Induced potential, generators, microphones and transformers (physics only, Higher) – GCSE Physics exam questions with mark scheme

1Grade 6–7HigherTriple only5 marks

A student pushes a bar magnet into a coil of wire connected to a sensitive voltmeter. The voltmeter shows a reading.

(a) Explain why a potential difference is induced across the coil. [1]

(b) The magnet is then held still inside the coil. State and explain the reading on the voltmeter. [2]

(a) the magnet moves relative to the coil / the magnetic field through the coil changes (the coil 'cuts' magnetic field lines)(1)
(b) zero(1)
(b) there is no relative movement / the magnetic field through the coil is not changing(1)
(c) any two from: move the magnet faster / use a stronger magnet / use a coil with more turns (allow: put an iron core in the coil)(1)
(c) second way from the list above(1)
Answer: (a) the field through the coil changes (relative motion) (b) zero — no relative motion (c) e.g. move magnet faster, stronger magnet, more turns
Examiner tip: A potential difference is induced only while the field is CHANGING — a stationary magnet gives zero.

2Grade 6–7HigherTriple only5 marks

A transformer is used to step down the potential difference from 230 V to 11.5 V. The primary coil has 2000 turns. Use the Physics Equations Sheet.

(a) Calculate the number of turns on the secondary coil. [3]

(b) The current in the secondary coil is 2.0 A. Assume the transformer is 100% efficient.

Calculate the current in the primary coil. [2]

(a) $\dfrac{230}{11.5} = \dfrac{2000}{n_s}$(1)
(a) $n_s = \dfrac{2000 \times 11.5}{230}$(1)
(a) 100 turns(1)
(b) $230 \times I_p = 11.5 \times 2.0$(1)
(b) 0.10 A(1)
Answer: (a) 100 turns (b) 0.10 A
Examiner tip: For a step-down transformer the current goes UP as the p.d. goes down — check your answer makes sense.

3Grade 8–9HigherTriple only6 marks

(a) Explain how a transformer produces a potential difference across its secondary coil. [4]

(b) The National Grid uses step-up transformers. Explain why transmitting electricity at a high potential difference reduces energy losses. [2]

(a) an alternating current in the primary coil (allow: alternating p.d. across the primary coil)(1)
(a) produces a changing magnetic field(1)
(a) the iron core is easily magnetised, so the changing field passes through the core to the secondary coil(1)
(a) the changing magnetic field (through the secondary coil) induces an (alternating) potential difference across the secondary coil(1)
(b) for the same power, a higher potential difference means a lower current (P = VI)(1)
(b) a lower current means less heating of the cables (allow: lower I²R losses), so less energy is wasted(1)
Answer: (a) a.c. in primary → changing field → carried by iron core through secondary → p.d. induced across secondary (b) higher p.d. → lower current for same power → less heating in cables
Examiner tip: The key word is CHANGING magnetic field — that is why transformers only work with a.c.

4Grade 4–5HigherTriple only4 marks

A transformer is made from two coils of insulated wire wound on a core.

(a) Name the metal that the core of a transformer is usually made from. [1]

(b) Give the property of this metal that makes it suitable for a transformer core. [1]

Is this a step-up transformer or a step-down transformer? Give a reason for your answer. [2]

(a) iron allow soft iron; do not accept steel(1)
(b) it is easily magnetised allow it is easily magnetised and demagnetised; ignore it is magnetic / it is a conductor(1)
(c) step-down(1)
(c) (because) the secondary coil has fewer turns than the primary coil (so the secondary p.d. is lower than the primary p.d.) allow the ratio of turns is 20 : 1; reason only scores if step-down is chosen(1)
Answer: (a) iron (b) easily magnetised (c) step-down: fewer turns on the secondary (60) than on the primary (1200)
Examiner tip: In (c) the reason must compare the two coils: 'it has 60 turns' on its own does not score.

5Grade 6–7HigherTriple only6 marks

The diagram shows a view from above of a straight wire placed between the poles of two magnets. The two ends of the wire are connected to a centre-zero ammeter.

When the wire is moved quickly in direction A, the ammeter needle moves to the right.

(a) Name the effect that causes a current in the wire. [1]

(b) The wire is then moved towards the S pole, along the direction of the field lines.

