Concept of a magnetic field — 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

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 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

8 (a) Describe what is meant by a magnetic field. For Examiner’s … Use … … … … [3] (b) A small mass is placed in a field of force that is either electric or magnetic or gravitational. State the nature of the field of force when the mass is (i) charged and the force is opposite to the direction of the field, … [1] (ii) uncharged and the force is in the direction of the field, … [1] (iii) charged and there is a force only when the mass is moving, … [1] (iv) charged and there is no force on the mass when it is stationary or moving in a particular direction. … [1]

7 marks

Mark scheme: 8 (a) region (of space) / area where B1 a force is experienced by M1 current-carrying conductor / moving charge / permanent magnet A1 [3] (b) (i) electric B1 [1] (ii) gravitational B1 [1] (iii) magnetic B1 [1] (iv) magnetic B1 [1] GCE A/AS LEVEL – October/November 2008 9702 04 Section B

This question in 9702/41 Oct/Nov 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

6 The current in a long, straight vertical wire is in the direction XY, as shown in Fig. 6.1. For Examiner’s Y Use D C A B X Fig. 6.1 (a) On Fig. 6.1, sketch the pattern of the magnetic flux in the horizontal plane ABCD due to the current-carrying wire. Draw at least four flux lines. [3] (b) The current-carrying wire is within the Earth’s magnetic field. As a result, the pattern drawn in Fig. 6.1 is superposed with the horizontal component of the Earth’s magnetic field. Fig. 6.2 shows a plan view of the plane ABCD with the current in the wire coming out of the plane. D C magnetic field current out of of Earth plane ABCD A B Fig. 6.2 The horizontal component of the Earth’s magnetic field is also shown. (i) On Fig. 6.2, mark with the letter P a point where the magnetic field due to the For current-carrying wire could be equal and opposite to that of the Earth. [1] Examiner’s Use (ii) For a long, straight wire carrying current I, the magnetic flux density B at distance r from the centre of the wire is given by the expression I B = μ0 2πr where μ0 is the permeability of free space. The point P in (i) is found to be 1.9 cm from the centre of the wire for a current of 1.7 A. Calculate a value for the horizontal component of the Earth’s magnetic flux density. flux density = … T [2] (c) The current in the wire in (b)(ii) is increased. The point P is now found to be 2.8 cm from the wire. Determine the new current in the wire. current = … A [2]

8 marks

Mark scheme: 6 (a) concentric circles …(at least three lines) … M1 with increasing separation … A1 correct direction clear … B1 [3] (b) (i) correct position to left of wire … B1 [1] (ii) B = (4π × 10-7 × 1.7) / (2π × 1.9 × 10-2) … C1 = 1.8 × 10-5 T … A1 [2] (c) distance ∝ current … C1 current = (2.8 / 1.9) × 1.7 = 2.5 A … A1 [2] [Total: 8]

This question in 9702/41 Oct/Nov 2009

11 Outline briefly the main principles of the use of magnetic resonance to obtain diagnostic For information about internal body structures. Examiner’s Use … … … … … … … … … … … … [6]

6 marks

Mark scheme: 11 large / 1 T magnetic field applied along body (allow ‘across’) (1) r.f. pulse applied … (1) causes hydrogen nuclei / protons … (1) to resonate … (1) (nuclei) return to equilibrium state / after relaxation time … (1) r.f. (pulse) emitted … (1) pulses detected, processed and displayed … (1) resonant frequency depends on magnetic field strength … (1) calibrated non-uniform field enables nuclei to be located … (1) any six points, one mark each … B6 [6] [Total: 6] GCE A/AS LEVEL – October/November 2009 9702 41

This question in 9702/41 Oct/Nov 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 (a) A constant current is maintained in a long straight vertical wire. A Hall probe is positioned For a distance r from the centre of the wire, as shown in Fig. 5.1. Examiner’s Use current-carrying wire Hall probe X Y terminals to r Hall probe circuitry and voltmeter Fig. 5.1 (i) Explain why, when the Hall probe is rotated about the horizontal axis XY, the Hall voltage varies between a maximum positive value and a maximum negative value. … … … [2] (ii) The maximum Hall voltage VH is measured at different distances r. Data for VH and the corresponding values of r are shown in Fig. 5.2. VH / V r / cm 0.290 1.0 0.190 1.5 0.140 2.0 0.097 3.0 0.073 4.0 0.060 5.0 Fig. 5.2 It is thought that VH and r are related by an expression of the form k VH = r where k is a constant. 1. Without drawing a graph, use data from Fig. 5.2 to suggest whether the For expression is valid. Examiner’s Use [2] 1 2. A graph showing the variation with of VH is plotted. r State the features of the graph that suggest that the expression is valid. … … [1] (b) The Hall probe in (a) is now replaced with a small coil of wire connected to a sensitive voltmeter. The coil is arranged so that its plane is normal to the magnetic field of the wire. (i) State Faraday’s law of electromagnetic induction and hence explain why the voltmeter indicates a zero reading. … … … … [3] (ii) State three different ways in which an e.m.f. may be induced in the coil. 1. … … 2. … … 3. … … [3]

11 marks

Mark scheme: 5 (a) (i) VH depends on angle between (plane of) probe and B-field B1 either VH max when plane and B-field are normal to each other or VH zero when plane and B-field are parallel or VH depends on sine of angle between plane and B-field B1 [2] (ii) 1 calculates VHr at least three times M1 to 1 s.f. constant so valid or approx constant so valid or to 2 s.f., not constant so invalid A1 [2] 2 straight line passes through origin B1 [1] (b) (i) e.m.f. induced is proportional / equal to M1 rate of change of (magnetic) flux (linkage) A1 constant field in coil / flux (linkage) of coil does not change B1 [3] (ii) e.g. vary current (in wire) / switch current on or off / use a.c. current rotate coil move coil towards / away from wire (1 mark each, max 3) B3 [3]

This question in 9702/41 May/June 2010

6 (a) A uniform magnetic field has constant flux density B. A straight wire of fixed length For carries a current I at an angle θ to the magnetic field, as shown in Fig. 6.1. Examiner’s Use magnetic field flux density B I current-carrying wire Fig. 6.1 (i) The current I in the wire is changed, keeping the angle θ constant. On Fig. 6.2, sketch a graph to show the variation with current I of the force F on the wire. F 0 0 I Fig. 6.2 [2] (ii) The angle θ between the wire and the magnetic field is now varied. The current I is For kept constant. Examiner’s On Fig. 6.3, sketch a graph to show the variation with angle θ of the force F on the Use wire. F 0 0 30 60 90 / ° Fig. 6.3 [3] (b) A uniform magnetic field is directed at right-angles to the rectangular surface PQRS of a slice of a conducting material, as shown in Fig. 6.4. uniform magnetic field Q R direction of movement P S of electrons Fig. 6.4 Electrons, moving towards the side SR, enter the slice of conducting material. The electrons enter the slice at right-angles to side SR. (i) Explain why, initially, the electrons do not travel in straight lines across the slice from side SR to side PQ. … … … [2] (ii) Explain to which side, PS or QR, the electrons tend to move. … … … [2]

9 marks

Mark scheme: 6 (a) (i) straight line with positive gradient M1 through origin A1 [2] (ii) maximum force shown at θ = 90° M1 zero force shown at θ = 0° M1 reasonable curve with F about ½ max at 30° A1 [3] (b) (i) force on electron due to magnetic field B1 force on electron normal to magnetic field and direction of electron B1 [2] (ii) quote / mention of (Fleming’s) left hand rule M1 electron moves towards QR A1 [2]

