Force on a current-carrying conductor — Paper 4 · A Level Physics 9702

6 Two long, straight, current-carrying conductors, PQ and XY, are held a constant distance apart, as shown in Fig. 6.1. Q Y I I P X Fig. 6.1 The conductors each carry the same magnitude current in the same direction. A plan view from above the conductors is shown in Fig. 6.2. current out current out of paper of paper Q Y Fig. 6.2 (a) On Fig. 6.2 draw arrows, one in each case, to show the direction of (i) the magnetic field at Q due to the current in wire XY (label this arrow B), [1] (ii) the force at Q as a result of the magnetic field due to the current in wire XY (label this arrow F). [1] Use (b) (i) State Newton’s third law of motion. … … … [1] (ii) Use this law and your answer in (a)(ii) to state the direction of the force on wire XY. … … [1] (c) The magnetic flux density B at a distance d from a long straight wire carrying a current I is given by I B = 2.0 × 10–7 × . d Use this expression to explain why, under normal circumstances, wires carrying alternating current are not seen to vibrate. Make reasonable estimates of the magnitudes of the quantities involved. … … … … … [4]

8 marks

Mark scheme: 6 (a) (i) arrow B in correct direction (down the page) B1 (ii) arrow F in correct direction (towards Y) B1 [2] (b) (i) When two bodies interact, force on one body is equal but opposite in direction to force on the other body. B1 [1] (ii) direction opposite to that in (a)(ii) B1 [1] (c) suggested reasonable values of I and d B1 mention of expression F = BIL B1 force between wires is small M1 compared to weight of wire A1 [4]

This question in 9702/41 May/June 2006

6 (a) A straight conductor carrying a current I is at an angle θ to a uniform magnetic field of flux density B, as shown in Fig. 6.1. magnetic field, flux density B θθ θ θ current I Fig. 6.1 The conductor and the magnetic field are both in the plane of the paper. State (i) an expression for the force per unit length acting on the conductor due to the magnetic field, force per unit length = … [1] (ii) the direction of the force on the conductor. … [1] Examiner’s Use (b) A coil of wire consisting of two loops is suspended from a fixed point as shown in Fig. 6.2. 0.75 cm 9.4 cm Fig. 6.2 Each loop of wire has diameter 9.4 cm and the separation of the loops is 0.75 cm. The coil is connected into a circuit such that the lower end of the coil is free to move. (i) Explain why, when a current is switched on in the coil, the separation of the loops of the coil decreases. … … … … … [4] (ii) Each loop of the coil may be considered as being a long straight wire. In SI units, the magnetic flux density B at a distance x from a long straight wire carrying a current I is given by the expression I B = 2.0 × 10–7 . x When the current in the coil is switched on, a mass of 0.26 g is hung from the free end of the coil in order to return the loops of the coil to their original separation. Calculate the current in the coil. current = … A [4]

10 marks

Mark scheme: 6 (a) (i) BI sinθ … B1 [1] (ii) (downwards) into (the plane of) the paper … B1 [1] (b) (i) magnetic field (due to current) in one loop OR each loop acts as a coil … B1 cuts/is normal to current in second loop OR produces magnetic field … B1 causing force on second loop OR fields in same direction … M1 either Newton’s 3rd discussed or vice versa clear gives rise to attraction OR so attracts … A1 [4] (ii) B = 2 × 10-7 I/0.75 × 10-2 (= 2.67 × 10-5 I) … C1 force = 0.26 × 10-3 × 9.81 (= 2.55 × 10-3 N) … C1 F = BIL 2.55 × 10-3 = 2.67 × 10-5 × I2 × 2π × 4.7 × 10-2 … C1 I = 18 A … A1 [4] GCE A/AS LEVEL – October/November 2007 9702 04

This question in 9702/41 Oct/Nov 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) Define the tesla. For Examiner’s … Use … … … [3] (b) A large horseshoe magnet produces a uniform magnetic field of flux density B between its poles. Outside the region of the poles, the flux density is zero. The magnet is placed on a top-pan balance and a stiff wire XY is situated between its poles, as shown in Fig. 6.1. Y pole P magnet X top-pan balance Fig. 6.1 The wire XY is horizontal and normal to the magnetic field. The length of wire between the poles is 4.4 cm. A direct current of magnitude 2.6 A is passed through the wire in the direction from X to Y. The reading on the top-pan balance increases by 2.3 g. (i) State and explain the polarity of the pole P of the magnet. … … … … [3] (ii) Calculate the flux density between the poles. For Examiner’s Use flux density = … T [3] (c) The direct current in (b) is now replaced by a very low frequency sinusoidal current of r.m.s. value 2.6 A. Calculate the variation in the reading of the top-pan balance. variation in reading = … g [2]

