Electromagnetic induction — Paper 4 · A Level Physics 9702

6 (a) Define magnetic flux density. … … … … [3] (b) A flat coil consists of N turns of wire and has area A. The coil is placed so that its plane is at an angle θ to a uniform magnetic field of flux density B, as shown in Fig. 6.1. flat coil area A θ magnetic field flux density B θ Fig. 6.1 Using the symbols A, B, N and θ and making reference to the magnetic flux in the coil, derive an expression for the magnetic flux linkage through the coil. [2] Use (c) (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) The magnetic flux density B in the coil is now made to vary with time t as shown in Fig. 6.2. B 0 0 t T 2T 3T Fig. 6.2 E 0 0 t T 2T 3T Fig. 6.3 On Fig. 6.3, sketch the variation with time t of the e.m.f. E induced in the coil. [3]

10 marks

Mark scheme: 6 (a) (numerically equal to) force per unit length …………………….…….… M1 on straight conductor carrying unit current ……………………………. A1 normal to the field ………………………………………………………… A1 [3] (b) flux through coil = BA sinθ ……………………………………………….. B1 flux linkage = BAN sinθ ………………..………………………………… B1 [2] (c) (i) (induced) e.m.f. proportional to ………………………………..…..…..… M1 rate of change of flux (linkage) …………………….……………………. A1 [2] (ii) graph: two square sections in correct positions, zero elsewhere ….. B1 pulses in opposite directions …………………………………… B1 amplitude of second about twice amplitude of first ………….. B1 [3]

This question in 9702/41 Oct/Nov 2005

5 A metal disc is swinging freely between the poles of an electromagnet, as shown in Fig. 5.1. metal disc pole-piece of electromagnet Fig. 5.1 When the electromagnet is switched on, the disc comes to rest after a few oscillations. (a) (i) State Faraday’s law of electromagnetic induction and use the law to explain why an e.m.f. is induced in the disc. … … … … [2] (ii) Explain why eddy currents are induced in the metal disc. … … … [2] (b) Use energy principles to explain why the disc comes to rest after a few oscillations. … … … … [3]

7 marks

Mark scheme: 5 (a) (i) (induced) e.m.f proportional/equal to rate of change of flux (linkage) B1 (allow ‘induced voltage, induced p.d.) flux is cust as the disc moves M1 hence inducing an e.m.f A0 [2] (ii) field in disc is not uniform/rate of cutting not same/speed of disc not same (over whole disc) B1 so different e.m.f.’s in different parts of disc M1 lead to eddy currents A0 [2] (b) eddy currents dissipate thermal energy in disc B1 energy derived from oscillation of disc B1 energy of disc depends on amplitude of oscillations B1 [3] GCE A/AS LEVEL - OCT/NOV 2006 9702 04

This question in 9702/41 Oct/Nov 2006

7 A magnet is suspended vertically from a fixed point by means of a spring, as shown in Fig. 7.1. spring magnet R coil Fig. 7.1 One end of the magnet hangs inside a coil of wire. The coil is connected in series with a resistor R. (a) The magnet is displaced vertically a small distance D and then released. Fig. 7.2 shows the variation with time t of the vertical displacement d of the magnet from its equilibrium position. +D d 0 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 t /s –D Fig. 7.2 Examiner’s Use (i) State and explain, by reference to electromagnetic induction, the nature of the oscillations of the magnet. … … … … … … [5] (ii) Calculate the angular frequency ω0 of the oscillations. ω0 = …………….……………… rad s–1 [2] (b) The resistance of the resistor R is increased. The magnet is again displaced a vertical distance D and released. On Fig. 7.2, sketch the variation with time t of the displacement d of the magnet. [2] Examiner’s Use (c) The resistor R in Fig. 7.1 is replaced by a variable-frequency signal generator of constant r.m.s. output voltage. The angular frequency ω of the generator is gradually increased from about 0.7ω0 to about 1.3ω0, where ω0 is the angular frequency calculated in (a)(ii). (i) On the axes of Fig. 7.3, sketch a graph to show the variation with ω of the amplitude A of the oscillations of the magnet. [2] A 0 0.7ω0 1.0ω0 1.3ω0 ω Fig. 7.3 (ii) State the name of the phenomenon illustrated in the graph of Fig. 7.3. … [1] (iii) Briefly describe one situation where the phenomenon named in (ii) is useful and one situation where it should be avoided. useful: … … avoid: … … [2]

14 marks

Mark scheme: 7 (a) (i) oscillations are damped/amplitude decreases B1 as magnet moves, flux is cut by coil B1 e.m.f./current is induced in the coil B1 causing energy loss in load OR force on magnet B1 energy is derived from oscillations of magnet OR force opposes motion of magnet B1 [5] (ii) T = 0.60 s C1 ω0 (= 2π/T) = 10.5 rad s–1 A1 [2] (b) sketch: sinusoidal wave with period unchanged or slightly smaller M1 same initial displacement, less damping A1 [2] (c) (i) sketch: general shape – peaked curve M1 peak at ω0 and amplitude never zero A1 [2] (ii) resonance B1 [1] (iii) useful: e.g. child on swing, microwave oven heating B1 avoid: e.g. vibrating panels, vibrating bridges B1 [2] (for credit, stated example must be put in context) Section B

This question in 9702/41 May/June 2007

6 A small rectangular coil ABCD contains 140 turns of wire. The sides AB and BC of the coil For are of lengths 4.5 cm and 2.8 cm respectively, as shown in Fig. 6.1. Examiner’s Use pole-piece of magnet 2.8cm B 4.5cm C A D axis of rotation Fig. 6.1 The coil is held between the poles of a large magnet so that the coil can rotate about an axis through its centre. The magnet produces a uniform magnetic field of flux density B between its poles. When the current in the coil is 170 mA, the maximum torque produced in the coil is 2.1 × 10–3 N m. (a) For the coil in the position for maximum torque, state whether the plane of the coil is parallel to, or normal to, the direction of the magnetic field. … [1] (b) For the coil in the position shown in Fig. 6.1, calculate the magnitude of the force on (i) side AB of the coil, force = … N [2] (ii) side BC of the coil. For Examiner’s Use force = … N [1] (c) Use your answer to (b)(i) to show that the magnetic flux density B between the poles of the magnet is 70 mT. [2] (d) (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) The current in the coil in (a) is switched off and the coil is positioned as shown in Fig. 6.1. The coil is then turned through an angle of 90° in a time of 0.14 s. Calculate the average e.m.f. induced in the coil. e.m.f. = … V [3]

11 marks

Mark scheme: 6 (a) parallel (to the field) B1 [1] (b) (i) torque = F × d 2.1 × 10–3 = F × 2.8 × 10–2 C1 F = 0.075 N A1 [2] (use of 4.5 cm scores no marks) (ii) zero A1 [1] (c) F = BILN(sinθ) C1 0.075 = B × 0.170 × 4.5 × 10–2 × 140 M1 B = 7.0 × 10–2 T = 70 mT A0 [2] (d) (i) (induced) e.m.f. is proportional to / equal to rate of change of M1 (magnetic) flux (linkage) A1 [2] (ii) change in flux linkage = BAN = 0.070 × 4.5 × 10–2 × 2.8 × 10–2 × 140 C1 = 0.0123 Wb turns induced e.m.f = 0.0123 / 0.14 C1 = 88 mV A1 [3] (Note: This is a simplified treatment. A full treatment would involve the averaging of B cosθ leading to a √2 factor)

This question in 9702/41 May/June 2008

6 A simple iron-cored transformer is illustrated in Fig. 6.1. For Examiner’s Use laminated soft-iron core primary secondary coil coil Fig. 6.1 (a) Suggest why the core is (i) a continuous loop, … … [1] (ii) laminated. … … … [2] (b) (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) Use Faraday’s law to explain the operation of the transformer. … … … … [3] (c) State two advantages of the use of alternating voltages for the transmission and use of For electrical energy. Examiner’s Use 1. … … 2. … … [2]

10 marks

Mark scheme: 6 (a) (i) either prevent loss of magnetic flux or improves flux linkage with secondary B1 [1] (ii) reduces eddy current (losses) B1 reduces losses of energy (in core) B1 [2] (b) (i) (induced) e.m.f. proportional to / equal to M1 rate of change of (magnetic) flux (linkage) A1 [2] (ii) changing current in primary gives rise to (1) changing flux in core (1) flux links with the secondary coil (1) changing flux in secondary coil, inducing e.m.f. (1) GCE A/AS LEVEL – October/November 2008 9702 04 (any three, 1 each to max 3) B3 [3] (c) e.g. can change voltage easily / efficiently high voltage transmission reduces power losses (any two sensible suggestions, 1 each) B2 [2]

This question in 9702/41 Oct/Nov 2008

7 You are provided with a coil of wire, a bar magnet and a sensitive ammeter. For Examiner’s Outline an experiment to verify Lenz’s law. Use … … … … … … … … [6]

6 marks

Mark scheme: 7 coil in series with meter (do not allow inclusion of a cell) B1 push known pole into coil B1 observe current direction (not reading) B1 (induced) field / field from coil repels magnet B1 either states rule to determine direction of magnetic field in coil or reversing magnet direction gives opposite deflection on meter B1 direction of induced current such as to oppose the change producing it B1 [6]

This question in 9702/41 May/June 2009

6 An ideal transformer is illustrated in Fig. 6.1. For Examiner’s Use primary coil secondary coil input output soft-iron core Fig. 6.1 (a) (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) Use the law to explain why a transformer will not operate using a direct current input. … … … [2] (b) (i) State Lenz’s law. … … … [2] (ii) Use Lenz’s law to explain why the input potential difference and the output e.m.f. are not in phase. … … … [2] (c) Electrical energy is usually transmitted using alternating high voltages. For Examiner’s Suggest one advantage, for the transmission of electrical energy, of using Use (i) alternating voltage, … … [1] (ii) high voltage. … … [1]

10 marks

Mark scheme: 6 (a) (i) e.m.f. induced proportional / equal to … M1 rate of change of (magnetic) flux (linkage) … A1 [2] (ii) e.m.f. (induced) only when flux is changing / cut … B1 direct current gives constant flux … B1 [2] (b) (i) (induced) e.m.f. / current acts in such a direction to produce effects … B1 to oppose the change causing it … B1 [2] (ii) (induced) current in secondary produces magnetic field … M1 opposes (changing) field produced in primary … M1 so not in phase … A0 [2] (c) (i) alternating means that voltage / current is easy to change … B1 [1] (ii) high voltage means less power / energy loss (during transmission) … B1 [1] [Total: 10]

This question in 9702/42 Oct/Nov 2009

6 A simple iron-cored transformer is illustrated in Fig. 6.1. For Examiner’s iron Use core input output primary secondary coil coil Fig. 6.1 (a) (i) State why the primary and secondary coils are wound on a core made of iron. … … … [1] (ii) Suggest why thermal energy is generated in the core when the transformer is in use. … … … … [3] (b) The root-mean-square (r.m.s.) voltage and current in the primary coil are VP and IP For respectively. Examiner’s The r.m.s. voltage and current in the secondary coil are VS and IS respectively. Use (i) Explain, by reference to direct current, what is meant by the root-mean-square value of an alternating current. … … … [2] (ii) Show that, for an ideal transformer, VS IP = . VP IS [2]

8 marks

Mark scheme: 6 (a) (i) e.g. prevent flux losses / improve flux linkage B1 [1] (ii) flux in core is changing B1 e.m.f. / current (induced) in core B1 induced current in core causes heating B1 [3] (b) (i) that value of the direct current producing same (mean) power / heating M1 in a resistor A1 [2] (ii) power in primary = power in secondary M1 VP IP = VS IS A1 [2]

This question in 9702/41 Oct/Nov 2010

6 A simple iron-cored transformer is illustrated in Fig. 6.1. For Examiner’s iron Use core input output primary secondary coil coil Fig. 6.1 (a) (i) State why the primary and secondary coils are wound on a core made of iron. … … … [1] (ii) Suggest why thermal energy is generated in the core when the transformer is in use. … … … … [3] (b) The root-mean-square (r.m.s.) voltage and current in the primary coil are VP and IP For respectively. Examiner’s The r.m.s. voltage and current in the secondary coil are VS and IS respectively. Use (i) Explain, by reference to direct current, what is meant by the root-mean-square value of an alternating current. … … … [2] (ii) Show that, for an ideal transformer, VS IP = . VP IS [2]

8 marks

Mark scheme: 6 (a) (i) e.g. prevent flux losses / improve flux linkage B1 [1] (ii) flux in core is changing B1 e.m.f. / current (induced) in core B1 induced current in core causes heating B1 [3] (b) (i) that value of the direct current producing same (mean) power / heating M1 in a resistor A1 [2] (ii) power in primary = power in secondary M1 VP IP = VS IS A1 [2]

This question in 9702/42 Oct/Nov 2010

5 The poles of a horseshoe magnet measure 5.0 cm × 2.4 cm, as shown in Fig. 5.1. For Examiner’s Use direction of A movement of wire copper wire 5.0 cm pole piece 2.4 cm of magnet Fig. 5.1 The uniform magnetic flux density between the poles of the magnet is 89 mT. Outside the region of the poles, the magnetic flux density is zero. A stiff copper wire is connected to a sensitive ammeter of resistance 0.12 Ω. A student moves the wire at a constant speed of 1.8 m s–1 between the poles in a direction parallel to the faces of the poles. (a) Calculate the magnetic flux between the poles of the magnet. magnetic flux = … Wb [2] (b) (i) Use your answer in (a) to determine, for the wire moving between the poles of the magnet, the e.m.f. induced in the wire. e.m.f. = … V [3] (ii) Show that the reading on the ammeter is approximately 70 mA. For Examiner’s Use [1] (c) By reference to Lenz’s law, a force acts on the wire to oppose the motion of the wire. The student who moved the wire between the poles of the magnet claims not to have felt this force. Explain quantitatively a reason for this claim. … … [3]

9 marks

Mark scheme: 5 (a) magnetic flux = BA = 89 × 10–3 × 5.0 × 10–2 × 2.4 × 10–2 C1 = 1.07 × 10–4 Wb A1 [2] (b) (i) e.m.f. = ∆φ / ∆t C1 (for ∆φ = 1.07 × 10–4 Wb), ∆t = 2.4 × 10–2 / 1.8 = 1.33 × 10–2 s C1 e.m.f. = (1.07 × 10–4) / (1.33 × 10–2) = 8.0 × 10–3 V A1 [3] (ii) current = 8.0 × 10–3 / 0.12 M1 ≈ 70 mA A0 [1] (c) force on wire = BIL = 89 × 10–3 × 70 × 10–3 × 5.0 × 10–2 C1 ≈ 3 × 10–4 (N) M1 suitable comment e.g. this force is too / very small (to be felt) A1 [3] 2

This question in 9702/43 Oct/Nov 2010

6 A transformer is illustrated in Fig. 6.1. For Examiner’s laminated iron Use core load primary secondary coil coil Fig. 6.1 (a) (i) Explain why the coils are wound on a core made of iron. … … [1] (ii) Suggest why thermal energy is generated in the core. … … … [2] (b) (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) Use Faraday’s law to explain why the potential difference across the load and the e.m.f. of the supply are not in phase. … … … … [2] (c) Electrical energy is usually transmitted using alternating current. Suggest why the For transmission is achieved using Examiner’s Use (i) high voltages, … … … [2] (ii) alternating current. … … [1]

10 marks

Mark scheme: 6 (a) (i) to concentrate the (magnetic) flux / reduce flux losses B1 [1] (ii) changing flux (in core) induces current in core M1 currents in core give rise to a heating effect A1 [2] (b) (i) e.m.f. induced proportional to M1 rate of change of (magnetic) flux (linkage) A1 [2] (ii) magnetic flux in phase with / proportional to e.m.f. / current in primary coil M1 e.m.f. / p.d. across secondary proportional to rate of change of flux M1 so e.m.f. of supply not in phase with p.d. across secondary A0 [2] (c) (i) for same power (transmission), high voltage with low current B1 with low current, less energy losses in transmission cables B1 [2] (ii) voltage is easily / efficiently changed B1 [1] GCE AS/A LEVEL – May/June 2011 9702 41

