S5 Physics Paper: Electromagnetic Induction, Transformers and AC (100 marks)

2026-11-12

S5 Physics – electromagnetism

S5 Physics paper: 6 questions, 24 parts, 100 marks, 75 minutes. Covers Faraday induction law, magnetic flux, transformers, AC rms values, generators and transmission loss.

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Paper details　S5 Physics practice paper with six questions on Faraday induction law, magnetic flux change, transformer turns ratio, AC peak and rms values, generator emf and high voltage transmission loss.

Section A: Faraday Law of Electromagnetic Induction (16 marks)

A coil of 200 turns and area 0.020 m² is placed in a uniform magnetic field perpendicular to its plane. The flux density falls uniformly from 0.50 T to 0.10 T in 0.20 s. Give units in every answer.

#	Question	Answer and marking notes
1	Find the change in flux density.	Answer: the change is 0.50 minus 0.10, that is 0.40 T. (Marking: 1 mark for subtracting, 2 marks for the value with unit.)
2	Find the change in magnetic flux through the coil.	Answer: flux change equals area times flux density change, that is 0.020 times 0.40, so 8.0 times 10 to the power minus 3 Wb. (Marking: 1 mark for the formula, 1 mark for substitution, 2 marks for the value with unit.)
3	Find the magnitude of the induced emf in the coil.	Answer: emf equals turns times flux change divided by time, that is 200 times 8.0 times 10 to the power minus 3, divided by 0.20, so 8.0 V. (Marking: 2 marks for the formula, 2 marks for substitution, 1 mark for the value with unit.)
4	If the field instead increases at the same rate, how does the direction of the induced current change? Explain with Lenz law.	Answer: the direction reverses. Lenz law states that the induced current produces a field that opposes the change in flux, so when the flux increases the coil must produce an opposing field and the current flows the other way. (Marking: 2 marks for the direction, 2 marks for the explanation.)

Section B: A Rod Cutting Magnetic Field Lines (16 marks)

A metal rod of length 0.60 m moves at 4.0 m/s perpendicular to a uniform field of 0.25 T. Its ends are connected to a 2.0 ohm resistor. The field, the rod and the motion are mutually perpendicular.

#	Question	Answer and marking notes
5	Find the induced emf across the rod.	Answer: emf equals field times length times speed, that is 0.25 times 0.60 times 4.0, so 0.60 V. (Marking: 1 mark for the formula, 2 marks for substitution, 1 mark for the value with unit.)
6	Find the current in the circuit.	Answer: current equals emf divided by resistance, that is 0.60 divided by 2.0, so 0.30 A. (Marking: 1 mark for Ohm law, 1 mark for substitution, 2 marks for the value with unit.)
7	Find the power dissipated in the resistor.	Answer: power equals current squared times resistance, that is 0.30 squared times 2.0, so 0.18 W. (Marking: 1 mark for the formula, 2 marks for substitution, 1 mark for the value with unit.)
8	Find the external force needed to keep the rod moving at constant speed.	Answer: force equals field times current times length, that is 0.25 times 0.30 times 0.60, so 0.045 N. (Marking: 2 marks for the formula, 1 mark for substitution, 1 mark for the value with unit.)

Section C: Transformers and Turns Ratio (18 marks)

An ideal transformer is connected to a 220 V AC supply. The primary has 1100 turns and the secondary 55 turns. The secondary delivers 44 W.

#	Question	Answer and marking notes
9	Find the primary to secondary turns ratio.	Answer: the turns ratio is 1100 divided by 55, that is 20 to 1. (Marking: 2 marks for the division, 2 marks for stating the ratio.)
10	Find the secondary voltage.	Answer: secondary voltage equals primary voltage divided by 20, that is 220 divided by 20, so 11 V. (Marking: 2 marks for the formula, 2 marks for substitution, 1 mark for the value with unit.)
11	Find the secondary current.	Answer: secondary current equals power divided by voltage, that is 44 divided by 11, so 4.0 A. (Marking: 1 mark for the formula, 2 marks for substitution, 1 mark for the value with unit.)
12	If the efficiency is only 92 per cent, what is the actual output power? Where does the rest of the energy go?	Answer: output equals 44 times 0.92, that is 40.48 W, about 40.5 W. The remaining 3.5 W is lost as heat in the coil resistance and the iron core. (Marking: 3 marks for the calculation, 2 marks for the energy loss.)

