ExaminerPrep › Question bank › IGCSE Physics (Edexcel) › 6 Magnetism and electromagnetism
International GCSE Physics · Pearson Edexcel 4PH1 · Topic 6
123 free, high-quality past-paper-style questions on Magnetism and electromagnetism for Edexcel IGCSE Physics, each with a full mark scheme. Covers: (b) Magnetism; (d) Electromagnetic Induction; (c) Electromagnetism. For when you have finished the past papers, or want to practise one topic at a time.Notes 6(b) →Notes 6(c) →Notes 6(d) →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
A thin copper wire hangs between the poles of a horseshoe magnet in a model of an electric bell. When there is a current in the wire, the wire is pushed sideways out of the magnet.
Which change would make the wire be pushed in the opposite direction?
| Marking point | Mark | Notes |
|---|---|---|
| The direction of the force depends on the directions of both the current and the magnetic field. Reversing only one of them reverses the force, so reversing the current pushes the wire the other way. A larger current only makes the force bigger. Reversing both the current and the field reverses the force twice, so its direction is unchanged. A wire parallel to the field has no force on it at all. | ✓ 1 | Answer B |
Answer: B
Common error: Reversing the current and the field together gives two reversals, so the force ends up in the same direction as before.
A geologist uses a bar magnet to test rock samples for iron ore.
Which statement about the magnetic field lines around the bar magnet is correct?
| Marking point | Mark | Notes |
|---|---|---|
| By convention, a field line shows the direction of the force on a north pole. A north pole is repelled by the magnet's N pole and attracted to its S pole, so outside the magnet the lines run from N to S. The field is strongest where the lines are closest together, not furthest apart. Field lines never cross, because the field can only have one direction at any point. | ✓ 1 | Answer C |
Answer: C
Common error: Spacing shows strength: closer lines mean a stronger field. Students often reverse this relationship.
A crane in a scrap yard uses a large electromagnet to lift old cars. The electromagnet is a coil of insulated copper wire wound around an iron core.
Which change would make the electromagnet stronger?
| Marking point | Mark | Notes |
|---|---|---|
| The strength of an electromagnet increases with the number of turns, the size of the current and the use of an iron core. Adding more turns gives a stronger field, so C is correct. Reversing the current only swaps the poles and leaves the strength unchanged. Copper is not a magnetic material, so a copper core would make the magnet much weaker. Reducing the current weakens the field. | ✓ 1 | Answer C |
Answer: C
Common error: Reversing the current swaps the north and south poles. It does not change how strong the electromagnet is.
A security door in a hospital is held shut by an electromagnetic lock. An electromagnet fixed to the door frame attracts an iron plate fixed to the door. The electromagnet is a coil of wire wound around a soft iron core.
State two changes that would make the electromagnet hold the door more strongly.
The door is designed to open automatically during a power cut. Explain why the door can open when the current is switched off.
The electromagnet operates at 12 V and draws a current of 0.50 A. Calculate the energy it transfers if it stays switched on for 24 hours.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Any two different changes, one mark each, maximum 2: | — | |
| increase the current in the coil, for example by increasing the p.d. across it | ✓ 1 | |
| increase the number of turns on the coil | ✓ 1 | |
| Also accept as either mark: use a thicker (wider) soft iron core, or reduce the gap between the core and the plate. | — | |
| Increasing the current and increasing the p.d. count as one change only. | — | |
| Part (b) | ||
| With no current there is no magnetic field from the coil, and the soft iron core loses its magnetism . | ✓ 1 | |
| The iron plate is therefore no longer attracted, so the door can be pushed open . | ✓ 1 | |
| Part (c) | ||
| Use E = I × V × t . | ✓ 1 | |
| t = 24 × 3600 = 86 400 s . | ✓ 1 | |
| E = 0.50 × 12 × 86 400 = 5.2 × 105 J (518 400 J) . | ✓ 1 | |
Common error: Always convert the time into seconds before using E = IVt. Using 24 directly gives an answer 3600 times too small.
The main magnet in an MRI scanner is a very long solenoid. The patient lies inside the solenoid.
Describe the magnetic field inside a long solenoid carrying a steady current.
State the effect on the magnetic field of reversing the direction of the current in the solenoid.
Steel objects such as oxygen cylinders are not allowed in the scanner room. Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The field lines are straight and parallel to the axis of the solenoid . | ✓ 1 | |
| The lines are equally spaced, so the field is uniform and has the same strength throughout . | ✓ 1 | |
| Part (b) | ||
| The direction of the field reverses, so the north and south poles swap . | ✓ 1 | |
| Part (c) | ||
| Steel is a magnetic material, so the field induces magnetism in it . | ✓ 1 | |
| The object is then strongly attracted towards the scanner and could fly into it and injure someone . | ✓ 1 | |
Exam tip: To get full marks when describing a uniform field, mention both that the lines are parallel and that they are equally spaced.
A bicycle dynamo has a small magnet that spins inside a fixed coil when the wheel turns. The dynamo powers the bicycle's front lamp.
Which change would increase the size of the voltage induced in the coil?
| Marking point | Mark | Notes |
|---|---|---|
| The induced voltage depends on how quickly the magnetic field through the coil changes. Spinning the magnet faster makes the field change more rapidly, so a larger voltage is induced. Reversing the rotation only reverses the direction of the induced voltage, not its size. Fewer turns gives a smaller voltage. An unmagnetised iron cylinder has no magnetic field of its own, so it would induce no voltage. | ✓ 1 | Answer C |
Answer: C
Common error: Changing the direction of motion changes the direction of the induced voltage, not its size.
A student investigates how the number of turns on a coil affects the peak voltage induced when a magnet falls through it. She drops the same magnet from the same height each time.
Her results:
50 turns: 0.12 V
100 turns: 0.24 V
150 turns: 0.36 V
200 turns: 0.48 V
State the independent variable in this investigation.
Describe the relationship between the number of turns and the peak induced voltage.
Predict the peak voltage for a coil with 250 turns.
State two variables the student kept the same to make this a fair test.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The number of turns on the coil . | ✓ 1 | |
| Part (b) | ||
| The peak voltage is directly proportional to the number of turns, for example doubling the turns doubles the voltage . | ✓ 1 | |
| Part (c) | ||
| 0.60 V | ✓ 1 | |
| Part (d) | ||
| First valid control variable, e.g. the same magnet, so the field strength is the same . | ✓ 1 | |
| Second valid control variable, e.g. the same drop height, so the magnet's speed is the same . | ✓ 1 | |
| Accept any two from: same magnet or magnet strength; same drop height; same orientation of the magnet; same coil diameter or area. | — | |
Exam tip: To claim 'directly proportional', show that the ratio stays constant. Here voltage ÷ turns = 0.0024 V per turn every time.
A sorting machine uses a magnet to pull metal parts off a conveyor belt. The parts must be attracted to the magnet, but they must not stay magnetised after they drop off.
Which material should the parts be made from?
| Marking point | Mark | Notes |
|---|---|---|
| Soft iron is a magnetic material. It becomes an induced magnet when it is near a magnet, and it loses its magnetism easily when it is removed. Steel is also attracted, but it keeps its magnetism and would become a permanent magnet. Aluminium and copper are metals but are not magnetic materials, so a magnet does not attract them. | ✓ 1 | Answer B |
Answer: B
Common error: Not all metals are magnetic. Of the common metals, only iron, steel, nickel and cobalt are attracted to a magnet.
A recycling centre uses a crane with a large electromagnet to lift steel cans out of a pile of mixed waste. The pile also contains aluminium drinks cans.
State why the electromagnet does not lift the aluminium cans.
Explain why an electromagnet is used rather than a permanent magnet.
State two changes that would make the electromagnet stronger.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Aluminium is not a magnetic material / is not attracted to magnets | ✓ 1 | |
| Part (b) | ||
| An electromagnet can be switched off (by turning off the current) | ✓ 1 | |
| so the steel cans can be dropped/released where they are needed | ✓ 1 | |
| Part (c) | ||
| Acceptable changes: increase the current; increase the number of turns on the coil; add a (soft) iron core / use a larger iron core. | — | |
| One valid change from the list | ✓ 1 | |
| A second, different valid change from the list | ✓ 1 | |
Common error: 'Use a bigger battery' is too vague to earn credit. Say 'increase the current' or 'increase the voltage' instead.
A student wants to map the magnetic field around a bar magnet. She places the magnet on a sheet of paper and has a small plotting compass and a pencil.
Describe how she can use the plotting compass to draw one field line.
State what the spacing of the field lines shows about the magnetic field.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Place the compass near one pole of the magnet and mark a dot at the position of the north end of the compass needle | ✓ 1 | |
| Move the compass so that the south end (tail) of the needle is on the last dot, then mark a new dot at the north end; repeat until the other pole is reached | ✓ 1 | |
| Join the dots with a smooth curve and add an arrow pointing from the magnet's N pole towards its S pole | ✓ 1 | |
| Part (b) | ||
| The closer together the lines, the stronger the magnetic field (or: wider spacing means a weaker field) | ✓ 1 | |
Exam tip: Outside a magnet, field-line arrows point from the north pole to the south pole. This is the direction the north end of a plotting compass needle points.
A solenoid is used inside an electric door chime. The solenoid is a coil of wire, and a current passes through it.
Describe the shape of the magnetic field inside and outside the solenoid.
When end X of the solenoid is viewed head-on, the current flows anticlockwise. State which pole forms at end X.
State the effect of placing a soft iron rod inside the solenoid.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Inside: strong, uniform field with parallel, equally spaced lines along the axis | ✓ 1 | |
| Outside: similar to the field of a bar magnet | ✓ 1 | |
| Part (b) | ||
| North pole | ✓ 1 | |
| Part (c) | ||
| The magnetic field becomes (much) stronger | ✓ 1 | |
Exam tip: To find the poles of a solenoid, look at one end. Current flowing anticlockwise gives a North pole, and clockwise gives a South pole.