This question in 9702/42 May/June 2010

6 (a) A uniform magnetic field has constant flux density B. A straight wire of fixed length For carries a current I at an angle θ to the magnetic field, as shown in Fig. 6.1. Examiner’s Use magnetic field flux density B I current-carrying wire Fig. 6.1 (i) The current I in the wire is changed, keeping the angle θ constant. On Fig. 6.2, sketch a graph to show the variation with current I of the force F on the wire. F 0 0 I Fig. 6.2 [2] (ii) The angle θ between the wire and the magnetic field is now varied. The current I is For kept constant. Examiner’s On Fig. 6.3, sketch a graph to show the variation with angle θ of the force F on the Use wire. F 0 0 30 60 90 / ° Fig. 6.3 [3] (b) A uniform magnetic field is directed at right-angles to the rectangular surface PQRS of a slice of a conducting material, as shown in Fig. 6.4. uniform magnetic field Q R direction of movement P S of electrons Fig. 6.4 Electrons, moving towards the side SR, enter the slice of conducting material. The electrons enter the slice at right-angles to side SR. (i) Explain why, initially, the electrons do not travel in straight lines across the slice from side SR to side PQ. … … … [2] (ii) Explain to which side, PS or QR, the electrons tend to move. … … … [2]

9 marks

Mark scheme: 6 (a) (i) straight line with positive gradient M1 through origin A1 [2] (ii) maximum force shown at θ = 90° M1 zero force shown at θ = 0° M1 reasonable curve with F about ½ max at 30° A1 [3] (b) (i) force on electron due to magnetic field B1 force on electron normal to magnetic field and direction of electron B1 [2] (ii) quote / mention of (Fleming’s) left hand rule M1 electron moves towards QR A1 [2]

This question in 9702/43 May/June 2010

5 Positive ions are travelling through a vacuum in a narrow beam. The ions enter a region of For uniform magnetic field of flux density B and are deflected in a semi-circular arc, as shown in Examiner’s Fig. 5.1. Use detector uniform magnetic field 12.8 cm beam of positive ions Fig. 5.1 The ions, travelling with speed 1.40 × 105 m s–1, are detected at a fixed detector when the diameter of the arc in the magnetic field is 12.8 cm. (a) By reference to Fig. 5.1, state the direction of the magnetic field. … [1] (b) The ions have mass 20 u and charge +1.6 × 10–19 C. Show that the magnetic flux density is 0.454 T. Explain your working. [3] (c) Ions of mass 22 u with the same charge and speed as those in (b) are also present in For the beam. Examiner’s Use (i) On Fig. 5.1, sketch the path of these ions in the magnetic field of magnetic flux density 0.454 T. [1] (ii) In order to detect these ions at the fixed detector, the magnetic flux density is changed. Calculate this new magnetic flux density. magnetic flux density = … T [2]

7 marks

Mark scheme: 5 (a) field into (the plane of) the paper B1 [1] (b) force due to magnetic field provides the centripetal force B1 mv2 / r = Bqv C1 B = (20 × 1.66 × 10–27 × 1.40 × 105) / (1.6 × 10–19 × 6.4 × 10–2) B1 = 0.454 T A0 [3] (c) (i) semicircle with diameter greater than 12.8 cm B1 [1] 22 (ii) new flux density = × 0.454 C1 20 B = 0.499 T A1 [2]

This question in 9702/41 Oct/Nov 2010

5 Positive ions are travelling through a vacuum in a narrow beam. The ions enter a region of For uniform magnetic field of flux density B and are deflected in a semi-circular arc, as shown in Examiner’s Fig. 5.1. Use detector uniform magnetic field 12.8 cm beam of positive ions Fig. 5.1 The ions, travelling with speed 1.40 × 105 m s–1, are detected at a fixed detector when the diameter of the arc in the magnetic field is 12.8 cm. (a) By reference to Fig. 5.1, state the direction of the magnetic field. … [1] (b) The ions have mass 20 u and charge +1.6 × 10–19 C. Show that the magnetic flux density is 0.454 T. Explain your working. [3] (c) Ions of mass 22 u with the same charge and speed as those in (b) are also present in For the beam. Examiner’s Use (i) On Fig. 5.1, sketch the path of these ions in the magnetic field of magnetic flux density 0.454 T. [1] (ii) In order to detect these ions at the fixed detector, the magnetic flux density is changed. Calculate this new magnetic flux density. magnetic flux density = … T [2]

7 marks

Mark scheme: 5 (a) field into (the plane of) the paper B1 [1] (b) force due to magnetic field provides the centripetal force B1 mv2 / r = Bqv C1 B = (20 × 1.66 × 10–27 × 1.40 × 105) / (1.6 × 10–19 × 6.4 × 10–2) B1 = 0.454 T A0 [3] (c) (i) semicircle with diameter greater than 12.8 cm B1 [1] 22 (ii) new flux density = × 0.454 C1 20 B = 0.499 T 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

5 (a) State what is meant by a magnetic field. For Examiner’s … Use … … [2] (b) A charged particle of mass m and charge +q is travelling with velocity v in a vacuum. It enters a region of uniform magnetic field of flux density B, as shown in Fig. 5.1. region of magnetic field path of charged particle Fig. 5.1 The magnetic field is normal to the direction of motion of the particle. The path of the particle in the field is the arc of a circle of radius r. (i) Explain why the path of the particle in the field is the arc of a circle. … … … … [2] (ii) Show that the radius r is given by the expression mv r = . Bq [2] (c) A thin metal foil is placed in the magnetic field in (b). For A second charged particle enters the region of the magnetic field. It loses kinetic energy Examiner’s as it passes through the foil. The particle follows the path shown in Fig. 5.2. Use region of uniform magnetic field foil Fig. 5.2 (i) On Fig. 5.2, mark with an arrow the direction of travel of the particle. [1] (ii) The path of the particle has different radii on each side of the foil. The radii are 7.4 cm and 5.7 cm. Determine the ratio final momentum of particle initial momentum of particle for the particle as it passes through the foil. ratio = … [2]

9 marks

Mark scheme: 5 (a) region (of space) where there is a force M1 either on / produced by magnetic pole or on / produced by current carrying conductor / moving charge A1 [2] (b) (i) force on particle is (always) normal to velocity / direction of travel B1 speed of particle is constant B1 [2] (ii) magnetic force provides the centripetal force B1 mv2 / r = Bqv M1 r = mv / Bq A0 [2] (c) (i) direction from ‘bottom to top’ of diagram B1 [1] (ii) radius proportional to momentum C1 ratio = 5.7 / 7.4 = 0.77 A1 [2] (answer must be consistent with direction given in (c)(i))

This question in 9702/41 May/June 2011

12 Many television receivers are connected to an aerial using a coaxial cable. Such a cable is For illustrated in Fig. 12.1. Examiner’s Use copper wire polythene plastic insulator covering copper braid Fig. 12.1 (a) State two functions of the copper braid. 1. … … 2. … … [2] (b) Suggest two reasons why a coaxial cable is used, rather than a wire pair, to connect the aerial to the receiver. 1. … … 2. … … [2] (c) A coaxial cable has an attenuation per unit length of 200 dB km–1. The length of the co-axial cable between an aerial and the receiver is 12 m. Calculate the ratio input signal power to coaxial cable . output signal power from coaxial cable

4 marks

Mark scheme: 12 (a) e.g. acts as ‘return’ for the signal shields inner core from noise / interference / cross-talk (any two sensible answers, 1 each, max 2) B2 [2] (b) e.g. greater bandwidth less attenuation (per unit length) less noise / interference (any two sensible answers, 1 each, max 2) B2 [2] (c) attenuation is 2.4 dB C1 attenuation = 10 lg(P1 / P2) C1 ratio = 1.7 A1 [3]

This question in 9702/41 May/June 2011

10 Explain briefly the main principles of the use of magnetic resonance to obtain diagnostic For information about internal body structures. Examiner’s Use … … … … … … … … … … … [8]

