Magnetic fields due to currents — Paper 4 · A Level Physics 9702

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

7 Two long straight parallel copper wires A and B are clamped vertically. The wires pass For through holes in a horizontal sheet of card PQRS, as shown in Fig. 7.1. Examiner’s Use wire A wire B S R P Q Fig. 7.1 (a) There is a current in wire A in the direction shown on Fig. 7.1. On Fig. 7.1, draw four field lines in the plane PQRS to represent the magnetic field due to the current in wire A. [3] (b) A direct current is now passed through wire B in the same direction as that in wire A. The current in wire B is larger than the current in wire A. (i) On Fig. 7.1, draw an arrow in the plane PQRS to show the direction of the force on wire B due to the magnetic field produced by the current in wire A. [1] (ii) Wire A also experiences a force. State and explain which wire, if any, will experience the larger force. … … … [2] (c) The direct currents in wires A and B are now replaced by sinusoidal alternating currents of equal peak values. The currents are in phase. Describe the variation, if any, of the force experienced by wire B. … … … … [3]

9 marks

Mark scheme: 7 (a) sketch: concentric circles (minimum of 3 circles) M1 separation increasing with distance from wire A1 correct direction B1 [3] (b) (i) arrow direction from wire B towards wire A B1 [1] (ii) either reference to Newton’s third law or force on each wire proportional to product of the two currents M1 so forces are equal A1 [2] (c) force always towards wire A/always in same direction B1 varies from zero (to a maximum value) (1) variation is sinusoidal / sin2 (1) (at) twice frequency of current (1) (any two, one each) B2 [3]

This question in 9702/41 May/June 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

5 (a) Define the tesla. For Examiner’s … Use … … [2] (b) Two long straight vertical wires X and Y are separated by a distance of 4.5 cm, as illustrated in Fig. 5.1. 4.5 cm wire X wire Y Q R P S 6.3 A Fig. 5.1 The wires pass through a horizontal card PQRS. The current in wire X is 6.3 A in the upward direction. Initially, there is no current in wire Y. (i) On Fig. 5.1, sketch, in the plane PQRS, the magnetic flux pattern due to the current in wire X. Show at least four flux lines. [3] (ii) The magnetic flux density B at a distance x from a long straight current-carrying For wire is given by the expression Examiner’s Use μ0I B = 2πx where I is the current in the wire and μ0 is the permeability of free space. Calculate the magnetic flux density at wire Y due to the current in wire X. flux density = … T [2] (iii) A current of 9.3 A is now switched on in wire Y. Use your answer in (ii) to calculate the force per unit length on wire Y. force per unit length = … N m–1 [2] (c) The currents in the two wires in (b)(iii) are not equal. Explain whether the force per unit length on the two wires will be the same, or different. … … … [2]

11 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) sketch: concentric circles M1 increasing separation (must show more than 3 circles) A1 correct direction (anticlockwise, looking down) B1 [3] (ii) B = (4π × 10–7 × 6.3) / (2π × 4.5 × 10–2) C1 = 2.8 × 10–5 T A1 [2] (iii) F = BIL (sinθ) C1 = 2.8 × 10–5 × 9.3 × 1 F/L = 2.6 × 10–4 N m–1 A1 [2] (c) force per unit length depends on product IXIY / by Newton’s third law / action and reaction are equal and opposite M1 so same for both A1 [2] GCE AS/A LEVEL – May/June 2013 9702 42

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

4 A coaxial cable is frequently used to connect an aerial to a television receiver. Such a cable is illustrated in Fig. 4.1. plastic insulator covering copper core copper braid Fig. 4.1 (a) Suggest two functions of the copper braid. 1. … … 2. … … [2] (b) Suggest two reasons why a wire pair is not usually used to connect the aerial to the receiver. 1. … … 2. … … [2] (c) The coaxial cable connecting an aerial to a receiver has length 14 m. The cable has an attenuation per unit length of 190 dB km−1. Calculate the fractional loss in signal power during transmission of the signal along the cable. fractional loss = … [4]

