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Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)
Question
A student pushes the N pole of a bar magnet into a 200-turn coil connected to a centre-zero galvanometer. Describe what happens to the galvanometer needle, and explain why.
Solution
As the magnet moves toward the coil, the magnetic field through the coil increases. The changing field induces a p.d. across the coil's ends.
Because the coil is connected to a complete circuit (the galvanometer), an induced current flows. The galvanometer needle deflects to one side.
The induced current creates a magnetic field that opposes the change — the near end of the coil becomes a N pole, repelling the incoming magnet (the student must do work pushing it in).
When the magnet stops moving, the field is constant; no induced current flows and the needle returns to zero.
Answer
The needle deflects while the magnet is moving (greater deflection for faster pushes, stronger magnets, or more turns). When the magnet stops, the needle returns to zero.
Question
State four ways a student could increase the induced p.d. when pushing a magnet into a coil.
Solution
Push the magnet in faster (faster change of field through the coil).
Use a stronger magnet (larger flux density B in the coil's region).
Use a coil with more turns.
Use a coil with a larger cross-sectional area.
Answer
Faster motion, stronger magnet, more turns, larger coil area.
Question
After pushing the magnet into the coil, the student pulls it out at the same speed. Describe and explain what happens to the galvanometer needle.
Solution
Pulling the magnet out decreases the magnetic field through the coil — the field is changing in the opposite sense to before.
The induced p.d. therefore reverses, and so does the induced current.
The galvanometer needle deflects to the opposite side. The induced current makes the near end of the coil a S pole (to attract the receding N pole and oppose the change).
Answer
The needle deflects to the OPPOSITE side. The induced current's direction has reversed because the field through the coil is now decreasing instead of increasing.
Question
Explain why no current flows when a strong magnet is held motionless inside a coil.
Solution
For an induced p.d. to appear, the magnetic field threading the coil must be changing.
A stationary magnet produces a constant magnetic field inside the coil — no change with time.
With no changing field, there is no induced p.d., and no current flows.
Answer
No induced current flows because the magnetic field through the coil is not changing. Induction requires a changing field.
Question
A magnet is dropped through a vertical copper tube and is observed to fall much more slowly than the same magnet dropped through a plastic tube. Explain in terms of the generator effect.
Solution
As the magnet falls through the copper tube, the magnetic field through each cross-section of the tube changes.
The copper is a conductor, so the changing field induces circular currents in the tube wall (Lenz's law).
These induced currents create their own magnetic field that opposes the change — i.e. it tries to slow the magnet's fall.
The plastic tube is an insulator, so no current can flow → no opposing field → the magnet falls under gravity unimpeded.
Answer
The changing field of the falling magnet induces currents in the copper tube. These currents produce a magnetic field that opposes the magnet's motion (Lenz's law), slowing its fall. Plastic does not conduct, so no opposing currents form.
Question
An alternator and a dynamo both contain a coil rotating in a magnetic field. State the key difference in their construction and the resulting difference in their output.
Solution
The alternator uses two complete slip rings to connect each end of the coil to the external circuit.
The dynamo uses a split-ring commutator (one ring divided into two halves) to connect the coil to the external circuit.
Result: the alternator's output is alternating current (a sine wave that reverses every half-cycle). The dynamo's output is one-direction pulsing DC (positive humps that never go negative).
Answer
Alternator: two slip rings → AC output (sine wave). Dynamo: split-ring commutator → DC output (one direction, pulsing).
Question
A student doubles the rotational speed of an alternator. Describe what happens to the output trace on an oscilloscope.
Solution
Faster motion → faster change in field through the coil → larger induced p.d. The amplitude (peak height) of the sine wave doubles.
More rotations per second → higher frequency. The period halves and twice as many cycles fit on the same time axis.
Answer
The amplitude doubles AND the frequency doubles (the period halves). The waveform shows taller peaks crammed twice as close together.
