/0.50=80\,\text{Hz}$ . The new period is $T=1/80=0.0125\,\text{s}$ . The cycles are closer together in time because the period is shorter. The graph has a larger peak potential difference because the coil cuts magnetic field lines at a greater rate.	For this alternator, one rotation produces one output cycle, so rotations per second give frequency and $T=1/f$ gives period. Faster rotation shortens the time scale and increases the rate at which the coil cuts magnetic field lines, increasing the peak induced potential difference.	6
Total Question 4			6
05.1	$Q=It=0.40\times30=12\,\text{C}$ . $E=Pt=2.4\times30=72\,\text{J}$ . The induced potential difference in the rotating coil reverses every half-turn. The split-ring commutator swaps the external connections at the same time, so the external polarity stays in one direction although its magnitude varies.	Use the stated mean current and mean power over the interval, then distinguish varying d.c. from steady d.c. The split ring preserves the external polarity even though the induced potential difference and its magnitude vary as the coil rotates.	4
Total Question 5			4

4.7.3.3 · Microphones (physics only) (HT only)

Tier 1 · Easy

Mark scheme for 4.7.3.3 Tier 1 · Easy
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01.1	The electrical signal has a greater amplitude. Its frequency is unchanged because the pitch is unchanged.	Greater sound amplitude drives larger diaphragm and coil motion, increasing the induced signal amplitude. The unchanged sound frequency keeps the signal frequency unchanged.	2
Total Question 1			2
02.1	The sound itself moves the diaphragm and attached coil relative to the magnetic field. The generator effect induces the electrical signal, so a driving battery is not required.	Follow the microphone's input-to-output direction. Mechanical motion supplied by the sound changes the field through the coil and induces a potential difference.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.3.3 Tier 2 · Standard
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01.1	Sound-pressure variations make the diaphragm vibrate. The attached coil moves relative to a magnetic field. The changing magnetic field through the coil induces a potential difference by the generator effect. A complete circuit carries a current that varies with the sound.	Trace energy and cause in order: sound moves the diaphragm, the diaphragm moves the coil through the field, and the generator effect produces the varying electrical signal.	4
Total Question 1			4
02.1	Trace A has frequency $500\,\text{Hz}$ and trace B has frequency $250\,\text{Hz}$ . A came from the higher-pitched sound because its frequency is greater. B came from the louder sound because its signal amplitude is greater.	Convert the periods: $T_A=2.0\,\text{ms}=0.0020\,\text{s}$ and $T_B=4.0\,\text{ms}=0.0040\,\text{s}$ . Then $f_A=1/T_A=1/0.0020=500\,\text{Hz}$ and $f_B=1/T_B=1/0.0040=250\,\text{Hz}$ . Microphone signal frequency follows pitch, while signal amplitude follows sound amplitude or loudness.	4
Total Question 2			4
03.1	Use the signal generator to drive the loudspeaker at several known frequencies and record the microphone output on the oscilloscope. Keep the microphone-loudspeaker distance and the signal amplitude or volume setting constant. Measure the period of several microphone cycles for each setting and calculate frequency using $f=1/T$ . Compare each calculated microphone frequency with the corresponding signal-generator frequency, repeating readings where possible.	Vary only input frequency, measure the output period over several cycles, and compare input with output. Fixed geometry and amplitude isolate the frequency response; repeat readings allow a mean and reduce the effect of random error.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.7.3.3 Tier 3 · Hard
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01.1	P produces the larger signal amplitude because the louder sound drives a larger diaphragm and coil vibration than Q. R produces the higher signal frequency because its higher pitch makes the diaphragm vibrate more frequently than S. Moving the coil relative to the field induces the potential difference. Faster change of magnetic field gives a larger instantaneous induced potential difference.	Use the controlled pairs separately: loudness changes vibration and signal amplitude when pitch is fixed, while pitch changes vibration and signal frequency when loudness is fixed. The generator effect converts the coil's motion through the field into voltage.	5
Total Question 1			5
02.1	Increasing the coil from 200 to 400 turns increases the peak signal from $24\,\text{mV}$ to $47\,\text{mV}$ . Using a stronger magnet increases it further from $47\,\text{mV}$ to $81\,\text{mV}$ . Both changes make the microphone more sensitive because the same sound produces a larger electrical output. The tone fixes how often the diaphragm and coil vibrate each second. The generator effect changes signal amplitude with coil and field design, but it does not change that vibration frequency.	Use controlled comparisons: the first pair changes turns only and the second changes field strength only. Those factors change induced-voltage amplitude; the unchanged input tone fixes the common repetition rate.	5
Total Question 2			5
03.1	M is a moving-coil microphone. Motion of M's diaphragm moves its coil relative to the magnetic field, so the generator effect induces a potential difference. M transfers energy from sound or mechanical vibration to an electrical signal. N is a moving-coil loudspeaker. The supplied alternating current in N's coil experiences a reversing force by the motor effect, making the diaphragm vibrate. N transfers electrical energy to mechanical vibration and sound in the surrounding air.	Classify by cause and output. Motion producing voltage is the generator effect and identifies a microphone; supplied current producing motion is the motor effect and identifies a loudspeaker.	6
Total Question 3			6
04.1	The trace amplitude decreases at $2.0\,\text{m}$ because the sound is quieter there. The quieter sound makes the diaphragm and attached coil vibrate with a smaller amplitude. The smaller coil motion changes the magnetic field through the coil more slowly, inducing a smaller potential difference. The spacing between cycles is unchanged because the note's frequency is unchanged. Using a coil with more turns or a stronger permanent magnet could restore the trace amplitude.	Separate amplitude from frequency on the trace. Greater distance reduces the sound amplitude and hence the coil motion and induced-potential amplitude, but the same note retains its frequency; increasing turns or field strength increases the induced output.	5
Total Question 4			5
05.1	M produces no induced signal because its detached coil does not move relative to the magnetic field. N can have an induced potential difference across the coil because the coil still moves through the field. N has no current in the external circuit because that circuit is incomplete. P produces a smaller induced-potential amplitude because the same coil motion occurs in a weaker magnetic field. P's signal frequency is unchanged because the same sound fixes the diaphragm and coil vibration frequency.	Test each link in the microphone chain independently: coil motion is required for induction, a complete circuit is required for current, field strength affects amplitude, and the input vibration fixes frequency.	5
Total Question 5			5