A tidal stream turbine is fixed to the sea bed. Moving water turns blades, which rotate a coil of wire inside a magnetic field to generate electricity. The ends of the coil are connected to the output cable through slip rings.
State two ways the induced voltage from the generator could be increased.
Explain why the output of this generator is alternating current (a.c.).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Any two of: rotate the coil faster; use stronger magnets; use more turns of wire on the coil; use a coil with a larger area. | — | |
| First correct way | ✓ 1 | |
| Second correct way | ✓ 1 | |
| Part (b) | ||
| Each side of the coil moves one way through the magnetic field, then the opposite way, every half turn (accept: the flux linkage of the coil increases then decreases, or the field lines are cut in alternating directions) | ✓ 1 | |
| So the direction of the induced voltage / current reverses every half turn, and the slip rings keep the coil connected to the same output terminals, so the output is not rectified | ✓ 1 | |
Common error: A stronger tide spins the coil faster, but the answer must name the physics factor, such as the speed of rotation, not 'more water'.
A long straight cable carries a steady current to a water pumping station. Which statement describes the magnetic field around the cable?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: the field forms concentric circles around the wire and weakens with distance, so the lines spread out. A describes the field of a solenoid inside, not a wire. C ignores that the field gets weaker with distance. D confuses the pattern with an electric field around a point charge. | ✓ 1 | Answer B |
Answer: B
Common error: The field around a straight wire is circular, not radial, and it is strongest closest to the wire.
A student places a pair of magnets on a top-pan balance. A straight horizontal wire is clamped so that 0.050 m of it lies between the poles, where the magnetic flux density is 0.15 T. The wire is at right angles to the field. A current of 4.0 A is passed through the wire. Use g = 10 N/kg.
Explain why the reading on the balance changes when the current is switched on.
Calculate the force on the wire. Use the equation F = BIl.
Calculate the change in the balance reading in grams.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The wire carrying a current in a magnetic field experiences a force | ✓ 1 | |
| there is an equal and opposite force on the magnets, which changes the force on the balance | ✓ 1 | |
| Part (b) | ||
| F = BIl = 0.15 × 4.0 × 0.050 | ✓ 1 | |
| F = 0.030 N | ✓ 1 | |
| Part (c) | ||
| mass = F ÷ g = 0.030 ÷ 10 (ECF from (b)) | ✓ 1 | |
| 0.0030 kg = 3.0 g | ✓ 1 | |
Exam tip: Balances read in grams: convert to kilograms before using weight = mass × g.
A long straight cable carries a current of 25 A to a pump in a water supply station.
Describe the pattern of the magnetic field around the cable.
State what happens to the strength of the field 5 cm from the cable when the current is doubled.
A plotting compass is placed beside the cable. Explain why the needle turns round when the current is reversed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the field lines are concentric circles centred on the cable | ✓ 1 | |
| the lines are closer together near the cable, where the field is stronger | ✓ 1 | |
| Part (b) | ||
| the field strength increases (it doubles) | ✓ 1 | |
| Part (c) | ||
| reversing the current reverses the direction of the magnetic field | ✓ 1 | |
| the compass needle lines up with the field, so it points the opposite way | ✓ 1 | |
Exam tip: Magnetic field direction is the direction a north pole would be pushed, which is the way a compass needle points.
A recycling plant uses an electromagnet to lift steel cans from a conveyor belt. Which pair of changes would both make the electromagnet stronger?
| Marking point | Mark | Notes |
|---|---|---|
| C is correct: more turns and a larger current each increase the field strength. A reduces the turns, which weakens the field. B reduces the current, which weakens the field. D removes the iron core, which greatly reduces the field. | ✓ 1 | Answer C |
Answer: C
Exam tip: Remember the three ways to strengthen an electromagnet: more turns, larger current, and an iron core.
A student connects a coil of wire to a sensitive voltmeter and pushes a bar magnet into the coil.
State what the student sees on the voltmeter while the magnet is moving into the coil.
State two ways the student could increase the voltage induced.
Explain why the voltmeter reads zero when the magnet is held still inside the coil.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A small reading or deflection on the voltmeter / a voltage is induced | ✓ 1 | |
| Part (b) | ||
| Any two from: move the magnet faster; use a stronger magnet; use a coil with more turns | ✓ 1 | |
| Second different correct way | ✓ 1 | |
| Part (c) | ||
| The magnetic field through the coil is not changing / no field lines are being cut | ✓ 1 | |
| So no voltage is induced | ✓ 1 | |
Exam tip: Induction needs a change: the magnet and coil must move relative to each other for a voltage to appear.
A student investigates the magnetic field around a bar magnet using a plotting compass.
State the direction of the magnetic field lines outside a bar magnet.
Describe how the student can use a plotting compass to plot a field line.
Describe how the field lines show where the magnetic field is strongest.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| From the north pole to the south pole | ✓ 1 | |
| Part (b) | ||
| Place the compass near the magnet and mark the position of the compass needle ends with dots | ✓ 1 | |
| Move the compass so one end is at the previous dot, repeat, then join the dots to draw the line | ✓ 1 | |
| Part (c) | ||
| Field lines are closest together where the field is strongest | ✓ 1 | |
Exam tip: Magnetic field lines always point from north to south outside the magnet.
A shop's public address system uses a loudspeaker. A coil attached to the cone is placed in the field of a permanent magnet. An alternating current from the amplifier flows through the coil.
State the effect that makes the coil move when a current flows through it.
Explain why the cone vibrates backwards and forwards.
State what change to the current makes the sound louder, and what change makes the pitch higher.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The motor effect / a force on a current-carrying wire in a magnetic field | ✓ 1 | |
| Part (b) | ||
| The current changes direction each half cycle | ✓ 1 | |
| So the force on the coil reverses direction each half cycle | ✓ 1 | |
| Part (c) | ||
| Louder: increase the size of the current | ✓ 1 | |
| Higher pitch: increase the frequency of the alternating current | ✓ 1 | |
Common error: Saying a bigger frequency makes the sound louder; frequency affects pitch, while the size of the current affects loudness.
Which statement about the magnetic field lines around a bar magnet is correct?
| Marking point | Mark | Notes |
|---|---|---|
| Field lines point away from a north pole and towards a south pole, and the closer the lines, the stronger the field, so the third statement is correct. The first has the direction reversed and the spacing wrong. The second is wrong because field lines never cross. The fourth has the direction right but links close lines to a weak field. | ✓ 1 | Answer C |
Answer: C
Common error: Field lines never cross, and a crowded set of lines always means a strong field.
A shake torch is charged by shaking. It then passes a constant current of 25 mA through its LED, and a total of 12 C of charge flows through the LED before the torch goes out. For how long does the LED stay lit?
| Marking point | Mark | Notes |
|---|---|---|
| Charge = current × time, so t = Q ÷ I = 12 ÷ 0.025 = 480 s, after converting 25 mA to 0.025 A. 0.48 s comes from using 25 instead of 0.025. 300 s comes from multiplying 12 by 25. 48 s comes from converting 25 mA to 0.25 A. | ✓ 1 | Answer A |
Answer: A
Common error: Convert milliamperes to amperes before using Q = It.
A proton moves at right angles to a uniform magnetic field. Which statement about the magnetic force on the proton is correct?
| Marking point | Mark | Notes |
|---|---|---|
| A moving charge in a magnetic field feels a force at right angles to both its velocity and the field. A force at right angles to the motion does no work, so the speed stays constant while the direction changes. Speeding up and slowing down both need a force along the line of motion. Saying there is no force ignores that a moving charge is itself equivalent to a current. | ✓ 1 | Answer C |
Answer: C
Common error: A magnetic force on a moving charge changes its direction, never its speed.
A bicycle dynamo produces an alternating p.d. When the speed of the bicycle doubles, what happens to the peak p.d. and to the frequency of the output?
| Marking point | Mark | Notes |
|---|---|---|
| Faster rotation makes the magnetic field change faster in the coil, which gives a larger induced p.d., and it also makes the magnet complete more turns each second, which raises the frequency. Both double. Peak p.d. doubling with an unchanged frequency ignores the extra turns per second. An unchanged peak p.d. ignores the faster rate of change of field. Both halving reverses the effect of speeding up. | ✓ 1 | Answer D |
Answer: D
Common error: Faster rotation increases both the size and the frequency of the output of a generator.
A bar magnet is pushed into a coil that is connected to a sensitive voltmeter. This is a demonstration of how a sensor on a bridge could detect vibration.
State what the voltmeter shows while the magnet moves into the coil.
The magnet is then held still inside the coil. State the reading and explain it.
State two ways to increase the p.d. induced.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a small p.d. (a reading) is induced across the coil | ✓ 1 | |
| Part (b) | ||
| the reading is zero | ✓ 1 | |
| there is no change in the magnetic field linking the coil (no field lines are cut) | ✓ 1 | |
| Part (c) | ||
| move the magnet faster | ✓ 1 | |
| use a stronger magnet or add more turns to the coil (accept any two valid ways) | ✓ 1 | |
Common error: A p.d. is induced only while the magnetic field through the coil is changing, not while the magnet is simply inside it.
A straight wire carrying a current lies at right angles to the magnetic field between the poles of a magnet. The wire is pushed upwards.
Name the effect that causes the wire to move.
State the direction of the force if only the current is reversed.
State the direction of the force if both the current and the magnetic field are reversed.