Question
A student replaces the magnets in an alternator with stronger neodymium magnets but keeps the rotation speed the same. Describe what happens to the output trace.
Solution
Stronger B → bigger induced p.d. for the same rate of motion. Peak amplitude increases.
Rotation speed unchanged → period and frequency unchanged.
Answer
Peaks become bigger; frequency and period are unchanged.
Question
Sketch the output of a DC dynamo and explain the shape of the trace using the position of the coil.
Solution
The trace is a series of positive humps — like the absolute value of a sine wave.
Peaks occur when the coil's sides move at 90° to B (maximum rate of cutting field lines).
Zeros occur when the coil's plane is parallel to B (no field lines being cut at that instant). The trace touches zero between peaks but never goes negative because the commutator reverses the connection just as the internal p.d. would reverse.
Answer
All-positive humps. Peak height = peak induced p.d.; humps touch zero between cycles. Two humps per coil rotation.
Question
An oscilloscope trace of an alternator's output shows a sine wave with a period of 20 ms. Calculate the frequency of rotation of the coil and the AC frequency of the output.
Solution
Period T = 20 ms = 0.020 s.
Evaluate.
Coil rotates 50 times per second (matches UK mains).
Answer
f = 50 Hz; the coil rotates 50 times per second.
Question
Explain how a moving-coil microphone converts sound waves into an alternating current.
Solution
Sound waves are pressure variations that push and pull a thin diaphragm at the front of the microphone, making it vibrate.
A voice coil is fixed to the back of the diaphragm. The coil oscillates with the diaphragm in the radial magnetic field of a permanent magnet.
As the coil's wires move through the magnetic field, they cut field lines. By the generator effect, an alternating p.d. is induced across the coil.
Connected to an external circuit, this induces an alternating current with the same frequency as the sound and an amplitude that depends on the sound's loudness.
Answer
Sound vibrates the diaphragm → coil vibrates in the magnetic field → generator effect induces an alternating p.d. → AC flows in the circuit. Frequency matches the sound; amplitude matches loudness.
Question
State two similarities and two differences between a moving-coil microphone and a loudspeaker.
Solution
Similarity 1: both contain a coil of wire in the radial magnetic field of a permanent magnet, fixed to a diaphragm or cone.
Similarity 2: both are transducers — they convert energy between mechanical (sound/vibration) and electrical forms.
Difference 1: the loudspeaker uses the motor effect (current → force on the coil → cone vibration → sound). The microphone uses the generator effect (sound → diaphragm vibration → coil motion → induced p.d.).
Difference 2: energy flow is opposite. Loudspeaker takes in electrical energy and gives out sound; microphone takes in sound and gives out electrical energy.
Answer
Similarities: same hardware (coil, magnet, diaphragm); both are transducers. Differences: loudspeaker uses motor effect, microphone uses generator effect; energy flow is reversed.
Question
A 440 Hz tone plays at a moderate volume into a moving-coil microphone. (a) State the frequency of the induced AC. (b) Describe what would happen to the induced AC if the volume of the sound is doubled.
Solution
(a) The diaphragm vibrates at the same frequency as the sound wave (440 Hz). The coil oscillates at 440 Hz and the induced AC has frequency 440 Hz.
(b) Doubling the loudness means larger pressure variations on the diaphragm → larger diaphragm displacement → larger coil velocity → larger induced p.d.
The frequency stays at 440 Hz; the amplitude of the AC roughly doubles.
Answer
(a) 440 Hz. (b) Same frequency (440 Hz); approximately double the amplitude.
Question
A pupil notices that their microphone has no battery yet still produces a signal. Explain how this is possible.
Solution
A moving-coil microphone generates its own signal by electromagnetic induction. No external power supply is needed.
The sound waves themselves provide the energy: they do work on the diaphragm, which transfers that energy to the moving coil; the coil's motion in the magnetic field induces a p.d.
Energy chain: sound (kinetic energy of air) → kinetic energy of diaphragm + coil → electrical energy in the induced current.