State the reading on the ammeter. Give a reason for your answer. [2]

(d) A student notices that it is harder to move the wire when there is a current in it. Explain why. [1]

(a) generator effect allow electromagnetic induction; do not accept motor effect(1)
(b) zero allow the needle does not move(1)
(b) the wire is not cutting (across) the magnetic field lines allow there is no change in the magnetic field around the wire; mark independently(1)
(c) move the wire in direction B / in the opposite direction(1)
(c) reverse the magnetic field / swap the poles of the magnets allow turn the magnets round; do not accept both changes made together as one answer; ignore move the wire faster / slower(1)
(d) the (induced) current produces a magnetic field that opposes the movement of the wire allow opposes the change that induced it; allow there is a force on the current-carrying wire in the opposite direction to its motion(1)
Answer: (a) generator effect (b) zero, the wire does not cut any field lines (c) move the wire in direction B; reverse the poles of the magnets (d) the induced current's magnetic field opposes the motion
Examiner tip: Reversing the motion AND the field together gives a right deflection again, so the two changes in (c) must be made separately.

6Grade 6–7HigherTriple only6 marks

A coil of wire is rotated at a steady rate in a magnetic field. This produces an alternating potential difference across the ends of the coil. The graph shows how the potential difference varies with time.

(a) Give the name of a generator that produces an alternating potential difference. [1]

(b) Calculate the frequency of the potential difference. [2]

Describe two ways the graph would change. [2]

(d) The generator is changed into a dynamo by connecting the coil to a split-ring commutator. The coil is rotated at the original rate.

Describe how the graph would be different. [1]

(a) alternator allow a.c. generator(1)
(b) period = 40 ms / 0.040 s (read from graph)(1)
(b) frequency = 1 ÷ 0.040 = 25 Hz correct answer with no working scores 2 marks; allow ecf from an incorrect period read from the graph(1)
(c) the peak / maximum potential difference is larger allow peak p.d. doubles to 12 V(1)
(c) the time for one cycle is shorter / there are more cycles in the same time allow period halves to 20 ms; allow frequency doubles to 50 Hz(1)
(d) the potential difference is never negative / does not reverse direction (the negative halves are flipped to be positive) allow it is d.c.; allow all of the graph is above the time axis(1)
Answer: (a) alternator (b) 25 Hz (c) higher peak p.d. (12 V) and shorter period (20 ms) / higher frequency (d) p.d. stays positive: it does not change direction
Examiner tip: Students often forget to convert 40 ms into 0.040 s before using f = 1/T, giving 0.025 Hz.

7Grade 6–7HigherTriple only6 marks

The diagram shows a cross-section through a moving-coil microphone.

(a) Explain how the microphone produces an electrical signal when sound waves reach it. [4]

(b) The microphone detects a louder sound with the same pitch.

Explain how the electrical signal changes. [2]

(a) the pressure variations in the sound wave make the diaphragm vibrate / move backwards and forwards(1)
(a) the coil (attached to the diaphragm) moves (backwards and forwards) in the magnetic field (of the permanent magnet)(1)
(a) a potential difference is induced across the coil allow a current is induced in the coil / circuit; allow the coil cuts the field lines so a p.d. is induced(1)
(a) the direction of the induced p.d. / current reverses each time the coil changes direction, so the signal is an alternating current that matches the pressure variations of the sound allow the signal has the same frequency as the sound wave(1)
(b) the diaphragm / coil vibrates with a larger amplitude allow moves further / faster(1)
(b) so the induced p.d. / current has a larger amplitude (but the same frequency) do not accept a higher frequency(1)
Answer: (a) sound pressure variations vibrate the diaphragm; the attached coil moves in the magnet's field; a p.d. is induced; it reverses as the coil changes direction, giving an a.c. signal matching the sound (b) coil vibrates with larger amplitude, so the signal has a larger amplitude (same frequency)
Examiner tip: Microphones use the generator effect (not the motor effect, which is the loudspeaker): marks are lost when students say a current makes the coil move.

8Grade 8–9HigherTriple only6 marks

A student holds a bar magnet vertically above a long coil of wire, with the north pole at the bottom. The coil is connected to a datalogger that records the potential difference across the coil.

The student releases the magnet. It falls through the middle of the coil and out of the bottom. The graph shows the potential difference recorded.

Explain the shape of the graph. [6]

as the magnet falls into the coil the magnetic field through the coil changes, so a potential difference is induced (generator effect)(1)
there is no reading before about 0.05 s because the magnet is not yet close enough to change the field in the coil(1)
the first peak is positive while the north pole enters the coil / the field through the coil increases(1)
while the magnet is inside the coil the field through it is (almost) not changing, so the p.d. falls to (about) zero between the peaks(1)
as the magnet leaves the coil the field through it decreases / the south pole leaves last, so the p.d. is in the opposite direction (negative)(1)
the magnet accelerates as it falls, so it is moving faster when it leaves the coil than when it enters(1)
so the field changes more quickly when it leaves, giving a larger maximum p.d. (0.45 V compared with 0.30 V)(1)
and the magnet takes less time to leave the coil, so the second peak is narrower(1)
(stretch) the same amount of field passes in and out, so the areas under the two peaks are equal(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logically linked explanation of the whole graph: why a p.d. is induced, why it is zero while the magnet is inside the coil, why the second peak is negative, and why it is both larger and narrower, linked to the magnet speeding up as it falls.
Level 2	3–4	Explains that a changing field induces a p.d. and that the p.d. reverses as the magnet leaves. Attempts to explain the different sizes of the peaks, but the link to the magnet's increasing speed may be missing or incomplete.
Level 1	1–2	Simple relevant statements, e.g. 'the moving magnet induces a voltage' or 'it goes negative when the magnet comes out', with little explanation.
0	0	No relevant content.