8 marks

Mark scheme: 4(a) acts as ‘return’ (conductor) for signal • shielding from noise/crosstalk/interference Two sensible suggestions, 1 mark each. B2 4(b) • small bandwidth • (there is) noise/interference/crosstalk • large attenuation/energy loss • reflections due to poor impedance matching Two sensible suggestions, 1 mark each. B2 4(c) attenuation = 190 × 14 × 10–3 (= 2.66 dB) C1 ratio / dB = (–)10 lg(P2 / P1) C1 2.66 = –10 lg (POUT / PIN) POUT/ PIN = 0.54 C1 fractional loss = 1 – (POUT / PIN) = 1 – 0.54 = 0.46 A1 or 2.66 = 10 lg (PIN / POUT) PIN/ POUT = 1.85 (C1) fractional loss = (PIN – POUT) / PIN = (1.85 – 1) / 1.85 = 0.46 (A1)

This question in 9702/43 Oct/Nov 2017

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) A long straight vertical wire carries a current I. The wire passes through a horizontal card EFGH, as shown in Fig. 8.1 and Fig. 8.2. current out of plane of paper H G I wire H G E F E F Fig. 8.1 Fig. 8.2 (view from above) On Fig. 8.2, draw the pattern of the magnetic field produced by the current-carrying wire on the plane EFGH. [3] (b) Two long straight parallel wires P and Q are situated a distance 3.1 cm apart, as illustrated in Fig. 8.3. 6.2 A 8.5 A wire P wire Q 3.1cm Fig. 8.3 The current in wire P is 6.2 A. The current in wire Q is 8.5 A. The magnetic flux density B at a distance x from a long straight wire carrying current I is given by the expression μ0I B = 2πx where μ0 is the permeability of free space. Calculate: (i) the magnetic flux density at wire Q due to the current in wire P flux density = … T [2] (ii) the force per unit length, in N m–1, acting on wire Q due to the current in wire P. force per unit length = … N m–1 [2] (c) The currents in wires P and Q are different in magnitude. State and explain whether the forces per unit length on the two wires will be different. … … … [2] [Total: 9]

9 marks

Mark scheme: 8(a) concentric circles (around the wire) M1 at least 3 circles shown, all with increasing separation A1 direction anticlockwise B1 8(b)(i) B = (4π × 10–7 × 6.2) / (2π × 3.1 × 10–2) C1 B = 4.0 × 10–5 T A1 8(b)(ii) F = BIL or F / L = BI C1 F / L = 4.0 × 10–5 × 8.5F / L = 3.4 × 10–4 N m–1 A1 8(c) correct application of Newton’s 3rd law to the forces or F / L is proportional to the product of the two currents M1 so same magnitude A1

This question in 9702/41 Oct/Nov 2019

8 (a) A long straight vertical wire carries a current I. The wire passes through a horizontal card EFGH, as shown in Fig. 8.1 and Fig. 8.2. current out of plane of paper H G I wire H G E F E F Fig. 8.1 Fig. 8.2 (view from above) On Fig. 8.2, draw the pattern of the magnetic field produced by the current-carrying wire on the plane EFGH. [3] (b) Two long straight parallel wires P and Q are situated a distance 3.1 cm apart, as illustrated in Fig. 8.3. 6.2 A 8.5 A wire P wire Q 3.1cm Fig. 8.3 The current in wire P is 6.2 A. The current in wire Q is 8.5 A. The magnetic flux density B at a distance x from a long straight wire carrying current I is given by the expression μ0I B = 2πx where μ0 is the permeability of free space. Calculate: (i) the magnetic flux density at wire Q due to the current in wire P flux density = … T [2] (ii) the force per unit length, in N m–1, acting on wire Q due to the current in wire P. force per unit length = … N m–1 [2] (c) The currents in wires P and Q are different in magnitude. State and explain whether the forces per unit length on the two wires will be different. … … … [2] [Total: 9]

9 marks

Mark scheme: 8(a) concentric circles (around the wire) M1 at least 3 circles shown, all with increasing separation A1 direction anticlockwise B1 8(b)(i) B = (4π × 10–7 × 6.2) / (2π × 3.1 × 10–2) C1 B = 4.0 × 10–5 T A1 8(b)(ii) F = BIL or F / L = BI C1 F / L = 4.0 × 10–5 × 8.5F / L = 3.4 × 10–4 N m–1 A1 8(c) correct application of Newton’s 3rd law to the forces or F / L is proportional to the product of the two currents M1 so same magnitude A1