8 marks

Mark scheme: 10 strong / large (uniform) magnetic field B1 nuclei precess / rotate about field direction (1) radio frequency pulse B1 at Larmor frequency (1) causes resonance / nuclei absorb energy B1 on relaxation / de-excitation, nuclei emit r.f. pulse B1 pulse detected and processed (1) non-uniform field superposed on uniform field B1 allows position of resonating nuclei to be determined B1 allows for location of detection to be changed (1) (six points, 1 each plus any two extra – max 8) [8]

This question in 9702/42 May/June 2011

10 Explain briefly the main principles of the use of magnetic resonance to obtain diagnostic For information about internal body structures. Examiner’s Use … … … … … … … … … … … [8]

8 marks

Mark scheme: 10 strong / large (uniform) magnetic field B1 nuclei precess / rotate about field direction (1) radio frequency pulse B1 at Larmor frequency (1) causes resonance / nuclei absorb energy B1 on relaxation / de-excitation, nuclei emit r.f. pulse B1 pulse detected and processed (1) non-uniform field superposed on uniform field B1 allows position of resonating nuclei to be determined B1 allows for location of detection to be changed (1) (six points, 1 each plus any two extra – max 8) [8]

This question in 9702/43 May/June 2011

6 (a) Define the tesla. For Examiner’s … Use … … … [3] (b) A charged particle of mass m and charge +q is travelling with velocity v in a vacuum. It enters a region of uniform magnetic field of flux density B as shown in Fig. 6.1. particle mass m, charge +q uniform magnetic field flux density B Fig. 6.1 The magnetic field is normal to the direction of motion of the particle. The path of the particle in the field is the arc of a circle of radius r. (i) Explain why the path of the particle in the field is the arc of a circle. … … … … [2] (ii) Show that the radius r is given by the expression mv r = . Bq [1] (c) A uniform magnetic field is produced in the region PQRS, as shown in Fig. 6.2. For Examiner’s Use P Q X uniform magnetic field S R Fig. 6.2 The magnetic field is normal to the page. At point X, a gamma-ray photon interaction causes two particles to be formed. The paths of these particles are shown in Fig. 6.2. (i) Suggest, with a reason, why each of the paths is a spiral, rather than the arc of a circle. … … … [2] (ii) State and explain what can be deduced from the paths about 1. the charges on the two particles, … … … [2] 2. the initial speeds of the two particles. … … … [2]

12 marks

Mark scheme: 6 (a) unit of magnetic flux density B1 field normal to (straight) conductor carrying current of 1 A M1 force per unit length is 1 N m–1 A1 [3] (b) (i) force on particle always normal to direction of motion M1 (and speed of particle is constant) magnetic force provides the centripetal force A1 [2] (ii) mv 2/r = Bqv M1 r = mv/Bq A0 [1] (c) (i) the momentum/speed is becoming less M1 so the radius is becoming smaller A1 [2] (ii) 1. spirals are in opposite directions M1 so oppositely charged A1 [2] 2. equal initial radii M1 so equal (initial) speeds A1 [2] GCE AS/A LEVEL – October/November 2011 9702 43

This question in 9702/43 Oct/Nov 2011

12 In a cellular phone network, a region is divided into a number of cells, each with its own base For station. Examiner’s Use (a) Suggest and explain two reasons why a region is divided into a number of cells. 1. … … … 2. … … … [4] (b) A passenger in a car is using a mobile phone as the car moves across several cells. Outline how it is ensured that the phone call is continuous. … … … … … [4]

8 marks

Mark scheme: 12 (a) e.g. carrier frequencies can be re-used (without interference) (M1) so increased number of handsets can be used (A1) e.g. lower power transmitters (M1) so less interference (A1) e.g. UHF used (M1) so must be line-of-sight/short handset aerial (A1) (any two sensible suggestions with explanation, max 4) B4 [4] (b) computer at cellular exchange B1 monitors the signal power B1 relayed from several base stations B1 switches call to base station with strongest signal B1 [4]

This question in 9702/43 Oct/Nov 2011

12 A person is to be investigated using a magnetic resonance (MR) scanner. For Examiner’s (a) This technique involves the use of two superimposed magnetic fields. Use Describe the functions of these two magnetic fields. … … … … … … [4] (b) The frequency f of the electromagnetic waves emitted by protons on relaxation in an MR scanner is given by the equation f = 2cB where B is the total magnetic flux density and c is a constant equal to 1.34 × 108 s–1 T–1. The magnetic flux density changes by 2.0 × 10–4 T for each 1.0 cm thickness of tissue in a section. The scanner is adjusted so that the thickness of each section is 3.0 mm. Calculate, for corresponding points in neighbouring sections, (i) the difference in magnetic flux density, difference in flux density = … T [1] (ii) the change in emitted frequency. frequency change = … Hz [2]

7 marks

Mark scheme: 12 (a) strong uniform (magnetic) field M1 either aligns nuclei or gives rise to Larmor/resonant frequency in r.f. region A1 non-uniform (magnetic) field M1 either enables nuclei to be located or changes the Larmor/resonant frequency A1 [4] (b) (i) difference in flux density = 2.0 × 10–2 × 3.0 × 10–3 = 6.0 × 10–5 T A1 [1] (ii) ∆f = 2 × c × ∆B C1 = 2 × 1.34 × 108 × 6.0 × 10–5 = 1.6 × 104 Hz A1 [2] GCE AS/A LEVEL – May/June 2012 9702 41

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

12 A person is to be investigated using a magnetic resonance (MR) scanner. For Examiner’s (a) This technique involves the use of two superimposed magnetic fields. Use Describe the functions of these two magnetic fields. … … … … … … [4] (b) The frequency f of the electromagnetic waves emitted by protons on relaxation in an MR scanner is given by the equation f = 2cB where B is the total magnetic flux density and c is a constant equal to 1.34 × 108 s–1 T–1. The magnetic flux density changes by 2.0 × 10–4 T for each 1.0 cm thickness of tissue in a section. The scanner is adjusted so that the thickness of each section is 3.0 mm. Calculate, for corresponding points in neighbouring sections, (i) the difference in magnetic flux density, difference in flux density = … T [1] (ii) the change in emitted frequency. frequency change = … Hz [2]

7 marks

Mark scheme: 12 (a) strong uniform (magnetic) field M1 either aligns nuclei or gives rise to Larmor/resonant frequency in r.f. region A1 non-uniform (magnetic) field M1 either enables nuclei to be located or changes the Larmor/resonant frequency A1 [4] (b) (i) difference in flux density = 2.0 × 10–2 × 3.0 × 10–3 = 6.0 × 10–5 T A1 [1] (ii) ∆f = 2 × c × ∆B C1 = 2 × 1.34 × 108 × 6.0 × 10–5 = 1.6 × 104 Hz A1 [2] GCE AS/A LEVEL – May/June 2012 9702 43

This question in 9702/43 May/June 2012

6 (a) (i) State the condition for a charged particle to experience a force in a magnetic field. For Examiner’s … Use … … [2] (ii) State an expression for the magnetic force F acting on a charged particle in a magnetic field of flux density B. Explain any other symbols you use. … … … [2] (b) A sample of a conductor with rectangular faces is situated in a magnetic field, as shown in Fig. 6.1. direction of magnetic field B C direction of F G movement A D of electrons E H Fig. 6.1 The magnetic field is normal to face ABCD in the downward direction. Electrons enter face CDHG at right-angles to the face. As the electrons pass through the conductor, they experience a force due to the magnetic field. (i) On Fig. 6.1, shade the face to which the electrons tend to move as a result of this force. [1] (ii) The movement of the electrons in the magnetic field causes a potential difference between two faces of the conductor. Using the lettering from Fig. 6.1, state the faces between which this potential difference will occur. face … and face … [1] (c) Explain why the potential difference in (b) causes an additional force on the moving electrons in the conductor. … … … [2]