Name the rule used to find the direction of the force.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the motor effect | ✓ 1 | |
| Part (b) | ||
| downwards | ✓ 1 | |
| Part (c) | ||
| upwards (unchanged) | ✓ 1 | |
| Part (d) | ||
| Fleming's left-hand rule | ✓ 1 | |
Common error: Reversing one of the current or the field reverses the force; reversing both leaves it unchanged.
A straight horizontal tram cable carries a direct current towards the north. At this location the Earth's magnetic field acts vertically downwards.
In which direction is the force on the cable?
| Marking point | Mark | Notes |
|---|---|---|
| Use Fleming's left-hand rule. Point the first finger along the field (downwards) and the second finger along the current (north). The thumb then points west, so B is correct. East is the answer you get if the field or the current is reversed, or if the right hand is used. The force is always at right angles to both the current and the field. It therefore cannot be vertical (parallel to the field) or north (parallel to the current). | ✓ 1 | Answer B |
Answer: B
Exam tip: In Fleming's left-hand rule, the First finger is the Field, the seCond finger is the Current and the thuMb is the Motion (force). The force is always perpendicular to both the current and the field.
A wireless earbud contains a tiny loudspeaker. The loudspeaker has a coil of wire attached to a thin plastic cone. The coil sits in the field of a small permanent magnet. An alternating current from the music player passes through the coil.
Explain how the loudspeaker produces a sound wave. Include how the loudness and the pitch of the sound are controlled.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The coil carries a current inside the magnetic field of the permanent magnet . | ✓ 1 | |
| A current-carrying conductor in a magnetic field experiences a force, which is the motor effect . | ✓ 1 | |
| The current is alternating, so the direction of the force keeps reversing . | ✓ 1 | |
| The coil and cone therefore vibrate back and forth, pushing the air to make compressions and rarefactions, which form a sound wave . | ✓ 1 | |
| The frequency of the vibration equals the frequency of the current, so a higher-frequency current gives a higher pitch . | ✓ 1 | |
| A larger current gives a larger force and a larger amplitude of vibration, so the sound is louder . | ✓ 1 | |
Exam tip: Link each step to the next: current in a field gives a force; alternating current gives a reversing force; a reversing force gives vibration; vibration produces sound. Name the motor effect explicitly.
In proton beam therapy, a beam of protons is steered towards a tumour using magnets. Protons are positively charged.
State the condition needed for a magnetic field to exert a force on a moving charged particle.
The proton beam travels horizontally towards the east. The steering magnet provides a field acting vertically upwards. Determine the direction of the force on the protons.
A beam of electrons travels east through the same field. State the direction of the force on the electrons.
Suggest why the protons follow a curved path in the field rather than speeding up.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The particle must move in a direction that is not parallel to the magnetic field . | ✓ 1 | |
| Part (b) | ||
| The conventional current is in the same direction as the moving protons, so the current is towards the east . | ✓ 1 | |
| Using Fleming's left-hand rule, the force is towards the south . | ✓ 1 | |
| Part (c) | ||
| North, because electrons are negative, so their conventional current is towards the west . | ✓ 1 | |
| Part (d) | ||
| The force is always at right angles to the direction of motion . | ✓ 1 | |
| So the force changes the direction of the protons but not their speed . | ✓ 1 | |
Common error: For negative particles such as electrons, the conventional current is opposite to the direction of motion. Reverse the direction of the second finger before applying the left-hand rule.
An electric guitar pickup is a coil of wire wound around a permanent magnet. A steel string is magnetised by the magnet. When the string vibrates close to the pickup, a voltage is induced in the coil.
The guitarist places a hand on the string so that it stops vibrating completely.
What is the voltage induced in the coil now?
| Marking point | Mark | Notes |
|---|---|---|
| A voltage is induced only when the magnetic field through the coil changes. A vibrating string makes the field through the coil change continually. A string at rest still has a magnetic field, but that field is now constant, so the induced voltage is zero. A stationary magnet or magnetised object induces nothing, however strong it is. | ✓ 1 | Answer B |
Answer: B
Exam tip: Always link induction to a change in the magnetic field or to relative movement. A strong but steady field induces no voltage.
A mobile phone charger contains a transformer that changes 230 V a.c. into 5.0 V a.c. The primary coil has 2300 turns.
How many turns does the secondary coil have?
| Marking point | Mark | Notes |
|---|---|---|
| Use Vp/Vs = np/ns. Rearranging gives ns = np × Vs/Vp = 2300 × 5.0/230 = 50 turns. This is a step-down transformer, so the secondary coil must have fewer turns than the primary. 46 is only the voltage ratio. 460 comes from dividing 2300 by 5.0. 105 800 comes from using the ratio the wrong way up. | ✓ 1 | Answer B |
Answer: B
Common error: Check your answer makes sense. A step-down transformer must have fewer secondary turns than primary turns.
An offshore wind turbine generator produces an alternating voltage of 690 V. A transformer in the base of the turbine increases this to 34 500 V before the electricity is sent to shore. The secondary coil of the transformer has 12 000 turns.
State the equation linking the voltages across, and the numbers of turns on, the primary and secondary coils of a transformer.
Calculate the number of turns on the primary coil.
The current in the primary coil is 3000 A.
Calculate the current in the secondary coil. Assume the transformer is 100% efficient.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Vp/Vs = np/ns | ✓ 1 | |
| Part (b) | ||
| np = 12 000 × 690/34 500 ; np = 240 turns | ✓ 2 | |
| Part (c) | ||
| VpIp = VsIs ; 690 × 3000 = 34 500 × Is ; Is = 60 A | ✓ 3 | |
Exam tip: For a step-up transformer, the voltage goes up by the same factor that the current goes down. Here both factors are 50, which is a quick check.
A hospital X-ray machine needs a very high voltage. A transformer increases the 240 V a.c. mains supply to 120 000 V. The primary coil has 400 turns.
Describe the structure of a transformer.
Calculate the number of turns on the secondary coil.
The current in the secondary circuit is 5.0 mA.
Calculate the current in the primary coil. Assume the transformer is 100% efficient.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| It has a primary coil and a secondary coil of insulated wire . Both coils are wound on a soft iron core . The two coils are not electrically connected to each other . | ✓ 3 | |
| Part (b) | ||
| ns = 400 × 120 000/240 ; ns = 200 000 turns | ✓ 2 | |
| Part (c) | ||
| VpIp = VsIs, so 240 × Ip = 120 000 × 0.0050 ; Ip = 600/240 ; Ip = 2.5 A | ✓ 3 | |
Common error: Convert mA to A before substituting: 5.0 mA = 0.0050 A. Leaving the current in mA gives an answer 1000 times too big.
A cycle computer measures speed using a small magnet fixed to a spoke of the front wheel and a coil fixed to the bicycle fork. The magnet passes the coil once for every complete turn of the wheel. Each time it does so, a short voltage pulse is produced in the coil. The wheel has a circumference of 2.1 m.
Explain why a voltage pulse is produced each time the magnet passes the coil.
Explain why the pulses are larger when the cyclist rides faster.
The computer detects 5.0 pulses every second.
Calculate the speed of the bicycle.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The magnetic field through the coil changes as the magnet moves past it (or the field lines are cut by the coil) . A changing magnetic field through a coil induces a voltage across it . | ✓ 2 | |
| Part (b) | ||
| The magnet moves past the coil more quickly . The magnetic field through the coil changes at a greater rate, so a larger voltage is induced . | ✓ 2 | |
| Part (c) | ||
| Distance travelled per second = 5.0 × 2.1 . Speed = 10.5 m/s, accept 11 m/s . | ✓ 2 | |
Exam tip: One pulse means one turn of the wheel, which moves the bicycle forward by one circumference.
A horizontal wire in a factory carries a current towards the east. It passes through a magnetic field that points vertically downwards.
In which direction is the force on the wire?
| Marking point | Mark | Notes |
|---|---|---|
| Use Fleming's left-hand rule. Point the first finger along the field (down) and the second finger along the current (east). The thumb then points north, so the force acts towards the north. 'South' is the answer you get if you wrongly use the right hand. 'Vertically upwards' comes from treating the force as opposite to the field, but the force is always at right angles to both the field and the current. 'West' is opposite to the current, and the force is never along the line of the current. | ✓ 1 | Answer B |
Answer: B
Exam tip: Fleming's left-hand rule: First finger = Field, seCond finger = Current, thuMb = Motion (force). Always use your LEFT hand for motors.
An office door is held shut by a magnetic lock. A coil wrapped around an iron core is fixed to the door frame and attracts a steel plate on the door. The coil has a resistance of 48 Ω and is connected to a 12 V supply.
Calculate the current in the coil.
The supply connections are reversed. Explain whether the lock still holds the door shut.
The lock is switched on for 24 hours a day. Calculate the energy it transfers in one day.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = V / R = 12 / 48 | ✓ 1 | |
| = 0.25 A | ✓ 1 | |
| Part (b) | ||
| Yes, the lock still holds the door: the poles of the electromagnet swap but the steel plate is still attracted | ✓ 1 | |
| because magnetism is induced in the steel plate with the opposite pole facing the coil | ✓ 1 | |
| Part (c) | ||
| E = I × V × t = 0.25 × 12 × (24 × 3600) | ✓ 1 | |
| = 2.6 × 105 J | ✓ 1 | |
Common error: Time must be in seconds when you use E = I × V × t. One day is 86 400 s, not 24 s.
In a demonstration, a light aluminium rod rests across two horizontal metal rails that run north to south. The rod lies east to west. A large flat magnet under the rails produces a magnetic field pointing vertically upwards through the rod. When the switch is closed, the current in the rod flows from west to east.
State the direction in which the rod starts to roll.
Explain why there is a force on the rod when there is a current in it.
Suggest two changes that would make the rod accelerate more quickly.