11 marks

Mark scheme: 6 (a) unit of magnetic flux density / magnetic field strength B1 (uniform) field normal to wire carrying current of 1 A M1 giving force (per unit length) of 1 N m–1 A1 [3] (b) (i) force on magnet / balance is downwards (so by Newton’s third law) B1 force on wire is upwards M1 pole P is a north pole A1 [3] (ii) F = BIL and F = mg (g missing, then 0/3 in (ii)) C1 2.3 × 10–3 × 9.8 = B × 2.6 × 4.4 × 10–2 (g = 10, loses this mark) C1 B = 0.20 T A1 [3] (c) reading for maximum current = 2.3 × √2 C1 total variation = 2 × 2.3 × √2 = 6.5 g A1 [2]

This question in 9702/41 May/June 2009

5 Two long straight vertical wires X and Y pass through a horizontal card, as shown in For Fig. 5.1. Examiner’s Use wire X wire Y horizontal card Fig. 5.1 The current in each wire is in the upward direction. The top view of the card, seen by looking vertically downwards at the card, is shown in Fig. 5.2. card wire X wire Y current out current out of card of card Fig. 5.2 (not to scale) (a) On Fig. 5.2, For Examiner’s (i) draw four field lines to represent the pattern of the magnetic field around wire X due Use solely to the current in wire X, [2] (ii) draw an arrow to show the direction of the force on wire Y due to the magnetic field of wire X. [1] (b) The magnetic flux density B at a distance x from a long straight wire due to a current I in the wire is given by the expression B = 0I , 2 x where is the permeability of free space. 0 The current in wire X is 5.0 A and that in wire Y is 7.0 A. The separation of the wires is 2.5 cm. (i) Calculate the force per unit length on wire Y due to the current in wire X. force per unit length = … N m–1 [4] (ii) The currents in the wires are not equal. State and explain whether the forces on the two wires are equal in magnitude. … … … [2]

9 marks

Mark scheme: 5 (a) (i) concentric circles, anticlockwise ……(minimum 3 circles) … M1 separation of lines increases with distance from wire … A1 [2] (ii) direction from Y towards X … A1 [1] (b) (i) flux density at wire Y = (4π × 10-7 × 5.0) / (2π × 2.5 × 10-2) … C1 = 4.0 × 10-5 T … C1 force per unit length = BI = 4.0 × 10-5 × 7.0 … C1 = 2.8 × 10-4 N … A1 [4] (ii) either force depends on product of the currents in the two wires … M1 so equal … A1 or (isolated system so) Newton’s 3rd law applies … (M1) so equal … (A1) [2] [Total: 9] GCE A/AS LEVEL – October/November 2009 9702 42

This question in 9702/42 Oct/Nov 2009

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

5 Positively charged particles are travelling in a vacuum through three narrow slits S1, S2 and For Examiner’s S3, as shown in Fig. 5.1. Use S1 S2 S3 beam of charged particles direction of electric field Fig. 5.1 Each particle has speed v and charge q. There is a uniform magnetic field of flux density B and a uniform electric field of field strength E in the region between the slits S2 and S3. (a) State the expression for the force F acting on a charged particle due to (i) the magnetic field, … [1] (ii) the electric field. … [1] (b) The electric field acts downwards in the plane of the paper, as shown in Fig. 5.1. State and explain the direction of the magnetic field so that the positively charged particles may pass undeviated through the region between slits S2 and S3. … … … [2]

4 marks

Mark scheme: 5 (a) (i) Bqv(sinθ) or Bqv(cosθ) B1 [1] (ii) qE B1 [1] (b) FB must be opposite in direction to FE B1 so magnetic field into plane of paper B1 [2] GCE AS/A LEVEL – October/November 2011 9702 41

This question in 9702/41 Oct/Nov 2011

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 … … … [3] (b) A horseshoe magnet is placed on a balance. A stiff metal wire is clamped horizontally between the poles, as illustrated in Fig. 5.1. horseshoe magnet stiff metal wire balance pan Fig. 5.1 The magnetic flux density in the space between the poles of the magnet is uniform and is zero outside this region. The length of the metal wire normal to the magnetic field is 6.4 cm. When a current in the wire is switched on, the reading on the balance increases by 2.4 g. The current in the wire is 5.6 A. (i) State and explain the direction of the force on the wire due to the current. … … … … [3] (ii) Calculate the magnitude of the magnetic flux density between the poles of the For magnet. Examiner’s Use flux density = … T [2] (c) A low frequency alternating current is now passed through the wire in (b). The root-mean-square (r.m.s.) value of the current is 5.6 A. Describe quantitatively the variation of the reading seen on the balance. … … … … [2]