This question in 9702/43 Oct/Nov 2019

10 (a) A long straight vertical wire A carries a current in an upward direction. The wire passes through the centre of a horizontal card, as illustrated in Fig. 10.1. card current-carrying wire A Fig. 10.1 The card is viewed from above. The card is shown from above in Fig. 10.2. card wire A carrying current out of plane of paper Fig. 10.2 On Fig. 10.2, draw four lines to represent the magnetic field produced by the current-carrying wire. [3] (b) Two wires A and B are now placed through a card. The two wires are parallel and carrying currents in the same direction, as illustrated in Fig. 10.3. wire B wire A card Fig. 10.3 (i) Explain why a magnetic force is exerted on each wire. … … … … [2] (ii) State the directions of the forces. … … [1] (c) The currents in the two wires are not equal. Explain whether the magnetic forces on the two wires are equal in magnitude. … … … [1] [Total: 7]

7 marks

Mark scheme: 10(a) concentric circles centred on the wire B1 separation of lines increasing with distance from wire B1 arrows show anti-clockwise direction B1 10(b)(i) current in (each) wire creates a magnetic field (at the other wire) B1 current (in wire) at 90° to field causes force B1 10(b)(ii) force on each wire towards other wire/attractive B1 10(c) Newton’s third law pair of forces so yes (forces are equal) or force proportional to product of both currents so yes (forces are equal) B1

This question in 9702/42 Oct/Nov 2020

8 (a) Two long straight wires P and Q are parallel to each other, as shown in Fig. 8.1. There is a current in each wire in the direction shown. The pattern of the magnetic field lines in a plane normal to wire P due to the current in the wire is also shown. wire P wire Q plane direction of current magnetic field pattern Fig. 8.1 (i) Draw arrows on the magnetic field lines in Fig. 8.1 around wire P to show the direction of the field. [1] (ii) Determine the direction of the force on wire Q due to the magnetic field from wire P. … [1] (iii) The current in wire Q is less than the current in wire P. State and explain whether the magnitude of the force on wire P is less than, equal to, or greater than the magnitude of the force on wire Q. … … … [2] (b) Nuclear magnetic resonance imaging (NMRI) is used to obtain diagnostic information about internal structures in the human body. Radio waves are produced and directed towards the body. The radio waves affect the protons within the body. (i) Explain why radio waves are used. … … … [2] (ii) Explain why the radio waves are applied in pulses. … … … [2] [Total: 8]

8 marks

Mark scheme: 8(a)(i) at least one anticlockwise arrow and no clockwise arrows B1 8(a)(ii) (force is to the) left B1 8(a)(iii) force is the same B1 Newton’s third law (of motion) or force depends on the product of the two currents B1 Question Answer Marks 8(b)(i) frequency of radio waves is equal to natural frequency of protons B1 resonance of protons occurs / protons absorb energy B1 8(b)(ii) in between pulses / when pulse stops B1 Any 1 from: • protons de-excite • protons emit r.f. pulses • emitted (r.f.) pulse (from proton) detected B1

This question in 9702/42 Feb/March 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

8 Two long straight parallel wires P and Q carry currents into the plane of the paper, as shown in Fig. 8.1. P Q current I current 2I Fig. 8.1 The current in P is I and the current in Q is 2I. (a) (i) On Fig. 8.1, draw an arrow to show the direction of the magnetic field at wire Q due to the current in wire P. Label this arrow B. [1] (ii) On Fig. 8.1, draw another arrow to show the direction of the force acting on wire Q due to the current in wire P. Label this arrow F. [1] (b) (i) State, with a reason, how the magnitude of the force acting on wire P compares with the magnitude of the force acting on wire Q. … … … … [2] (ii) State how the direction of the force on wire P compares with the direction of the force on wire Q. … … [1] [Total: 5]

5 marks

Mark scheme: 8(a)(i) arrow from Q pointing downwards, labelled B B1 8(a)(ii) arrow from Q pointing towards P, labelled F B1 8(b)(i) force is proportional to product of both currents (I and 2I) or Newton’s third law B1 forces are equal B1 8(b)(ii) opposite B1