Answer: Changing field as magnet enters induces a positive p.d.; about zero inside the coil (field hardly changing); negative peak as it leaves (field decreasing, opposite direction); magnet has accelerated so it leaves faster: the field changes faster, so the peak is larger (0.45 V vs 0.30 V) and shorter.
Examiner tip: Most answers explain why the p.d. reverses but forget to explain why the second peak is larger and narrower; for that you need the magnet speeding up as it falls.

9Grade 4–5HigherTriple only6 marks

A student builds a model transformer. The primary coil has 100 turns and is connected to a 4.0 V alternating supply. She changes the number of turns on the secondary coil and measures the secondary potential difference each time.

Number of turns on secondary coil	50	100	200	300	400
Secondary potential difference in V	1.9	3.8	7.7	9.6	15.3

(a) Explain why the transformer would not work if the primary coil were connected to a direct (d.c.) supply. [2]

(b) Identify the anomalous result in the table. [1]

(c) Calculate the secondary potential difference the student would expect for 400 turns on the secondary coil. Use the Physics Equations Sheet. [2]

(d) Suggest why the measured values are slightly smaller than the values calculated using the equation. [1]

(a) a direct current produces a magnetic field (in the core) that does not change(1)
(a) so no potential difference is induced across the secondary coil allow a changing magnetic field is needed to induce a p.d.; allow induced p.d. is only produced while the d.c. is switched on or off(1)
(b) 9.6 V (for 300 turns)(1)
(c) $\dfrac{4.0}{V_s} = \dfrac{100}{400}$ allow $V_s = 4.0 \times \dfrac{400}{100}$(1)
(c) $V_s = 16$ V correct answer with no working scores 2 marks(1)
(d) the transformer is not 100% efficient allow energy is dissipated (by heating) in the coils / core; allow not all of the magnetic field from the primary passes through the secondary; ignore 'energy is lost' without further detail; ignore human error(1)
Answer: (a) d.c. gives a constant field in the core, so no p.d. is induced in the secondary (b) 9.6 V at 300 turns (c) 16 V (d) the transformer is not 100% efficient: some energy is dissipated heating the coils and core
Examiner tip: For the anomaly, compare each reading with the pattern (about 0.038 V per turn): 9.6 V at 300 turns should be about 11.5 V.

10Grade 8–9HigherTriple only7 marks

A power station generator produces a potential difference of 25 kV and a current of 20 000 A.

A transformer is used to increase the potential difference to 400 kV for the transmission cables of the National Grid. The transformer is 100% efficient.

(a) Calculate the current in the transmission cables. Use the Physics Equations Sheet. [3]

(b) The total resistance of the transmission cables is 0.40 Ω.

Calculate the power dissipated in the transmission cables. [2]

(c) Show that the power dissipated in the cables would be 256 times greater if they carried the 20 000 A current from the generator directly. [2]

(a) $400\,000 \times I_s = 25\,000 \times 20\,000$ allow substitution with both p.d.s in kV: $400 \times I_s = 25 \times 20\,000$(1)
(a) $I_s = \dfrac{25\,000 \times 20\,000}{400\,000}$(1)
(a) $I_s = 1250$ A correct answer with no working scores 3 marks; allow 1300 A(1)
(b) $P = 1250^2 \times 0.40$ allow ecf from (a)(1)
(b) $P = 625\,000$ W allow 6.3 × 10⁵ W / 625 kW / 0.625 MW(1)
(c) the current would be 16 times greater ($20\,000 \div 1250 = 16$)(1)
(c) power dissipated depends on current squared ($P = I^2R$), so it would be $16^2 = 256$ times greater allow calculation of both powers: 20 000² × 0.40 = 1.6 × 10⁸ W and 1.6 × 10⁸ ÷ 6.25 × 10⁵ = 256(1)
Answer: (a) 1250 A (b) 625 000 W (625 kW) (c) current is 16 times larger; P = I²R so power is 16² = 256 times larger
Examiner tip: In (a), convert both kV values consistently (or leave both in kV) and remember the current goes DOWN when the p.d. goes up.

Still losing marks on this topic?

Book a free introductory session and we'll find exactly which step is costing you.

Book a free intro