8 marks

Mark scheme: 6 (a) (i) particle must be moving M1 with component of velocity normal to magnetic field A1 [2] (ii) F = Bqv sin θ M1 q, v and θ explained A1 [2] (b) (i) face BCGF shaded A1 [1] (ii) between face BCGF and face ADHE A1 [1] (c) potential difference gives rise to an electric field M1 either FE = qE (no need to explain symbols) or electric field gives rise to force (on an electron) A1 [2] GCE AS/A LEVEL – October/November 2012 9702 41

This question in 9702/41 Oct/Nov 2012

11 (a) In modern communications systems, the majority of data is transmitted in digital form For rather than analogue form. Examiner’s Suggest three advantages of the transmission of data in digital form. Use 1. … … 2. … … 3. … … [3] (b) A recording is made of some music. For this recording, the music is sampled at a rate of 44.1 kHz and each sample consists of a 16-bit word. (i) Suggest the effect on the quality of the recording of 1. sampling at a high frequency rather than a lower frequency, … … [1] 2. using a long word length rather than a shorter word length. … … [1] (ii) The recording lasts for a total time of 5 minutes 40 seconds. Calculate the number of bits generated during the recording. number = … [2]

7 marks

Mark scheme: 11 (a) e.g. noise can be eliminated / filtered / signal can be regenerated extra bits can be added to check for errors multiplexing possible digital circuits are more reliable / cheaper data can be encrypted for security any sensible advantages, 1 each, max. 3 B3 [3] (b) (i) 1. higher frequencies can be reproduced B1 [1] 2. smaller changes in loudness / amplitude can be detected B1 [1] (ii) bit rate = 44.1 × 103 × 16 C1 = 7.06 × 105 s–1 number = 7.06 × 106 × 340 = 2.4 × 108 A1 [2]

This question in 9702/41 Oct/Nov 2012

6 (a) (i) State the condition for a charged particle to experience a force in a magnetic field. For Examiner’s … Use … … [2] (ii) State an expression for the magnetic force F acting on a charged particle in a magnetic field of flux density B. Explain any other symbols you use. … … … [2] (b) A sample of a conductor with rectangular faces is situated in a magnetic field, as shown in Fig. 6.1. direction of magnetic field B C direction of F G movement A D of electrons E H Fig. 6.1 The magnetic field is normal to face ABCD in the downward direction. Electrons enter face CDHG at right-angles to the face. As the electrons pass through the conductor, they experience a force due to the magnetic field. (i) On Fig. 6.1, shade the face to which the electrons tend to move as a result of this force. [1] (ii) The movement of the electrons in the magnetic field causes a potential difference between two faces of the conductor. Using the lettering from Fig. 6.1, state the faces between which this potential difference will occur. face … and face … [1] (c) Explain why the potential difference in (b) causes an additional force on the moving electrons in the conductor. … … … [2]

8 marks

Mark scheme: 6 (a) (i) particle must be moving M1 with component of velocity normal to magnetic field A1 [2] (ii) F = Bqv sin θ M1 q, v and θ explained A1 [2] (b) (i) face BCGF shaded A1 [1] (ii) between face BCGF and face ADHE A1 [1] (c) potential difference gives rise to an electric field M1 either FE = qE (no need to explain symbols) or electric field gives rise to force (on an electron) A1 [2] GCE AS/A LEVEL – October/November 2012 9702 42

This question in 9702/42 Oct/Nov 2012

4 A proton of mass m and charge +q is travelling through a vacuum in a straight line with For speed v. Examiner’s It enters a region of uniform magnetic field of magnetic flux density B, as shown in Fig. 4.1. Use region of uniform magnetic field proton mass m charge +q v Fig. 4.1 The magnetic field is normal to the direction of motion of the proton. (a) Explain why the path of the proton in the magnetic field is an arc of a circle. … … … [2] (b) The angular speed of the proton in the magnetic field is ω. Derive an expression for ω in terms of B, q and m. [4]

6 marks

Mark scheme: 4 (a) force on proton is normal to velocity and field M1 provides centripetal force (for circular motion) A1 [2] (b) magnetic force = Bqv B1 centripetal force = mrω2 or mv2/r B1 v = rω B1 Bqv = Bqrω = mrω2 ω = Bq/m A1 [4]

This question in 9702/43 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

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

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

6 A charged particle of mass m and charge –q is travelling through a vacuum at constant For speed v. Examiner’s It enters a uniform magnetic field of flux density B. The initial angle between the direction of Use motion of the particle and the direction of the magnetic field is 90°. (a) Explain why the path of the particle in the magnetic field is the arc of a circle. … … … … [3] (b) The radius of the arc in (a) is r. q Show that the ratio for the particle is given by the expression m q v = . m Br [1] (c) The initial speed v of the particle is 2.0 × 107 m s–1. The magnetic flux density B is 2.5 × 10–3 T. The radius r of the arc in the magnetic field is 4.5 cm. q (i) Use these data to calculate the ratio m. ratio = … C kg–1 [2] (ii) The path of the negatively-charged particle before it enters the magnetic field is For shown in Fig. 6.1. Examiner’s Use magnetic field into plane of paper path of particle Fig. 6.1 The direction of the magnetic field is into the plane of the paper. On Fig. 6.1, sketch the path of the particle in the magnetic field and as it emerges from the field. [2]

8 marks

Mark scheme: 6 (a) force due to magnetic field is constant B1 force is (always) normal to direction of motion this force provides the centripetal force A1 [3] (b) mv2 / r = Bqv M1 hence q / m = v / Br A0 [1] (c) (i) q / m = (2.0 × 107) / (2.5 × 10–3 × 4.5 × 10–2) C1 q / m = 1.8 × 1011 C kg–1 A1 [2] (ii) sketch: curved path, constant radius, in direction towards bottom of page M1 tangent to curved path on entering and on leaving the field A1 [2]

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

6 A charged particle of mass m and charge –q is travelling through a vacuum at constant For speed v. Examiner’s It enters a uniform magnetic field of flux density B. The initial angle between the direction of Use motion of the particle and the direction of the magnetic field is 90°. (a) Explain why the path of the particle in the magnetic field is the arc of a circle. … … … … [3] (b) The radius of the arc in (a) is r. q Show that the ratio for the particle is given by the expression m q v = . m Br [1] (c) The initial speed v of the particle is 2.0 × 107 m s–1. The magnetic flux density B is 2.5 × 10–3 T. The radius r of the arc in the magnetic field is 4.5 cm. q (i) Use these data to calculate the ratio m. ratio = … C kg–1 [2] (ii) The path of the negatively-charged particle before it enters the magnetic field is For shown in Fig. 6.1. Examiner’s Use magnetic field into plane of paper path of particle Fig. 6.1 The direction of the magnetic field is into the plane of the paper. On Fig. 6.1, sketch the path of the particle in the magnetic field and as it emerges from the field. [2]

8 marks

Mark scheme: 6 (a) force due to magnetic field is constant B1 force is (always) normal to direction of motion this force provides the centripetal force A1 [3] (b) mv2 / r = Bqv M1 hence q / m = v / Br A0 [1] (c) (i) q / m = (2.0 × 107) / (2.5 × 10–3 × 4.5 × 10–2) C1 q / m = 1.8 × 1011 C kg–1 A1 [2] (ii) sketch: curved path, constant radius, in direction towards bottom of page M1 tangent to curved path on entering and on leaving the field A1 [2]