This question in 9702/42 Oct/Nov 2021

6 (a) State the two conditions that must be satisfied for a copper wire, placed in a magnetic field, to experience a magnetic force. 1 … … 2 … … [2] (b) A long air-cored solenoid is connected to a power supply, so that the solenoid creates a magnetic field. Fig. 6.1 shows a cross-section through the middle of the solenoid. Z section through solenoid wires Y W X Fig. 6.1 The direction of the magnetic field at point W is indicated by the arrow. Three other points are labelled X, Y and Z. (i) On Fig. 6.1, draw arrows to indicate the direction of the magnetic field at each of the points X, Y and Z. [3] (ii) Compare the magnitude of the flux density of the magnetic field: ● at X and at W … … ● at Y and at Z. … … [2] (c) Two long parallel current-carrying wires are placed near to each other in a vacuum. Explain why these wires exert a magnetic force on each other. You may draw a labelled diagram if you wish. … … … … [3] [Total: 10]

10 marks

Mark scheme: 6(a) there must be a current (in the wire) B1 (wire) must be at a non-zero angle to the magnetic field B1 6(b)(i) arrow from X pointing horizontally to the left B1 arrow from Y pointing diagonally upwards and to the left at about 45° B1 arrow from Z pointing horizontally to the right B1 6(b)(ii) (flux densities at W and X are approximately) equal B1 (flux density at) Y greater than (flux density at) Z B1 6(c) current in wire creates magnetic field around wire B1 (each) wire sits in the magnetic field created by the other B1 (for each wire,) current / wire is perpendicular to magnetic field (due to other wire), (so) experiences a (magnetic) force B1

This question in 9702/42 May/June 2022

6 (a) State what is meant by a magnetic field. … … … [2] (b) A long, straight wire P carries a current into the page, as shown in Fig. 6.1. wire P current into page Fig. 6.1 On Fig. 6.1, draw four field lines to represent the magnetic field around wire P due to the current in the wire. [3] (c) A second long, straight wire Q, carrying a current of 5.0 A out of the page, is placed parallel to wire P, as shown in Fig. 6.2. wire P wire Q current current 5.0 A into page out of page Fig. 6.2 The flux density of the magnetic field at wire Q due to the current in wire P is 2.6 mT. (i) Calculate the magnetic force per unit length exerted on wire Q by wire P. force per unit length = … N m–1 [2] (ii) State the direction of the force exerted on wire Q by wire P. … [1] (iii) The flux density of the magnetic field at wire P due to the current in wire Q is 1.5 mT. Determine the magnitude of the current in wire P. Explain your reasoning. current = … A [2] [Total: 10]

10 marks

Mark scheme: 6(a) a region where a force acts on M1 a current-carrying conductor or a moving charge or a magnetic material / magnetic pole A1 6(b) concentric circles around the wire B1 spacing between circles increases with distance from wire B1 arrows showing direction of field is clockwise B1 6(c)(i) F = BIL C1 force per unit length = BI = 2.6  10–3  5.0 = 0.013 N m–1 A1 6(c)(ii) to the right B1 6(c)(iii) force (per unit length) has the same magnitude due to Newton’s 3rd law B1 0.013 = 1.5  10–3  I current = 8.7 A A1

This question in 9702/41 May/June 2023

6 (a) State what is meant by a magnetic field. … … … [2] (b) A long, straight wire P carries a current into the page, as shown in Fig. 6.1. wire P current into page Fig. 6.1 On Fig. 6.1, draw four field lines to represent the magnetic field around wire P due to the current in the wire. [3] (c) A second long, straight wire Q, carrying a current of 5.0 A out of the page, is placed parallel to wire P, as shown in Fig. 6.2. wire P wire Q current current 5.0 A into page out of page Fig. 6.2 The flux density of the magnetic field at wire Q due to the current in wire P is 2.6 mT. (i) Calculate the magnetic force per unit length exerted on wire Q by wire P. force per unit length = … N m–1 [2] (ii) State the direction of the force exerted on wire Q by wire P. … [1] (iii) The flux density of the magnetic field at wire P due to the current in wire Q is 1.5 mT. Determine the magnitude of the current in wire P. Explain your reasoning. current = … A [2] [Total: 10]