Suggest why the rod is made of aluminium and not steel.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| South (along the rails) | ✓ 1 | |
| Part (b) | ||
| The current in the rod produces its own magnetic field | ✓ 1 | |
| This field interacts with the field of the magnet, giving a force on the rod | ✓ 1 | |
| Part (c) | ||
| One valid change, e.g. increase the current / use a stronger magnet / use a rod of smaller mass / reduce friction between the rod and the rails | ✓ 1 | |
| A second, different valid change from the same list | ✓ 1 | |
| Part (d) | ||
| Aluminium is not magnetic, so it is not attracted to the magnet and is free to roll / a steel rod would be pulled down onto the rails | ✓ 1 | |
Exam tip: Use Fleming's left-hand rule. Point the First finger along the Field, the seCond finger along the Current, and your thuMb gives the Motion.
An electric guitar has a pickup under each steel string. A pickup is a small permanent magnet with a coil of fine wire wound around it. The magnet magnetises the part of the string just above it. The coil is connected to an amplifier.
Explain why a voltage is induced in the coil when the string vibrates.
Suggest why the induced voltage is alternating.
Suggest why this pickup does not work with nylon strings.
The guitarist plucks the string harder. It vibrates at the same frequency but with a larger amplitude.
Suggest and explain how this affects the peak induced voltage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The magnetised string moves, so the magnetic field through the coil changes | ✓ 1 | |
| A changing magnetic field through the coil (or field lines cutting the coil) induces a voltage | ✓ 1 | |
| Part (b) | ||
| The string moves towards and then away from the coil, so the change in the field keeps reversing direction | ✓ 1 | |
| Part (c) | ||
| Nylon is not magnetic, so it is not magnetised and gives no changing field through the coil | ✓ 1 | |
| Part (d) | ||
| The peak voltage is larger | ✓ 1 | |
| The string moves faster, so the magnetic field through the coil changes more quickly; OR the string moves further / closer to the coil, so the field through the coil changes by more in the same time | ✓ 1 | |
Exam tip: A voltage is induced only while something is changing. Always link the induced voltage to the magnetic field through the coil changing, and link a bigger voltage to a faster change.
A straight wire inside a games-controller vibration unit lies at right angles to a uniform magnetic field and experiences a force of 0.12 N. The current in the wire is doubled and the length of wire in the field is halved. The field strength does not change. What is the new force on the wire?
| Marking point | Mark | Notes |
|---|---|---|
| F = BIl. Doubling I multiplies F by 2 and halving l multiplies it by 0.5, so the force is unchanged at 0.12 N. 0.24 N ignores the change in length. 0.030 N treats both changes as halving. 0.48 N treats both changes as doubling. | ✓ 1 | Answer C |
Answer: C
Exam tip: When several quantities change in a formula, find the factor for each change and multiply the factors together.
In an electric guitar, a coil is wound around a magnet placed just below a steel string. When the string is plucked, a voltage is induced in the coil. Which change would increase the size of the induced voltage?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: the induced voltage is larger when the coil has more turns. A nylon string is not magnetic, so it would not change the magnetic field and no voltage would be induced. Plucking more gently reduces the vibration and the rate of change of field. Moving the pickup away weakens the change in field at the coil. | ✓ 1 | Answer A |
Answer: A
Exam tip: Induced voltage increases with a faster change in field, a stronger magnet and more turns on the coil.
In a practical, a student uses a top-pan balance to investigate how the force on a wire depends on the current. The wire lies at right angles to a uniform field of flux density 0.20 T. The increase in balance reading, Δm, is recorded for each current, I.
I = 1.0 A: Δm = 0.8 g
I = 2.0 A: Δm = 1.6 g
I = 3.0 A: Δm = 2.4 g
I = 4.0 A: Δm = 3.2 g
Use g = 10 N/kg.
Describe the relationship between Δm and I shown by the data.
Calculate the force on the wire when the current is 3.0 A.
Calculate the length of wire in the magnetic field. Use F = BIl.
Suggest one variable the student must keep constant in this investigation.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Δm increases as I increases | ✓ 1 | |
| Δm is directly proportional to I (doubling the current doubles Δm) | ✓ 1 | |
| Part (b) | ||
| Δm = 2.4 g = 0.0024 kg; F = mg = 0.0024 × 10 | ✓ 1 | |
| F = 0.024 N | ✓ 1 | |
| Part (c) | ||
| rearranges to l = F ÷ (BI) | ✓ 1 | |
| l = 0.024 ÷ (0.20 × 3.0) (ECF from (b)) | ✓ 1 | |
| l = 0.040 m | ✓ 1 | |
| Part (d) | ||
| the magnetic flux density (or the length of wire in the field, or the angle between wire and field) | ✓ 1 | |
Common error: Do not leave the balance change in grams when you substitute into F = mg; the mass must be in kilograms.
A student places magnets on a top-pan balance, with 0.040 m of a clamped horizontal wire between the poles in a field of flux density 0.14 T. With no current the reading is 182.4 g. With the current switched on the reading is 185.2 g. Use g = 10 N/kg.
Calculate the increase in the reading.
Calculate the force on the wire.
Calculate the current in the wire. Use F = BIl.
The current is reversed. Predict the new reading and explain your prediction.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 185.2 − 182.4 = 2.8 g | ✓ 1 | |
| Part (b) | ||
| 2.8 g = 0.0028 kg; F = mg = 0.0028 × 10 (ECF from (a)) | ✓ 1 | |
| F = 0.028 N | ✓ 1 | |
| Part (c) | ||
| I = F ÷ (Bl) = 0.028 ÷ (0.14 × 0.040) (ECF from (b)) | ✓ 1 | |
| I = 5.0 A | ✓ 1 | |
| Part (d) | ||
| the force on the wire reverses (it is now downwards), so the force on the magnets is upwards | ✓ 1 | |
| the reading decreases by 2.8 g, to 179.6 g | ✓ 1 | |
Common error: The force on the balance is the reaction to the force on the wire: when the wire is pushed one way, the magnets are pushed the other way.
A guitar string vibrates at a frequency of 330 Hz above a pickup coil that has 6000 turns. The peak voltage induced in the coil is 12 mV.
State the frequency of the alternating voltage induced in the coil.
Calculate the time for one cycle of the voltage, in milliseconds. Give your answer to 2 significant figures.
A second pickup has 7500 turns and is otherwise identical. The same string is played in the same way. The induced voltage is proportional to the number of turns. Calculate the peak voltage from the second pickup.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 330 Hz | ✓ 1 | |
| Part (b) | ||
| T = 1 ÷ f = 1 ÷ 330 | ✓ 1 | |
| T = 3.0 ms | ✓ 1 | |
| Part (c) | ||
| V = 12 × (7500 ÷ 6000) | ✓ 1 | |
| V = 15 mV | ✓ 1 | |
Common error: One cycle takes 1 ÷ f seconds; multiply by 1000 to give milliseconds.
Protons from the Sun travel towards the Earth and pass through the Earth's magnetic field. Which statement describes the effect of the magnetic field on a proton moving across the field?
| Marking point | Mark | Notes |
|---|---|---|
| C is correct: a moving charge in a magnetic field feels a force at right angles to its velocity, which changes direction without changing speed. A and B wrongly put the force along the line of motion. D is wrong because a proton is a moving positive charge and does experience a force. | ✓ 1 | Answer C |
Answer: C
Common error: Thinking a magnetic force speeds up or slows down a moving charge; the force is perpendicular to the motion.
A straight wire between the poles of a magnet carries a current and experiences an upward force. The direction of the current is reversed and, at the same time, the two magnetic poles are swapped so that the field direction is also reversed. What happens to the force on the wire?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: reversing the current reverses the force, and reversing the field reverses it again, so the direction is back to the original. A assumes only one reversal. C wrongly adds the two effects. D wrongly assumes the changes cancel the force itself. | ✓ 1 | Answer B |
Answer: B
Exam tip: Each reversal (current or field) flips the force direction, so two reversals together cancel out.
A horizontal cable in a wind turbine carries a current towards the east. The cable lies in a uniform magnetic field that points vertically downwards. Using Fleming's left-hand rule, in which direction is the force on the cable?
| Marking point | Mark | Notes |
|---|---|---|
| D is correct: with the first finger pointing down (field) and the second finger pointing east (current), the thumb points north (force). A is the result of using the right hand. B and C are wrong because the force must be perpendicular to both the field and the current, so it cannot be vertical. | ✓ 1 | Answer D |
Answer: D
Common error: Using the right hand instead of the left; the left-hand rule is for the force on a current in a field.
A technician in an epidemiology laboratory uses a centrifuge to separate blood samples. The centrifuge is spun by a d.c. motor, which contains a rectangular coil between the poles of two magnets.
State the rule used to find the direction of the force on a current-carrying wire in a magnetic field.
Explain why the coil turns when a current flows.
State two ways of making the motor spin faster.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Fleming's left-hand rule | ✓ 1 | |
| Part (b) | ||
| The current flows in opposite directions in the two sides of the coil | ✓ 1 | |
| Each side experiences a force from the magnetic field | ✓ 1 | |
| The forces are in opposite directions so they make the coil turn | ✓ 1 | |
| Part (c) | ||
| Any two from: increase the current; use stronger magnets; use more turns on the coil | ✓ 1 | |
| Second different correct way | ✓ 1 | |
Exam tip: In explanation questions, link each step: current, then force on each side, then the turning effect.
At a hydroelectric power station, flowing water turns a coil in a magnetic field inside a generator, which produces an alternating voltage.
Explain how turning the coil produces a voltage.
State two changes that would increase the size of the voltage produced.