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

6 A particle has mass m and charge +q and is travelling with speed v through a vacuum. For The initial direction of travel is parallel to the plane of two charged horizontal metal plates, as Examiner’s shown in Fig. 6.1. Use + V metal plate path of particle metal plate Fig. 6.1 The uniform electric field between the plates has magnitude 2.8 × 104 V m–1 and is zero outside the plates. The particle passes between the plates and emerges beyond them, as illustrated in Fig. 6.1. (a) Explain why the path of the particle in the electric field is not an arc of a circle. … … … [1] (b) A uniform magnetic field is now formed in the region between the metal plates. The magnetic field strength is adjusted so that the positively charged particle passes undeviated between the plates, as shown in Fig. 6.2. + V region of uniform electric and magnetic fields path of particle path of particle Fig. 6.2 (i) State and explain the direction of the magnetic field. For Examiner’s … Use … … [2] (ii) The particle has speed 4.7 × 105 m s–1. Calculate the magnitude of the magnetic flux density. Explain your working. magnetic fl ux density = … T [3] (c) The particle in (b) has mass m, charge +q and speed v. Without any further calculation, state the effect, if any, on the path of a particle that has (i) mass m, charge –q and speed v, … [1] (ii) mass m, charge +q and speed 2v, … [1] (iii) mass 2m, charge +q and speed v. … [1]

9 marks

Mark scheme: 6 (a) either constant speed parallel to plate or accelerated motion / force normal to plate / in direction field B1 so not circular A0 [1] (b) (i) direction of force due to magnetic field opposite to that due to electric field B1 magnetic field into plane of page B1 [2] (ii) force due to magnetic field = force due to electric field B1 Bqv = qE B = E / v C1 = (2.8 × 104) / (4.7 × 105) = 6.0 × 10–2 T A1 [3] (c) (i) no change / not deviated B1 [1] (ii) deviated upwards B1 [1] (iii) no change / not deviated B1 [1]

This question in 9702/43 Oct/Nov 2013

10 Magnetic resonance imaging (MRI) requires the use of a non-uniform magnetic field For superimposed on a large uniform magnetic field. Examiner’s Use State and explain the purpose of (a) the large uniform magnetic field, … … … … … [3] (b) the non-uniform magnetic field. … … … … … [3]

6 marks

Mark scheme: 10 (a) nuclei spin / precess B1 spin / precess about direction of magnetic field B1 either frequency of precession depends on magnetic field strength or large field means frequency in radio frequency range B1 [3] (b) non-uniform field means frequency of precession different in different regions of subject B1 enables location of precessing nuclei to be determined B1 enables thickness of slice to be varied / location of slice to be changed B1 [3]

This question in 9702/43 Oct/Nov 2013

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

6 (a) State the type of field, or fields, that may cause a force to be exerted on a particle that is (i) uncharged and moving, … [1] (ii) charged and stationary, … [1] (iii) charged and moving at right-angles to the field. … [2] (b) A particle X has mass 3.32 × 10−26 kg and charge +1.60 × 10−19 C. The particle is travelling in a vacuum with speed 7.60 × 104 m s−1. It enters a region of uniform magnetic field that is normal to the direction of travel of the particle. The particle travels in a semicircle of diameter 12.2 cm, as shown in Fig. 6.1. region of uniform magnetic field 12.2 cm path of particle X Fig. 6.1 For the uniform magnetic field, (i) state its direction, … … [1] (ii) calculate the magnetic flux density. magnetic flux density = … T [3] (c) A second particle Y has mass less than that of particle X in (b) and the same charge. It enters the region of uniform magnetic field in (b) with the same speed and along the same initial path as particle X. On Fig. 6.1, draw the path of particle Y in the region of the magnetic field. [1]

9 marks

Mark scheme: 6 (a) (i) gravitational B1 [1] (ii) gravitational and electric B1 [1] (iii) magnetic and one other field given B1 magnetic, graviational and electric B1 [2] (b) (i) out of (plane of) paper/page (not “upwards”) B1 [1] (ii) B = mv / qr C1 = (3.32 × 10–26 × 7.6 × 104) / (1.6 × 10–19 × 6.1 × 10–2) C1 = 0.26 T A1 [3] (c) sketch: semicircle with diameter < 12.2 cm B1 [1]

This question in 9702/41 May/June 2015

13 During magnetic resonance imaging to obtain information about internal body structures, a large constant magnetic field is used with a calibrated non-uniform magnetic field superimposed on it. (a) State and explain the purpose of (i) the large constant magnetic field, … … … [2] (ii) the non-uniform magnetic field. … … … … [3] (b) The de-excitation energy E (measured in joule) of a proton in magnetic resonance imaging is given by the expression E = 2.82 × 10−26 B where B is the magnetic flux density measured in tesla. The energy E is emitted as a photon of electromagnetic radiation in the radio-frequency range. Calculate the magnetic flux density required for the radio frequency to be 42 MHz. magnetic flux density = … T [2]

7 marks

Mark scheme: 13 (a) (i) to align nuclei/protons B1 to cause Larmor/precessional frequency to be in r.f. region B1 [2] (ii) Larmor/precessional frequency depends on (applied magnetic) field strength B1 knowing field strength enables (region of precessing) nuclei to be located M1 by knowing the frequency A1 [3] (b) E = 2.82 × 10–26 × B 6.63 × 10–34 × 42 × 106 = 2.82 × 10–26 × B C1 B = 0.99 T A1 [2]

This question in 9702/41 May/June 2015

6 (a) State the type of field, or fields, that may cause a force to be exerted on a particle that is (i) uncharged and moving, … [1] (ii) charged and stationary, … [1] (iii) charged and moving at right-angles to the field. … [2] (b) A particle X has mass 3.32 × 10−26 kg and charge +1.60 × 10−19 C. The particle is travelling in a vacuum with speed 7.60 × 104 m s−1. It enters a region of uniform magnetic field that is normal to the direction of travel of the particle. The particle travels in a semicircle of diameter 12.2 cm, as shown in Fig. 6.1. region of uniform magnetic field 12.2 cm path of particle X Fig. 6.1 For the uniform magnetic field, (i) state its direction, … … [1] (ii) calculate the magnetic flux density. magnetic flux density = … T [3] (c) A second particle Y has mass less than that of particle X in (b) and the same charge. It enters the region of uniform magnetic field in (b) with the same speed and along the same initial path as particle X. On Fig. 6.1, draw the path of particle Y in the region of the magnetic field. [1]

9 marks

Mark scheme: 6 (a) (i) gravitational B1 [1] (ii) gravitational and electric B1 [1] (iii) magnetic and one other field given B1 magnetic, graviational and electric B1 [2] (b) (i) out of (plane of) paper/page (not “upwards”) B1 [1] (ii) B = mv / qr C1 = (3.32 × 10–26 × 7.6 × 104) / (1.6 × 10–19 × 6.1 × 10–2) C1 = 0.26 T A1 [3] (c) sketch: semicircle with diameter < 12.2 cm B1 [1]

This question in 9702/43 May/June 2015

13 During magnetic resonance imaging to obtain information about internal body structures, a large constant magnetic field is used with a calibrated non-uniform magnetic field superimposed on it. (a) State and explain the purpose of (i) the large constant magnetic field, … … … [2] (ii) the non-uniform magnetic field. … … … … [3] (b) The de-excitation energy E (measured in joule) of a proton in magnetic resonance imaging is given by the expression E = 2.82 × 10−26 B where B is the magnetic flux density measured in tesla. The energy E is emitted as a photon of electromagnetic radiation in the radio-frequency range. Calculate the magnetic flux density required for the radio frequency to be 42 MHz. magnetic flux density = … T [2]

7 marks

Mark scheme: 13 (a) (i) to align nuclei/protons B1 to cause Larmor/precessional frequency to be in r.f. region B1 [2] (ii) Larmor/precessional frequency depends on (applied magnetic) field strength B1 knowing field strength enables (region of precessing) nuclei to be located M1 by knowing the frequency A1 [3] (b) E = 2.82 × 10–26 × B 6.63 × 10–34 × 42 × 106 = 2.82 × 10–26 × B C1 B = 0.99 T A1 [2]