This question in 9702/41 May/June 2011

5 A bar magnet is suspended vertically from the free end of a helical spring, as shown in For Fig. 5.1. Examiner’s Use helical spring magnet coil V Fig. 5.1 One pole of the magnet is situated in a coil. The coil is connected in series with a high-resistance voltmeter. The magnet is displaced vertically and then released. The variation with time t of the reading V of the voltmeter is shown in Fig. 5.2. V 0 0 0.5 1.0 1.5 2.0 2.5 t / s Fig. 5.2 (a) (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) Use Faraday’s law to explain why For Examiner’s 1. there is a reading on the voltmeter, Use … … [1] 2. this reading varies in magnitude, … … [1] 3. the reading has both positive and negative values. … … [1] (b) Use Fig. 5.2 to determine the frequency f0 of the oscillations of the magnet. f0 = … Hz [2] (c) The magnet is now brought to rest and the voltmeter is replaced by a variable frequency alternating current supply that produces a constant r.m.s. current in the coil. The frequency of the supply is gradually increased from 0.7 f0 to 1.3 f0, where f0 is the frequency calculated in (b). On the axes of Fig. 5.3, sketch a graph to show the variation with frequency f of the amplitude A of the new oscillations of the bar magnet. A 0 0.7 f0 f0 1.3 f0 f [2] Fig. 5.3 (d) (i) Name the phenomenon illustrated on your completed graph of Fig. 5.3. For Examiner’s … [1] Use (ii) State one situation where the phenomenon named in (i) is useful. … … [1]

11 marks

Mark scheme: 5 (a) (i) (induced) e.m.f. proportional to M1 rate of change of (magnetic) flux (linkage) / rate of flux cutting A1 [2] (ii) 1. moving magnet causes change of flux linkage B1 [1] 2. speed of magnet varies so varying rate of change of flux B1 [1] 3. magnet changes direction of motion (so current changes direction) B1 [1] (b) period = 0.75 s C1 frequency = 1.33 Hz A1 [2] (c) graph: smooth correctly shaped curve with peak at f0 M1 A never zero A1 [2] (d) (i) resonance B1 [1] (ii) e.g. quartz crystal for timing / production of ultrasound A1 [1]

This question in 9702/42 May/June 2011

5 A bar magnet is suspended vertically from the free end of a helical spring, as shown in For Fig. 5.1. Examiner’s Use helical spring magnet coil V Fig. 5.1 One pole of the magnet is situated in a coil. The coil is connected in series with a high-resistance voltmeter. The magnet is displaced vertically and then released. The variation with time t of the reading V of the voltmeter is shown in Fig. 5.2. V 0 0 0.5 1.0 1.5 2.0 2.5 t / s Fig. 5.2 (a) (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) Use Faraday’s law to explain why For Examiner’s 1. there is a reading on the voltmeter, Use … … [1] 2. this reading varies in magnitude, … … [1] 3. the reading has both positive and negative values. … … [1] (b) Use Fig. 5.2 to determine the frequency f0 of the oscillations of the magnet. f0 = … Hz [2] (c) The magnet is now brought to rest and the voltmeter is replaced by a variable frequency alternating current supply that produces a constant r.m.s. current in the coil. The frequency of the supply is gradually increased from 0.7 f0 to 1.3 f0, where f0 is the frequency calculated in (b). On the axes of Fig. 5.3, sketch a graph to show the variation with frequency f of the amplitude A of the new oscillations of the bar magnet. A 0 0.7 f0 f0 1.3 f0 f [2] Fig. 5.3 (d) (i) Name the phenomenon illustrated on your completed graph of Fig. 5.3. For Examiner’s … [1] Use (ii) State one situation where the phenomenon named in (i) is useful. … … [1]

11 marks

Mark scheme: 5 (a) (i) (induced) e.m.f. proportional to M1 rate of change of (magnetic) flux (linkage) / rate of flux cutting A1 [2] (ii) 1. moving magnet causes change of flux linkage B1 [1] 2. speed of magnet varies so varying rate of change of flux B1 [1] 3. magnet changes direction of motion (so current changes direction) B1 [1] (b) period = 0.75 s C1 frequency = 1.33 Hz A1 [2] (c) graph: smooth correctly shaped curve with peak at f0 M1 A never zero A1 [2] (d) (i) resonance B1 [1] (ii) e.g. quartz crystal for timing / production of ultrasound A1 [1]

This question in 9702/43 May/June 2011

7 (a) State Lenz’s law. For Examiner’s … Use … … [2] (b) A simple transformer with a soft-iron core is illustrated in Fig. 7.1. laminated soft-iron core input output primary coil secondary coil Fig. 7.1 (i) Explain why the core is 1. made of iron, … … [1] 2. laminated. … … … [2] (ii) An e.m.f. is induced in the secondary coil of the transformer. Explain how a current in the primary coil gives rise to this induced e.m.f. … … … … … … [4]

9 marks

Mark scheme: 7 (a) induced e.m.f./current produces effects / acts in such a direction / tends M1 to oppose the change causing it A1 [2] (b) (i) 1. to reduce flux losses / increase flux linkage / easily magnetised and demagnetised B1 [1] 2. to reduce energy / heat losses (do not allow ‘to prevent energy losses’) M1 caused by eddy currents A1 [2] (allow 1 mark for ‘reduce eddy currents’) (ii) alternating current / voltage B1 gives rise to (changing) flux in core B1 flux links the secondary coil M1 (by Faraday’s law) changing flux induces e.m.f. (in secondary coil) A1 [4]

This question in 9702/41 Oct/Nov 2012

7 (a) State Lenz’s law. For Examiner’s … Use … … [2] (b) A simple transformer with a soft-iron core is illustrated in Fig. 7.1. laminated soft-iron core input output primary coil secondary coil Fig. 7.1 (i) Explain why the core is 1. made of iron, … … [1] 2. laminated. … … … [2] (ii) An e.m.f. is induced in the secondary coil of the transformer. Explain how a current in the primary coil gives rise to this induced e.m.f. … … … … … … [4]

9 marks

Mark scheme: 7 (a) induced e.m.f./current produces effects / acts in such a direction / tends M1 to oppose the change causing it A1 [2] (b) (i) 1. to reduce flux losses / increase flux linkage / easily magnetised and demagnetised B1 [1] 2. to reduce energy / heat losses (do not allow ‘to prevent energy losses’) M1 caused by eddy currents A1 [2] (allow 1 mark for ‘reduce eddy currents’) (ii) alternating current / voltage B1 gives rise to (changing) flux in core B1 flux links the secondary coil M1 (by Faraday’s law) changing flux induces e.m.f. (in secondary coil) A1 [4]

This question in 9702/42 Oct/Nov 2012

5 (a) State the relation between magnetic flux density B and magnetic flux Φ, explaining any For other symbols you use. Examiner’s Use … … … [2] (b) A large horseshoe magnet has a uniform magnetic field between its poles. The magnetic field is zero outside the space between the poles. A small Hall probe is moved at constant speed along a line XY that is midway between, and parallel to, the faces of the poles of the magnet, as shown in Fig. 5.1. Hall probe pole of X magnet pole of magnet Y Fig. 5.1 An e.m.f. is produced by the Hall probe when it is in the magnetic field. For The angle between the plane of the probe and the direction of the magnetic field is not Examiner’s varied. Use On the axes of Fig. 5.2, sketch a graph to show the variation with time t of the e.m.f. VH produced by the Hall probe. VH 0 t probe enters probe leaves magnetic magnetic field field Fig. 5.2 [2] (c) (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) The Hall probe in (b) is replaced by a small flat coil of wire. The coil is moved at constant speed along the line XY. The plane of the coil is parallel to the faces of the poles of the magnet. On the axes of Fig. 5.3, sketch a graph to show the variation with time t of the e.m.f. E induced in the coil. E 0 t coil enters coil leaves magnetic magnetic field field Fig. 5.3 [3]

9 marks

Mark scheme: 5 (a) either φ = BA sinθ M1 where A is the area (through which flux passes) θ is the angle between B and (plane of) A A1 or φ = BA (M1) where A is area normal to B (A1) [2] (b) graph: VH constant and non zero between the poles and zero outside M1 sharp increase/decrease at ends of magnet A1 [2] GCE AS/A LEVEL – October/November 2012 9702 43 (c) (i) (induced) e.m.f. proportional to M1 rate of change of (magnetic) flux (linkage) A1 [2] (ii) short pulse on entering and on leaving region between poles M1 pulses approximately the same shape but opposite polarities A1 e.m.f. zero between poles and outside A1 [3]

This question in 9702/43 Oct/Nov 2012

5 (a) Define the tesla. For Examiner’s … Use … … [2] (b) A long solenoid has an area of cross-section of 28 cm2, as shown in Fig. 5.1. solenoid area of cross-section 28 cm2 coil C 160 turns Fig. 5.1 A coil C consisting of 160 turns of insulated wire is wound tightly around the centre of the solenoid. The magnetic flux density B at the centre of the solenoid is given by the expression B = μ0n I where I is the current in the solenoid, n is a constant equal to 1.5 × 103 m–1 and is μ0 the permeability of free space. Calculate, for a current of 3.5 A in the solenoid, (i) the magnetic flux density at the centre of the solenoid, flux density = … T [2] (ii) the flux linkage in the coil C. For Examiner’s Use flux linkage = … Wb [2] (c) (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) The current in the solenoid in (b) is reversed in direction in a time of 0.80 s. Calculate the average e.m.f. induced in coil C. e.m.f. = … V [2]

10 marks

Mark scheme: 5 (a) (uniform magnetic) flux normal to long (straight) wire carrying a current of 1 A M1 (creates) force per unit length of 1 N m–1 A1 [2] (b) (i) flux density = 4π × 10–7 × 1.5 × 103 × 3.5 C1 = 6.6 × 10–3 T A1 [2] (ii) flux linkage = 6.6 × 10–3 × 28 × 10–4 × 160 C1 = 3.0 × 10–3 Wb A1 [2] (c) (i) (induced) e.m.f. proportional to rate of M1 change of (magnetic) flux (linkage) A1 [2] (ii) e.m.f. = (2 × 3.0 × 10–3) / 0.80 C1 = 7.4 × 10–3 V A1 [2]

This question in 9702/41 May/June 2013

6 A simple transformer is illustrated in Fig. 6.1. For Examiner’s Use load input resistor primary secondary coil coil laminated iron core Fig. 6.1 (a) State (i) why the iron core is laminated, … … … [2] (ii) what is meant by an ideal transformer. … … [1] (b) An ideal transformer has 300 turns on the primary coil and 8100 turns on the secondary coil. The root-mean-square input voltage to the primary coil is 9.0 V. Calculate the peak voltage across the load resistor connected to the secondary coil. peak voltage = … V [2]

5 marks

Mark scheme: 6 (a) (i) to reduce power loss in the core B1 due to eddy currents/induced currents B1 [2] (ii) either no power loss in transformer or input power = output power B1 [1] (b) either r.m.s. voltage across load = 9.0 × (8100 / 300) C1 peak voltage across load = √2 × 243 = 340 V A1 [2] or peak voltage across primary coil = 9.0 × √2 (C1) peak voltage across load = 12.7 × (8100/300) = 340 V (A1)

This question in 9702/41 May/June 2013

6 (a) State Faraday’s law of electromagnetic induction. For Examiner’s … Use … … [2] (b) The output of an ideal transformer is connected to a bridge rectifier, as shown in Fig. 6.1. 240 V r.m.s. load resistor Fig. 6.1 The input to the transformer is 240 V r.m.s. and the maximum potential difference across the load resistor is 9.0 V. (i) On Fig. 6.1, mark with the letter P the positive output from the rectifier. [1] (ii) Calculate the ratio number of turns on primary coil . number of turns on secondary coil ratio = … [3] (c) The variation with time t of the potential difference V across the load resistor in (b) is For shown in Fig. 6.2. Examiner’s Use V 0 t Fig. 6.2 A capacitor is now connected in parallel with the load resistor to produce some smoothing. (i) Explain what is meant by smoothing. … … [1] (ii) On Fig. 6.2, draw the variation with time t of the smoothed output potential difference. [2]

9 marks

Mark scheme: 6 (a) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 [2] (b) (i) positive terminal identified (upper connection to load) B1 [1] (ii) VP = √2 × VRMS C1 ratio = 240 √2 / 9 C1 ratio = 38 A1 [3] (VP = VRMS / √2 gives ratio = 18.9 and scores 1/3) (ratio = 240 / 9 = 26.7 scores 1/3) (ratio = 9 / (240 / √2) = 0.0265 is inverted ratio and scores 1/3) (c) (i) e.g. (output) p.d. / voltage / current does not fall to zero e.g. range of (output) p.d. / voltage / current is reduced (any sensible answer) B1 [1] (ii) sketch: same peak value at start of discharge M1 correct shape between one peak and the next A1 [2]

This question in 9702/42 May/June 2013

5 (a) Define the tesla. For Examiner’s … Use … … [2] (b) A long solenoid has an area of cross-section of 28 cm2, as shown in Fig. 5.1. solenoid area of cross-section 28 cm2 coil C 160 turns Fig. 5.1 A coil C consisting of 160 turns of insulated wire is wound tightly around the centre of the solenoid. The magnetic flux density B at the centre of the solenoid is given by the expression B = μ0n I where I is the current in the solenoid, n is a constant equal to 1.5 × 103 m–1 and is μ0 the permeability of free space. Calculate, for a current of 3.5 A in the solenoid, (i) the magnetic flux density at the centre of the solenoid, flux density = … T [2] (ii) the flux linkage in the coil C. For Examiner’s Use flux linkage = … Wb [2] (c) (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) The current in the solenoid in (b) is reversed in direction in a time of 0.80 s. Calculate the average e.m.f. induced in coil C. e.m.f. = … V [2]

10 marks

Mark scheme: 5 (a) (uniform magnetic) flux normal to long (straight) wire carrying a current of 1 A M1 (creates) force per unit length of 1 N m–1 A1 [2] (b) (i) flux density = 4π × 10–7 × 1.5 × 103 × 3.5 C1 = 6.6 × 10–3 T A1 [2] (ii) flux linkage = 6.6 × 10–3 × 28 × 10–4 × 160 C1 = 3.0 × 10–3 Wb A1 [2] (c) (i) (induced) e.m.f. proportional to rate of M1 change of (magnetic) flux (linkage) A1 [2] (ii) e.m.f. = (2 × 3.0 × 10–3) / 0.80 C1 = 7.4 × 10–3 V A1 [2]

This question in 9702/43 May/June 2013

6 A simple transformer is illustrated in Fig. 6.1. For Examiner’s Use load input resistor primary secondary coil coil laminated iron core Fig. 6.1 (a) State (i) why the iron core is laminated, … … … [2] (ii) what is meant by an ideal transformer. … … [1] (b) An ideal transformer has 300 turns on the primary coil and 8100 turns on the secondary coil. The root-mean-square input voltage to the primary coil is 9.0 V. Calculate the peak voltage across the load resistor connected to the secondary coil. peak voltage = … V [2]

5 marks

Mark scheme: 6 (a) (i) to reduce power loss in the core B1 due to eddy currents/induced currents B1 [2] (ii) either no power loss in transformer or input power = output power B1 [1] (b) either r.m.s. voltage across load = 9.0 × (8100 / 300) C1 peak voltage across load = √2 × 243 = 340 V A1 [2] or peak voltage across primary coil = 9.0 × √2 (C1) peak voltage across load = 12.7 × (8100/300) = 340 V (A1)

This question in 9702/43 May/June 2013

5 (a) An incomplete diagram for the magnetic flux pattern due to a current-carrying solenoid For is illustrated in Fig. 5.1. Examiner’s Use direction of current Fig. 5.1 (i) On Fig. 5.1, draw arrows on the field lines to show the direction of the magnetic field. [1] (ii) State the feature of Fig. 5.1 that indicates that the magnetic field strength at each end of the solenoid is less than that at the centre. … [1] (b) A Hall probe is placed near one end of the solenoid in (a), as shown in Fig. 5.2. Y to circuit Hall probe for Hall probe X Fig. 5.2 The Hall probe is rotated about the axis XY. State and explain why the magnitude of the Hall voltage varies. … … … [2] (c) (i) State Faraday’s law of electromagnetic induction. For Examiner’s … Use … … [2] (ii) The Hall probe in (b) is replaced by a small coil of wire connected to a sensitive voltmeter. State three different ways in which an e.m.f. may be induced in the coil. 1. … … 2. … … 3. … … [3]