The coil is now turned at half the original speed. State what happens to the frequency of the alternating voltage and to the size of the voltage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The coil cuts the magnetic field lines / the magnetic field through the coil changes | ✓ 1 | |
| This induces a voltage across the coil | ✓ 1 | |
| The direction of the voltage reverses every half turn as the sides move in opposite directions through the field | ✓ 1 | |
| Part (b) | ||
| Any two from: rotate the coil faster; use stronger magnets; use more turns on the coil | ✓ 1 | |
| Second different correct way | ✓ 1 | |
| Part (c) | ||
| The frequency halves | ✓ 1 | |
| The size of the voltage is smaller / decreases | ✓ 1 | |
Common error: Writing that the generator "makes" electricity from nothing; the voltage is induced by a changing magnetic field through the coil.
A wind turbine rotor turns at 0.25 revolutions per second. A gearbox makes the generator coil turn 200 times faster than the rotor. The generator produces one complete cycle of alternating voltage for each revolution of its coil.
Calculate the number of revolutions per second of the generator coil.
State the frequency of the alternating voltage produced.
Calculate the time taken for one complete cycle of the voltage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 0.25 × 200 | ✓ 1 | |
| A1 50 revolutions per second | ✓ 1 | |
| Part (b) | ||
| 50 Hz (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 T = 1 ÷ f = 1 ÷ 50 | ✓ 1 | |
| A1 0.02 s (ECF from (b)) | ✓ 1 | |
Exam tip: Period and frequency are reciprocals: T = 1 ÷ f.
A hospital lamp uses an ideal transformer. The primary coil has 1150 turns and is connected to a 230 V a.c. supply. The lamp needs 20 V from the secondary coil. How many turns must the secondary coil have?
| Marking point | Mark | Notes |
|---|---|---|
| For an ideal transformer Vp/Vs = Np/Ns, so Ns = 1150 × 20 ÷ 230 = 100 turns. 13 225 comes from inverting the ratio. 940 comes from subtracting the voltage difference from the turns. 4 comes from multiplying the voltages and dividing by the primary turns. | ✓ 1 | Answer B |
Answer: B
Exam tip: Check the direction: a step-down transformer must have fewer turns on the secondary than on the primary.
A beekeeper builds a hive monitor in which a bar magnet is pushed into a coil to generate a small voltage. Which change gives the largest induced voltage?
| Marking point | Mark | Notes |
|---|---|---|
| The induced voltage is proportional to the number of turns and to the rate at which the field is cut. Doubling both gives four times the voltage. Halving the turns and doubling the speed leaves it unchanged. Turning the magnet round reverses the direction but not the size. Doubling the turns and halving the speed also leaves it unchanged. | ✓ 1 | Answer D |
Answer: D
Common error: A change that raises one factor and lowers another may leave the voltage the same, so combine the changes before comparing.
A beekeeper uses a shake torch for inspecting hives at night. Shaking the torch stores enough energy for 45 C of charge to pass through its LED, at a potential difference of 3.0 V. The LED current is 0.15 A.
Calculate how long the LED stays lit, in seconds.
Calculate the energy transferred by the LED, in joules.
Calculate the power of the LED.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 t = Q ÷ I = 45 ÷ 0.15 | ✓ 1 | |
| A1 300 s | ✓ 1 | |
| Part (b) | ||
| M1 E = QV = 45 × 3.0 | ✓ 1 | |
| A1 135 J | ✓ 1 | |
| Part (c) | ||
| M1 P = E ÷ t = 135 ÷ 300 or P = IV = 0.15 × 3.0 | ✓ 1 | |
| A1 0.45 W (ECF from (a) and (b)) | ✓ 1 | |
Exam tip: Write the equation, substitute with correct units, then give the answer; power can be found two ways, which makes a good check.
A hospital keeps a shake torch for power cuts. In one use the torch lights its lamp for 12 minutes with a constant current of 0.050 A. The potential difference across the lamp is 2.5 V. The person shaking the torch does 150 J of work.
Calculate the charge that flows through the lamp, in coulombs.
Calculate the energy transferred by the lamp, in joules.
Explain why shaking the torch produces a potential difference across a coil inside it.
Calculate the percentage of the work done by the person that is transferred by the lamp.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 12 × 60 = 720 s and Q = It | ✓ 1 | |
| A1 36 C | ✓ 1 | |
| Part (b) | ||
| M1 E = QV = 36 × 2.5 (ECF from (a)) | ✓ 1 | |
| A1 90 J | ✓ 1 | |
| Part (c) | ||
| B1 a magnet moves through/past the coil so the magnetic field through the coil changes (field lines are cut) | ✓ 1 | |
| B1 this induces a potential difference across the coil | ✓ 1 | |
| Part (d) | ||
| M1 90 ÷ 150 (ECF from (b)) | ✓ 1 | |
| A1 60 % | ✓ 1 | |
Exam tip: For an explanation of induction, mention a changing magnetic field or field lines being cut, not just 'a magnet near a coil'.
The control unit of a coffee roaster needs 9.0 V a.c. from a 230 V mains supply. An ideal step-down transformer is used with 1840 turns on the primary coil. How many turns are needed on the secondary coil?
| Marking point | Mark | Notes |
|---|---|---|
| Ns = Np × Vs ÷ Vp = 1840 × 9.0 ÷ 230 = 72 turns. 1831 subtracts the voltages instead of using a ratio. 47000 inverts the ratio, which would be a step-up transformer. 26 is 230 ÷ 9.0, ignoring the turns on the primary. | ✓ 1 | Answer A |
Answer: A
Exam tip: For a step-down transformer, the secondary coil must have fewer turns than the primary; check your answer is smaller.
A graph of the output p.d. of a bicycle dynamo against time is a sine wave with a peak of 6.0 V. One complete cycle of the graph takes 0.040 s. The dynamo produces one cycle of output for each complete turn of its magnet, and the magnet turns 10 times for every one turn of the bicycle wheel. The circumference of the wheel is 2.0 m.
Calculate the frequency of the output of the dynamo.
Calculate the speed of the bicycle.
Describe how the graph changes when the cyclist goes faster.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| frequency = 1 ÷ period = 1 ÷ 0.040 | ✓ 1 | |
| 25 Hz | ✓ 1 | |
| Part (b) | ||
| wheel turns per second = 25 ÷ 10 = 2.5 (ECF from (a)) | ✓ 1 | |
| distance per turn = 2.0 m | ✓ 1 | |
| speed = 2.5 × 2.0 = 5.0 m/s | ✓ 1 | |
| Part (c) | ||
| the peak of the wave is higher | ✓ 1 | |
| one cycle takes less time, so the waves are closer together | ✓ 1 | |
Common error: Frequency is cycles per second, which is 1 divided by the period measured in seconds.
A power station supplies 5.0 MW of electrical power to a transmission line of total resistance 4.0 Ω. Initially the power is supplied at 25 kV.
Calculate the current in the line.
Calculate the power wasted as heat in the line.
A transformer steps the supply up to 100 kV with the same power delivered. Calculate the power now wasted in the line.
Explain why stepping up the voltage reduces the power wasted.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P ÷ V = 5000000 ÷ 25000 | ✓ 1 | |
| 200 A | ✓ 1 | |
| Part (b) | ||
| P = I2R = 2002 × 4.0 | ✓ 1 | |
| 160000 W | ✓ 1 | |
| Part (c) | ||
| current = 5000000 ÷ 100000 = 50 A | ✓ 1 | |
| P = 502 × 4.0 | ✓ 1 | |
| 10000 W | ✓ 1 | |
| Part (d) | ||
| for the same power a higher voltage means a smaller current | ✓ 1 | |
| the power wasted depends on I2, so heating in the line is much smaller | ✓ 1 | |
Exam tip: Power loss in a cable is I²R, so halving the current cuts the loss to a quarter.
An electromagnet on a crane lifts steel girders for a bridge. It is made from a coil of insulated wire wound around an iron core.
State two ways to make the electromagnet stronger.
Explain why a soft iron core is used.
State what happens to the magnetic poles of the electromagnet when the current is reversed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| increase the current in the coil | ✓ 1 | |
| increase the number of turns of the coil (accept any two valid ways) | ✓ 1 | |
| Part (b) | ||
| the iron becomes magnetised, which makes the magnetic field much stronger | ✓ 1 | |
| soft iron loses its magnetism when the current is switched off, so the girder can be released | ✓ 1 | |
| Part (c) | ||
| the poles swap (the magnetic field direction reverses) | ✓ 1 | |
Exam tip: Say why the core is soft (loses magnetism easily), not just that it is iron.
| Level 0 (0) | No rewardable content. |
| Level 1 (1–2) | Isolated points e.g. testing pairs or stating aluminium is not magnetic; no method to distinguish the magnet from the iron. |
| Level 2 (3–4) | Identifies the aluminium and recognises that the magnet and iron attract each other, with some explanation (induced magnetism) OR a partial end-to-middle method lacking reasoning. |
| Level 3 (5–6) | Logically sequenced method identifying all three bars, including the end-to-middle test explained by the field being strongest at the poles, and why attraction alone is not conclusive. |
| Level 0 (0) | No rewardable content. |
| Level 1 (1–2) | Isolated points, e.g. steel is magnetic, or a voltage is induced when something moves near a coil; no linked explanation. |
| Level 2 (3–4) | Explains that the vibrating string changes the field through the coil and induces a voltage, with some reference to induced magnetism OR to the alternating nature/frequency OR to the nylon string; reasoning partly linked. |
| Level 3 (5–6) | Coherent linked explanation: induced magnetism in the steel string → changing field through the coil → induced voltage that alternates at the string's frequency, with size linked to the amplitude/speed of vibration and a correct reason why nylon does not work. |
A car's starter motor runs from the 12 V battery and has a power of 1.5 kW. The driver's ignition switch is connected in a separate circuit to the coil of a relay. The relay coil carries a current of 0.15 A. When the relay coil is energised, it attracts an iron armature, which closes a pair of heavy-duty contacts in the starter motor circuit.