10 marks

Mark scheme: 6(a) a region where a force acts on M1 a current-carrying conductor or a moving charge or a magnetic material / magnetic pole A1 6(b) concentric circles around the wire B1 spacing between circles increases with distance from wire B1 arrows showing direction of field is clockwise B1 6(c)(i) F = BIL C1 force per unit length = BI = 2.6  10–3  5.0 = 0.013 N m–1 A1 6(c)(ii) to the right B1 6(c)(iii) force (per unit length) has the same magnitude due to Newton’s 3rd law B1 0.013 = 1.5  10–3  I current = 8.7 A A1

This question in 9702/43 May/June 2023

7 (a) Define magnetic flux density. … … … [2] (b) A long, straight wire carries a current into the page, as shown in Fig. 7.1. Fig. 7.1 On Fig. 7.1, draw four field lines to represent the magnetic field around the wire due to the current in it. [3] (c) Two identical wires X and Y are placed parallel to each other. The wires both carry current into the page, as shown in Fig. 7.2. X Y Fig. 7.2 (i) Explain why the two wires exert a magnetic force on each other. … … … … [2] (ii) On Fig. 7.2, draw an arrow to show the direction of the magnetic force exerted on wire X. Label your arrow F. [1] (iii) The current in X is double the current in Y. State how the magnetic force exerted on wire Y compares with the magnetic force exerted on wire X. … … … [2] (iv) The direction of the current in both wires is now reversed. State, with a reason, the effect of this change on the direction of the force on wire X. … … [1] [Total: 11]

11 marks

Mark scheme: 7(a) • force per unit length B2 • force per unit current • length / current perpendicular to field 1 mark for any two points, 2 marks for all three points 7(b) concentric circles around the wire (at least two circles needed) B1 spacing between circles increases with distance from wire (at least four circles needed) B1 arrows showing direction of field is clockwise B1 7(c)(i) (each) wire sits in the (magnetic) field created by the other B1 current (in one wire) is perpendicular to (magnetic) field (due to other wire) so (magnetic) force acts (on wire) B1 7(c)(ii) arrow drawn, starting from X and pointing towards Y, labelled F B1 7(c)(iii) (forces have) equal magnitudes B1 (forces are in) opposite directions B1 7(c)(iv) no change (in the direction of the force) since both the current in X and the field due to Y have reversed B1

This question in 9702/41 Oct/Nov 2024

7 (a) Define magnetic flux density. … … … [2] (b) A long, straight wire carries a current into the page, as shown in Fig. 7.1. Fig. 7.1 On Fig. 7.1, draw four field lines to represent the magnetic field around the wire due to the current in it. [3] (c) Two identical wires X and Y are placed parallel to each other. The wires both carry current into the page, as shown in Fig. 7.2. X Y Fig. 7.2 (i) Explain why the two wires exert a magnetic force on each other. … … … … [2] (ii) On Fig. 7.2, draw an arrow to show the direction of the magnetic force exerted on wire X. Label your arrow F. [1] (iii) The current in X is double the current in Y. State how the magnetic force exerted on wire Y compares with the magnetic force exerted on wire X. … … … [2] (iv) The direction of the current in both wires is now reversed. State, with a reason, the effect of this change on the direction of the force on wire X. … … [1] [Total: 11]

11 marks

Mark scheme: 7(a) • force per unit length B2 • force per unit current • length / current perpendicular to field 1 mark for any two points, 2 marks for all three points 7(b) concentric circles around the wire (at least two circles needed) B1 spacing between circles increases with distance from wire (at least four circles needed) B1 arrows showing direction of field is clockwise B1 7(c)(i) (each) wire sits in the (magnetic) field created by the other B1 current (in one wire) is perpendicular to (magnetic) field (due to other wire) so (magnetic) force acts (on wire) B1 7(c)(ii) arrow drawn, starting from X and pointing towards Y, labelled F B1 7(c)(iii) (forces have) equal magnitudes B1 (forces are in) opposite directions B1 7(c)(iv) no change (in the direction of the force) since both the current in X and the field due to Y have reversed B1

This question in 9702/43 Oct/Nov 2024