4.7.3.4 · Transformers (physics only) (HT only)

Tier 1 · Easy

Mark scheme for 4.7.3.4 Tier 1 · Easy
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01.1	A steady direct current produces an unchanging magnetic field in the core. Without a changing magnetic field, no continuous potential difference is induced in the secondary.	A transformer relies on the generator effect. Alternating current continually changes the core field; steady direct current does not, apart from a brief change when switched.	2
Total Question 1			2
02.1	The alternating primary current produces a changing magnetic field in the iron core. That changing field passes through the secondary coil and induces a potential difference across it.	Trace the coupling through the core rather than through a wire connection: alternating current creates a changing core field, and the generator effect acts on the secondary coil.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.3.4 Tier 2 · Standard
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01.1	$15.3\,\text{V}$	Use $V_{\mathrm{p}}/V_{\mathrm{s}}=n_{\mathrm{p}}/n_{\mathrm{s}}$ . Thus $V_{\mathrm{s}}=V_{\mathrm{p}}n_{\mathrm{s}}/n_{\mathrm{p}}=230\times80/1200=15.3\,\text{V}$ .	3
Total Question 1			3
02.1	Row R is inconsistent. Its corrected secondary potential difference is $11.5\,\text{V}$ .	For an ideal transformer, $V_s=V_pn_s/n_p$ . P gives $240\times40/800=12\,\text{V}$ and Q gives $240\times150/600=60\,\text{V}$ . R gives $230\times50/1000=11.5\,\text{V}$ , not $13.8\,\text{V}$ .	3
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03.1	The normal output power is $12\times5.0=60\,\text{W}$ , giving primary current $60/240=0.25\,\text{A}$ . The fuse does not melt during normal operation because $0.25\,\text{A}<0.40\,\text{A}$ . The fault output power is $12\times10.0=120\,\text{W}$ , giving primary current $120/240=0.50\,\text{A}$ . The fuse melts during the fault because $0.50\,\text{A}>0.40\,\text{A}$ .	For an ideal transformer, input power equals output power. Normally, $P=12\times5.0=60\,\text{W}$ and $I_p=60/240=0.25\,\text{A}$ . During the fault, $P=12\times10.0=120\,\text{W}$ and $I_p=120/240=0.50\,\text{A}$ , which exceeds the $0.40\,\text{A}$ fuse rating by $0.10\,\text{A}$ .	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.7.3.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Secondary current $=1.50\,\text{A}$ . Primary current $=75.0\,\text{A}$ . $n_{\mathrm{s}}:n_{\mathrm{p}}=50:1$ . The smaller transmission current reduces heating losses in cables.	Use $P=VI$ . The secondary current is $I_{\mathrm{s}}=18000/12000=1.50\,\text{A}$ and, for equal ideal power, $I_{\mathrm{p}}=18000/240=75.0\,\text{A}$ . The turn ratio equals the potential-difference ratio: $n_{\mathrm{s}}/n_{\mathrm{p}}=12000/240=50$ . For the same transmitted power, a larger potential difference means a smaller current, reducing cable heating.	6