This question in 9702/43 May/June 2015

6 (a) A particle has mass m, charge +q and speed v. State the magnitude and direction of the force, if any, on the particle when the particle is travelling along the direction of (i) a uniform gravitational field of field strength g, … … [2] (ii) a uniform magnetic field of flux density B. … … [1] (b) Two charged horizontal metal plates, situated in a vacuum, produce a uniform electric field of field strength E between the plates. The field strength outside the region between the plates is zero. The particle in (a) enters the region of the electric field at right-angles to the direction of the field, as illustrated in Fig. 6.1. particle, v charge +q mass m E Fig. 6.1 A uniform magnetic field is to be applied in the same region as the electric field so that the particle passes undeviated through the region between the plates. (i) State and explain the direction of the magnetic field. … … [2] (ii) Derive, with explanation, the relation between the speed v and the magnitudes of the electric field strength E and the magnetic flux density B. [3] (c) A second particle has the same mass m and charge +q as that in (b) but its speed is 2v. This particle enters the region between the plates along the same direction as the particle in (b). On Fig. 6.1, sketch the path of this particle in the region between the plates. [2]

10 marks

Mark scheme: 6 (a) (i) force = mg M1 along the direction of the field/of the motion A1 [2] (ii) no force B1 [1] (b) (i) force due to E-field downwards so force due to B-field upwards B1 into the plane of the paper B1 [2] (ii) force due to magnetic field = Bqv B1 force due to electric field = Eq B1 (use of FB and FE not explained, allow 1/2) forces are equal (and opposite) so Bv = E or Eq = Bqv so E = Bv B1 [3] (c) sketch: smooth curved path M1 in ‘upward’ direction A1 [2]

This question in 9702/41 Oct/Nov 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

9 A magnetic field of flux density B is normal to face PQRS of a slice of a conducting material, as shown in Fig. 9.1. magnetic field flux density % S R Z FXUUHQW I P Q X Y Fig. 9.1 A current I in the slice is normal to face QRZY of the slice. The Hall voltage VH across the slice is given by the expression BI VH = . ntq (a) (i) State what is represented by the symbol n. … … [1] (ii) The symbol t represents the length of one side of the slice. Use letters from Fig. 9.1 to identify t. … [1] (b) (i) In general, the Hall voltage produced in a slice of a metal is very small. For a slice of the same dimensions with the same current and magnetic flux density, the Hall voltage produced in a semiconductor material is much larger. Suggest and explain why. … … … [2] (ii) In some semiconducting materials, electrons are mainly responsible for conduction. In other semiconducting materials, holes are mainly responsible for conduction. Suggest and explain the difference, if any, that conduction by electrons or by holes will have on the Hall voltage. … … … … [3] [Total: 7]

7 marks

Mark scheme: 9 (a) (i) number density of charge carriers/free electrons or number per unit volume of charge carriers/free electrons B1 [1] (ii) PX or QY or RZ B1 [1] (b) (i) VH is inversely proportional to n B1 for semiconductors, n is (much) smaller than for metals B1 [2] (ii) magnetic field would deflect holes and electrons in same direction B1 (because) electrons are (–)ve, holes are (+)ve M1 so VH has opposite polarity/opposite sign A1 [3]

This question in 9702/42 May/June 2016

10 Explain the function of the non-uniform magnetic field that is superimposed on a large uniform magnetic field in diagnosis using nuclear magnetic resonance imaging (NMRI). … … … … … … … … [4] [Total: 4]

4 marks

Mark scheme: 10 nuclei emitting r.f. (pulse) B1 Larmor frequency/r.f. frequency emitted/detected depends on magnitude of magnetic field B1 nuclei can be located (within a slice) B1 changing field enables position of detection (slice) to be changed B1 [4]

This question in 9702/42 Oct/Nov 2016

8 (a) State what is meant by a magnetic field. … … … [2] (b) A particle of charge +q and mass m is travelling in a vacuum with speed v. The particle enters, at a right angle, a uniform magnetic field of flux density B, as shown in Fig. 8.1. uniform magnetic field flux density B d particle charge +q mass m speed v Fig. 8.1 The particle leaves the field after following a semi-circular path of diameter d. (i) State the direction of the magnetic field. … [1] (ii) Explain why the speed of the particle is not affected by the magnetic field. … … … [2] (iii) Show that the diameter d of the semi-circular path is given by the expression 2 mv d = . Bq [2] (iv) Use the expression in (b)(iii) to show that the time TF spent in the field by the particle is independent of its speed v. [2] [Total: 9]

9 marks

Mark scheme: 8(a) region (of space) where there is a force M1 produced by / on a magnet / magnetic pole / moving charge / current-carrying conductor A1 8(b)(i) out of (the plane of) the paper / page B1 8(b)(ii) the force on the particle is (always) perpendicular to the velocity / perpendicular to the direction of travel / towards the centre of path B1 no work is done by the force on the particle / there is no acceleration in the direction of the velocity / the acceleration is (always) perpendicular to the velocity B1 8(b)(iii) F = Bqv or F = mv 2 /r C1 mv 2 / (d / 2) = Bqv so d = 2mv / Bq A1 8(b)(iv) time = distance / speed T(F) = πd / 2v C1 T(F) = (π / 2v) × (2mv / Bq) T(F) = πm / Bq and so T(F) independent of v A1

This question in 9702/42 Feb/March 2017

8 (a) Explain what is meant by a magnetic field. … … … [2] (b) A particle has mass m, charge +q and speed v. The particle enters a uniform magnetic field of flux density B such that, on entry, it is moving normal to the magnetic field, as shown in Fig. 8.1. path of particle mass m charge +q speed v region of magnetic field Fig. 8.1 The direction of the magnetic field is perpendicular to, and into, the plane of the paper. (i) On Fig. 8.1, draw the path of the particle through, and beyond, the region of the magnetic field. [3] (ii) There is a force acting on the particle, causing it to accelerate. Explain why the speed of the particle on leaving the magnetic field is v. … … … [1] (c) The particle in (b) loses an electron so that its charge becomes +2q. Its change in mass is negligible. Determine, in terms of v, the initial speed of the particle such that its path through the magnetic field is unchanged. Explain your working. speed = … [3] [Total: 9]

9 marks

Mark scheme: 8(a) region where there is a force M1 experienced by a current-carrying conductor/moving charge/(permanent) magnet A1 8(b)(i) single path, deflection in ‘upward’ direction B1 acceptable circular arc in whole field B1 no ‘kinks’ at start or end of curvature, and straight outside region of field B1 8(b)(ii) force (on particle) is normal to velocity/direction of motion/direction of speed B1 8(c) magnetic force provides/is the centripetal force B1 Bqv = mv2 / r or r = mv / Bq C1 (if q is doubled), new speed = 2v A1

This question in 9702/41 Oct/Nov 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

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

8 (a) Explain what is meant by a magnetic field. … … … [2] (b) A particle has mass m, charge +q and speed v. The particle enters a uniform magnetic field of flux density B such that, on entry, it is moving normal to the magnetic field, as shown in Fig. 8.1. path of particle mass m charge +q speed v region of magnetic field Fig. 8.1 The direction of the magnetic field is perpendicular to, and into, the plane of the paper. (i) On Fig. 8.1, draw the path of the particle through, and beyond, the region of the magnetic field. [3] (ii) There is a force acting on the particle, causing it to accelerate. Explain why the speed of the particle on leaving the magnetic field is v. … … … [1] (c) The particle in (b) loses an electron so that its charge becomes +2q. Its change in mass is negligible. Determine, in terms of v, the initial speed of the particle such that its path through the magnetic field is unchanged. Explain your working. speed = … [3] [Total: 9]