10 marks

Mark scheme: 5 (a) (long) straight conductor carrying current of 1 A M1 current/wire normal to magnetic field M1 (for flux density 1 T,) force per unit length is 1 N m–1 A1 [3] (b) (i) (originally) downward force on magnet (due to current) B1 by Newton’s third law (allow “N3”) M1 upward force on wire A1 [3] (ii) F = BIL 2.4 × 10–3 × 9.8 = B × 5.6 × 6.4 × 10–2 C1 B = 0.066 T (need 2 SF) A1 [2] (g missing scores 0/2, but g = 10 leading to 0.067T scores 1/2) (c) new reading is 2.4√2 g C1 either changes between +3.4 g and –3.4 g or total change is 6.8 g A1 [2]

This question in 9702/42 May/June 2012

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 9702/43 May/June 2012

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

6 A stiff straight copper wire XY is held fixed in a uniform magnetic field of flux density 2.6 × 10−3 T, as shown in Fig. 6.1. stiff wire X uniform magnetic field 34° flux density 2.6 × 10–3 T 4.7 cm 34° Y Fig. 6.1 The wire XY has length 4.7 cm and makes an angle of 34° with the magnetic field. (a) Calculate the force on the wire due to a constant current of 5.4 A in the wire. force = … N [2] (b) The current in the wire is now changed to an alternating current of r.m.s. value 1.7 A. Determine the total variation in the force on the wire due to the alternating current. variation in force = … N [3]

5 marks

Mark scheme: 6 (a) F = BIL sinθ C1 = 2.6 × 10–3 × 5.4 × 4.7 × 10–2 × sin 34° = 3.69 × 10–4 N A1 [2] (allow 1 mark for use of cos 34°) (b) peak current = 1.7 × √2 C1 = 2.4 A max. force = 2.6 × 10–3 × 2.4 × 4.7 × 10–2 × sin 34° = 1.64 × 10–4 N C1 variation = 2 × 1.64 × 10–4 = 3.3 × 10–4 N A1 [3] 2

This question in 9702/41 Oct/Nov 2014

6 A stiff straight copper wire XY is held fixed in a uniform magnetic field of flux density 2.6 × 10−3 T, as shown in Fig. 6.1. stiff wire X uniform magnetic field 34° flux density 2.6 × 10–3 T 4.7 cm 34° Y Fig. 6.1 The wire XY has length 4.7 cm and makes an angle of 34° with the magnetic field. (a) Calculate the force on the wire due to a constant current of 5.4 A in the wire. force = … N [2] (b) The current in the wire is now changed to an alternating current of r.m.s. value 1.7 A. Determine the total variation in the force on the wire due to the alternating current. variation in force = … N [3]

5 marks

Mark scheme: 6 (a) F = BIL sinθ C1 = 2.6 × 10–3 × 5.4 × 4.7 × 10–2 × sin 34° = 3.69 × 10–4 N A1 [2] (allow 1 mark for use of cos 34°) (b) peak current = 1.7 × √2 C1 = 2.4 A max. force = 2.6 × 10–3 × 2.4 × 4.7 × 10–2 × sin 34° = 1.64 × 10–4 N C1 variation = 2 × 1.64 × 10–4 = 3.3 × 10–4 N A1 [3] 2

This question in 9702/42 Oct/Nov 2014

9 A stiff wire is held horizontally between the poles of a magnet, as illustrated in Fig. 9.1. S N stiff wire Fig. 9.1 When a constant current of 6.0 A is passed through the wire, there is an additional downwards force on the magnet of 0.080 N. (a) On Fig. 9.1, draw an arrow on the wire to show the direction of the current in the wire. Explain your answer. … … … … [3] (b) The constant current of 6.0 A is now replaced by a low-frequency sinusoidal current. The root-mean-square (r.m.s.) value of this current is 2.5 A. Calculate the difference between the maximum and the minimum forces now acting on the magnet. difference = … N [4] [Total: 7]

7 marks

Mark scheme: 9 (a) (by Newton’s third law) force on wire is up(wards) M1 by (Fleming’s) left-hand rule/right-hand slap rule to give current A1 in direction left to right shown on diagram A1 [3] (b) force ∝ current or F = BIL or B (= 0.080 / 6.0L) = 1 / 75L C1 maximum current = 2.5 × √2 C1 = 3.54 A maximum force in one direction = (3.54 / 6.0) × 0.080 C1 = 0.047 N difference (= 2 × 0.047) = 0.094 N or force varies from 0.047 N upwards to 0.047 N downwards A1 [4]