Describe how turning the ignition switch makes the starter motor turn.
Suggest two reasons why a relay is used rather than connecting the ignition switch directly into the starter motor circuit.
Calculate the current in the starter motor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Closing the ignition switch causes a current in the relay coil | ✓ 1 | |
| The coil becomes an electromagnet / produces a magnetic field | ✓ 1 | |
| The iron armature is attracted to the coil and moves | ✓ 1 | |
| This pushes the contacts together, completing the starter circuit so there is a (large) current in the motor | ✓ 1 | |
| Part (b) | ||
| Acceptable reasons: the ignition switch only carries a small current, so it and its wiring can be small/cheap/thin; the large current is kept away from the driver, which is safer; the thick cables for the large current can be kept short, running directly between the battery and the motor. | — | |
| First valid reason | ✓ 1 | |
| Second, different valid reason | ✓ 1 | |
| Part (c) | ||
| I = P / V = 1500 / 12 | ✓ 1 | |
| = 125 A | ✓ 1 | |
Common error: Convert kW to W before you calculate. 1.5 kW = 1500 W, so using 1.5 gives an answer that is 1000 times too small.
A student wants to find out how the height from which a magnet is dropped affects the peak voltage induced in a coil. She has a vertical plastic tube with a coil of wire wound around its middle, a small strong magnet, a datalogger with a voltage sensor, and a metre rule.
Describe how the student could carry out this investigation.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Connect the coil to the voltage sensor / datalogger | ✓ 1 | |
| Use the metre rule to measure the drop height from a consistent reference point (e.g. bottom of the magnet to the top of the coil), then release the magnet from rest (not pushed) so it falls through the tube | ✓ 1 | |
| Read the peak (maximum) induced voltage from the datalogger voltage–time trace | ✓ 1 | |
| Repeat for a range of at least five different drop heights | ✓ 1 | |
| Repeat each height (e.g. three times) and calculate a mean peak voltage, ignoring anomalous results | ✓ 1 | |
| Keep at least one named control variable the same, e.g. the same magnet, the same pole entering first, the same coil / number of turns, or the same coil position on the tube; accept any other valid control | ✓ 1 | |
Exam tip: A strong method answer says how the independent variable is changed and measured. It also says how the dependent variable is recorded, names a control variable and mentions repeats. Plotting peak voltage against drop height then shows the relationship.
In a hospital physics laboratory, a student drops a bar magnet so that it falls vertically through the middle of a coil of 125 turns connected to a voltmeter. The voltage shows a pulse in one direction as the magnet enters the coil and a pulse in the opposite direction as it leaves. The largest voltage in the second pulse is 0.46 V.
Explain why the two pulses are in opposite directions.
The second pulse is larger and shorter than the first. Explain why.
The student drops the magnet from a greater height. State and explain the effect on the largest voltage.
The student repeats the original drop with a coil of 250 turns. Calculate the largest voltage in the second pulse.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 as the magnet enters, the magnetic field through the coil is increasing | ✓ 1 | |
| B1 as it leaves the field through the coil is decreasing, so the induced potential difference is reversed | ✓ 1 | |
| Part (b) | ||
| B1 the magnet is accelerating under gravity so it moves faster when it leaves the coil | ✓ 1 | |
| B1 the field changes faster so a larger potential difference is induced (for less time) | ✓ 1 | |
| Part (c) | ||
| B1 the largest voltage increases | ✓ 1 | |
| B1 the magnet is moving faster through the coil so the field changes faster | ✓ 1 | |
| Part (d) | ||
| M1 0.46 × 250 ÷ 125 (voltage proportional to number of turns) | ✓ 1 | |
| A1 0.92 V | ✓ 1 | |
Common error: A moving magnet gives a voltage only while the field through the coil is changing; there is no voltage when the magnet is at rest.
A coffee-roasting oven has a heating element that needs 24 V and 20 A. It is supplied from 240 V mains through a transformer whose primary coil has 1200 turns.
Assume the transformer is ideal. Calculate the number of turns on the secondary coil.
Assume the transformer is ideal. Calculate the current in the primary coil.
The transformer is in fact only 80% efficient. Calculate the current in the primary coil.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Ns = 1200 × 24 ÷ 240 | ✓ 1 | |
| 120 turns | ✓ 1 | |
| Part (b) | ||
| VpIp = VsIs, so Ip = 24 × 20 ÷ 240 | ✓ 1 | |
| 2.0 A | ✓ 1 | |
| Part (c) | ||
| output power = 24 × 20 = 480 W | ✓ 1 | |
| input power = 480 ÷ 0.80 = 600 W | ✓ 1 | |
| current = 600 ÷ 240 = 2.5 A | ✓ 1 | |
Common error: For an efficiency below 100%, the input power is larger than the output power, so divide by the efficiency.
The peak output p.d. of a cycle dynamo and the frequency of its output are both directly proportional to the speed of the bicycle. At 6.0 m/s the dynamo gives a peak p.d. of 4.5 V at a frequency of 30 Hz. A rear lamp needs a peak p.d. of at least 6.0 V to light at full brightness.
Calculate the lowest speed at which the lamp lights at full brightness.
Calculate the frequency of the output at this speed.
Calculate the time for one complete cycle of the output at this speed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| speed is proportional to peak p.d., so the speed ratio equals the p.d. ratio | ✓ 1 | |
| speed = 6.0 × (6.0 ÷ 4.5) | ✓ 1 | |
| 8.0 m/s | ✓ 1 | |
| Part (b) | ||
| frequency = 30 × (8.0 ÷ 6.0) | ✓ 1 | |
| 40 Hz (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| period = 1 ÷ frequency = 1 ÷ 40 | ✓ 1 | |
| 0.025 s (ECF from (b)) | ✓ 1 | |
Exam tip: If two quantities are directly proportional, their ratio stays fixed, so scale one by the same factor as the other.
A transformer in a microwave oven steps up the voltage from 230 V to 2300 V. Assume the transformer is 100% efficient.
How does the current in the secondary coil compare with the current in the primary coil?
| Marking point | Mark | Notes |
|---|---|---|
| For a 100% efficient transformer, VpIp = VsIs, so the input power equals the output power. The voltage rises by a factor of 10, so the current must fall by a factor of 10. C is correct. A wrongly assumes the current rises with the voltage. B wrongly assumes current is the same in both coils. D wrongly squares the turns ratio. | ✓ 1 | Answer C |
Answer: C
Exam tip: A step-up transformer increases voltage but decreases current by the same factor, because power in = power out.
An electric guitar pickup is a coil of thin wire wound around a small permanent magnet. The steel guitar string lies just above the pickup. The magnet magnetises the string. When the string is plucked, it vibrates up and down close to the coil.
State what is produced in the coil when the string vibrates.
State two changes to the pickup that would increase the size of the output.
Explain why the output from the pickup is an alternating voltage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| An induced voltage (induced emf / induced potential difference) . | ✓ 1 | |
| Part (b) | ||
| Any two different changes from: use more turns of wire on the coil; use a stronger magnet; mount the pickup closer to the string. | — | |
| First valid change | ✓ 1 | |
| Second, different valid change | ✓ 1 | |
| Do not credit changes to the string or the playing, such as plucking harder, because these are not changes to the pickup. | — | |
| Part (c) | ||
| The magnetised string moves towards the coil and then away from it, so the magnetic field through the coil increases and then decreases . | ✓ 1 | |
| The direction of the change keeps reversing, so the direction of the induced voltage keeps reversing . | ✓ 1 | |
Exam tip: A voltage is induced only while the magnetic field through the coil is changing. A stationary string produces no output.
A furniture restorer has a tub of mixed old screws. Some are made of steel and some are made of brass. She separates them by holding a strong permanent magnet over the tub.
Name one metal, other than iron and steel, that is attracted to a magnet.
A steel screw is attracted to the magnet whichever pole is held towards it, even though the screw was not a magnet before.
Explain why.
State why the brass screws stay in the tub.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Nickel or cobalt | ✓ 1 | |
| Part (b) | ||
| The magnet induces magnetism in the steel screw | ✓ 1 | |
| The end of the screw nearest the magnet becomes the opposite pole, so the screw is always attracted | ✓ 1 | |
| Part (c) | ||
| Brass is not a magnetic material / brass cannot be magnetised | ✓ 1 | |
Exam tip: Only repulsion proves that an object is a permanent magnet. Attraction could just be induced magnetism in a magnetic material.
A portable arc-welding set contains a transformer. Its primary coil has 460 turns and is connected to a 230 V a.c. supply. The secondary coil gives an output of 40 V.
How many turns are on the secondary coil?
| Marking point | Mark | Notes |
|---|---|---|
| Use Vp/Vs = np/ns, so ns = np × Vs/Vp = 460 × 40/230 = 80 turns. The output voltage is smaller, so this is a step-down transformer and it must have fewer secondary turns. 2645 comes from using the ratio upside down. 420 comes from subtracting the numbers. 18 400 comes from multiplying 460 by 40 and not dividing by 230. | ✓ 1 | Answer A |
Answer: A
Exam tip: Check your answer with a quick sense test. A step-down transformer must have fewer turns on the secondary coil than on the primary coil.
A laptop charger contains a transformer. The primary coil is connected to the 230 V mains and has 1840 turns. The output voltage from the secondary coil is 20 V.
Explain why a transformer would not work if the primary coil were connected to a d.c. supply.
Calculate the number of turns on the secondary coil.