Section D: Peak and rms Values of AC (16 marks)

The HK mains supply has an rms value of 220 V at 50 Hz. An electric kettle of rated power 1000 W is connected to it.

#	Question	Answer and marking notes
13	Find the peak voltage.	Answer: peak voltage equals square root of 2 times 220, that is 311 V to three significant figures. (Marking: 2 marks for the square root of 2 relation, 2 marks for the value with unit.)
14	Find the period of the supply.	Answer: period equals 1 divided by frequency, that is 1 divided by 50, so 0.020 s. (Marking: 2 marks for the formula, 2 marks for the value with unit.)
15	Find the resistance of the kettle.	Answer: resistance equals voltage squared divided by power, that is 220 squared divided by 1000, so 48.4 ohm. (Marking: 1 mark for the formula, 2 marks for substitution, 1 mark for the value with unit.)
16	Find the rms current through the kettle.	Answer: current equals power divided by voltage, that is 1000 divided by 220, so 4.55 A. (Marking: 1 mark for the formula, 2 marks for substitution, 1 mark for the value with unit.)

Section E: Rotating Coil Generator (17 marks)

The coil of an AC generator has 100 turns and area 0.015 m². It rotates uniformly at 60 revolutions per second in a uniform field of 0.80 T.

#	Question	Answer and marking notes
17	Find the maximum flux through the coil.	Answer: maximum flux equals field times area, that is 0.80 times 0.015, so 0.012 Wb. (Marking: 2 marks for the formula, 2 marks for the value with unit.)
18	Find the angular speed of the coil.	Answer: angular speed equals 2 pi times frequency, that is 2 pi times 60, so 377 rad per second. (Marking: 2 marks for the formula, 2 marks for substitution, 1 mark for the value with unit.)
19	Find the peak induced emf.	Answer: peak emf equals turns times maximum flux times angular speed, that is 100 times 0.012 times 377, so 452 V to three significant figures. (Marking: 2 marks for the formula, 1 mark for substitution, 1 mark for the value with unit.)
20	Why is the induced emf zero when the coil plane is perpendicular to the field?	Answer: the flux is then at its maximum but its rate of change is zero, and the induced emf depends on the rate of change rather than on the flux itself, so the instantaneous emf is zero. (Marking: 2 marks for the zero rate of change, 2 marks for the explanation.)

Section F: High Voltage Transmission and Power Loss (17 marks)

A power station transmits 50 kW to a town at 10 kV. The total resistance of the transmission line is 8.0 ohm. Assume a power factor of 1.

#	Question	Answer and marking notes
21	Find the current in the transmission line.	Answer: current equals power divided by voltage, that is 50000 divided by 10000, so 5.0 A. (Marking: 1 mark for the formula, 2 marks for substitution, 1 mark for the value with unit.)
22	Find the power lost in the line.	Answer: loss equals current squared times resistance, that is 5.0 squared times 8.0, so 200 W. (Marking: 1 mark for the formula, 2 marks for substitution, 1 mark for the value with unit.)
23	Find the percentage of the transmitted power lost.	Answer: the percentage equals 200 divided by 50000, times 100 per cent, so 0.40 per cent. (Marking: 2 marks for the formula, 2 marks for substitution, 1 mark for the answer.)
24	If the same power were sent at 400 V, how would the loss change? Use a formula to explain the benefit of high voltage transmission.	Answer: the voltage falls to one twenty-fifth, so the current rises to 125 A and the loss becomes 125 squared times 8.0, that is 125000 W, far more than the power sent, so it cannot work in practice. Since loss equals power over voltage, squared, times resistance, a higher voltage means a smaller current and the loss falls with the square of the voltage. (Marking: 2 marks for the larger current, 2 marks for the formula relation.)

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