This question in 9702/42 Oct/Nov 2016

8 A thin slice of conducting material is placed normal to a uniform magnetic field of flux density B, as shown in Fig. 8.1. magnetic field flux density B F E S R C D P Q current I Fig. 8.1 The magnetic field is normal to face CDEF and to face PQRS. A current I passes through the slice and is normal to the faces CDQP and FERS. A potential difference, the Hall voltage VH, is developed across the slice. (a) State the faces between which the Hall voltage VH is developed. … and … [1] (b) The current I is produced by charge carriers, each of charge +q moving at speed v in the direction of the current. The number density of the charge carriers is n. (i) Derive an expression relating the Hall voltage VH to v, B and d, where d is one of the dimensions of the slice. [3] (ii) Use your answer in (b)(i) and an expression for the current I in the slice to derive the expression BI VH = ntq. Explain your working. [2] (c) Suggest why the Hall voltage is difficult to detect in a thin slice of copper. … … … [2] [Total: 8]

8 marks

Mark scheme: 8(a) DERQ and CFSP B1 8(b)(i) force (on charge) due to magnetic field = force due to electric field or Bqv = Eq or v = E / B B1 E = VH / d B1 VH = Bvd B1 8(b)(ii) use of I = nAqv and A = dt M1 algebra clear leading to VH = BI / ntq A1 8(c) (in metal,) n is very large M1 (therefore) VH is small A1

This question in 9702/41 Oct/Nov 2017

3 (a) A mass is undergoing simple harmonic motion with amplitude x0. The maximum velocity of the mass has magnitude v0. On Fig. 3.1, show the variation with displacement x of the velocity v of the mass. v v0 0 −x0 0 x0 x −v0 Fig. 3.1 [2] (b) A straight stiff wire carries a constant current in a region of uniform magnetic flux density. The angle θ between the direction of the current and the direction of the magnetic field is varied. The maximum force on the wire is F0. On Fig. 3.2, show the variation with angle θ of the force F on the wire for values of θ between 0° and 90°. F0 F 0 0 90 θ/° Fig. 3.2 [2] (c) A sinusoidal supply has frequency 250 Hz and r.m.s. potential difference 2.8 V. On the axes of Fig. 3.3, show quantitatively the variation with time t of the voltage V for one cycle of the varying voltage. 8 V / V 6 4 2 00 1 2 3 4 5 t / ms −2 −4 −6 −8 Fig. 3.3 [2] (d) One particular fission reaction may be represented by the equation 23 9 52U + 10n 14516Ba + 9326Kr + 310n The variation with nucleon number A of the binding energy per nucleon BE is shown in Fig. 3.4. BE 0 0 A Fig. 3.4 On Fig. 3.4, mark on the line the position of (i) the nucleus 23952U (label this point U), (ii) the nucleus 14516Ba (label this point Ba), (iii) the nucleus 9326Kr (label this point Kr). [2] [Total: 8]

8 marks

Mark scheme: 3(a) reasonably shaped circle or oval surrounding the origin B1 closed loop passing through (0,±v0) and (±x0,0) B1 3(b) line from (0,0) to (90, F0) B1 curve with decreasing positive gradient, zero gradient at θ = 90 B1 3(c) reasonable sinusoidal wave, one cycle, period 4.0 ms B1 amplitude at 4.0 V B1 3(d) U near right-hand end of line with Ba between U and peak of graph B1 Ba on right hand side of peak and Kr between Ba and peak of graph B1

This question in 9702/42 Feb/March 2018

9 A rigid copper wire is held horizontally between the pole pieces of two magnets, as shown in Fig. 9.1. 8.5 cm 8.5 cm copper wire S S direction of current θ N N side view top view Fig. 9.1 The width of each pole piece is 8.5 cm. The uniform magnetic flux density B in the region between the poles of the magnets is 3.7 mT and is zero outside this region. The angle between the wire and the direction of the magnetic field is θ. The current in the wire is in the direction shown on Fig. 9.1. (a) By reference to the side view of Fig. 9.1, state and explain the direction of the force on the magnets. … … … … [2] (b) The constant current in the wire is 5.1 A. (i) For angle θ equal to 90°, calculate the force on the wire. force = … N [2] (ii) The angle θ is changed to 60°. 8 .5 The length of wire in the magnetic field is cm. c sin60 ° m Calculate the force on the wire. force = … N [1] (c) The constant current in the wire is now changed to an alternating current of frequency 20 Hz and root-mean-square (r.m.s.) value 5.1 A. The angle between the wire and the direction of the magnetic field is 90°. On Fig. 9.2, sketch a graph to show the variation with time t of the force F on the wire for two cycles of the alternating current. F / N 0 0 t / s Fig. 9.2 [3] [Total: 8]