The secondary current is 3.25 A. Assuming the transformer is 100% efficient, calculate the current in the primary coil.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A direct current produces a constant magnetic field in the iron core . | ✓ 1 | |
| The field in the secondary coil does not change, so no voltage is induced . | ✓ 1 | |
| Part (b) | ||
| Use Vp/Vs = Np/Ns . | ✓ 1 | |
| Ns = 1840 × 20 ÷ 230 . | ✓ 1 | |
| Ns = 160 turns . | ✓ 1 | |
| Part (c) | ||
| Use Vp × Ip = Vs × Is . | ✓ 1 | |
| Ip = 20 × 3.25 ÷ 230 . | ✓ 1 | |
| Ip = 0.28 A . | ✓ 1 | |
Common error: Do not use the turns ratio the same way up for current as for voltage. In a step-down transformer the primary current is smaller than the secondary current.
An emergency wind-up radio contains a small a.c. generator. When the handle is turned, a coil of wire rotates between the poles of a permanent magnet.
Describe how turning the handle produces a voltage across the ends of the coil.
State two changes to the design of the generator that would increase the peak output voltage at the same turning speed.
The handle is now turned twice as fast. Describe how the alternating output voltage changes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The coil rotates in the magnetic field of the magnet . | ✓ 1 | |
| The wires of the coil cut the magnetic field lines, so the magnetic field through the coil changes . | ✓ 1 | |
| This induces a voltage (potential difference) across the ends of the coil . | ✓ 1 | |
| Part (b) | ||
| Any two from: use a stronger magnet / increase the number of turns on the coil / increase the area of the coil. | — | |
| First correct change . | ✓ 1 | |
| Second correct change . | ✓ 1 | |
| Do not credit turning the handle faster, because the speed is fixed. | — | |
| Part (c) | ||
| The peak voltage increases (it doubles) . | ✓ 1 | |
| The frequency doubles, so the voltage changes direction twice as often . | ✓ 1 | |
Common error: Turning a generator faster does not only increase the frequency. The peak voltage also increases, because the field lines are cut at a faster rate.
An engineer needs a region of uniform magnetic field to test a small sensor. She has two flat permanent magnets.
Which arrangement gives a uniform field in the gap between them?
| Marking point | Mark | Notes |
|---|---|---|
| When opposite poles face each other, the field lines run straight across the gap from N to S. If the pole faces are flat, parallel and close together, these lines are parallel and equally spaced, so the field is uniform. When like poles face each other, they repel and leave a neutral point between them. When opposite poles are far apart or tilted, the field lines spread out and curve, so the field is not uniform. | ✓ 1 | Answer B |
Answer: B
Exam tip: A uniform field is shown by parallel, equally spaced field lines. Say both 'parallel' and 'equally spaced' when you describe one.
A shop sign uses a gas-discharge tube that needs a high voltage. A transformer converts the 230 V a.c. mains supply to 6900 V. The transformer is 100% efficient. Its primary coil has 150 turns.
State the type of transformer used.
Calculate the number of turns on the secondary coil.
The current in the secondary coil is 0.020 A.
Calculate the current in the primary coil.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Step-up (transformer) | ✓ 1 | |
| Part (b) | ||
| Vp/Vs = np/ns | ✓ 1 | |
| ns = 150 × 6900/230 | ✓ 1 | |
| ns = 4500 (turns) | ✓ 1 | |
| Part (c) | ||
| Vp × Ip = Vs × Is | ✓ 1 | |
| Ip = 6900 × 0.020/230 | ✓ 1 | |
| Ip = 0.60 A | ✓ 1 | |
Exam tip: If the transformer steps the voltage up, it must step the current down by the same factor. Use this to check that your current answer makes sense.
A small offshore wind farm sends its electrical output along a cable on the sea bed to a village on a remote island. The power sent is 2.4 MW, and it is transmitted at a voltage of 33 000 V.
Calculate the current in the cable.
The generators produce a much lower voltage, so step-up transformers are used at the wind farm.
Explain why transmitting at a high voltage reduces energy losses in the cable.
Suggest why a step-down transformer is needed before the electricity reaches the homes in the village.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P/V = 2 400 000 / 33 000 | ✓ 1 | |
| I = 73 A (accept 72.7 A) | ✓ 1 | |
| Part (b) | ||
| For the same power (P = IV), a higher voltage means a smaller current | ✓ 1 | |
| The cable has resistance, so energy is transferred as heat; the rate of heating is P = I2R, so it depends on the square of the current | ✓ 1 | |
| A smaller current therefore means much less energy is wasted as thermal energy in the cable | ✓ 1 | |
| Part (c) | ||
| 33 000 V is far too high and dangerous for homes, so it must be reduced to a safe mains voltage that suits appliances (about 230 V) | ✓ 1 | |
Common error: Do not write that a high voltage 'loses less voltage'. The key idea is that the current is smaller. Because the heating is I²R, a smaller current gives much less heating in the cable.
A retail charging station for electric scooters uses a transformer. The mains input is 230 V. The transformer has a secondary coil of 150 turns and gives an output of 23 V.
Explain why the transformer will only work with an alternating current input.
Calculate the number of turns on the primary coil.
State whether the transformer is a step-up or a step-down transformer, and give a reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| An alternating current produces a changing magnetic field in the iron core | ✓ 1 | |
| This induces a voltage in the secondary coil; a steady direct current would not | ✓ 1 | |
| Part (b) | ||
| M1 Np = 150 × 230 ÷ 23 | ✓ 1 | |
| A1 1500 turns | ✓ 1 | |
| Part (c) | ||
| Step-down, because the secondary voltage is lower / there are fewer turns on the secondary | ✓ 1 | |
Common error: Swapping the primary and secondary values in the turns-ratio equation; keep each voltage paired with its own coil.
A student tests the transformer equation using an iron-cored transformer with 100 turns on the primary coil. The primary coil is connected to a 4.0 V a.c. supply. The student measures the secondary voltage for different numbers of secondary turns.
Secondary turns: 50, 100, 200, 300
Secondary voltage in V: 1.9, 3.9, 7.8, 11.6
Calculate the secondary voltage predicted by the transformer equation for 300 turns.
Suggest one reason why every measured voltage is slightly lower than the predicted value.
Explain why the supply must be a.c. and not d.c.
With 300 secondary turns the secondary coil supplies a current of 0.50 A at 12 V. Assuming the transformer is 100 % efficient, calculate the current in the primary coil.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 Vs = 4.0 × 300 ÷ 100 | ✓ 1 | |
| A1 12 V | ✓ 1 | |
| Part (b) | ||
| B1 energy is lost, e.g. heating of the coils, eddy currents in the core, or not all the magnetic field passing through the secondary coil | ✓ 1 | |
| Part (c) | ||
| B1 an alternating current produces a changing magnetic field in the core | ✓ 1 | |
| B1 d.c. gives a constant field so no potential difference is induced in the secondary coil | ✓ 1 | |
| Part (d) | ||
| M1 VpIp = VsIs or Ip = 12 × 0.50 ÷ 4.0 | ✓ 1 | |
| A1 1.5 A | ✓ 1 | |
Exam tip: For a 100 % efficient transformer, power in equals power out, so the side with the lower voltage carries the larger current.
In a Formula 1 car, a motor-generator unit contains conductors in a magnetic field. A straight conductor of length 0.12 m carries a current of 85 A at right angles to a uniform magnetic field of flux density 0.45 T.
Calculate the force on the conductor. Give your answer to 2 significant figures.
State what happens to the force when the current is reversed.
Name the rule that can be used to find the direction of the force.
The length of conductor in the field is doubled and the current is halved. State the new force and explain your answer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 F = BIl | ✓ 1 | |
| M1 0.45 × 85 × 0.12 | ✓ 1 | |
| A1 4.6 N | ✓ 1 | |
| Part (b) | ||
| B1 the force has the same size but acts in the opposite direction | ✓ 1 | |
| Part (c) | ||
| B1 Fleming's left-hand rule | ✓ 1 | |
| Part (d) | ||
| B1 4.6 N (unchanged) | ✓ 1 | |
| B1 F depends on the product I × l, which is unchanged | ✓ 1 | |
Common error: The force depends on the current multiplied by the length, so one change can cancel the other.
A straight wire carries a current at right angles to a uniform magnetic field. The current is tripled and the magnetic flux density is halved. By what factor does the force on the wire change?
| Marking point | Mark | Notes |
|---|---|---|
| The force is BIL, so it is proportional to the product of B and I. The factor is 3 × 0.5 = 1.5. Unchanged would need the factors to cancel. Multiplying by 6 treats halving the field as doubling it. Multiplying by 2.5 subtracts 0.5 from 3 rather than multiplying. | ✓ 1 | Answer B |
Answer: B
Exam tip: When two quantities in a product change, multiply the two factors of change together.
Protons from the Sun travel towards the Earth. Near the equator the Earth's magnetic field is horizontal and points towards geographic north. A proton arrives travelling vertically downwards towards the surface.
State the direction in which the proton is deflected, and explain how you decided.
Explain why the magnetic force does not change the speed of the proton.
Suggest how the Earth's magnetic field reduces the number of protons reaching the surface.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a moving positive charge is equivalent to a conventional current in the direction of motion (downwards) | ✓ 1 | |
| use Fleming's left-hand rule with the field pointing north and the current pointing down | ✓ 1 | |
| the deflection is towards the east | ✓ 1 | |
| Part (b) | ||
| the force is at right angles to the velocity of the proton | ✓ 1 | |
| so no work is done on the proton and its speed stays the same while its direction changes | ✓ 1 | |
| Part (c) | ||
| the field bends the paths of many protons so that they are turned away from the Earth or guided towards the poles | ✓ 1 | |
Exam tip: Treat a moving positive charge as a current in the direction of motion, then apply Fleming's left-hand rule.