9 marks

Mark scheme: 8(a) region where there is a force M1 experienced by a current-carrying conductor/moving charge/(permanent) magnet A1 8(b)(i) single path, deflection in ‘upward’ direction B1 acceptable circular arc in whole field B1 no ‘kinks’ at start or end of curvature, and straight outside region of field B1 8(b)(ii) force (on particle) is normal to velocity/direction of motion/direction of speed B1 8(c) magnetic force provides/is the centripetal force B1 Bqv = mv2 / r or r = mv / Bq C1 (if q is doubled), new speed = 2v A1

This question in 9702/43 Oct/Nov 2018

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

9 Nuclear magnetic resonance imaging (NMRI) is used to obtain diagnostic information about internal body structures. State, during the use of NMRI, the function of: (a) the large constant magnetic field … … … … … [3] (b) the non-uniform magnetic field. … … … … … [2] [Total: 5]

5 marks

Mark scheme: 9(a) nuclei precess B1 precession is about direction of magnetic field B1 frequency of precession depends on field strength or frequency of precession is in radio-frequency range B1 9(b) Any two points from: • frequency (of precession) depends on position • to locate position of (spinning) nuclei • to change region where nuclei are detected B2

This question in 9702/41 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

9 Nuclear magnetic resonance imaging (NMRI) is used to obtain diagnostic information about internal body structures. State, during the use of NMRI, the function of: (a) the large constant magnetic field … … … … … [3] (b) the non-uniform magnetic field. … … … … … [2] [Total: 5]

5 marks

Mark scheme: 9(a) nuclei precess B1 precession is about direction of magnetic field B1 frequency of precession depends on field strength or frequency of precession is in radio-frequency range B1 9(b) Any two points from: • frequency (of precession) depends on position • to locate position of (spinning) nuclei • to change region where nuclei are detected B2

This question in 9702/43 May/June 2019

9 Diagnosis using nuclear magnetic resonance imaging (NMRI) requires the use of a non-uniform magnetic field superimposed on a constant magnetic field of large magnitude. Explain the purpose of: (a) the large constant magnetic field … … … … … [2] (b) the non-uniform magnetic field. … … … … … [2] [Total: 4]

4 marks

Mark scheme: 9(a) nuclei precess B1 precession is about (direction of magnetic) field or frequency of precession is in radio-frequency range B1 9(b) • frequency (of precession) depends on field strength • to locate/find position of (spinning) nuclei • to change region where nuclei are detected any two points, one mark each B2

This question in 9702/41 Oct/Nov 2019

9 Diagnosis using nuclear magnetic resonance imaging (NMRI) requires the use of a non-uniform magnetic field superimposed on a constant magnetic field of large magnitude. Explain the purpose of: (a) the large constant magnetic field … … … … … [2] (b) the non-uniform magnetic field. … … … … … [2] [Total: 4]

4 marks

Mark scheme: 9(a) nuclei precess B1 precession is about (direction of magnetic) field or frequency of precession is in radio-frequency range B1 9(b) • frequency (of precession) depends on field strength • to locate/find position of (spinning) nuclei • to change region where nuclei are detected any two points, one mark each B2

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

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) State what is meant by a magnetic field. … … … [2] (b) A rectangular piece of aluminium foil is situated in a uniform magnetic field of flux density B, as shown in Fig. 9.1. magnetic field, flux density B Q R T aluminium movement foil of electrons P S V W Fig. 9.1 The magnetic field is normal to the face PQRS of the foil. Electrons, each of charge −q, enter the foil at right angles to the face PQTV. (i) On Fig. 9.1, shade the face of the foil on which electrons initially accumulate. [1] (ii) Explain why electrons do not continuously accumulate on the face you have shaded. … … … … [3] (c) The Hall voltage VH developed across the foil in (b) is given by the expression BI VH = ntq where I is the current in the foil. (i) State the meaning of the quantity n. … … [1] (ii) Using the letters on Fig. 9.1, identify the distance t. … [1] (d) Suggest why, in practice, Hall probes are usually made using a semiconductor material rather than a metal. … … [1] [Total: 9]

9 marks

Mark scheme: 9(a) region where there is a force exerted on M1 a current-carrying conductor or a moving charge or a magnetic material/magnetic pole A1 9(b)(i) face PSWV shaded B1 9(b)(ii) accumulating electrons cause an electric field (between the faces) B1 force due to electric field opposes force due to magnetic field B1 accumulation stops when magnetic force equals electric force B1 9(c)(i) number density of charge carriers B1 9(c)(ii) PV or QT or SW B1 9(d) (for semiconductor,) n is (much) smaller so VH (much) larger B1

This question in 9702/41 May/June 2021

8 (a) Define magnetic flux density. … … … [2] (b) Electrons, each of mass m and charge q, are accelerated from rest in a vacuum through a potential difference V. Derive an expression, in terms of m, q and V, for the final speed v of the electrons. Explain your working. [2] (c) The accelerated electrons in (b) are injected at point S into a region of uniform magnetic field of flux density B, as illustrated in Fig. 8.1. region of uniform magnetic field, flux density B S path of electrons, radius r Fig. 8.1 The electrons move at right angles to the direction of the magnetic field. The path of the electrons is a circle of radius r. q (i) Show that the specific charge of the electrons is given by the expression m q 2V = 2 2. m B r Explain your working. [2] (ii) Electrons are accelerated through a potential difference V of 230 V. The electrons are injected normally into the magnetic field of flux density 0.38 mT. The radius r of the circular orbit of the electrons is 14 cm. Use this information to calculate a value for the specific charge of an electron. specific charge = … C kg−1 [2] (iii) Suggest why the arrangement outlined in (ii), using the same values of B and V, is not practical for the determination of the specific charge of α-particles. … … … [2] [Total: 10]

10 marks

Mark scheme: 8(a) • force per unit length • force per unit current • length/current perpendicular to field 1 mark for any two points, 2 marks for all three points B2 8(b) change in potential energy = change in kinetic energy or qV = ½mv2 B1 v = √(2qV / m) A1 8(c)(i) magnetic force = centripetal force or Bqv = mv2 / r M1 clear substitution of expression for v and correct algebra leading to q / m = 2V / B2r2 A1 8(c)(ii) q / m = (2 × 230) / [(0.38 × 10–3)2 × 0.142] C1 = 1.6 × 1011 C kg–1 A1 8(c)(iii) (for α-particle,) q / m is (much) smaller B1 r would be much larger B1

This question in 9702/42 May/June 2021

9 (a) State what is meant by a magnetic field. … … … [2] (b) A rectangular piece of aluminium foil is situated in a uniform magnetic field of flux density B, as shown in Fig. 9.1. magnetic field, flux density B Q R T aluminium movement foil of electrons P S V W Fig. 9.1 The magnetic field is normal to the face PQRS of the foil. Electrons, each of charge −q, enter the foil at right angles to the face PQTV. (i) On Fig. 9.1, shade the face of the foil on which electrons initially accumulate. [1] (ii) Explain why electrons do not continuously accumulate on the face you have shaded. … … … … [3] (c) The Hall voltage VH developed across the foil in (b) is given by the expression BI VH = ntq where I is the current in the foil. (i) State the meaning of the quantity n. … … [1] (ii) Using the letters on Fig. 9.1, identify the distance t. … [1] (d) Suggest why, in practice, Hall probes are usually made using a semiconductor material rather than a metal. … … [1] [Total: 9]

9 marks

Mark scheme: 9(a) region where there is a force exerted on M1 a current-carrying conductor or a moving charge or a magnetic material/magnetic pole A1 9(b)(i) face PSWV shaded B1 9(b)(ii) accumulating electrons cause an electric field (between the faces) B1 force due to electric field opposes force due to magnetic field B1 accumulation stops when magnetic force equals electric force B1 9(c)(i) number density of charge carriers B1 9(c)(ii) PV or QT or SW B1 9(d) (for semiconductor,) n is (much) smaller so VH (much) larger B1

This question in 9702/43 May/June 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 9702/41 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