9 marks

Mark scheme: 5 (a) (i) field shown as right to left B1 [1] (ii) lines are more spaced out at ends B1 [1] (b) Hall voltage depends on angle M1 either between field and plane of probe or maximum when field normal to plane of probe or zero when field parallel to plane of probe A1 [2] (c) (i) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 [2] (allow rate of cutting of flux) (ii) e.g. move coil towards/away from solenoid e.g. rotate coil e.g. vary current in solenoid e.g. insert iron core into solenoid (any three sensible suggestions, 1 each) B3 [3] GCE A LEVEL – October/November 2013 9702 41

This question in 9702/41 Oct/Nov 2013

5 (a) An incomplete diagram for the magnetic flux pattern due to a current-carrying solenoid For is illustrated in Fig. 5.1. Examiner’s Use direction of current Fig. 5.1 (i) On Fig. 5.1, draw arrows on the field lines to show the direction of the magnetic field. [1] (ii) State the feature of Fig. 5.1 that indicates that the magnetic field strength at each end of the solenoid is less than that at the centre. … [1] (b) A Hall probe is placed near one end of the solenoid in (a), as shown in Fig. 5.2. Y to circuit Hall probe for Hall probe X Fig. 5.2 The Hall probe is rotated about the axis XY. State and explain why the magnitude of the Hall voltage varies. … … … [2] (c) (i) State Faraday’s law of electromagnetic induction. For Examiner’s … Use … … [2] (ii) The Hall probe in (b) is replaced by a small coil of wire connected to a sensitive voltmeter. State three different ways in which an e.m.f. may be induced in the coil. 1. … … 2. … … 3. … … [3]

9 marks

Mark scheme: 5 (a) (i) field shown as right to left B1 [1] (ii) lines are more spaced out at ends B1 [1] (b) Hall voltage depends on angle M1 either between field and plane of probe or maximum when field normal to plane of probe or zero when field parallel to plane of probe A1 [2] (c) (i) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 [2] (allow rate of cutting of flux) (ii) e.g. move coil towards/away from solenoid e.g. rotate coil e.g. vary current in solenoid e.g. insert iron core into solenoid (any three sensible suggestions, 1 each) B3 [3] GCE A LEVEL – October/November 2013 9702 42

This question in 9702/42 Oct/Nov 2013

5 A uniform magnetic field of flux density B makes an angle θ with a flat plane PQRS, as For shown in Fig. 5.1. Examiner’s Use Q P e e magnetic field flux density B e e R S Fig. 5.1 The plane PQRS has area A. (a) State (i) what is meant by a magnetic field, … … [1] (ii) an expression, in terms of A, B and θ, for the magnetic flux Φ through the plane PQRS. … [1] (b) A vertical aluminium window frame DEFG has width 52 cm and length 95 cm, as shown in Fig. 5.2. 52 cm E hinge D 95 cm hinge F G Fig. 5.2 The frame is hinged along the vertical edge DG. The horizontal component BH of the Earth’s magnetic field is 1.8 × 10–5 T. For the closed window, the frame is normal to the horizontal component BH. The window is opened so that the plane of the window rotates through 90°. (i) Explain why, when the window is opened, the change in magnetic flux linkage due For to the vertical component of the Earth’s magnetic field is zero. Examiner’s Use … … [1] (ii) Calculate, for the window opening through an angle of 90°, the change in magnetic flux linkage. change in flux linkage = … Wb [2] (c) (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) The window in (b) is opened in a time of 0.30 s. Use your answer in (b)(ii) to calculate the average e.m.f. induced in the window frame. e.m.f. = … V [1] (iii) State the sides of the window frame between which the e.m.f. is induced. between side … and side … [1]

9 marks

Mark scheme: 5 (a) (i) region (of space) either where a moving charge (may) experience a force or around a magnet where another magnet experiences a force B1 [1] (ii) (Φ =) BA sinθ A1 [1] (b) (i) plane of frame is always parallel to BV / flux linkage always zero B1 [1] (ii) ∆Φ = 1.8 × 10–5 × 52 × 10–2 × 95 × 10–2 C1 = 8.9 × 10–6 Wb A1 [2] (c) (i) (induced) e.m.f. proportional to rate of M1 change of (magnetic) flux (linkage) A1 [2] (allow rate of cutting of flux) (ii) e.m.f. = (8.9 × 10–6) / 0.30 = 3.0 × 10–5 V A1 [1] (iii) This question part was removed from the assessment. All candidates were awarded 1 mark. B1 [1] GCE A LEVEL – October/November 2013 9702 43

This question in 9702/43 Oct/Nov 2013

7 A solenoid is connected in series with a battery and a switch. A Hall probe is placed close to one end of the solenoid, as illustrated in Fig. 7.1. solenoid Hall probe Fig. 7.1 The current in the solenoid is switched on. The Hall probe is adjusted in position to give the maximum reading. The current is then switched off. (a) The current in the solenoid is now switched on again. Several seconds later, it is switched off. The Hall probe is not moved. On the axes of Fig. 7.2, sketch a graph to show the variation with time t of the Hall voltage VH. VH 0 t current current switched on switched off Fig. 7.2 [3] (b) The Hall probe is now replaced by a small coil. The plane of the coil is parallel to the end of the solenoid. (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) On the axes of Fig. 7.3, sketch a graph to show the variation with time t of the e.m.f. E induced in the coil when the current in the solenoid is switched on and then switched off. E 0 t current current switched on switched off Fig. 7.3 [3]

8 marks

Mark scheme: 7 (a) graph: VH increases from zero when current switched on B1 VH then non-zero constant B1 VH returns to zero when current switched off B1 [3] (b) (i) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 [2] (ii) pulse as current is being switched on B1 zero e.m.f. when current in coil B1 pulse in opposite direction when switching off B1 [3]

This question in 9702/41 May/June 2014

5 A Hall probe is placed a distance d from a long straight current-carrying wire, as illustrated in Fig. 5.1. current-carrying 4.0 A wire Hall probe X Y d Fig. 5.1 The direct current in the wire is 4.0 A. Line XY is normal to the wire. The Hall probe is rotated about the line XY to the position where the reading VH of the Hall probe is maximum. (a) The Hall probe is now moved away from the wire, along the line XY. On the axes of Fig. 5.2, sketch a graph to show the variation of the Hall voltage VH with distance x of the probe from the wire. Numerical values are not required on your sketch. VH 0 0 d x Fig. 5.2 [2] (b) The Hall probe is now returned to its original position, a distance d from the wire. At this point, the magnetic flux density due to the current in the wire is proportional to the current. For a direct current of 4.0 A in the wire, the reading of the Hall probe is 3.5 mV. The direct current is now replaced by an alternating current of root-mean-square (r.m.s.) value 4.0 A. The period of this alternating current is T. On the axes of Fig. 5.3, sketch the variation with time t of the reading of the Hall voltage VH for two cycles of the alternating current. Give numerical values for VH, where appropriate. 6 VH / mV 4 2 0 0 T 2T t –2 –4 –6 Fig. 5.3 [3] (c) A student suggests that the Hall probe in (a) is replaced with a small coil connected in series with a millivoltmeter. The constant current in the wire is 4.0 A. In order to obtain data to plot a graph showing the variation with distance x of the magnetic flux density, the student suggests that readings of the millivoltmeter are taken when the coil is held in position at different values of x. Comment on this suggestion. … … … … [2]

7 marks

Mark scheme: 5 (a) only curve with decreasing gradient M1 acceptable value near xR 0 and does not reach zero A1 [2] (if graph line less than 4.0 cm do not allow A1 mark) (no credit if graph line has positive and negative values of VH) (b) graph: from 0 to 2T, two cycles of a sinusoidal wave M1 all peaks above 3.5 mV C1 peaks at 4.95 / 5.0 mV (allow 4.8 mV to 5.2 mV) A1 [3] (c) e.m.f. induced in coil when magnetic field / flux is changing / cutting B1 either at each position, magnetic field does not vary so no e.m.f. is induced in the coil / no reading on the millivoltmeter or at each position, switch off current and take millivoltmeter reading or at each position, rapidly remove coil from field and take meter reading B1 [2]

This question in 9702/42 May/June 2014

7 A solenoid is connected in series with a battery and a switch. A Hall probe is placed close to one end of the solenoid, as illustrated in Fig. 7.1. solenoid Hall probe Fig. 7.1 The current in the solenoid is switched on. The Hall probe is adjusted in position to give the maximum reading. The current is then switched off. (a) The current in the solenoid is now switched on again. Several seconds later, it is switched off. The Hall probe is not moved. On the axes of Fig. 7.2, sketch a graph to show the variation with time t of the Hall voltage VH. VH 0 t current current switched on switched off Fig. 7.2 [3] (b) The Hall probe is now replaced by a small coil. The plane of the coil is parallel to the end of the solenoid. (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) On the axes of Fig. 7.3, sketch a graph to show the variation with time t of the e.m.f. E induced in the coil when the current in the solenoid is switched on and then switched off. E 0 t current current switched on switched off Fig. 7.3 [3]

8 marks

Mark scheme: 7 (a) graph: VH increases from zero when current switched on B1 VH then non-zero constant B1 VH returns to zero when current switched off B1 [3] (b) (i) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 [2] (ii) pulse as current is being switched on B1 zero e.m.f. when current in coil B1 pulse in opposite direction when switching off B1 [3]

This question in 9702/43 May/June 2014

1 A light spring is suspended from a fixed point. A bar magnet is attached to the end of the spring, as shown in Fig. 1.1. spring bar magnet cardboard cup Fig. 1.1 In order to shield the magnet from draughts, a cardboard cup is placed around the magnet but does not touch it. The magnet is displaced vertically and then released. The variation with time t of the vertical displacement y of the magnet is shown in Fig. 1.2. 2.0 y / cm 1.0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 t / s –1.0 –2.0 Fig. 1.2 The mass of the magnet is 130 g. (a) For the oscillations of the magnet, use Fig. 1.2 to (i) determine the angular frequency ω, ω = … rad s−1 [2] (ii) show that the maximum kinetic energy of the oscillating magnet is 6.4 mJ. [2] (b) The cardboard cup is now replaced with a cup made of aluminium foil. During 10 complete oscillations of the magnet, the amplitude of vibration is seen to decrease to 0.75 cm from that shown in Fig. 1.2. The change in angular frequency is negligible. (i) Use Faraday’s law of electromagnetic induction to explain why the amplitude of the oscillations decreases. … … … … … [3] (ii) Show that the loss in energy of the oscillating magnet is 4.8 mJ. [2] (c) The mass of the aluminium cup in (b) is 6.2 g. The specific heat capacity of aluminium is 910 J kg−1 K−1. The energy in (b)(ii) is transferred to the cup as thermal energy. Calculate the mean rise in temperature of the cup. temperature rise = … K [2] Please turn over for Question 2.

11 marks

Mark scheme: 1 (a) (i) either ω = 2π / T or ω = 2πf and f = 1 / T C1 ω = 2π / 0.30 = 20.9 rad s–1 (accept 2 s.f.) A1 [2] (ii) kinetic energy = ½mω 2x02 or v = ωx0 and ½mv2 C1 = ½ × 0.130 × 20.92 × (1.5 × 10–2)2 = 6.4 × 10–3 J A1 [2] (b) (i) as magnet moves, flux is cut by cup / aluminium giving rise to induced e.m.f. (in cup) B1 induced e.m.f. gives rise to currents and heating of the cup B1 thermal energy derived from oscillations of magnet so amplitude decreases B1 or induced e.m.f. gives rise to currents which generate a magnetic field (B1) the magnetic field opposes the motion of the magnet so amplitude decreases (B1) [3] (ii) either use of ½mω 2x02 and x0 = 0.75 cm or x0 is halved so ¼ energy C1 to give new energy = 1.6 mJ either loss in energy = 6.4 – 1.6 or loss = ¾ × 6.4 giving loss = 4.8 mJ A1 [2] (c) q = mc∆θ 4.8 × 10–3 = 6.2 × 10–3 × 910 × ∆θ C1 ∆θ = 8.5 × 10–4 K A1 [2]

This question in 9702/43 Oct/Nov 2014

7 (a) A solenoid is connected in series with a resistor, as shown in Fig. 7.1. N S motion of magnet Fig. 7.1 As the magnet is being moved into the solenoid, thermal energy is transferred in the resistor. Use laws of electromagnetic induction to explain the origin of this thermal energy. … … … … … … … [4] (b) Explain why the alternating current in the primary coil of a transformer is not in phase with the alternating e.m.f. induced in the secondary coil. … … … … … … … [4]

8 marks

Mark scheme: 7 (a) moving magnet gives rise to/causes/induces e.m.f./current in solenoid/coil B1 (induced current) creates field/flux in solenoid that opposes (motion of) magnet B1 work is done/energy is needed to move magnet (into solenoid) B1 (induced) current gives heating effect (in resistor) which comes from the work done B1 [4] (b) current in primary coil give rise to (magnetic) flux/field B1 (magnetic) flux / field (in core) is in phase with current (in primary coil) B1 (magnetic) flux threads/links/cuts secondary coil inducing e.m.f. in secondary coil B1 (there must be a mention of secondary coil) e.m.f. induced proportional to rate of change/cutting of flux/field so not in phase B1 [4] 13

This question in 9702/42 May/June 2015

6 Suggest an explanation for each of the following observations. (a) Two wires are laid side-by-side and carry equal currents I in opposite directions, as shown in Fig. 6.1. I I Fig. 6.1 The total magnetic flux density due to the current in the wires is negligible. … … … … … [3] (b) An air-cored solenoid is connected in series with a battery, as shown in Fig. 6.2. solenoid iron core Fig. 6.2 As an iron core is inserted into the solenoid, an e.m.f. that opposes the e.m.f. of the battery is induced in the solenoid. … … … … … [4]

7 marks

Mark scheme: 6 (a) magnetic fields are equal in magnitude/strength/flux density M1 magnetic fields are opposite in direction M1 fields superpose/add/cancel to give zero/negligible resultant field A1 [3] (b) core causes increase in magnetic flux in the solenoid/induced poles in core or field induced in core B1 changing flux threads/cuts the turns on the solenoid M1 (by Faraday’s law) an e.m.f. is induced in the solenoid A1 by Lenz’s law, this e.m.f. opposes the battery e.m.f. A1 [4]

This question in 9702/43 Oct/Nov 2015

10 A small coil of wire is situated in a non-uniform magnetic field, as shown in Fig. 10.1. coil, constant 40 turns non-uniform velocity magnetic field P x Fig. 10.1 The coil consists of 40 turns of wire and moves with a constant speed in a straight line. The coil has displacement x from a fixed point P. The variation with x of the magnetic flux Φ in the coil is shown in Fig. 10.2. 8 7 \ / 10–6 Wb 6 5 4 3 2 0 1 2 3 4 5 6 x / cm Fig. 10.2 (a) The coil is moved at constant speed between point P and the point where x = 3.0 cm. (i) Calculate the change in magnetic flux linkage of the coil. change in flux linkage = … Wb [1] (ii) The e.m.f. induced in the coil is 5.0 × 10–4 V. Determine the speed of the coil. speed = … m s–1 [2] (b) On Fig. 10.3, sketch the variation with x of the e.m.f. E induced in the coil for values of x from x = 0 to x = 6.0 cm. E 0 0 1 2 3 4 5 6 x / cm Fig. 10.3 [2] [Total: 5]

5 marks

Mark scheme: 10 (a) (i) change in flux linkage = 40 × (5.0 – 3.0) × 10–6 = 8(.0) × 10–5 Wb A1 [1] (ii) time taken = 8.0 × 10–5 / 5.0 × 10–4 = 0.16 (s) C1 speed = 3.0 × 10–2 / 0.16 = 0.19 (0.188) m s–1 A1 or E = (∆Φ / ∆x) × speed speed = 5.0 × 10–4 / (8.0 ×10–5 / 3.0 ×10–2) (C1) = 0.19 (0.188) m s–1 (A1) [2] (b) a constant non-zero value of E from 0 to 3 cm and a different constant non-zero value of E from 3 to 6 cm M1 E from 3–6 cm has the opposite sign to and larger value than E from 0–3 cm A1 [2]