8 marks

Mark scheme: 9(a) using Fleming’s left-hand rule force on wire is upwards B1 by Newton’s third law, force on magnet is downwards B1 9(b)(i) F = BIL C1 = 3.7 × 10–3 × 5.1 × 8.5 × 10–2 = 1.6 × 10–3 N A1 9(b)(ii) F = 1.6 × 10–3 N A1 9(c) sketch: sinusoidal wave with two cycles B1 amplitude 2.3 × 10–3 N B1 period 0.05 s B1

This question in 9702/41 May/June 2018

9 A rigid copper wire is held horizontally between the pole pieces of two magnets, as shown in Fig. 9.1. 8.5 cm 8.5 cm copper wire S S direction of current θ N N side view top view Fig. 9.1 The width of each pole piece is 8.5 cm. The uniform magnetic flux density B in the region between the poles of the magnets is 3.7 mT and is zero outside this region. The angle between the wire and the direction of the magnetic field is θ. The current in the wire is in the direction shown on Fig. 9.1. (a) By reference to the side view of Fig. 9.1, state and explain the direction of the force on the magnets. … … … … [2] (b) The constant current in the wire is 5.1 A. (i) For angle θ equal to 90°, calculate the force on the wire. force = … N [2] (ii) The angle θ is changed to 60°. 8 .5 The length of wire in the magnetic field is cm. c sin60 ° m Calculate the force on the wire. force = … N [1] (c) The constant current in the wire is now changed to an alternating current of frequency 20 Hz and root-mean-square (r.m.s.) value 5.1 A. The angle between the wire and the direction of the magnetic field is 90°. On Fig. 9.2, sketch a graph to show the variation with time t of the force F on the wire for two cycles of the alternating current. F / N 0 0 t / s Fig. 9.2 [3] [Total: 8]

8 marks

Mark scheme: 9(a) using Fleming’s left-hand rule force on wire is upwards B1 by Newton’s third law, force on magnet is downwards B1 9(b)(i) F = BIL C1 = 3.7 × 10–3 × 5.1 × 8.5 × 10–2 = 1.6 × 10–3 N A1 9(b)(ii) F = 1.6 × 10–3 N A1 9(c) sketch: sinusoidal wave with two cycles B1 amplitude 2.3 × 10–3 N B1 period 0.05 s B1

This question in 9702/43 May/June 2018

8 (a) Define magnetic flux density. … … … … [3] (b) A stiff copper wire is balanced horizontally on a pivot, as shown in Fig. 8.1. 7.5 cm P pivot stiff wire S Q R Fig. 8.1 Sections PQ, QR and RS of the wire are situated in a uniform magnetic field of flux density B produced between the poles of a permanent magnet. The perpendicular distance of PQRS from the pivot is 7.5 cm. When a current of 2.7 A is passed through the wire, a small mass of 45 mg is placed a distance 8.8 cm from the pivot in order to restore the balance of the wire, as shown in Fig. 8.2. small mass 7.5 cm 8.8 cm 2.7 A 2.7 A P pivot stiff wire S Q R pole pieces of magnet Fig. 8.2 (i) Explain why, when the current is switched on, the current in the sections PQ and RS of the wire does not affect the balance of the wire. … … … [2] (ii) The length of section QR of the wire is 1.2 cm. Calculate the magnetic flux density B. B = … T [3] [Total: 8]

8 marks

Mark scheme: 8(a) force per unit current B1 force per unit length (of wire) B1 current normal to (magnetic) field B1 8(b)(i) forces (on PQ and RS) are horizontal B1 (hence they create) no moment about the pivot B1 or forces (on PQ and RS) are equal and opposite (B1) (hence there is) no net force (on the two sections) (B1) 8(b)(ii) realisation of the need to apply moments C1 BILx = mgy B × 2.7 × 1.2 × 10–2 × 7.5 = 45 × 10–6 × 9.81 × 8.8 C1 B = 1.6 × 10–2 T A1

This question in 9702/42 Oct/Nov 2018

8 A horseshoe magnet is placed on a top pan balance. A rigid copper wire is fixed between the poles of the magnet, as illustrated in Fig. 8.1. A rigid copper wire balance pan horseshoe magnet B Fig. 8.1 The wire is clamped at ends A and B. (a) When a direct current is switched on in the wire, the reading on the balance is seen to decrease. State and explain the direction of: (i) the force acting on the wire … … … … [3] (ii) the current in the wire. … … … [2] (b) A direct current of 4.6 A in the wire causes the reading on the balance to change by 4.5 × 10–3 N. The direct current is now replaced by an alternating current of frequency 40 Hz and root-mean-square (r.m.s.) value 4.6 A. On the axes of Fig. 8.2, sketch a graph to show the change in balance reading over a time of 50 ms. 8 6 4 change in 2 balance reading 0 / 10–3 N 0 10 20 30 40 50 time / ms –2 –4 –6 –8 Fig. 8.2 [3] [Total: 8]