Answer
The microphone generates its own signal by the generator effect; the energy comes from the sound waves themselves doing work on the diaphragm. No battery required.
Question
A pupil shouts into a small loudspeaker that is connected to a sensitive voltmeter. The voltmeter deflects. Explain why.
Solution
A loudspeaker contains the same components as a microphone — a coil, a magnet and a cone.
When sound waves hit the cone, the cone vibrates. The coil is attached to the cone and vibrates with it, moving through the radial magnetic field.
By the generator effect, an alternating p.d. is induced across the coil. The voltmeter registers this induced p.d.
Answer
The shout vibrates the cone, the coil moves in the magnetic field, and the generator effect induces an alternating p.d. that the voltmeter reads. The loudspeaker is acting as a poor-quality microphone.
Question
A step-up transformer has 200 turns on its primary and 5000 turns on its secondary. The primary voltage is 11 kV. Calculate the secondary voltage.
Solution
Write the turns-ratio equation.
Rearrange for V_s.
Substitute V_p = 11 000 V, n_s = 5000, n_p = 200.
Evaluate.
Answer
V_s = 275 kV (matches the National Grid transmission voltage).
Question
A phone-charger transformer must step 230 V down to 5.0 V. The primary has 4600 turns. How many turns are needed on the secondary?
Solution
Rearrange the turns-ratio equation for n_s.
Substitute n_p = 4600, V_s = 5.0, V_p = 230.
Evaluate.
Answer
100 turns on the secondary.
Question
A step-up transformer takes a primary input of 11 kV at 1000 A and outputs 275 kV. Assuming 100 % efficiency, calculate the output current.
Solution
Write the power-conservation equation.
Rearrange for I_s.
Substitute V_p = 11 000 V, I_p = 1000 A, V_s = 275 000 V.
Evaluate.
Answer
I_s = 40 A. Power in = power out = 11 MW.
Examiner note
Always check: did the current change in the OPPOSITE direction to the voltage? Voltage ×25 up, current ×25 down. Tick.
Question
Explain why electricity is transmitted across the National Grid at very high voltages (e.g. 400 kV) rather than at the 230 V used in homes.
Solution
Power transmitted: P = V × I. For a fixed P, higher V means lower I.
Power lost in cables due to resistance: P_loss = I² R. Doubling the current QUADRUPLES the cable losses.
By stepping voltage up (and current down) with a transformer, I² R losses in the transmission cables are made very small — most of the generated power reaches the consumer.
At 230 V the same power would need thousands of times more current → massive I² R losses, melted cables. Transmission would be impossible.
Answer
Stepping voltage up reduces current (P = VI). Cable losses are P_loss = I²R, so lower current means much smaller losses. The grid transmits at hundreds of kV and steps down at substations.
Question
A student connects a 12 V DC battery to the primary of a transformer and measures the secondary voltage on a voltmeter. Predict and explain what they will see.
Solution
A transformer works because a CHANGING magnetic field in the core induces an AC p.d. in the secondary (generator effect).
A DC battery produces a steady current in the primary → steady magnetic field in the core → no CHANGE → no induced p.d. in the secondary.
The voltmeter will read zero for most of the experiment. When the battery is first connected (or disconnected) the current is changing briefly, so a momentary pulse may register on a sensitive meter.
Answer
Voltmeter reads zero (apart from a brief pulse at switch-on/off). DC produces a steady field, which cannot induce a continuous secondary p.d.
Factors affecting the induced p.d. (qualitative — no equation required at GCSE)
When to use
Use as a memory aid for the four factors. AQA does NOT require a numerical formula at GCSE — only the qualitative dependence on these four variables.
Frequency and period of AC output
When to use
Use to convert between rotation period (from an oscilloscope trace) and the AC frequency. UK mains: f = 50 Hz, T = 20 ms.