This question in 9702/42 Feb/March 2016

10 (a) A coil of insulated wire is wound on a copper core, as illustrated in Fig. 10.1. insulated copper wire core Fig. 10.1 An alternating current is passed through the coil. The heating effect of the current in the coil is negligible. Explain why the temperature of the core rises. … … … … … … [4] (b) Two hollow tubes of equal length hang vertically as shown in Fig. 10.2. magnet A magnet B plastic aluminium tube tube Fig. 10.2 One tube is made of plastic and the other of aluminium. Two small similar bar magnets A and B are held above the tubes and then released simultaneously. The magnets do not touch the sides of the tubes. Explain why magnet B takes much longer than magnet A to fall through the tube. … … … … … … … [5] [Total: 9]

9 marks

Mark scheme: 10 (a) (non-uniform) magnetic flux in core is changing M1 induces (different) e.m.f. in (different parts of) the core A1 (eddy) currents form in the core M1 which give rise to heating A1 [4] (b) as magnet falls, tube cuts magnetic flux M1 e.m.f./(eddy) currents induced in metal/aluminium (tube) A1 (eddy) current heating of tube M1 with energy taken from falling magnet A1 or (eddy) currents produce magnetic field (M1) that opposes motion of magnet (A1) so magnet B has acceleration < g or magnet B has smaller acceleration/reaches terminal speed A1 [5]

This question in 9702/41 May/June 2016

10 Two coils P and Q are placed close to one another, as shown in Fig. 10.1. coil P coil Q V power supply Fig. 10.1 (a) The current in coil P is constant. An iron rod is inserted into coil P. Explain why, during the time that the rod is moving, there is a reading on the voltmeter connected to coil Q. … … … … [2] (b) The current in coil P is now varied as shown in Fig. 10.2. current 0 0 W1 W2 WLPH Fig. 10.2 On Fig. 10.3, show the variation with time of the reading of the voltmeter connected to coil Q for time t = 0 to time t = t2. voltmeter reading 0 0 W1 W2 time Fig. 10.3 [4] [Total: 6]

6 marks

Mark scheme: 10 (a) iron rod changes flux (density)/field B1 change of flux in coil Q causes induced e.m.f. B1 [2] (b) constant reading (either polarity) from time zero to near t1 B1 spike in one direction near t1 clearly showing a larger voltage M1 of opposite polarity A1 zero reading from near t1 to t2 B1 [4]

This question in 9702/42 May/June 2016

10 (a) A coil of insulated wire is wound on a copper core, as illustrated in Fig. 10.1. insulated copper wire core Fig. 10.1 An alternating current is passed through the coil. The heating effect of the current in the coil is negligible. Explain why the temperature of the core rises. … … … … … … [4] (b) Two hollow tubes of equal length hang vertically as shown in Fig. 10.2. magnet A magnet B plastic aluminium tube tube Fig. 10.2 One tube is made of plastic and the other of aluminium. Two small similar bar magnets A and B are held above the tubes and then released simultaneously. The magnets do not touch the sides of the tubes. Explain why magnet B takes much longer than magnet A to fall through the tube. … … … … … … … [5] [Total: 9]

9 marks

Mark scheme: 10 (a) (non-uniform) magnetic flux in core is changing M1 induces (different) e.m.f. in (different parts of) the core A1 (eddy) currents form in the core M1 which give rise to heating A1 [4] (b) as magnet falls, tube cuts magnetic flux M1 e.m.f./(eddy) currents induced in metal/aluminium (tube) A1 (eddy) current heating of tube M1 with energy taken from falling magnet A1 or (eddy) currents produce magnetic field (M1) that opposes motion of magnet (A1) so magnet B has acceleration < g or magnet B has smaller acceleration/reaches terminal speed A1 [5]

This question in 9702/43 May/June 2016

11 (a) State Faraday’s law of electromagnetic induction. … … … [2] (b) An alternating current is passed through an air-cored solenoid. An iron core is inserted into the solenoid and then held stationary within the solenoid. The current in the solenoid is now smaller. Explain why the root-mean-square (r.m.s.) value of the current in the solenoid is reduced as a result of inserting the core. … … … … … [3] (c) Practical transformers are very efficient. However, there are some power losses. State two sources of power loss within a transformer. 1. … … 2. … … [2] [Total: 7]

7 marks

Mark scheme: 11 (a) (induced) e.m.f. proportional/equal to rate M1 of change of (magnetic) flux (linkage) A1 [2] (b) (for same current) iron core gives large(r) (rates of change of) flux (linkage) B1 e.m.f induced in solenoid is greater (for same current) M1 induced e.m.f. opposes applied e.m.f. so current smaller/acts to reduce current A1 [3] or same supply so same induced e.m.f. balancing it (B1) (rate of change of) flux linkage is same (M1) smaller current for same flux when core present (A1) (c) e.g. (heating due to) eddy currents in core (heating due to current in) resistance of coils hysteresis losses/losses due to changing magnetic field in core Any two of the above marking points, 1 mark each B2 [2]

This question in 9702/42 Oct/Nov 2016

9 (a) State Faraday’s law of electromagnetic induction. … … … … [2] (b) The diameter of the cross-section of a long solenoid is 3.2 cm, as shown in Fig. 9.1. coil C long solenoid 85 turns 3.2 cm I I Fig. 9.1 A coil C, with 85 turns of wire, is wound tightly around the centre region of the solenoid. The magnetic flux density B, in tesla, at the centre of the solenoid is given by the expression B = π × 10−3 × I where I is the current in the solenoid in ampere. Show that, for a current I of 2.8 A in the solenoid, the magnetic flux linkage of the coil C is 6.0 × 10−4 Wb. [1] (c) The current I in the solenoid in (b) is reversed in 0.30 s. Calculate the mean e.m.f. induced in coil C. e.m.f. = … mV [2] (d) The current I in the solenoid in (b) is now varied with time t as shown in Fig. 9.2. 3.0 2.0 I/ A 1.0 0 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 t / s –1.0 –2.0 –3.0 Fig. 9.2 Use your answer to (c) to show, on Fig. 9.3, the variation with time t of the e.m.f. E induced in coil C. E / mV 00 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 t / s Fig. 9.3 [4]

9 marks

Mark scheme: 9 (a) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 [2] (b) flux linkage = BAN = π × 10–3 × 2.8 × π × (1.6 × 10–2)2 × 85 = 6.0 × 10–4 Wb B1 [1] (c) e.m.f. = ∆NΦ/ ∆t e.m.f. = (6.0 × 10–4 × 2) / 0.30 C1 e.m.f. = 4.0 mV A1 [2] (d) sketch: E = 0 for t = 0 → 0.3 s, 0.6 s → 1.0 s, 1.6 s → 2.0 s B1 E = 4 mV for t = 0.3 s → 0.6 s (either polarity) B1 E = 2 mV for t = 1.0 s → 1.6 s B1 with opposite polarity B1 [4]

This question in 9702/43 Oct/Nov 2016

9 An ideal transformer is shown in Fig. 9.1. core input output secondary primary coil coil 1200 turns Fig. 9.1 (a) Explain (i) why the core is made of iron, … … [1] (ii) why an electromotive force (e.m.f.) is not induced at the output when a constant direct voltage is at the input. … … … [2] (b) An alternating voltage of peak value 150 V is applied across the 1200 turns of the primary coil. The variation with time t of the e.m.f. E induced across the secondary coil is shown in Fig. 9.2. 60 E / V 40 20 0 0 5.0 10.0 15.0 20.0 25.0 t / ms –20 –40 –60 Fig. 9.2 Use data from Fig. 9.2 to (i) calculate the number of turns of the secondary coil, number = … [2] (ii) state one time when the magnetic flux linking the secondary coil is a maximum. time = … ms [1] (c) A resistor is connected between the output terminals of the secondary coil. The mean power dissipated in the resistor is 1.2 W. It may be assumed that the varying voltage across the resistor is equal to the varying e.m.f. E shown in Fig. 9.2. (i) Calculate the resistance of the resistor. resistance = … Ω [2] (ii) On Fig. 9.3, sketch the variation with time t of the power P dissipated in the resistor for t = 0 to t = 22.5 ms. 3.0 P / W 2.0 1.0 0 0 5.0 10.0 15.0 20.0 25.0 t / ms Fig. 9.3 [3] [Total: 11]

11 marks

Mark scheme: 9(a)(i) increase flux linkage (with secondary coil) / to reduce flux loss B1 9(a)(ii) e.m.f. (induced only) when flux (in core/coil) is changing B1 constant / direct voltage gives constant flux / field B1 9(b)(i) NS / NP = VS / VP C1 NS = (52 / 150) × 1200 = 416 turns A1 9(b)(ii) 0 ms or 7.5 ms or 15.0 ms or 22.5 ms A1 9(c)(i) either mean power = V 2 / 2R and V = 52 (V) C1 R = 522 / (2 × 1.2) = 1100 (1127) Ω A1 or mean power = V 2 / R and V = 52 / 2 (= 36.8 V) (C1) R = 36.82 / 1.2 = 1100 Ω (A1) 9(c)(ii) sinusoidal shape with troughs at zero power B1 only 3 ‘cycles’ B1 each ‘cycle’ is 2.4 W high and zero power at correct times B1

This question in 9702/42 Feb/March 2017

9 A simple transformer is illustrated in Fig. 9.1. laminated iron core input output Fig. 9.1 (a) (i) State why the transformer has an iron core, rather than having no core. … … [1] (ii) Explain why the core is laminated. … … … [2] (b) By reference to the action of a transformer, explain why the input to the transformer is an alternating voltage, rather than a constant voltage. … … … … … [3] [Total: 6]

6 marks

Mark scheme: 9(a)(i) core reduces loss of (magnetic) flux linkage/improves flux linkage B1 9(a)(ii) reduces (size of eddy) currents in core B1 (so that) heating of core is reduced B1 9(b) alternating voltage gives rise to changing magnetic flux in core M1 (changing) flux links the secondary coil A1 induced e.m.f. (in secondary) only when flux is changing/cut B1

This question in 9702/41 May/June 2017

9 A simple transformer is illustrated in Fig. 9.1. laminated iron core input output Fig. 9.1 (a) (i) State why the transformer has an iron core, rather than having no core. … … [1] (ii) Explain why the core is laminated. … … … [2] (b) By reference to the action of a transformer, explain why the input to the transformer is an alternating voltage, rather than a constant voltage. … … … … … [3] [Total: 6]

6 marks

Mark scheme: 9(a)(i) core reduces loss of (magnetic) flux linkage/improves flux linkage B1 9(a)(ii) reduces (size of eddy) currents in core B1 (so that) heating of core is reduced B1 9(b) alternating voltage gives rise to changing magnetic flux in core M1 (changing) flux links the secondary coil A1 induced e.m.f. (in secondary) only when flux is changing/cut B1

This question in 9702/43 May/June 2017

5 (a) (i) State Coulomb’s law for the force between two point charges. … … [1] (ii) Two point charges are situated in a vacuum and separated by a distance R. The force between the charges is FC. On Fig. 5.1, sketch a graph to show the variation of the force F between the charges with separation x for values of x from x = R to x = 4R. 1.0 Fc 0.8 Fc F 0.6 Fc 0.4 Fc 0.2 Fc 0 R 2R 3R 4R x [3] Fig. 5.1 (b) Two coils C and D are placed close to one another, as shown in Fig. 5.2. coil C coil D I e.m.f. E V Fig. 5.2 The variation with time t of the current I in coil C is shown in Fig. 5.3. On Fig. 5.4, show the variation with time t of the e.m.f. E induced in coil D for time t = 0 to time t = t5. I 0 0 t1 t2 t3 t4 t5 t Fig. 5.3 E 0 0 t1 t2 t3 t4 t5 t Fig. 5.4 [4] [Total: 8]

8 marks

Mark scheme: 5(a)(i) force proportional to product of charges and inversely proportional to square of separation A1 5(a)(ii) curve starting at (R, FC) B1 passing through (2R, 0.25FC) B1 passing through (4R, 0.06FC) B1 5(b) graph: E = 0 when current constant (0 to t1, t2 to t3, t4 to t5) B1 stepped from t1 to t2 and t3 to t4 B1 (steps) in opposite directions B1 later one larger in magnitude B1

This question in 9702/41 Oct/Nov 2017

5 (a) (i) State Coulomb’s law for the force between two point charges. … … [1] (ii) Two point charges are situated in a vacuum and separated by a distance R. The force between the charges is FC. On Fig. 5.1, sketch a graph to show the variation of the force F between the charges with separation x for values of x from x = R to x = 4R. 1.0 Fc 0.8 Fc F 0.6 Fc 0.4 Fc 0.2 Fc 0 R 2R 3R 4R x [3] Fig. 5.1 (b) Two coils C and D are placed close to one another, as shown in Fig. 5.2. coil C coil D I e.m.f. E V Fig. 5.2 The variation with time t of the current I in coil C is shown in Fig. 5.3. On Fig. 5.4, show the variation with time t of the e.m.f. E induced in coil D for time t = 0 to time t = t5. I 0 0 t1 t2 t3 t4 t5 t Fig. 5.3 E 0 0 t1 t2 t3 t4 t5 t Fig. 5.4 [4] [Total: 8]

8 marks

Mark scheme: 5(a)(i) force proportional to product of charges and inversely proportional to square of separation A1 5(a)(ii) curve starting at (R, FC) B1 passing through (2R, 0.25FC) B1 passing through (4R, 0.06FC) B1 5(b) graph: E = 0 when current constant (0 to t1, t2 to t3, t4 to t5) B1 stepped from t1 to t2 and t3 to t4 B1 (steps) in opposite directions B1 later one larger in magnitude B1

This question in 9702/43 Oct/Nov 2017

10 (a) (i) Define magnetic flux. … … … [2] (ii) State Faraday’s law of electromagnetic induction. … … … … [2] (b) A solenoid has a coil C of wire wound tightly about its centre, as shown in Fig. 10.1. coil C solenoid + – d.c. supply Fig. 10.1 The coil C has 96 turns. The uniform magnetic flux Φ (in weber) in the solenoid is given by the expression Φ = 6.8 × 10–6 × I where I is the current (in amperes) in the solenoid. Calculate the average electromotive force (e.m.f.) induced in coil C when a current of 3.5 A is reversed in the solenoid in a time of 2.4 ms. e.m.f. = … V [2] (c) The d.c. supply in Fig. 10.1 is now replaced with a sinusoidal alternating supply. Describe qualitatively the e.m.f. that is now induced in coil C. … … … [2] [Total: 8]

8 marks

Mark scheme: 10(a)(i) either product of flux density and area M1 direction of flux normal to area A1 or flux density × area × sinθ (M1) where θ is angle between direction of flux and area (A1) 10(a)(ii) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux linkage A1 10(b) e.m.f. = ∆( φN) / ∆t = (6.8 × 10–6 × 2 × 3.5 × 96) / (2.4 × 10–3) C1 = 1.9 V A1 Question Answer Marks 10(c) alternating C1 with same frequency as supply A1

This question in 9702/42 Feb/March 2018

10 (a) State Faraday’s law of electromagnetic induction. … … … … [2] (b) A coil of insulated wire is wound on to one end of a ferrous core and connected to a battery, as shown in Fig. 10.1. ferrous core aluminium ring coil of insulated wire Fig. 10.1 An aluminium ring is placed on the core. The ring can move freely along the length of the core. The switch is initially open. Use Faraday’s law and Lenz’s law to explain why the aluminium ring jumps upwards when the switch is closed. … … … … … … … [4] [Total: 6]

6 marks

Mark scheme: 10(a) induced e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) or of cutting (magnetic) flux A1 10(b) current in coil produces flux B1 (by Faraday’s law) changing flux induces e.m.f. in ring B1 current in ring causes field (around ring) B1 (by Lenz’s law) field around ring opposes field around coil B1

This question in 9702/41 May/June 2018

9 (a) State what is meant by the magnetic flux linkage of a coil. … … … … [3] (b) A coil of wire has 160 turns and diameter 2.4 cm. The coil is situated in a uniform magnetic field of flux density 7.5 mT, as shown in Fig. 9.1. magnetic field flux density 2.4 cm 7.5 mT coil 160 turns Fig. 9.1 The direction of the magnetic field is along the axis of the coil. The magnetic flux density is reduced to zero in a time of 0.15 s. Show that the average e.m.f. induced in the coil is 3.6 mV. [2] (c) The magnetic flux density B in the coil in (b) is now varied with time t as shown in Fig. 9.2. 10 B / mT 5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 t / s –5 –10 Fig. 9.2 Use data in (b) to show, on Fig. 9.3, the variation with time t of the e.m.f. E induced in the coil. 8 E / mV 6 4 2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 t / s –2 –4 –6 –8 Fig. 9.3 [4] [Total: 9]