8 marks

Mark scheme: 8(a)(i) Either Newton’s third law or equal and opposite forces B1 force on magnet is upwards B1 so force on wire downwards B1 8(a)(ii) using (Fleming’s) left-hand rule M1 current from B to A A1 8(b) sinusoidal wave with at least 1 cycle B1 peaks at +6.4 mN and –6.4 mN B1 time period 25 ms 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

8 (a) Explain what is meant by a magnetic field. … … … … [1] (b) The apparatus shown in Fig. 8.1 is used in an experiment to find the magnetic flux density B between the poles of a horseshoe magnet. Assume the magnetic field is uniform between the poles of the magnet and zero elsewhere. 45 mm horseshoe magnet 300 mm metal rod balance pan Fig. 8.1 The rigid metal rod of length 300 mm is fixed in position perpendicular to the direction of the magnetic field. The poles of the magnet are both 45 mm long. There is a current in the rod that causes a force on the rod. The balance is used to determine the magnitude of the force. The variation with current I of the force F on the rod is shown in Fig. 8.2. 10.0 8.0 F / mN 6.0 4.0 2.0 0 0 1.0 2.0 3.0 4.0 5.0 I / A Fig. 8.2 Calculate the magnetic flux density B. B = … T [2] (c) In a different experiment, electrons are accelerated through a potential difference and then enter a region of magnetic field. The magnetic field is into the plane of the paper and is perpendicular to the direction of travel of the electrons, as illustrated in Fig. 8.3. region of magnetic field into the plane of the paper electron beam Fig. 8.3 (i) Explain why the electrons follow a circular path when inside the region of the magnetic field. … … … … [3] (ii) State the measurements needed in order to determine the charge to mass ratio, e /me, of an electron. … … … [2] [Total: 8]

8 marks

Mark scheme: 8(a) a region where a magnet / magnetic material / moving charge / current carrying conductor experiences a force B1 8(b) B = F / Il e.g. = 9 × 10–3 / (5.0 × 0.045) C1 = 0.040 T A1 8(c)(i) force is (always) perpendicular to the velocity / direction of motion B1 magnetic force provides the centripetal force or force perpendicular to motion causes circular motion B1 magnitude of force (due to the magnetic field) is constant or no work done by force or the force does not change the speed B1 8(c)(ii) Applying the list rule, any 2 from: accelerating p.d. radius of path / radius of semicircle magnetic flux density B2

This question in 9702/42 Feb/March 2020

8 (a) Define the tesla. … … … … [3] (b) A magnet produces a uniform magnetic field of flux density B in the space between its poles. A rigid copper wire carrying a current is balanced on a pivot. Part PQLM of the wire is between the poles of the magnet, as illustrated in Fig. 8.1. 5.6 cm M P L weight W N S rigid copper Q wire pivot magnet Fig. 8.1 (not to scale) The wire is balanced horizontally by means of a small weight W. The section of the wire between the poles of the magnet is shown in Fig. 8.2. rigid copper wire M P L N S Q pole of magnet pole of magnet Fig. 8.2 (not to scale) Explain why: (i) section QL of the wire gives rise to a moment about the pivot … … … … [3] (ii) sections PQ and LM of the wire do not affect the equilibrium of the wire. … … … … [2] (c) Section QL of the wire has length 0.85 cm. The perpendicular distance of QL from the pivot is 5.6 cm. When the current in the wire is changed by 1.2 A, W is moved a distance of 2.6 cm along the wire in order to restore equilibrium. The mass of W is 1.3 × 10–4 kg. (i) Show that the change in moment of W about the pivot is 3.3 × 10–5 N m. [2] (ii) Use the information in (i) to determine the magnetic flux density B between the poles of the magnet. B = … T [3] [Total: 13]

13 marks

Mark scheme: 8(a) magnetic field normal to current B1 newton per ampere B1 newton per metre B1 8(b)(i) current in wire QL gives rise to a force or wire QL is perpendicular to the magnetic field B1 force on wire QL is vertical B1 force does not act through the pivot B1 8(b)(ii) forces act through the same line or forces are horizontal B1 forces are equal (in magnitude) and opposite (in direction) B1 8(c)(i) change = mg × (Δ)L C1 = 1.3 × 10–4 × 9.81 × 2.6 × 10–2 = 3.3 × 10–5 N m–1 A1 8(c)(ii) change = B × (Δ)I × L × x C1 3.3 × 10–5 = B × 1.2 × 0.85 × 10–2 × 5.6 × 10–2 C1 B = 0.058 T A1