Induced p.d. on the microphone coil (qualitative)
When to use
Qualitative use only. Bigger B, longer wire or faster motion → bigger induced p.d. AQA does not require numerical calculation at GCSE; just understand the proportionalities.
Turns-ratio equation
When to use
Use whenever you know three of (V_p, V_s, n_p, n_s) and want the fourth. Step-up: n_s > n_p; step-down: n_s < n_p. On the AQA equation sheet.
Power conservation (ideal transformer)
When to use
Use when you need to find the current in either coil, given the voltages and the other current. Assumes 100 % efficiency (no losses). On the AQA equation sheet.
The induction of a potential difference (and hence a current, if the circuit is complete) across a conductor when the magnetic field through it changes — either because the conductor moves relative to a field, or because the field itself changes.
The p.d. produced across the ends of a conductor by the generator effect. Larger for faster motion, stronger fields, more turns and larger coil area.
The current driven by an induced p.d. in a complete circuit. Its direction is such that its own magnetic field opposes the change in field that caused it.
The rule that an induced current always flows in a direction that creates a magnetic field opposing the change that produced it. This is a consequence of energy conservation.
A sensitive current-detecting meter (often centre-zero) used to show small induced currents. Deflects in opposite directions for currents of opposite signs.
A device that uses the generator effect to produce alternating current. A coil rotates in a magnetic field; the ends of the coil are connected to the external circuit via two slip rings.
A device that uses the generator effect to produce direct (pulsing) current. A coil rotates in a magnetic field; the ends of the coil are connected to the external circuit via a split-ring commutator.
Two complete metal rings on the axle of an AC generator, one connected to each end of the coil. They allow the alternating induced current to flow into the external circuit without reversal.
A ring split into two halves on the axle of a DC generator. As the coil rotates, the brushes swap which half they contact at the moment the internal p.d. reverses, so the output current always flows in the same direction.
Direct current whose magnitude varies with time but whose direction never reverses. The output of a dynamo: a series of positive humps from zero up to a peak and back.
A device that converts sound waves into electrical signals (AC) using the generator effect. A diaphragm vibrates a coil in a magnetic field, inducing a p.d.
A thin, light circular sheet at the front of the microphone that vibrates when sound waves hit it. Equivalent to the cone in a loudspeaker but much lighter.
A device that converts one form of energy or signal into another. A microphone is a sound-to-electrical transducer; a loudspeaker is an electrical-to-sound transducer.
A small coil of wire attached to the back of the diaphragm of a moving-coil microphone. When the diaphragm vibrates, the coil moves through a radial magnetic field, inducing an alternating p.d.
The alternating current generated in the microphone coil. Its frequency matches the sound's frequency; its amplitude depends on the loudness of the sound.
A device that uses electromagnetic induction to change the size of an alternating voltage. Consists of a primary coil, a secondary coil and a shared soft iron core.
A transformer in which the secondary coil has MORE turns than the primary coil (n_s > n_p). The secondary voltage is bigger than the primary; the secondary current is smaller.
A transformer in which the secondary coil has FEWER turns than the primary coil (n_s < n_p). The secondary voltage is smaller than the primary; the secondary current is bigger.
The input coil of a transformer, connected to the AC supply. The alternating current in the primary creates the alternating magnetic field in the iron core.
The output coil of a transformer, connected to the load. The alternating flux in the core induces an alternating p.d. across the secondary by the generator effect.
A continuous loop of soft (easily magnetised and demagnetised) iron that guides the magnetic field from the primary coil to the secondary coil with minimal energy loss.
Mistake
Saying a stationary magnet inside a coil produces a current.
Why it happens
Students see a strong magnet near a coil and assume something must happen.
How to avoid it
Induction requires a CHANGING field. A stationary magnet produces a constant field → no induced p.d. → no current.
Source: AQA Examiner Report Paper 2 2024.
Mistake
Stating that the induced current's magnetic field helps the change that caused it.
Why it happens
Confusion about Lenz's law direction.