9 marks

Mark scheme: 9(a) B1 magnetic flux density normal to area or reference to cross-sectional area or × sin (angle between B and A) B1 × number of turns on coil B1 9(b) e.m.f. = BAN / t or e.m.f = rate of change of flux linkage C1 = (7.5 × 10–3 × π × {1.2 × 10–2}2 × 160) / 0.15 = 3.6 × 10–3 V A1 9(c) sketch: zero for 0–0.10 s, 0.25–0.35 s, and 0.425–0.55 s, and non-zero outside these ranges B1 two horizontal steps, with zero voltage either side B1 with same polarity B1 correct values (1st step 3.6 mV and 2nd step 7.2 mV) B1

This question in 9702/42 May/June 2018

10 (a) State Faraday’s law of electromagnetic induction. … … … … [2] (b) A coil of insulated wire is wound on to one end of a ferrous core and connected to a battery, as shown in Fig. 10.1. ferrous core aluminium ring coil of insulated wire Fig. 10.1 An aluminium ring is placed on the core. The ring can move freely along the length of the core. The switch is initially open. Use Faraday’s law and Lenz’s law to explain why the aluminium ring jumps upwards when the switch is closed. … … … … … … … [4] [Total: 6]

6 marks

Mark scheme: 10(a) induced e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) or of cutting (magnetic) flux A1 10(b) current in coil produces flux B1 (by Faraday’s law) changing flux induces e.m.f. in ring B1 current in ring causes field (around ring) B1 (by Lenz’s law) field around ring opposes field around coil B1

This question in 9702/43 May/June 2018

9 (a) State Faraday’s law of electromagnetic induction. … … … [2] (b) A solenoid S is wound on a soft-iron core, as shown in Fig. 9.1. coil C V solenoid S soft-iron core V Hall probe Fig. 9.1 A coil C having 120 turns of wire is wound on to one end of the core. The area of cross- section of coil C is 1.5 cm2. A Hall probe is close to the other end of the core. When there is a constant current in solenoid S, the flux density in the core is 0.19 T. The reading on the voltmeter connected to the Hall probe is 0.20 V. The current in solenoid S is now reversed in a time of 0.13 s at a constant rate. (i) Calculate the reading on the voltmeter connected to coil C during the time that the current is changing. reading = … V [2] (ii) Complete Fig. 9.2 for the voltmeter readings for the times before, during and after the direction of the current is reversed. before current during current after current changes change when changes current is zero reading on voltmeter connected to coil C / V … … … reading on voltmeter connected to Hall probe / V 0.20 … … Fig. 9.2 [4] [Total: 8]

8 marks

Mark scheme: 9(a) (induced) e.m.f. proportional/equal to rate M1 of change of (magnetic) flux (linkage) A1 9(b)(i) induced e.m.f. = (∆B)AN / ∆t = (2 × 0.19 × 1.5 × 10–4 × 120) / 0.13 C1 = 0.053 V A1 9(b)(ii) reading on voltmeter connected to coil C / V: 0 0.053 0 (all three values required) A1 reading on voltmeter connected to Hall probe / V: zero in middle column B1 final column correct sign (negative) B1 final column correct magnitude (0.20) B1

This question in 9702/41 Oct/Nov 2018

9 (a) A Hall probe is placed near one end of a solenoid that has been wound on a soft-iron core, as shown in Fig. 9.1. soft-iron + – core solenoid Hall probe V Fig. 9.1 The current in the solenoid is switched on. The Hall probe is rotated until the reading VH on the voltmeter is maximum. The current in the solenoid is then varied, causing the magnetic flux density to change. The variation with time t of the magnetic flux density B at the Hall probe is shown in Fig. 9.2. 2 B / mT 1 0 0 t1 t2 t3 t4 t –1 –2 Fig. 9.2 At time t = 0, the Hall voltage is V0. On Fig. 9.3, draw a line to show the variation with time t of the Hall voltage VH for time t = 0 to time t = t4. V H V 0 0 0 t1 t2 t3 t4 t Fig. 9.3 [2] (b) The Hall probe in (a) is now replaced by a small coil of wire connected to a sensitive voltmeter, as shown in Fig. 9.4. soft-iron + – core solenoid small coil of wire V Fig. 9.4 The magnetic flux density, normal to the plane of the small coil, is again varied as shown in Fig. 9.2. On Fig. 9.5, draw a line to show the variation with time t of the e.m.f. E induced in the small coil for time t = 0 to time t = t4. E 0 0 t1 t2 t3 t4 t Fig. 9.5 [3] [Total: 5]

5 marks

Mark scheme: 9(a) and t3 → t4 horizontal straight line at different non-zero VH B1 t1 → t3 straight diagonal line with negative gradient and graph line starts at (0, V0) and ends at (t4, –2V0) B1 9(b) E = 0 for 0 → t1 and t3 → t4 B1 E is non-zero at all points between t1 → t3 M1 E has constant magnitude between t1 → t3 A1

This question in 9702/42 Oct/Nov 2018

9 (a) State Faraday’s law of electromagnetic induction. … … … [2] (b) A solenoid S is wound on a soft-iron core, as shown in Fig. 9.1. coil C V solenoid S soft-iron core V Hall probe Fig. 9.1 A coil C having 120 turns of wire is wound on to one end of the core. The area of cross- section of coil C is 1.5 cm2. A Hall probe is close to the other end of the core. When there is a constant current in solenoid S, the flux density in the core is 0.19 T. The reading on the voltmeter connected to the Hall probe is 0.20 V. The current in solenoid S is now reversed in a time of 0.13 s at a constant rate. (i) Calculate the reading on the voltmeter connected to coil C during the time that the current is changing. reading = … V [2] (ii) Complete Fig. 9.2 for the voltmeter readings for the times before, during and after the direction of the current is reversed. before current during current after current changes change when changes current is zero reading on voltmeter connected to coil C / V … … … reading on voltmeter connected to Hall probe / V 0.20 … … Fig. 9.2 [4] [Total: 8]

8 marks

Mark scheme: 9(a) (induced) e.m.f. proportional/equal to rate M1 of change of (magnetic) flux (linkage) A1 9(b)(i) induced e.m.f. = (∆B)AN / ∆t = (2 × 0.19 × 1.5 × 10–4 × 120) / 0.13 C1 = 0.053 V A1 9(b)(ii) reading on voltmeter connected to coil C / V: 0 0.053 0 (all three values required) A1 reading on voltmeter connected to Hall probe / V: zero in middle column B1 final column correct sign (negative) B1 final column correct magnitude (0.20) B1

This question in 9702/43 Oct/Nov 2018

10 (a) A cross-section through a current-carrying solenoid is shown in Fig. 10.1. current into page current out of page Fig. 10.1 On Fig. 10.1, draw field lines to represent the magnetic field inside the solenoid. [3] (b) State Faraday’s law of electromagnetic induction. … … … [2] (c) A coil of insulated wire is wound on to a soft-iron core. The coil is connected in series with a battery, a switch and an ammeter, as shown in Fig. 10.2. coil of soft-iron wire core A Fig. 10.2 Use laws of electromagnetic induction to explain why, when the switch is closed, the current increases gradually to its maximum value. … … … … … [3] [Total: 8]

8 marks

Mark scheme: 10(a) single straight line along full length of solenoid B1 at least two more parallel lines along full length of solenoid B1 correct direction – right to left B1 10(b) (induced) e.m.f. proportional / equal to rate M1 of change of (magnetic) flux (linkage) A1 10(c) increasing current causes increasing flux B1 increasing flux induces e.m.f. in coil B1 (induced) e.m.f. opposes growth of current B1

This question in 9702/42 Feb/March 2019

8 A solenoid is connected in series with a battery and a switch, as illustrated in Fig. 8.1. small coil A solenoid axis of solenoid Fig. 8.1 A small coil, connected to a sensitive ammeter, is situated near one end of the solenoid. As the current in the solenoid is switched on, there is a changing magnetic field inside the solenoid. (a) (i) State what is meant by a magnetic field. … … [1] (ii) On Fig. 8.1, draw an arrow on the axis of the solenoid to show the direction of the magnetic field inside the solenoid. Label this arrow P. [1] (b) As the current in the solenoid is switched on, there is a current induced in the small coil. This induced current gives rise to a magnetic field in the small coil. (i) State Lenz’s law. … … … [2] (ii) Use Lenz’s law to state and explain the direction of the magnetic field due to the induced current in the small coil. On Fig. 8.1, mark this direction with an arrow inside the small coil. … … … … [3] (c) The small coil has an area of cross-section 7.0 × 10–4 m2 and contains 75 turns of wire. A constant current in the solenoid produces a uniform magnetic flux of flux density 1.4 mT throughout the small coil. The direction of the current in the solenoid is reversed in a time of 0.12 s. Calculate the average e.m.f. induced in the small coil. e.m.f. = … V [3] [Total: 10]

10 marks

Mark scheme: 8(a)(i) region where a force is exerted on: a magnetic pole or a moving charge or a current-carrying wire B1 8(a)(ii) arrow on axis of solenoid pointing downwards labelled P B1 8(b)(i) direction of induced e.m.f./current M1 (tends to) oppose the change causing it A1 8(b)(ii) magnetic field in solenoid is increasing B1 field in coil in opposite direction to oppose increase B1 arrow inside or just above small coil pointing in opposite direction to P B1 8(c) e.m.f. = N∆φ / ∆t C1 = (75 × 1.4 × 10–3 × 2 × 7.0 × 10–4) / 0.12 C1 = 1.2 × 10–3 V A1

This question in 9702/41 May/June 2019

10 (a) State Faraday’s law of electromagnetic induction. … … … [2] (b) An ideal transformer is illustrated in Fig. 10.1. soft-iron core load E resistor primary coil secondary coil 2700 turns 450 turns Fig. 10.1 Explain why, when there is an alternating current in the primary coil, there is a current in the load resistor. … … … … … [3] (c) The primary coil in (b) has 2700 turns. The secondary coil has 450 turns. The e.m.f. E applied across the primary coil is given by the expression E = 220 sin(100πt ) where E is measured in volts and t is the time in seconds. Calculate the root-mean-square (r.m.s.) e.m.f. induced in the secondary coil. r.m.s. e.m.f. = … V [3] [Total: 8]

8 marks

Mark scheme: 10(a) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 10(b) current in primary coil gives rise to magnetic flux B1 changing (magnetic) flux in core links with secondary coil B1 induced e.m.f. (in secondary coil) causes current in load/resistor B1 10(c) correct application of turns ratio: to peak voltage ratio, giving (V0 / 220) = (450 / 2700) or to r.m.s. voltage ratio, giving (Vr.m.s. / 156) = (450 / 2700) C1 correct application of √2 factor: to peak applied e.m.f., giving 220 / √2 or to peak output em.f., giving 37 / √2 C1 Vr.m.s. = 26 V A1

This question in 9702/42 May/June 2019

8 A solenoid is connected in series with a battery and a switch, as illustrated in Fig. 8.1. small coil A solenoid axis of solenoid Fig. 8.1 A small coil, connected to a sensitive ammeter, is situated near one end of the solenoid. As the current in the solenoid is switched on, there is a changing magnetic field inside the solenoid. (a) (i) State what is meant by a magnetic field. … … [1] (ii) On Fig. 8.1, draw an arrow on the axis of the solenoid to show the direction of the magnetic field inside the solenoid. Label this arrow P. [1] (b) As the current in the solenoid is switched on, there is a current induced in the small coil. This induced current gives rise to a magnetic field in the small coil. (i) State Lenz’s law. … … … [2] (ii) Use Lenz’s law to state and explain the direction of the magnetic field due to the induced current in the small coil. On Fig. 8.1, mark this direction with an arrow inside the small coil. … … … … [3] (c) The small coil has an area of cross-section 7.0 × 10–4 m2 and contains 75 turns of wire. A constant current in the solenoid produces a uniform magnetic flux of flux density 1.4 mT throughout the small coil. The direction of the current in the solenoid is reversed in a time of 0.12 s. Calculate the average e.m.f. induced in the small coil. e.m.f. = … V [3] [Total: 10]

10 marks

Mark scheme: 8(a)(i) region where a force is exerted on: a magnetic pole or a moving charge or a current-carrying wire B1 8(a)(ii) arrow on axis of solenoid pointing downwards labelled P B1 8(b)(i) direction of induced e.m.f./current M1 (tends to) oppose the change causing it A1 8(b)(ii) magnetic field in solenoid is increasing B1 field in coil in opposite direction to oppose increase B1 arrow inside or just above small coil pointing in opposite direction to P B1 8(c) e.m.f. = N∆φ / ∆t C1 = (75 × 1.4 × 10–3 × 2 × 7.0 × 10–4) / 0.12 C1 = 1.2 × 10–3 V A1

This question in 9702/43 May/June 2019

11 (a) State Faraday’s law of electromagnetic induction. … … … [2] (b) A solenoid S has a small coil C placed near to one of its ends, as shown in Fig. 11.1. solenoid S 3.6 × 10–2 m coil C 63 turns Fig. 11.1 The coil C has a circular cross-section of diameter 3.6 × 10–2 m and contains 63 turns of wire. The solenoid S produces a uniform magnetic field of flux density B, in tesla, in the region of coil C given by the expression B = 9.4 × 10–4 I where I is the current, in ampere, in the solenoid S. The variation with time t of the current I in solenoid S is shown in Fig. 11.2. current I 0 0 t1 t2 t3 t4 t5 t6 t7 time t Fig. 11.2 State two times at which: (i) there is no electromotive force (e.m.f.) induced in coil C time … and time … [1] (ii) the induced e.m.f. in coil C is a maximum but with opposite polarities. time … and time … [1] (c) The alternating current in the solenoid S in (b) is replaced by a constant current of 5.0 A. Calculate the average e.m.f. induced in coil C when the current in solenoid S is reversed in a time of 6.0 ms. e.m.f. induced = … V [3] [Total: 7]

7 marks

Mark scheme: 11(a) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 11(b)(i) any two from t1, t3, t5, t7 A1 11(b)(ii) t2 and t4 or t4 and t6 A1 11(c) e.m.f. = N∆Φ / ∆t C1 = (2 × 9.4 × 10–4 × 5.0 × π × (1.8 × 10–2)2 × 63) / (6.0 × 10–3) C1 = 0.10 V A1

This question in 9702/42 Oct/Nov 2019

9 (a) A coil of wire is situated in a uniform magnetic field of flux density B. The coil has diameter 3.6 cm and consists of 350 turns of wire, as illustrated in Fig. 9.1. uniform magnetic field 3.6 cm flux density B coil, 350 turns Fig. 9.1 The variation with time t of B is shown in Fig. 9.2. 50 40 B / mT 30 20 10 0 0 0.2 0.4 0.6 0.8 t / s Fig. 9.2 (i) Show that, for the time t = 0 to time t = 0.20 s, the electromotive force (e.m.f.) induced in the coil is 0.080 V. [2] (ii) On the axes of Fig. 9.3, show the variation with time t of the induced e.m.f. E for time t = 0 to time t = 0.80 s. 0.2 E / V 0.1 0 0 0.2 0.4 0.6 0.8 t / s –0.1 –0.2 Fig. 9.3 (b) A bar magnet is held a small distance above the surface of an aluminium disc by means of a rod, as illustrated in Fig. 9.4. rotating magnet fixed aluminium disc Fig. 9.4 The aluminium disc is supported horizontally and held stationary. The magnet is rotated about a vertical axis at constant speed. Use laws of electromagnetic induction to explain why there is a torque acting on the aluminium disc. … … … … … … [4] [Total: 10]

10 marks

Mark scheme: 9(a)(i) e.m.f. = (Δ)B × AN / t C1 = 45 × 10–3 × π × (1.8 × 10–2)2 × 350 / 0.20 = 0.080 V A1 9(a)(ii) 0 to 0.2 s: straight horizontal line at 0.080 V or –0.080 V B1 0.2 s to 0.4 s: zero B1 0.4 s to 0.8 s: straight horizontal line at 0.040 V or –0.040 V B1 opposite polarity to 0 to 0.2 s line B1 9(b) either disc cuts flux lines (of the magnet) or there is a changing flux in the disc B1 (by Faraday’s law) e.m.f. is induced in the disc B1 e.m.f. causes (eddy) currents in the disc B1 current in the magnetic field (of the magnet) causes force on disc B1