This question in 9702/41 May/June 2020

8 (a) Define the tesla. … … … … [3] (b) A magnet produces a uniform magnetic field of flux density B in the space between its poles. A rigid copper wire carrying a current is balanced on a pivot. Part PQLM of the wire is between the poles of the magnet, as illustrated in Fig. 8.1. 5.6 cm M P L weight W N S rigid copper Q wire pivot magnet Fig. 8.1 (not to scale) The wire is balanced horizontally by means of a small weight W. The section of the wire between the poles of the magnet is shown in Fig. 8.2. rigid copper wire M P L N S Q pole of magnet pole of magnet Fig. 8.2 (not to scale) Explain why: (i) section QL of the wire gives rise to a moment about the pivot … … … … [3] (ii) sections PQ and LM of the wire do not affect the equilibrium of the wire. … … … … [2] (c) Section QL of the wire has length 0.85 cm. The perpendicular distance of QL from the pivot is 5.6 cm. When the current in the wire is changed by 1.2 A, W is moved a distance of 2.6 cm along the wire in order to restore equilibrium. The mass of W is 1.3 × 10–4 kg. (i) Show that the change in moment of W about the pivot is 3.3 × 10–5 N m. [2] (ii) Use the information in (i) to determine the magnetic flux density B between the poles of the magnet. B = … T [3] [Total: 13]

13 marks

Mark scheme: 8(a) magnetic field normal to current B1 newton per ampere B1 newton per metre B1 8(b)(i) current in wire QL gives rise to a force or wire QL is perpendicular to the magnetic field B1 force on wire QL is vertical B1 force does not act through the pivot B1 8(b)(ii) forces act through the same line or forces are horizontal B1 forces are equal (in magnitude) and opposite (in direction) B1 8(c)(i) change = mg × (Δ)L C1 = 1.3 × 10–4 × 9.81 × 2.6 × 10–2 = 3.3 × 10–5 N m–1 A1 8(c)(ii) change = B × (Δ)I × L × x C1 3.3 × 10–5 = B × 1.2 × 0.85 × 10–2 × 5.6 × 10–2 C1 B = 0.058 T A1

This question in 9702/43 May/June 2020

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) Define the tesla. … … … [2] (b) A stiff metal wire is used to form a rectangular frame measuring 8.0 cm × 6.0 cm. The frame is open at the top, and is suspended from a sensitive newton meter, as shown in Fig. 8.1. newton meter insulating thread 5.0 A 8.0 cm frame P Q 6.0 cm Fig. 8.1 The open ends of the frame are connected to a power supply so that there is a current of 5.0 A in the frame in the direction indicated in Fig. 8.1. The frame is slowly lowered into a uniform magnetic field of flux density B so that all of side PQ is in the field. The magnetic field lines are horizontal and at an angle of 50° to PQ, as shown in Fig. 8.2. B P Q view from above 50° Fig. 8.2 When side PQ of the frame first enters the magnetic field, the reading on the newton meter changes by 1.0 mN. (i) Determine the magnetic flux density B, in mT. B = … mT [2] (ii) State, with a reason, whether the change in the reading on the newton meter is an increase or a decrease. … … … [1] (iii) The frame is lowered further so that the vertical sides start to enter the magnetic field. Suggest what effect this will have on the frame. … … … [1] [Total: 6]

6 marks

Mark scheme: 8(a) newton per ampere per metre M1 where current/wire is perpendicular to magnetic field A1 8(b)(i) F = BIL sinθ C1 B = 1.0 / (5.0 × 0.060 × sin 50°) = 4.4 mT A1 8(b)(ii) (from Fleming’s left-hand rule) force on wire is upwards, so reading decreases B1 8(b)(iii) frame will rotate (so that PQ becomes perpendicular to the field) B1

This question in 9702/41 Oct/Nov 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

8 (a) Define the tesla. … … … [2] (b) A stiff metal wire is used to form a rectangular frame measuring 8.0 cm × 6.0 cm. The frame is open at the top, and is suspended from a sensitive newton meter, as shown in Fig. 8.1. newton meter insulating thread 5.0 A 8.0 cm frame P Q 6.0 cm Fig. 8.1 The open ends of the frame are connected to a power supply so that there is a current of 5.0 A in the frame in the direction indicated in Fig. 8.1. The frame is slowly lowered into a uniform magnetic field of flux density B so that all of side PQ is in the field. The magnetic field lines are horizontal and at an angle of 50° to PQ, as shown in Fig. 8.2. B P Q view from above 50° Fig. 8.2 When side PQ of the frame first enters the magnetic field, the reading on the newton meter changes by 1.0 mN. (i) Determine the magnetic flux density B, in mT. B = … mT [2] (ii) State, with a reason, whether the change in the reading on the newton meter is an increase or a decrease. … … … [1] (iii) The frame is lowered further so that the vertical sides start to enter the magnetic field. Suggest what effect this will have on the frame. … … … [1] [Total: 6]