Total Question 1			6
02.1	Output power $=144\,\text{W}$ . Input power $=180\,\text{W}$ . Primary current $=0.75\,\text{A}$ . The secondary has 150 turns. Power dissipated $=36\,\text{W}$ .	First $P_{\mathrm{out}}=V_sI_s=24\times6.0=144\,\text{W}$ . From $0.80=P_{\mathrm{out}}/P_{\mathrm{in}}$ , $P_{\mathrm{in}}=144/0.80=180\,\text{W}$ . Then $I_p=P_{\mathrm{in}}/V_p=180/240=0.75\,\text{A}$ . The ratio gives $n_s=n_pV_s/V_p=1500\times24/240=150$ . Finally, $180-144=36\,\text{W}$ is dissipated.	6
Total Question 2			6
03.1	The maximum cable current is $I_{\max}=\sqrt{(40\times10^3)/4.0}=100\,\text{A}$ . The minimum transmission potential difference is $V_{\min}=(2.0\times10^6)/100=2.0\times10^4\,\text{V}$ , or $20\,\text{kV}$ . The minimum secondary-to-primary turn ratio is $20000/500=40:1$ . Using at least this potential difference keeps the current no greater than $100\,\text{A}$ and therefore keeps $I^2R$ heating within the limit.	Use $P_{\text{loss}}=I^2R$ : $I_{\max}=\sqrt{(40\times10^3)/4.0}=100\,\text{A}$ . For transmitted power, $V_{\min}=P/I_{\max}=2.0\times10^6/100=2.0\times10^4\,\text{V}$ . The ideal turn ratio equals the voltage ratio, so $n_s/n_p=20000/500=40$ .	6
Total Question 3			6
04.1	$n_{\mathrm{s}}/n_{\mathrm{p}}=V_{\mathrm{s}}/V_{\mathrm{p}}=230/11000$ , so $n_{\mathrm{s}}=230$ turns. $P_{\mathrm{out}}=V_{\mathrm{s}}I_{\mathrm{s}}=230\times44=10\,120\,\text{W}$ . For an ideal transformer, $V_{\mathrm{p}}I_{\mathrm{p}}=V_{\mathrm{s}}I_{\mathrm{s}}$ . $I_{\mathrm{p}}=10120/11000=0.92\,\text{A}$ . It is step-down because the secondary potential difference ( $230\,\text{V}$ ) is lower than the primary ( $11\,\text{kV}$ ); this gives a value that is safe for use in homes.	Use the sheet turn equation with volts on both sides, calculate secondary power, then conserve power with $V_{\mathrm{p}}I_{\mathrm{p}}=V_{\mathrm{s}}I_{\mathrm{s}}$ . Compare the input and output potential differences to identify the transformer type.	5
Total Question 4			5
05.1	At $15\,\text{kV}$ , $I=P/V=600000/15000=40\,\text{A}$ . The cable-heating power is $I^2R=40^2\times5.0=8000\,\text{W}$ , or $8.0\,\text{kW}$ . At $3.0\,\text{kV}$ , the current would be $600000/3000=200\,\text{A}$ . The cable-heating power would be $200^2\times5.0=200000\,\text{W}$ , or $200\,\text{kW}$ . The current is $5$ times larger, so the heating is $5^2=25$ times larger because cable heating is proportional to $I^2$ .	Calculate current from transmitted power at each voltage, then use $P_{\mathrm{heating}}=I^2R$ . Comparing the currents predicts the heating ratio directly: a factor of $5$ in current produces a factor of $25$ in heating.	5
Total Question 5			5