This question in 9702/41 May/June 2020

10 (a) State Faraday’s law of electromagnetic induction. … … … [2] (b) A simple iron‑cored transformer is illustrated in Fig. 10.1. laminated soft-iron core input output primary coil secondary coil Fig. 10.1 (i) State one function of a transformer. … … [1] (ii) A sinusoidal alternating current in the primary coil gives rise to a varying magnetic flux linking the secondary coil. Use Faraday’s law to explain why the output from the transformer is an electromotive force (e.m.f.) that is alternating. … … … … … [3] (iii) State why the soft‑iron core of the transformer is laminated. … … [1] [Total: 7]

7 marks

Mark scheme: 10(a) (induced) electromotive force is proportional to rate M1 of change of (magnetic) flux (linkage) A1 10(b)(i) to change magnitude of potential difference B1 10(b)(ii) magnitude of e.m.f. varies as rate of change of flux changes B1 direction of e.m.f. changes when direction of change of flux reverses/when flux changes from increasing to decreasing B1 flux is continuously increasing and decreasing, so polarity of e.m.f. is continuously switching B1 10(b)(iii) to reduce energy/power losses or to reduce eddy currents B1

This question in 9702/42 May/June 2020

9 (a) A coil of wire is situated in a uniform magnetic field of flux density B. The coil has diameter 3.6 cm and consists of 350 turns of wire, as illustrated in Fig. 9.1. uniform magnetic field 3.6 cm flux density B coil, 350 turns Fig. 9.1 The variation with time t of B is shown in Fig. 9.2. 50 40 B / mT 30 20 10 0 0 0.2 0.4 0.6 0.8 t / s Fig. 9.2 (i) Show that, for the time t = 0 to time t = 0.20 s, the electromotive force (e.m.f.) induced in the coil is 0.080 V. [2] (ii) On the axes of Fig. 9.3, show the variation with time t of the induced e.m.f. E for time t = 0 to time t = 0.80 s. 0.2 E / V 0.1 0 0 0.2 0.4 0.6 0.8 t / s –0.1 –0.2 Fig. 9.3 (b) A bar magnet is held a small distance above the surface of an aluminium disc by means of a rod, as illustrated in Fig. 9.4. rotating magnet fixed aluminium disc Fig. 9.4 The aluminium disc is supported horizontally and held stationary. The magnet is rotated about a vertical axis at constant speed. Use laws of electromagnetic induction to explain why there is a torque acting on the aluminium disc. … … … … … … [4] [Total: 10]

10 marks

Mark scheme: 9(a)(i) e.m.f. = (Δ)B × AN / t C1 = 45 × 10–3 × π × (1.8 × 10–2)2 × 350 / 0.20 = 0.080 V A1 9(a)(ii) 0 to 0.2 s: straight horizontal line at 0.080 V or –0.080 V B1 0.2 s to 0.4 s: zero B1 0.4 s to 0.8 s: straight horizontal line at 0.040 V or –0.040 V B1 opposite polarity to 0 to 0.2 s line B1 9(b) either disc cuts flux lines (of the magnet) or there is a changing flux in the disc B1 (by Faraday’s law) e.m.f. is induced in the disc B1 e.m.f. causes (eddy) currents in the disc B1 current in the magnetic field (of the magnet) causes force on disc B1

This question in 9702/43 May/June 2020

9 (a) Define magnetic flux. … … … [2] (b) A simple transformer consists of two coils of wire wound on a soft-iron core, as illustrated in Fig. 9.1. soft-iron core primary coil secondary coil Fig. 9.1 There is a sinusoidal current in the primary coil. Explain: (i) how this current gives rise to an induced electromotive force (e.m.f.) in the secondary coil … … … … … [3] (ii) why the e.m.f. induced in the secondary coil is not constant. … … … … [2] (c) Explain why the soft-iron core in (b) is laminated. … … … [2] [Total: 9]

9 marks

Mark scheme: 9(a) flux density × area M1 where flux is normal to area A1 or flux density × area × sin θ (M1) where θ is angle between flux direction and (plane of) area (A1) 9(b)(i) (alternating) current creates changing (magnetic) flux B1 core links (magnetic) flux with secondary coil B1 changing flux (in secondary) causes induced e.m.f. B1 9(b)(ii) rate of change of flux is not constant B1 (induced) e.m.f. is proportional to rate of change of flux B1 9(c) reduces induced currents in core B1 hence reduces energy losses (in core) B1

This question in 9702/41 Oct/Nov 2020

9 (a) A small coil is placed close to one end of a solenoid connected to a power supply. The plane of the small coil is normal to the axis of the solenoid, as illustrated in Fig. 9.1. solenoid small coil power supply Fig. 9.1 The power supply causes the current I in the solenoid to vary with time t as shown in Fig. 9.2. current I 0 t1 t2 time t Fig. 9.2 (i) State Faraday’s law of electromagnetic induction. … … … [2] (ii) On the axes of Fig. 9.3, sketch a graph to show the variation with time t of the electromotive force (e.m.f.) induced in the small coil. e.m.f. 0 t1 t2 time t Fig. 9.3 [4] (b) The small coil in (a) is now replaced by a Hall probe. The Hall probe is positioned so that the reading for the probe is a maximum. The current I in the solenoid varies again as shown in Fig. 9.2. On the axes of Fig. 9.4, sketch a graph to show the variation with time t of the reading VH of the probe. VH 0 t1 t2 time t Fig. 9.4 [2] [Total: 8]

8 marks

Mark scheme: 9(a)(i) (induced) e.m.f. (directly) proportional to rate M1 of change of magnetic flux (linkage) A1 9(a)(ii) e.m.f. = 0 apart from thin pulses at t1 and t2 B1 rectangular pulses centred on t1 and t2, of widths 2 small squares and 1 small square respectively B1 e.m.fs. at t1 and t2 have opposite polarities B1 magnitude of e.m.f. at t2 double the magnitude of e.m.f. at t1 B1 9(b) VH shown as zero before (t1 – 2 squares) and after (t2 + 2 squares) and rises to a constant non-zero value between t1 and t2 M1 change at t1 shown as 2 small squares wide and change at t2 shown as 1 small square wide A1

This question in 9702/42 Oct/Nov 2020

9 (a) Define magnetic flux. … … … [2] (b) A simple transformer consists of two coils of wire wound on a soft-iron core, as illustrated in Fig. 9.1. soft-iron core primary coil secondary coil Fig. 9.1 There is a sinusoidal current in the primary coil. Explain: (i) how this current gives rise to an induced electromotive force (e.m.f.) in the secondary coil … … … … … [3] (ii) why the e.m.f. induced in the secondary coil is not constant. … … … … [2] (c) Explain why the soft-iron core in (b) is laminated. … … … [2] [Total: 9]

9 marks

Mark scheme: 9(a) flux density × area M1 where flux is normal to area A1 or flux density × area × sin θ (M1) where θ is angle between flux direction and (plane of) area (A1) 9(b)(i) (alternating) current creates changing (magnetic) flux B1 core links (magnetic) flux with secondary coil B1 changing flux (in secondary) causes induced e.m.f. B1 9(b)(ii) rate of change of flux is not constant B1 (induced) e.m.f. is proportional to rate of change of flux B1 9(c) reduces induced currents in core B1 hence reduces energy losses (in core) B1

This question in 9702/43 Oct/Nov 2020

9 (a) Define magnetic flux linkage. … … … [2] (b) A solenoid of diameter 6.0 cm and 540 turns is placed in a uniform magnetic field as shown in Fig. 9.1. solenoid 540 turns diameter 6.0 cm magnetic field Fig. 9.1 The variation with time t of the magnetic flux density is shown in Fig. 9.2. 250 200 flux density / mT 150 100 50 0 0 1 2 3 4 5 6 7 8 t / s Fig. 9.2 Calculate the maximum magnitude of the induced electromotive force (e.m.f.) in the solenoid. e.m.f. = … V [3] (c) A thin copper sheet X is supported on a rigid rod so that it hangs between the poles of a magnet as shown in Fig. 9.3. rod copper sheet X poles of magnet Fig. 9.3 Sheet X is displaced to one side and then released so that it oscillates. A motion sensor is used to record the displacement of X. A second thin copper sheet Y replaces sheet X. Sheet Y has the same overall dimensions as X but is cut into the shape shown in Fig. 9.4. copper sheet Y Fig. 9.4 The motion sensor is again used to record the displacement. The graph in Fig. 9.5 shows the variation with time t of the displacement s of each copper sheet. s 0 t Fig. 9.5 (i) State the name of the phenomenon illustrated by the gradual reduction in the amplitude of the dashed line. … [1] (ii) Deduce which copper sheet is represented by the dashed line. Explain your answer using the principles of electromagnetic induction. … … … … … … [4] [Total: 10]

10 marks

Mark scheme: 9(a) (magnetic) flux density × area × number of turns M1 area is perpendicular to (magnetic) field A1 9(b) use of t = 1.2 s C1 BAN t ε Δ = Δ 2 0.250 0.030 540 1.2 π × × × = C1 0.32 V = A1 9(c)(i) light damping B1 Question Answer Marks 9(c)(ii) sheet cuts (magnetic) flux and causes induced emf B1 (induced) emf causes (eddy) currents (in sheet) B1 either currents (in sheet) cause resistive force or currents (in sheet) dissipate energy B1 smaller currents in Y or larger currents in X, so dashed line is X B1

This question in 9702/42 Feb/March 2021

10 (a) State Lenz’s law. … … … [2] (b) A metal ring is suspended from a fixed point P by means of a thread, as shown in Fig. 10.1. P P metal magnet ring pole piece metal ring N S Fig. 10.1 Fig. 10.2 The ring is displaced a distance d and then released. The ring completes many oscillations before coming to rest. The poles of a magnet are now placed near to the ring so that the ring hangs midway between the poles of the magnet, as shown in Fig. 10.2. The ring is again displaced a distance d and then released. Explain why the ring completes fewer oscillations before coming to rest. … … … … … … [4] (c) The ring in (b) is now cut so that it has the shape shown in Fig. 10.3. Fig. 10.3 Explain why, when the procedure in (b) is repeated, the cut ring completes more oscillations than the complete ring when oscillating between the poles of the magnet. … … … … … [3] [Total: 9]

9 marks

Mark scheme: 10(a) direction of (induced) e.m.f. M1 is such as to oppose the change causing it A1 10(b) ring cuts (magnetic) flux and causes induced e.m.f. in ring B1 (induced) e.m.f. causes (eddy/induced) currents (in ring) B1 currents (in ring) cause magnetic field (around ring) M1 two fields interact to cause resistive/opposing force A1 or current (in ring) is in a magnetic field (M1) which causes resistive force (A1) or currents (in ring) dissipate thermal energy (M1) (thermal) energy comes from energy of oscillations (A1) 10(c) current cannot pass all the way around the ring B1 (induced) currents smaller B1 smaller resistive force (so more oscillations) or smaller rate of dissipation of energy (so more oscillations) B1

This question in 9702/41 May/June 2021

9 (a) State two situations in which a charged particle in a magnetic field does not experience a force. 1. … … 2. … … [2] (b) A loosely coiled metal spring is suspended from a fixed point, as shown in Fig. 9.1. fixed point spring small mass flexible lead Fig. 9.1 Electrical connections are made to the ends of the spring by means of a flexible lead. The length of the spring is measured before the switch is closed and then again after the switch is closed. When the switch is closed, a magnetic field is set up around each coil of the spring. By reference to these magnetic fields, explain why there is a change in length of the spring. State whether the spring extends or contracts. … … … … … … [4] (c) With the switch in (b) closed, the small mass on the free end of the spring is now made to oscillate vertically. Use the principles of electromagnetic induction to explain why small fluctuations in the current in the spring are found to occur. … … … … [3] [Total: 9]

9 marks

Mark scheme: 9(a) (particle is) stationary/not moving B1 (particle is) moving parallel to the (magnetic) field B1 9(b) magnetic field around each coil is circular or each coil is normal to magnetic field due to adjacent coils B1 current in coil interacts with (magnetic) field to exert force (on coil) B1 force is normal to both coil and magnetic field or force parallel to axis (of coil) B1 forces between coils are attractive so spring contracts B1 9(c) (oscillating) coils cut magnetic flux or as separation of coils changes, magnetic flux changes B1 cutting flux causes induced e.m.f. in coils B1 changing (induced) e.m.f. causes changing current (in coil) B1

This question in 9702/42 May/June 2021

10 (a) State Lenz’s law. … … … [2] (b) A metal ring is suspended from a fixed point P by means of a thread, as shown in Fig. 10.1. P P metal magnet ring pole piece metal ring N S Fig. 10.1 Fig. 10.2 The ring is displaced a distance d and then released. The ring completes many oscillations before coming to rest. The poles of a magnet are now placed near to the ring so that the ring hangs midway between the poles of the magnet, as shown in Fig. 10.2. The ring is again displaced a distance d and then released. Explain why the ring completes fewer oscillations before coming to rest. … … … … … … [4] (c) The ring in (b) is now cut so that it has the shape shown in Fig. 10.3. Fig. 10.3 Explain why, when the procedure in (b) is repeated, the cut ring completes more oscillations than the complete ring when oscillating between the poles of the magnet. … … … … … [3] [Total: 9]

9 marks

Mark scheme: 10(a) direction of (induced) e.m.f. M1 is such as to oppose the change causing it A1 10(b) ring cuts (magnetic) flux and causes induced e.m.f. in ring B1 (induced) e.m.f. causes (eddy/induced) currents (in ring) B1 currents (in ring) cause magnetic field (around ring) M1 two fields interact to cause resistive/opposing force A1 or current (in ring) is in a magnetic field (M1) which causes resistive force (A1) or currents (in ring) dissipate thermal energy (M1) (thermal) energy comes from energy of oscillations (A1) 10(c) current cannot pass all the way around the ring B1 (induced) currents smaller B1 smaller resistive force (so more oscillations) or smaller rate of dissipation of energy (so more oscillations) B1

This question in 9702/43 May/June 2021

9 (a) State, by reference to the power dissipated in a resistor, what is meant by the root-mean-square (r.m.s.) value of an alternating voltage. … … … … [2] (b) A coil is rotating freely, on frictionless bearings, at constant speed in a uniform magnetic field. This rotation causes an induced alternating electromotive force (e.m.f.) across the open terminals of the coil. The induced e.m.f. has r.m.s. value 12 V and frequency 50 Hz. The speed of rotation of the coil is now doubled. (i) State and explain, with reference to the principles of electromagnetic induction, the effect of the increased speed of rotation on the r.m.s. value of the induced e.m.f. … … … … [2] (ii) On Fig. 9.1, sketch the variation with time t of the induced e.m.f. E across the terminals of the coil at the increased speed of rotation. Your line should extend from time t = 0 to time t = 20 ms. Assume that E = 0 when t = 0. 40 E / V 20 0 0 5 10 15 20 t / ms –20 –40 Fig. 9.1 [3] (c) State and explain the effect on the motion of the coil in (b) of connecting a load resistor across its terminals. … … … … [2] [Total: 9]

9 marks

Mark scheme: 9(a) constant voltage M1 that produces/dissipates same power as (the mean power of) the alternating voltage A1 9(b)(i) (maximum) rate of cutting of (magnetic) flux doubles B1 (peak and hence) r.m.s. induced e.m.f. doubles B1 9(b)(ii) sketch: (sinusoidal) wave of period 10 ms B1 peak E shown as ± 34 V (1 mark out of 2 awarded if peak E shown as ± 17 V or ± 24 V) B2 9(c) current in the coil results in forces that oppose its rotation or current in the resistor dissipates the energy of rotation B1 coil stops rotating B1