A probe boom contains a straight wire of length 0.15 m placed at right angles to a uniform magnetic field of flux density 0.080 T. The current in the wire is 2.5 A.
Calculate the force on the wire.
The current is increased to 4.0 A and the flux density is decreased to 0.050 T. Show that the force on the wire is unchanged.
A student says that increasing the current always increases the force. Explain why this is not true here.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F = BIL = 0.080 × 2.5 × 0.15 | ✓ 1 | |
| 0.030 N | ✓ 1 | |
| Part (b) | ||
| F = 0.050 × 4.0 × 0.15 | ✓ 1 | |
| 0.030 N, the same as before | ✓ 1 | |
| Part (c) | ||
| the force depends on B × I, and the increase in current is cancelled by the decrease in flux density | ✓ 1 | |
Common error: The force depends on the product of flux density, current and length, so changing one quantity can be cancelled by changing another.
An electric car uses regenerative braking. When the driver brakes, the electric motor is disconnected from the battery's supply and used as a generator. The car's wheels turn the coil of the motor, and the electricity produced recharges the battery.
Explain how regenerative braking recharges the battery and slows the car down.
The car has a mass of 1500 kg and brakes from 20 m/s to rest. 60% of its kinetic energy is stored in the battery. Calculate the energy stored.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The moving wheels make the coil rotate inside the magnetic field of the motor's magnets . | ✓ 1 | |
| The coil cuts the field lines, so the magnetic field through it changes . | ✓ 1 | |
| This induces a voltage in the coil . | ✓ 1 | |
| The induced voltage drives a current through the battery, transferring energy to its chemical store . | ✓ 1 | |
| The induced current in the coil is in a magnetic field, so the coil experiences a force . | ✓ 1 | |
| This force opposes the rotation, which slows the wheels, as kinetic energy is transferred away from the car . | ✓ 1 | |
| Part (b) | ||
| KE = ½ × m × v2 = 0.5 × 1500 × 202 . | ✓ 1 | |
| KE = 300 000 J . | ✓ 1 | |
| Energy stored = 0.60 × 300 000 = 180 000 J, or 1.8 × 105 J . | ✓ 1 | |
Exam tip: To explain why the car slows down, link the induced current back to the motor effect. A current-carrying coil in a field feels a force that opposes its motion.
A transformer in the controller of a model railway steps the 230 V mains supply down to 16 V. The primary coil has 1150 turns. Assume the transformer is 100% efficient.
Explain why the transformer would not work if it were connected to a 230 V direct current supply.
Calculate the number of turns on the secondary coil.
The secondary coil supplies a current of 2.0 A to the train. Calculate the current in the primary coil.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A direct current produces a constant/unchanging magnetic field in the core | ✓ 1 | |
| so no voltage is induced in the secondary coil, because induction needs a changing field | ✓ 1 | |
| Part (b) | ||
| ns = np × Vs / Vp = 1150 × 16 / 230 | ✓ 1 | |
| = 80 turns | ✓ 1 | |
| Part (c) | ||
| Vp × Ip = Vs × Is | ✓ 1 | |
| Ip = (16 × 2.0) / 230 | ✓ 1 | |
| = 0.14 A | ✓ 1 | |
Exam tip: A step-down transformer lowers the voltage but raises the current. Check that your primary current is smaller than the secondary current.
An electric toothbrush is charged by standing it on a base unit. The base contains a coil connected to the a.c. mains supply. The handle of the toothbrush contains a second coil, connected through a circuit to a rechargeable battery. There is no electrical contact between the base and the handle.
Explain how a current is produced in the coil in the handle, and why the charger would not work if the base coil were connected to a steady d.c. supply.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Award one mark for each of the following points, up to 6 marks: | — | |
| There is an alternating current in the base coil | ✓ 1 | |
| This produces a magnetic field around the base coil | ✓ 1 | |
| The magnetic field is continuously changing / reversing because the current is alternating | ✓ 1 | |
| The changing field passes through the coil in the handle | ✓ 1 | |
| A (changing) voltage is induced in the handle coil, so a current flows in its complete circuit | ✓ 1 | |
| With d.c. the magnetic field would be constant, so no voltage would be induced (after switch-on) | ✓ 1 | |
Exam tip: Build a clear chain of points: a.c. in the first coil, then a changing field, then the field links the second coil, then an induced voltage. Each link earns a mark.
An electron and a proton enter the same uniform magnetic field at the same speed, moving at right angles to the field. Ignore gravity. Which row describes their motion?
| Marking point | Mark | Notes |
|---|---|---|
| D is correct: opposite charges are deflected in opposite directions, and the proton's much larger mass means it is harder to turn, so its circle is larger. A ignores the opposite charges. B ignores the difference in mass. C reverses the effect of mass. | ✓ 1 | Answer D |
Answer: D
Common error: The magnetic force depends on the sign of the charge, so particles of opposite sign curve in opposite directions.
In an old arcade game display, a beam of electrons travels horizontally from left to right across the page. A uniform magnetic field acts on the beam and points into the page.
State the direction of the conventional current represented by the electron beam.
Use Fleming's left-hand rule to determine the direction of the force on the electrons.
Explain why the electrons follow a curved path but their speed does not change.
State what happens to the beam when the direction of the magnetic field is reversed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| right to left (opposite to the motion of the electrons) | ✓ 1 | |
| Part (b) | ||
| first finger points into the page (field), second finger points to the left (conventional current) | ✓ 1 | |
| thumb gives the force: down the page | ✓ 1 | |
| Part (c) | ||
| the force is always at right angles to the motion, so it changes the direction of travel | ✓ 1 | |
| the force does no work on the electrons, so their kinetic energy and speed stay the same | ✓ 1 | |
| Part (d) | ||
| the beam is deflected the other way (up the page) | ✓ 1 | |
Common error: Electrons move opposite to the conventional current, so use the reversed direction in Fleming's left-hand rule.
In a pharmaceutical laboratory, singly charged ions of a drug molecule enter a region of uniform magnetic field of flux density 0.30 T at right angles to the field. Each ion has charge 1.6 × 10−19 C and speed 1.2 × 104 m/s. The radius, r, of the circular path is given by r = mv ÷ (qB).
Ion X has mass 3.0 × 10−25 kg. Ion Y has mass 6.0 × 10−25 kg.
Calculate the radius of the path of ion X.
Ion Z has the same mass and speed as X but a charge of −1.6 × 10−19 C. Describe how its path differs from that of X.
The field is changed so that ion Y moves in a circle of radius 0.075 m. Calculate the new flux density.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| r = (3.0 × 10−25 × 1.2 × 104) ÷ (1.6 × 10−19 × 0.30) | ✓ 1 | |
| r = 0.075 m | ✓ 1 | |
| Part (b) | ||
| it curves in the opposite direction | ✓ 1 | |
| its radius is the same as that of X (same size circle) | ✓ 1 | |
| Part (c) | ||
| B = mv ÷ (qr) = (6.0 × 10−25 × 1.2 × 104) ÷ (1.6 × 10−19 × 0.075) | ✓ 1 | |
| B = 0.60 T | ✓ 1 | |
Exam tip: In r = mv ÷ (qB), radius increases with mass and speed and decreases with charge and field strength; use this to check your answer is sensible.
A transformer at a wind farm steps up the voltage from the generator before it enters the grid. The primary coil has 100 turns and the input voltage is 400 V. The output voltage is 11 000 V. The transformer can be assumed to be 100% efficient and the output power is 2.2 MW.
Calculate the number of turns on the secondary coil.
Calculate the current in the primary coil.
Explain why electricity is transmitted at a high voltage rather than at 400 V.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 Ns = 100 × 11 000 ÷ 400 | ✓ 1 | |
| A1 2750 turns | ✓ 1 | |
| Part (b) | ||
| M1 input power = output power = 2 200 000 W | ✓ 1 | |
| M1 I = P ÷ V = 2 200 000 ÷ 400 | ✓ 1 | |
| A1 5500 A | ✓ 1 | |
| Part (c) | ||
| Higher voltage means a smaller current for the same power | ✓ 1 | |
| Less energy is wasted as heat in the cables | ✓ 1 | |
Common error: Using the output voltage with the input power; with 100% efficiency the power is the same on both sides, but the voltages are different.
A small beekeeping research station has a wind generator that produces 2400 W at 240 V. The power is sent along cables of total resistance 5.0 Ω. Assume that all transformers are 100 % efficient.
Calculate the current in the cables when the power is sent at 240 V.
Calculate the power lost as heat in the cables at this current.
A step-up transformer raises the voltage to 2400 V before the power enters the cables. Calculate the current in the cables.
Calculate the power lost in the cables at this current.
Explain why transmitting power at a higher voltage reduces the energy wasted in the cables.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P ÷ V = 2400 ÷ 240 | ✓ 1 | |
| A1 10 A | ✓ 1 | |
| Part (b) | ||
| M1 P = I2R = 102 × 5.0 (ECF from (a)) | ✓ 1 | |
| A1 500 W | ✓ 1 | |
| Part (c) | ||
| M1 I = 2400 ÷ 2400 | ✓ 1 | |
| A1 1.0 A | ✓ 1 | |
| Part (d) | ||
| M1 I2R = 1.02 × 5.0 (ECF from (c)) | ✓ 1 | |
| A1 5.0 W | ✓ 1 | |
| Part (e) | ||
| B1 for the same power, a higher voltage means a lower current | ✓ 1 | |
| B1 power lost depends on the square of the current, so the loss in the cables is much smaller | ✓ 1 | |
Exam tip: In transmission questions, find the current first, then use I²R for the power lost.
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