6 marks

Mark scheme: 8(a) newton per ampere per metre M1 where current/wire is perpendicular to magnetic field A1 8(b)(i) F = BIL sinθ C1 B = 1.0 / (5.0 × 0.060 × sin 50°) = 4.4 mT A1 8(b)(ii) (from Fleming’s left-hand rule) force on wire is upwards, so reading decreases B1 8(b)(iii) frame will rotate (so that PQ becomes perpendicular to the field) B1

This question in 9702/43 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

7 (a) Define magnetic flux density. … … … … [3] (b) An insulated rectangular coil of wire, consisting of 40 turns, is suspended in a cradle from a newton meter, as shown in Fig. 7.1. newton meter cradle coil 40 turns 5.00 cm 3.00 cm Fig. 7.1 The vertical sides of the coil have a length of 5.00 cm and the horizontal sides have a length of 3.00 cm. The initial reading on the newton meter is 0.563 N. A U-shaped magnet rests on a top-pan balance that is set to a reading of 0.00 g. The lower edge of the coil is lowered into the region between the poles of the U-shaped magnet, as shown in the side view in Fig. 7.2. newton meter initial reading 0.563 N coil (viewed from the side) poles of magnet top-pan balance initial reading 0.00 g Fig. 7.2 The magnetic field in the region between the poles is uniform. The lower edge of the coil is entirely within the uniform magnetic field. A current of 3.94 A is now passed through the coil. This causes the reading on the top-pan balance to change to 2.16 g. (i) Explain why the current causes a vertical force to act on the coil. … … … [2] (ii) Determine, to three significant figures, the flux density B of the uniform magnetic field. B = … T [3] (iii) Determine what is now the reading on the newton meter. Explain your reasoning. reading = … N [2] [Total: 10]

10 marks

Mark scheme: 7(a) force per unit current M1 force per unit length M1 current / wire is perpendicular to (magnetic) field (lines) A1 7(b)(i) current (in coil) is perpendicular to magnetic field (so force on wire) B1 force (on wire) is perpendicular to current and field (so is vertical) B1 or current and field are both horizontal (so force is vertical) 7(b)(ii) NBIL = mg C1 B = (2.16  10–3  9.81) / (40  3.94  0.0300) C1 = 4.48  10–3 T A1 7(b)(iii) (magnetic) forces (on balance and newton meter) are (equal and) opposite B1 reading = 0.563 – (2.16  10–3  9.81) A1 = 0.542 N

This question in 9702/42 Oct/Nov 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

6 A rectangular coil PQRS of wire is free to rotate about its axis XY, as shown in Fig. 6.1. Y magnetic R field magnetic S Q field force on PS θ θ force on QR Q P θ P 1.2 A END VIEW X Fig. 6.1 (not to scale) The coil has length QR of 5.4 cm, width PQ of 2.5 cm and has 190 turns of wire. The plane of the coil is at an angle θ to a uniform magnetic field of flux density 5.2 × 10–3 T. The axis XY of the coil is normal to the field. The current in the coil is 1.2 A. (a) (i) Calculate the magnitude of the force on side QR of the coil. force = … N [3] (ii) Use your answer in (a)(i) to show that the torque τ on the coil is given by τ = 1.6 × 10–3 cos θ N m. [2] (iii) Using the expression in (a)(ii) sketch, on the axes of Fig. 6.2, a graph to show the variation of the torque τ with angle θ for values of θ between 0 and 360°. Label the τ axis with an appropriate scale. τ / 10–3 N m 0 0 90 180 270 360 θ/ ° Fig. 6.2 [3] (b) The coil is now replaced by an identical coil wound on a ferrous core. Suggest, with a reason, how the torque on this coil compares with the torque on the original coil. … … … [2] [Total: 10]

10 marks

Mark scheme: 6(a)(i) F = BIL C1 force on QR = 5.2  10–3  1.2  0.054  190 C1 = 0.064 N A1 6(a)(ii) torque = force  perpendicular distance between forces C1 = 0.064  0.025 cos = (1.6  10–3) cos  N m A1 6(a)(iii) one complete cycle of a sinusoidal curve between 0 and 360° B1 τ axis labelled to show maximum and minimum torques at ± 1.6  10–3 N m B1 maximum at 0 and 360° and minimum at 180° (or vice versa), with torque shown as zero at 90° and 270° B1 6(b) (ferrous core) increases magnetic flux density B1 amplitude of torque increases B1

This question in 9702/44 May/June 2025