This question in 9702/41 Oct/Nov 2021

10 Fig. 10.1 shows a simple laminated iron-cored transformer consisting of a primary coil of 25 000 turns and a secondary coil of 625 turns. laminated iron core 25 000 625 VIN 640 Ω VOUT turns turns Fig. 10.1 The output potential difference (p.d.) VOUT is applied to a load resistor of resistance 640 Ω. (a) (i) State the function of the iron core. … … [1] (ii) Explain why the iron core is laminated. … … … [2] (b) The input p.d. VIN is a sinusoidal alternating voltage of peak value 12 kV and period 40 ms. (i) Calculate the maximum value of VOUT. maximum VOUT = … V [1] (ii) Calculate the root-mean-square (r.m.s.) current in the load resistor. r.m.s. current = … A [1] (iii) On Fig. 10.2, sketch the variation with time t of the power P dissipated in the load resistor for time t = 0 to t = 40 ms. Assume that P = 0 when t = 0. 200 P / W 100 0 0 10 20 30 40 t / ms –100 –200 Fig. 10.2 [3] (c) Explain, with reference to Fig.10.2, why the mean power in the load resistor is 70 W. … … … … [2] [Total: 10]

10 marks

Mark scheme: 10(a)(i) to increase the magnetic flux linkage (between the coils) B1 10(a)(ii) to reduce energy losses B1 by reducing induced currents B1 10(b)(i) maximum VOUT = 12 000 × (625 / 25 000) = 300 V A1 10(b)(ii) r.m.s. current = 300 / (640 × √2) = 0.33 A A1 10(b)(iii) sketch: sinusoidal shape in positive half of the graph, sitting with ‘minima’ resting on the time-axis (at P = 0) B1 each ‘cycle’ shown repeating every 20 ms B1 maximum P shown as 140 W B1 10(c) power curve is symmetrical about the midpoint (on the power axis) B1 mean power is half the peak power B1

This question in 9702/42 Oct/Nov 2021

9 (a) State, by reference to the power dissipated in a resistor, what is meant by the root-mean-square (r.m.s.) value of an alternating voltage. … … … … [2] (b) A coil is rotating freely, on frictionless bearings, at constant speed in a uniform magnetic field. This rotation causes an induced alternating electromotive force (e.m.f.) across the open terminals of the coil. The induced e.m.f. has r.m.s. value 12 V and frequency 50 Hz. The speed of rotation of the coil is now doubled. (i) State and explain, with reference to the principles of electromagnetic induction, the effect of the increased speed of rotation on the r.m.s. value of the induced e.m.f. … … … … [2] (ii) On Fig. 9.1, sketch the variation with time t of the induced e.m.f. E across the terminals of the coil at the increased speed of rotation. Your line should extend from time t = 0 to time t = 20 ms. Assume that E = 0 when t = 0. 40 E / V 20 0 0 5 10 15 20 t / ms –20 –40 Fig. 9.1 [3] (c) State and explain the effect on the motion of the coil in (b) of connecting a load resistor across its terminals. … … … … [2] [Total: 9]

9 marks

Mark scheme: 9(a) constant voltage M1 that produces/dissipates same power as (the mean power of) the alternating voltage A1 9(b)(i) (maximum) rate of cutting of (magnetic) flux doubles B1 (peak and hence) r.m.s. induced e.m.f. doubles B1 9(b)(ii) sketch: (sinusoidal) wave of period 10 ms B1 peak E shown as ± 34 V (1 mark out of 2 awarded if peak E shown as ± 17 V or ± 24 V) B2 9(c) current in the coil results in forces that oppose its rotation or current in the resistor dissipates the energy of rotation B1 coil stops rotating B1

This question in 9702/43 Oct/Nov 2021

6 A small solenoid of area of cross section 1.6 × 10–3 m2 is placed inside a larger solenoid of area of cross-section 6.4 × 10–3 m2, as shown in Fig. 6.1. smaller solenoid larger solenoid area of cross-section area of cross-section 1.6 × 10–3 m2 6.4 × 10–3 m2 3000 turns 600 turns d.c. Fig. 6.1 (not to scale) The larger solenoid has 600 turns and is attached to a d.c. power supply to create a magnetic field. The smaller solenoid has 3000 turns. (a) Compare the magnetic flux in the two solenoids. … … … [1] (b) Compare the magnetic flux linkage in the two solenoids. … … … [1] (c) (i) State Lenz’s law of electromagnetic induction. … … … [2] (ii) The terminals of the smaller solenoid are connected together. The smaller solenoid is then removed from inside the larger solenoid. With reference to magnetic fields, explain why a force is needed to remove the smaller solenoid. … … … … … … [3] [Total: 7]

7 marks

Mark scheme: 6(a) less in smaller solenoid B1 6(b) greater in smaller solenoid B1 6(c)(i) direction of (induced) e.m.f. M1 such as to (produce effects that) oppose the change that caused it A1 6(c)(ii) change of flux (linkage) in smaller solenoid induces e.m.f. in smaller solenoid B1 (induced) current in smaller solenoid causes field around it B1 the two fields (interact to) create an attractive force B1

This question in 9702/42 Feb/March 2022

6 (a) Define magnetic flux. … … … [2] (b) A square coil of wire of side length 12 cm consists of 8 insulated turns. The coil is stationary in a uniform magnetic field. The plane of the coil is perpendicular to the magnetic field, as shown in Fig. 6.1. magnetic field lines into the page 12 cm square coil 8 turns terminals Fig. 6.1 The flux density B of the magnetic field varies with time t as shown in Fig. 6.2. 400 B / mT 200 0 0 0.2 0.4 0.6 0.8 t / s Fig. 6.2 (i) Determine the magnetic flux linkage inside the coil at time t = 0.60 s. Give a unit with your answer. magnetic flux linkage = … unit … [3] (ii) State how Fig. 6.2 shows that the electromotive force (e.m.f.) E induced across the terminals between t = 0 and t = 0.60 s is constant. … [1] (iii) Calculate the magnitude of E. E = … V [2] (c) The procedure in (b) is repeated, but this time the terminals of the coil are connected together. State and explain the effect on the coil of connecting the terminals together during the change of magnetic flux density shown in Fig. 6.2. … … … … [3] [Total: 11]

11 marks

Mark scheme: 6(a) product of (magnetic) flux density and area M1 where area is perpendicular to the (magnetic) field A1 6(b)(i) N = BAN C1 = 400  10–3  0.122  8 C1 = 0.046 Wb A1 6(b)(ii) (line is a) straight line B1 6(b)(iii) (induced) e.m.f. = rate of change of flux linkage C1 e.m.f. = N / t = 0.046 / 0.60 = 0.077 V A1 6(c) (induced e.m.f. causes) current flow (in the coil) B1 either current (in magnetic field) causes forces to act on the coil B1 (opposite sides of) coil forced inwards B1 or current causes dissipation of energy in the resistance of the coil (B1) temperature of the coil rises (B1)

This question in 9702/41 May/June 2022

7 (a) State Faraday’s law of electromagnetic induction. … … … [2] (b) Two coils are wound on an iron bar, as shown in Fig. 7.1. coil 2 V iron bar coil 1 V1 Fig. 7.1 Coil 1 is connected to a potential difference (p.d.) V1 that gives rise to a magnetic field in the iron bar. Fig. 7.2 shows the variation with time t of the magnetic flux density B in the iron bar. B 0 0 0.1 0.2 0.3 t / s 0.4 Fig. 7.2 A voltmeter measures the electromotive force (e.m.f.) V2 that is induced across coil 2. On Fig. 7.3, sketch the variation with t of V2 between t = 0 and t = 0.40 s. V2 0 0 0.1 0.2 0.3 t / s 0.4 Fig. 7.3 [4] (c) Coil 2 in (b) is now replaced with a copper ring that rests loosely on top of coil 1. The supply to coil 1 is replaced with a cell and a switch that is initially open, as shown in Fig. 7.4. iron bar copper ring coil 1 Fig. 7.4 (i) The switch is now closed. As it is closed, the copper ring is observed to jump upwards. Explain why this happens. … … … … [3] (ii) Suggest, with a reason, what would be the effect of repeating the procedure in (c)(i) with the terminals of the cell reversed. … … [1] [Total: 10]

10 marks

Mark scheme: 7(a) induced e.m.f. is (directly) proportional to rate M1 of change of (magnetic) flux (linkage) A1 7(b) V2 stepped, all at non-zero values, between t = 0 and t = 0.40 s B1 V2 shown with same non-zero magnitude up to t = 0.15 s and after t = 0.25 s but with a different magnitude between these times B1 V2 shown with a magnitude between t = 0.15 s and t = 0.25 s that is three times the magnitude before t = 0.15 s and after t = 0.25 s B1 V2 shown with same sign up to t = 0.15 s and after t = 0.25 s, and opposite sign in between B1 7(c)(i) changing current in coil causes changing (magnetic) field or changing (magnetic) flux causes induced e.m.f. in ring B1 induced e.m.f. in ring causes current in ring B1 (magnetic) field due to (induced) current in ring interacts with (coil’s) field to cause upwards force (on ring) or (induced) current in ring perpendicular to (coil’s magnetic) field causes upwards force (on ring) B1 7(c)(ii) both magnetic fields reverse direction so ring still jumps up or current (in ring) and (coil’s) field both reverse so ring still jumps up B1

This question in 9702/42 May/June 2022

6 (a) Define magnetic flux. … … … [2] (b) A square coil of wire of side length 12 cm consists of 8 insulated turns. The coil is stationary in a uniform magnetic field. The plane of the coil is perpendicular to the magnetic field, as shown in Fig. 6.1. magnetic field lines into the page 12 cm square coil 8 turns terminals Fig. 6.1 The flux density B of the magnetic field varies with time t as shown in Fig. 6.2. 400 B / mT 200 0 0 0.2 0.4 0.6 0.8 t / s Fig. 6.2 (i) Determine the magnetic flux linkage inside the coil at time t = 0.60 s. Give a unit with your answer. magnetic flux linkage = … unit … [3] (ii) State how Fig. 6.2 shows that the electromotive force (e.m.f.) E induced across the terminals between t = 0 and t = 0.60 s is constant. … [1] (iii) Calculate the magnitude of E. E = … V [2] (c) The procedure in (b) is repeated, but this time the terminals of the coil are connected together. State and explain the effect on the coil of connecting the terminals together during the change of magnetic flux density shown in Fig. 6.2. … … … … [3] [Total: 11]

11 marks

Mark scheme: 6(a) product of (magnetic) flux density and area M1 where area is perpendicular to the (magnetic) field A1 6(b)(i) N = BAN C1 = 400  10–3  0.122  8 C1 = 0.046 Wb A1 6(b)(ii) (line is a) straight line B1 6(b)(iii) (induced) e.m.f. = rate of change of flux linkage C1 e.m.f. = N / t = 0.046 / 0.60 = 0.077 V A1 6(c) (induced e.m.f. causes) current flow (in the coil) B1 either current (in magnetic field) causes forces to act on the coil B1 (opposite sides of) coil forced inwards B1 or current causes dissipation of energy in the resistance of the coil (B1) temperature of the coil rises (B1)

This question in 9702/43 May/June 2022

5 A capacitor of capacitance 470 μF is connected to a battery of electromotive force (e.m.f.) 24 V in the circuit of Fig. 5.1. X Y S 24 V V 470 μF P Q 5.6 kΩ 5.6 kΩ Fig. 5.1 The two-way switch S is initially at position X. P and Q are identical long straight wires, each with a resistance of 5.6 kΩ. These wires are placed near to, and parallel to, each other. Wire Q is connected to a voltmeter. At time t = 0, switch S is moved to position Y so that the capacitor discharges through wire P. (a) (i) Calculate the charge Q0 on the capacitor at time t = 0. Q0 = … C [2] (ii) Calculate the current I0 in wire P at time t = 0. I0 = … A [1] (iii) Calculate the time constant τ of the discharge circuit. τ = … s [2] (iv) On Fig. 5.2, sketch a line to show the variation with t of the current I in wire P as the capacitor discharges. I0 I 0 0 t Fig. 5.2 [2] (b) (i) Explain why there is an induced e.m.f. across wire Q during the discharge of the capacitor. … … … … [3] (ii) On Fig. 5.3, sketch a line to suggest the variation with t of the voltmeter reading V. V 0 0 t Fig. 5.3 [1] [Total: 11]

11 marks

Mark scheme: 5(a)(i) Q = CV C1 Q0 = 24  470  10–6 A1 = 0.011 C 5(a)(ii) I0 = 24 / 5600 A1 = 4.3  10–3 A 5(a)(iii) = RC C1 = 5600  470  10–6 A1 = 2.6 s 5(a)(iv) line with negative gradient throughout passing through (0, I0) B1 exponential decay curve asymptotic to t-axis B1 5(b)(i) current in wire P gives rise to a magnetic field B1 as current (in P) changes, wire Q cuts (magnetic) flux (of wire P) B1 cutting magnetic flux causes induced e.m.f. (across Q) B1 5(b)(ii) sketch shows line with a negative gradient throughout B1

This question in 9702/41 Oct/Nov 2022

8 (a) State Lenz’s law of electromagnetic induction. … … … [2] (b) Two coils of insulated wire are wound on an iron bar, as shown in Fig. 8.1. coil 1 coil 2 iron bar I1 V V2 Fig. 8.1 There is a current I1 in coil 1 that varies with time t as shown in Fig. 8.2. 1.0 I1 / A 0.5 0 0 0.02 0.04 0.06 0.08 t / s – 0.5 – 1.0 Fig. 8.2 (i) The variation with t of I1 can be represented by the equation I1 = X sin Yt where X and Y are constants. Use Fig. 8.2 to determine the values of X and Y. Give units with your answers. X = … unit … Y = … unit … [3] (ii) The current in coil 1 gives rise to a magnetic field in the iron bar. Assume that the flux density of this magnetic field is proportional to I1. An alternating electromotive force (e.m.f.) is induced across coil 2. The p.d. across coil 2 is measured using the voltmeter and has a root-mean-square (r.m.s.) value of 4.6 V. On Fig. 8.3, sketch a line to show the variation with t of V2 between t = 0 and t = 0.08 s. 10 V2 / V 5 0 0 0.02 0.04 0.06 0.08 t / s – 5 – 10 Fig. 8.3 [3] (iii) Use the laws of electromagnetic induction to explain the shape of your line in (b)(ii). … … … … [3] [Total: 11]

11 marks

Mark scheme: 8(a) direction of induced e.m.f. M1 such as to (produce effects that) oppose the change that caused it A1 8(b)(i) X = 0.85 A A1 Y = 2 / 0.040 C1 = 160 rad s–1 A1 8(b)(ii) two cycles of a sinusoidal curve with a period of 0.040 s B1 correct phase (i.e. V2 max / min at t = 0, 0.02, 0.04, 0.06 and 0.08 s, and V2 zero at t = 0.01, 0.03, 0.05, 0.07 s) B1 maximum / minimum V2 shown (consistently) at ± 6.5 V B1 8(b)(iii) (magnitude of) V2 is proportional to rate of change of (magnetic) flux B1 • V2 is proportional to gradient of I1–t curve B2 • V2 has maximum magnitude when I1–t curve is steepest • V2 is zero when I1–t curve is horizontal / a maximum or minimum • V2 changes sign when sign of gradient of I1–t curve changes Any two points, 1 mark each

This question in 9702/42 Oct/Nov 2022

5 A capacitor of capacitance 470 μF is connected to a battery of electromotive force (e.m.f.) 24 V in the circuit of Fig. 5.1. X Y S 24 V V 470 μF P Q 5.6 kΩ 5.6 kΩ Fig. 5.1 The two-way switch S is initially at position X. P and Q are identical long straight wires, each with a resistance of 5.6 kΩ. These wires are placed near to, and parallel to, each other. Wire Q is connected to a voltmeter. At time t = 0, switch S is moved to position Y so that the capacitor discharges through wire P. (a) (i) Calculate the charge Q0 on the capacitor at time t = 0. Q0 = … C [2] (ii) Calculate the current I0 in wire P at time t = 0. I0 = … A [1] (iii) Calculate the time constant τ of the discharge circuit. τ = … s [2] (iv) On Fig. 5.2, sketch a line to show the variation with t of the current I in wire P as the capacitor discharges. I0 I 0 0 t Fig. 5.2 [2] (b) (i) Explain why there is an induced e.m.f. across wire Q during the discharge of the capacitor. … … … … [3] (ii) On Fig. 5.3, sketch a line to suggest the variation with t of the voltmeter reading V. V 0 0 t Fig. 5.3 [1] [Total: 11]