How to avoid it
The induced current ALWAYS opposes the change. If it helped, energy would be created from nothing.
Mistake
Listing only three of the four factors affecting induced p.d.
Why it happens
Students forget the area of the coil.
How to avoid it
Memorise the foursome: speed of motion, magnetic flux density, number of turns, area of coil.
Mistake
Confusing 'induced p.d.' with 'induced current' — saying current flows even when the circuit is open.
Why it happens
Students gloss over the difference.
How to avoid it
Induced p.d. appears whenever there is a changing field, regardless of circuit. Induced CURRENT requires a COMPLETE circuit.
Mistake
Mixing up slip rings (AC) and split-ring commutator (DC) in a generator.
Why it happens
Same shapes (rings on a shaft).
How to avoid it
Memorise: TWO COMPLETE rings = AC (alternator); ONE RING CUT IN HALF = DC (dynamo). The commutator's whole job is to flip the connection at exactly the moment the internal p.d. would reverse.
Source: AQA Examiner Report Paper 2 2024.
Mistake
Drawing the dynamo output as a flat horizontal line (smooth DC like a battery).
Why it happens
Students think 'DC' always means constant.
How to avoid it
Dynamo DC is ONE-DIRECTION but PULSING — draw positive humps that touch zero between them.
Mistake
Saying that faster rotation only makes the peaks bigger (not faster).
Why it happens
Students stop at the size change.
How to avoid it
Always state BOTH effects: bigger peaks AND higher frequency (period decreases).
Mistake
Saying a stronger magnet increases the frequency of the output.
Why it happens
Conflating field strength with rotation speed.
How to avoid it
Frequency depends ONLY on rotation speed. A stronger magnet increases peak p.d. but not frequency.
Mistake
Saying a microphone uses the motor effect.
Why it happens
Confusion with a loudspeaker, which contains the same parts.
How to avoid it
Microphone = generator effect (sound moves coil → AC induced). Loudspeaker = motor effect (AC moves coil → sound). Same hardware, opposite physics.
Source: AQA Examiner Report Paper 2 2024.
Mistake
Stating that a microphone needs a battery to produce a signal.
Why it happens
Students don't realise the sound itself provides the energy.
How to avoid it
A moving-coil microphone is self-powered — the kinetic energy of the air drives the diaphragm and coil, and the generator effect converts that to electrical energy. No battery required.
Mistake
Saying a microphone amplifies the sound.
Why it happens
Students confuse the microphone with the amplifier that follows it.
How to avoid it
A microphone CONVERTS sound to electrical signal — it doesn't amplify. Amplification is a separate stage handled by an amplifier circuit.
Mistake
Saying the amplitude of the AC sets the frequency of the sound.
Why it happens
Confusion between amplitude and frequency.
How to avoid it
Frequency ↔ frequency (sound and AC are the same). Amplitude ↔ loudness (bigger amplitude = louder).
Mistake
Flipping the turns-ratio equation upside down (e.g. writing ).
Why it happens
Forgetting which subscript goes where.
How to avoid it
Match the subscripts on both sides: . Subscripts p over s = p over s.
Source: AQA Examiner Report Paper 2 2023.
Mistake
Saying a transformer can work with DC (or being unsure).
Why it happens
Not connecting transformers to induction.
How to avoid it
Induction requires a CHANGING field. DC → constant field → no induction → no output. Transformers ONLY work with AC.
Mistake
Treating voltage and current as independent — e.g. saying both can be increased by adding more secondary turns.
Why it happens
Students forget power conservation.
How to avoid it
Use BOTH equations together: AND . If voltage goes up by a factor, current goes down by the same factor.
Mistake
Using steel (hard magnetic material) for the core to make it more efficient.
Why it happens
Students think 'stronger' or 'harder' = better.
How to avoid it
Soft iron is used precisely because it magnetises AND demagnetises easily. A hard material would lose lots of energy as heat (hysteresis) every cycle and the transformer would be inefficient.