11 marks

Mark scheme: 5(a)(i) Q = CV C1 Q0 = 24  470  10–6 A1 = 0.011 C 5(a)(ii) I0 = 24 / 5600 A1 = 4.3  10–3 A 5(a)(iii) = RC C1 = 5600  470  10–6 A1 = 2.6 s 5(a)(iv) line with negative gradient throughout passing through (0, I0) B1 exponential decay curve asymptotic to t-axis B1 5(b)(i) current in wire P gives rise to a magnetic field B1 as current (in P) changes, wire Q cuts (magnetic) flux (of wire P) B1 cutting magnetic flux causes induced e.m.f. (across Q) B1 5(b)(ii) sketch shows line with a negative gradient throughout B1

This question in 9702/43 Oct/Nov 2022

6 A heavy aluminium disc has a radius of 0.36 m. The disc rotates with the wheels of a vehicle and forms part of an electromagnetic braking system on the vehicle. In order to activate the braking system, a uniform magnetic field of flux density 0.17 T is switched on. This magnetic field is perpendicular to the plane of rotation of the disc, as shown in Fig. 6.1. aluminium disc, radius 0.36 m rim rotation of disc axle magnetic field, flux density 0.17 T Fig. 6.1 (a) (i) Define magnetic flux. … … … [2] (ii) Calculate the magnetic flux through the disc. Give a unit with your answer. magnetic flux = … unit … [2] (b) The disc is rotating at a rate of 25 revolutions per second. Calculate the magnitude of the electromotive force (e.m.f.) induced between the axle and the rim of the disc. e.m.f. = … V [3] (c) The axle and the rim are connected into an external circuit that enables the energy of the rotation of the disc to be stored for future use. The direction of rotation is shown in Fig. 6.1. Use Lenz’s law of electromagnetic induction to determine whether the current in the disc is from the rim to the axle or from the axle to the rim. Explain your reasoning. … … … … … [3] [Total: 10]

10 marks

Mark scheme: 6(a)(i) product of (magnetic) flux density and area M1 area perpendicular to the (magnetic) field A1 6(a)(ii) flux = B  r2 = 0.17    0.362 C1 = 6.9  10–2 Wb A1 6(b) time for one revolution = 1 / 25 s C1 e.m.f. = rate of cutting flux or  / t C1 = 0.069  25 = 1.7 V A1 6(c) current (in disc) is perpendicular to magnetic field or current causes force to act on disc B1 force opposes rotation of disc B1 left-hand rule indicates current is from rim to axle B1

This question in 9702/42 May/June 2023

7 (a) A Hall probe containing a thin slice of semiconducting material is placed in a uniform magnetic field of flux density B. The largest faces of the slice are perpendicular to the magnetic field, as shown in Fig. 7.1. 5.4 A semiconducting slice x magnetic field, flux density B Q 5.4 A P Fig. 7.1 The thickness x of the slice is 1.8 mm. The number density of charge carriers in the semiconducting material is 1.5 × 1016 m–3. A constant current of 5.4 A is passed through the slice between the shaded faces. The Hall voltage VH that is developed between the terminals PQ is recorded. Fig. 7.2 shows the variation with time t of B. 4 B / 10–6 T 2 0 0 0.02 0.04 0.06 0.08 t / s Fig. 7.2 (i) Show that, when B is equal to 4.0 × 10–6 T, the magnitude of VH is 5.0 V. [1] (ii) On Fig. 7.3, sketch the variation of VH with t between t = 0 and t = 0.080 s. 6 VH / V 4 2 0 0 0.02 0.04 0.06 0.08 t / s –2 – 4 –6 Fig. 7.3 [3] (b) The Hall probe in (a) is replaced with a small flat coil that has 3000 turns. The cross-sectional area of the coil is 3.4 × 10–4 m2. The plane of the coil is perpendicular to the magnetic field. The electromotive force (e.m.f.) E induced between the terminals of the coil is recorded as B varies as shown in Fig. 7.2. (i) Show that the magnitude of E at time t = 0.010 s is 2.0 × 10–4 V. [3] (ii) On Fig. 7.4, sketch the variation of E with t between t = 0 and t = 0.080 s. 4 E / 10–4 V 2 0 0 0.02 0.04 0.06 0.08 t / s –2 – 4 Fig. 7.4 [4] [Total: 11]

11 marks

Mark scheme: 7(a)(i) VH = BI / ntq A1 = (4.0  10–6  5.4) / (1.5  1016  1.8  10–3  1.60  10–19) = 5.0 V 7(a)(ii) sketch: straight diagonal line from (0, 0) to t = 0.020 s B1 and straight diagonal line between two non-zero VH values of same sign from t = 0.040 to 0.050 s horizontal straight line at VH = 5.0 V from t = 0.020 to 0.040 s B1 horizontal straight line at VH = 2.5 V from t = 0.050 to 0.080 s B1 7(b)(i) e.m.f. = rate of change of (magnetic) flux (linkage) C1 E = NA ΔB / Δt or E = NA  gradient (at t = 0.010 s) C1 E = 3000  3.4  10–4  (4.0  10–6) / (0.020) = 2.0  10–4 V A1 7(b)(ii) sketch: line showing non-zero E from t = 0 to t = 0.020 s and from t = 0.040 s to t = 0.050 s, and E = 0 at all other times B1 ‘top hats’ showing constant non-zero E from t = 0 to t = 0.020 s and from t = 0.040 s to t = 0.050 s B1 magnitude of E shown as 2.0  10–4 V in both non-zero sections B1 sign of E in the t = 0 to t = 0.020 s region opposite to the sign of E in the t = 0.040 s to t = 0.050 s region B1

This question in 9702/42 Oct/Nov 2023

7 (a) State Faraday’s law of electromagnetic induction. … … … [2] (b) Fig. 7.1 shows a coil at rest in a uniform magnetic field that is parallel to the axis of the coil. coil magnetic field V Fig. 7.1 The coil is connected to a centre-zero voltmeter. The flux density B of the uniform magnetic field varies with time t as shown in Fig. 7.2. 8 B / mT 4 0 0 1 2 3 4 5 6 t / ms Fig. 7.2 The coil consists of 340 turns, each of cross-sectional area 3.2 × 10–4 m2. (i) Calculate the maximum magnetic flux through one turn of the coil. maximum magnetic flux = … Wb [2] (ii) Determine the maximum rate of change of magnetic flux linkage in the coil. maximum rate of change of flux linkage = … Wb s–1 [3] (iii) State the maximum electromotive force (e.m.f.) V0 induced across the coil. V0 = … V [1] (iv) On Fig. 7.3, sketch the variation of the e.m.f. V induced across the coil with t from t = 0 to t = 6.0 ms. V0 V 0 0 1 2 3 4 5 6 t / ms –V0 Fig. 7.3 [3] (v) The variation of V with t can be described by V = A sin Bt where A and B are constants. Determine the values of A and B. Give units with your answers. A = … unit … B = … unit … [3] [Total: 14]

14 marks

Mark scheme: 7(a) (induced) e.m.f. is (directly) proportional to rate M1 of change of (magnetic) flux (linkage) A1 7(b)(i)  = BA C1 = 7.2  10–3  3.2  10–4 = 2.3  10–6 Wb A1 7(b)(ii) tangent drawn at steepest point on Fig. 7.2 C1 evidence of multiplication by 340 C1 maximum rate of change of flux = 0.82 Wb s–1 A1 7(b)(iii) V0 = 0.82 V or V0 given as identical numerical answer to the answer in (b)(ii) A1 7(b)(iv) sinusoidal curve of period 2.0 ms from t = 0 to t = 6.0 ms B1 all peaks at +V0 and all troughs at –V0 B1 line showing V = 0 at (and only at) t = 0, 1.0, 2.0, 3.0, 4.0, 5.0 and 6.0 ms B1 7(b)(v) A = 0.82 V or A has same numerical value as answer in (b)(iii), with unit V A1 B = 2 / (2.0  10–3) C1 = 3100 rad s–1 A1

This question in 9702/42 May/June 2024

1 A metal wheel consists of an axle A, eight spokes and a rim, as shown in Fig. 1.1. spoke axle A rim X Fig. 1.1 Point X is on the rim at the end of one of the spokes. The rim has a radius of 0.85 m. The wheel is rotating clockwise with an angular speed of 140 rad s–1. (a) For point X, determine: (i) the speed speed = … m s–1 [2] (ii) the centripetal acceleration. acceleration = … m s–2 [2] (b) There is a uniform magnetic field of flux density 0.18 T into the plane of the page. (i) State Lenz’s law of electromagnetic induction. … … … [2] (ii) Show that the time taken for point X to complete one revolution is 45 ms. [1] (iii) Calculate the magnetic flux cut by spoke AX during one revolution of the wheel. Give a unit with your answer. magnetic flux = … unit … [3] (iv) Determine the magnitude of the electromotive force (e.m.f.) induced across spoke AX. induced e.m.f. = … V [2] (v) Use Lenz’s law to explain whether the potential is higher at end A or end X of the spoke. … … … [1] [Total: 13]

13 marks

Mark scheme: Question Answer Marks 1(a)(i) v = r C1 = 0.85  140 A1 = 120 m s–1 1(a)(ii) a = r2 or a = v2 / r C1 a = 0.85  1402 or 1202 / 0.85 A1 = 1.7  104 m s–2 1(b)(i) direction of (induced) e.m.f. M1 is such as to (produce effects that) oppose the change that caused it A1 1(b)(ii) T = 2 /  A1 = 2 / 140 = 0.045 s = 45 ms 1(b)(iii)  = BA C1 = 0.18    0.852 C1 = 0.41 Wb A1 1(b)(iv) E =  / t C1 = 0.41 / 0.045 A1 = 9.1 V 1(b)(v) force (on spoke) must be anticlockwise, so current is from A to X (by Fleming’s left hand rule), so X is at the higher potential B1

This question in 9702/42 Oct/Nov 2024

7 (a) State Faraday’s law of electromagnetic induction. … … … [2] (b) A metal rod is accelerated uniformly from rest in a uniform magnetic field as shown in Fig. 7.1. magnetic field rod into page direction of acceleration Fig. 7.1 The rod has length l and the flux density of the magnetic field is B. An electromotive force (e.m.f.) is induced in the rod. The variation with time t of the induced e.m.f. E is shown in Fig. 7.2. 0.3 E / mV 0.2 0.1 0 0 1 2 t / s Fig. 7.2 (i) Explain how Fig. 7.2 shows that E is proportional to the velocity v of the rod. … … … [2] (ii) Use Faraday’s law to show that the variation of E with time t is given by E = Blat where a is the acceleration of the rod. [3] (iii) The length of the rod is 0.45 m. The acceleration a of the rod is 7.8 m s–2. Determine the value of B. B = … T [2] [Total: 9]

9 marks

Mark scheme: 7(a) (induced) e.m.f. is (directly) proportional to rate M1 of change of (magnetic) flux (linkage) A1 7(b)(i) (uniform acceleration so) velocity is (directly) proportional to time M1 (Fig. 7.2 shows) e.m.f. is (directly) proportional to time so E is proportional to v. A1 7(b)(ii) (v = at so) distance moved in time t = att C1 = BA C1 E = ( / t) = B  L  (att) / t = BLat A1 7(b)(iii) B = (0.30  10–3) / (0.45  7.8  2.0) C1 = 4.3  10–5 T A1

This question in 9702/42 Feb/March 2025

7 (a) State Faraday’s law of electromagnetic induction. … … … [2] (b) An aircraft is flying horizontally at constant speed v through the Earth’s magnetic field, as shown in Fig. 7.1. vertical component of Earth’s aircraft magnetic field, 38 μT P v v 68 m Q VIEW FROM SIDE VIEW FROM ABOVE Fig. 7.1 At the location of the aircraft, the vertical component of the Earth’s magnetic field is 38 μT towards the ground. The distance between the wingtips P and Q of the aircraft is 68 m. As the aircraft moves through the magnetic field, an electromotive force (e.m.f.) of 0.54 V is induced between the wingtips P and Q. (i) Calculate the magnetic flux cut by the wings of the aircraft in a time of 15 s. Give a unit with your answer. magnetic flux = …………………………………… unit ….……. [2] (ii) Determine the area of flux cut by the wings in a time of 15 s. area = … m2 [2] (iii) Use your answer in (b)(ii) to determine the speed v of the aircraft. v = … m s–1 [2] (iv) Use Lenz’s law of electromagnetic induction to explain which of the wingtips P and Q is at the higher induced potential. … … … … … [3] [Total: 11]

11 marks

Mark scheme: 7(a) (induced) e.m.f. is (directly) proportional to rate M1 of change of (magnetic) flux (linkage) A1 7(b)(i) flux = e.m.f.  time C1 flux = 0.54  15 A1 = 8.1 Wb 7(b)(ii) = BA C1 area = 8.1 / (38  10–6) A1 = 2.1  105 m2 7(b)(iii) area = speed  time  width C1 v = (2.1  105) / (15  68) A1 = 210 m s–1 7(b)(iv) opposing force (due to current in wings) must be backwards B1 from Fleming’s left-hand rule, current (in wings) must be from Q to P B1 current is from – to + inside an e.m.f. source so P is at higher potential B1

This question in 9702/41 Oct/Nov 2025

7 (a) State Faraday’s law of electromagnetic induction. … … … [2] (b) An aircraft is flying horizontally at constant speed v through the Earth’s magnetic field, as shown in Fig. 7.1. vertical component of Earth’s aircraft magnetic field, 38 μT P v v 68 m Q VIEW FROM SIDE VIEW FROM ABOVE Fig. 7.1 At the location of the aircraft, the vertical component of the Earth’s magnetic field is 38 μT towards the ground. The distance between the wingtips P and Q of the aircraft is 68 m. As the aircraft moves through the magnetic field, an electromotive force (e.m.f.) of 0.54 V is induced between the wingtips P and Q. (i) Calculate the magnetic flux cut by the wings of the aircraft in a time of 15 s. Give a unit with your answer. magnetic flux = …………………………………… unit ….……. [2] (ii) Determine the area of flux cut by the wings in a time of 15 s. area = … m2 [2] (iii) Use your answer in (b)(ii) to determine the speed v of the aircraft. v = … m s–1 [2] (iv) Use Lenz’s law of electromagnetic induction to explain which of the wingtips P and Q is at the higher induced potential. … … … … … [3] [Total: 11]

11 marks

Mark scheme: 7(a) (induced) e.m.f. is (directly) proportional to rate M1 of change of (magnetic) flux (linkage) A1 7(b)(i) flux = e.m.f.  time C1 flux = 0.54  15 A1 = 8.1 Wb 7(b)(ii) = BA C1 area = 8.1 / (38  10–6) A1 = 2.1  105 m2 7(b)(iii) area = speed  time  width C1 v = (2.1  105) / (15  68) A1 = 210 m s–1 7(b)(iv) opposing force (due to current in wings) must be backwards B1 from Fleming’s left-hand rule, current (in wings) must be from Q to P B1 current is from – to + inside an e.m.f. source so P is at higher potential B1

This question in 9702/43 Oct/Nov 2025

7 (a) State Lenz’s law of electromagnetic induction. … … … [2] (b) A helicopter hovering in stationary equilibrium has four rotors, each of length 12 m, as shown in the view from above in Fig. 7.1. direction of rotation rotors X 12 m O Fig. 7.1 The vertical component of the Earth’s magnetic field at the helicopter is downwards with a flux density of 0.047 mT. The rotors each rotate in a horizontal plane in the direction shown with a frequency of 85 Hz. (i) Calculate the magnetic flux Φ cut by rotor OX during one complete rotation. Give a unit with your answer. Φ = … unit … [3] (ii) Determine the magnitude of the electromotive force (e.m.f.) induced across the length of rotor OX. e.m.f. = … V [2] (iii) Use Lenz’s law to explain whether end O or end X of the rotor is at the higher potential. … … … … [2] [Total: 9]

9 marks

Mark scheme: 7(a) direction of induced e.m.f. M1 such as to (produce effects that) oppose the change that caused it A1 7(b)(i)  = BA C1 = 0.047  10–3    122 C1 = 0.021 Wb A1 7(b)(ii) e.m.f. = flux cut / time C1 = 0.021 / (1 / 85) A1 = 1.8 V 7(b)(iii) force (due to current) cause anticlockwise moment (to oppose rotation) B1 (from the LH rule) current is from O to X, so X is at the higher potential B1

This question in 9702/44 Oct/Nov 2025