2 A sphere of mass 1.6 × 10–10 kg has a charge of +0.27 nC. The sphere is in a uniform electric field that acts vertically upwards, as shown in the side view in Fig. 2.1. SIDE VIEW electric field lines plane in which sphere moves sphere Fig. 2.1 The force exerted on the sphere by the electric field causes the sphere to remain at a fixed vertical height in a horizontal plane. There is a uniform magnetic field in the region of the electric field. The sphere moves at a speed of 0.78 m s–1 in the horizontal plane. The magnetic field causes the sphere to move in a circular path of radius 3.4 m, as shown in the view from above in Fig. 2.2. VIEW FROM ABOVE electric field lines 3.4 m out of the page path of sphere sphere Fig. 2.2 (a) (i) Determine the direction of the uniform magnetic field. … [1] (ii) Explain why the motion of the sphere in the horizontal plane is circular. … … … [2] (b) Calculate the strength of the uniform electric field. electric field strength = … N C–1 [2] (c) By considering the magnetic force on the sphere, show that the flux density of the uniform magnetic field is 0.14 T. [3] [Total: 8]

8 marks

Mark scheme: 2(a)(i) (vertically) downwards B1 2(a)(ii) magnetic force (on sphere) is perpendicular to its velocity B1 magnetic force perpendicular to velocity is the centripetal force or magnetic force perpendicular to velocity causes centripetal acceleration or acceleration perpendicular to velocity is centripetal (acceleration) or magnetic force does not change the speed of the sphere or magnetic force has constant magnitude B1 2(b) mg = Eq C1 E = (1.6  10–10  9.81) / (0.27  10–9) = 5.8 N C–1 A1 2(c) centripetal force = magnetic force or Bqv = mv2 / r B1 B = mv / qr C1 = (1.6  10–10  0.78) / (0.27  10–9  3.4) = 0.14 T A1

This question in 9702/41 May/June 2022

2 A sphere of mass 1.6 × 10–10 kg has a charge of +0.27 nC. The sphere is in a uniform electric field that acts vertically upwards, as shown in the side view in Fig. 2.1. SIDE VIEW electric field lines plane in which sphere moves sphere Fig. 2.1 The force exerted on the sphere by the electric field causes the sphere to remain at a fixed vertical height in a horizontal plane. There is a uniform magnetic field in the region of the electric field. The sphere moves at a speed of 0.78 m s–1 in the horizontal plane. The magnetic field causes the sphere to move in a circular path of radius 3.4 m, as shown in the view from above in Fig. 2.2. VIEW FROM ABOVE electric field lines 3.4 m out of the page path of sphere sphere Fig. 2.2 (a) (i) Determine the direction of the uniform magnetic field. … [1] (ii) Explain why the motion of the sphere in the horizontal plane is circular. … … … [2] (b) Calculate the strength of the uniform electric field. electric field strength = … N C–1 [2] (c) By considering the magnetic force on the sphere, show that the flux density of the uniform magnetic field is 0.14 T. [3] [Total: 8]

8 marks

Mark scheme: 2(a)(i) (vertically) downwards B1 2(a)(ii) magnetic force (on sphere) is perpendicular to its velocity B1 magnetic force perpendicular to velocity is the centripetal force or magnetic force perpendicular to velocity causes centripetal acceleration or acceleration perpendicular to velocity is centripetal (acceleration) or magnetic force does not change the speed of the sphere or magnetic force has constant magnitude B1 2(b) mg = Eq C1 E = (1.6  10–10  9.81) / (0.27  10–9) = 5.8 N C–1 A1 2(c) centripetal force = magnetic force or Bqv = mv2 / r B1 B = mv / qr C1 = (1.6  10–10  0.78) / (0.27  10–9  3.4) = 0.14 T A1

This question in 9702/43 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) A Hall probe is placed in a magnetic field. The Hall voltage is zero. The Hall probe is rotated to a new position in the magnetic field. The Hall voltage is now maximum. Explain these observations. … … … [2] (b) The formula for calculating the Hall voltage VH as measured by a Hall probe is BI VH = ntq. Table 6.1 shows the value of n for two materials. Table 6.1 material n / m–3 silicon 9.65 × 1015 copper 8.49 × 1028 (i) State the meaning of n. … … [1] (ii) Explain why a Hall probe is made from silicon rather than copper. … … [1] (c) A Hall probe gives a maximum reading of 24 mV when placed in a uniform magnetic field of flux density 32 mT. The same Hall probe is then placed in a magnetic field of fixed direction and varying flux density. The Hall probe is in a fixed position so that the angle between the Hall probe and the magnetic field is the same as when the Hall voltage was 24 mV. The variation of the reading VH on the Hall probe with time t from time t = 0 to time t = 8.6 s is shown in Fig. 6.1. 40 30 VH / mV 20 10 0 0 1 2 3 4 5 6 7 8 9 t / s Fig. 6.1 A coil with 780 turns and a diameter of 3.6 cm is placed in this varying magnetic field. The plane of the coil is perpendicular to the field lines. Calculate the magnitude of the maximum electromotive force (e.m.f.) induced in the coil in the time between t = 0 and t = 8.6 s. e.m.f. = … V [4] [Total: 8]

8 marks

Mark scheme: 6(a) it is zero when (plane of) probe is parallel to the (magnetic) field (lines) B1 it is maximum when (plane of) probe is perpendicular to (magnetic) field (lines) B1 6(b)(i) number density of charge carriers B1 6(b)(ii) smaller value of n so greater Hall voltage / VH B1 6(c) (36 mV corresponds to) 48 mT C1 use of 1.4 s or (8.6 – 7.2) s C1 E = BAN / t C1 48  10 −3  0.018 2  780 A1 = 1.4 = 0.027 V

This question in 9702/42 Feb/March 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 9702/41 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) Define magnetic flux density. … … … … [2] (b) Electrons are moving in a vacuum with speed 1.7 × 107 m s–1. The electrons enter a uniform magnetic field of flux density 4.8 mT. Fig. 6.1 shows the path of the electrons. magnetic field, flux density 4.8 mT electrons, speed 1.7 × 107 m s–1 X d Fig. 6.1 The path of the electrons remains in the plane of the page. (i) State the direction of the magnetic field. … … [1] (ii) Show that the magnitude of the force exerted on each electron by the magnetic field is 1.3 × 10–14 N. [2] (iii) On Fig. 6.1, draw an arrow to indicate the direction of the centripetal acceleration of the electron where it enters the magnetic field at point X. [1] (iv) Use the information in (b)(ii) to calculate the distance d between the path of the electrons entering the magnetic field and the path of the electrons leaving it. d = … m [3] (c) The electrons in (b) are replaced with positrons that are moving with speed 3.4 × 107 m s–1 along the same initial path as the electrons. The positrons enter the magnetic field at point X on Fig. 6.1. On Fig. 6.1, draw a line to show the path of the positrons through the magnetic field. [3] [Total: 12]

12 marks

Mark scheme: 6(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 6(b)(i) into the page B1 6(b)(ii) F = Bqv C1 = 4.8  10–3  1.6  10–19  1.7  107 = 1.3  10–14 N A1 6(b)(iii) arrow at point X pointing down the page B1 6(b)(iv) F = mv2 / r C1 1.3  10–14 = (9.11  10–31)  (1.7  107)2 / r C1 (r = 0.020 m) A1 d = 2r d = 0.040 m 6(c) path shows upwards deflection such that the curvature is always anticlockwise within the field B1 circular path with larger radius B1 line enters field at X and leaves field at distance 2d vertically from X B1

This question in 9702/41 Oct/Nov 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

6 (a) Define magnetic flux density. … … … … [2] (b) Electrons are moving in a vacuum with speed 1.7 × 107 m s–1. The electrons enter a uniform magnetic field of flux density 4.8 mT. Fig. 6.1 shows the path of the electrons. magnetic field, flux density 4.8 mT electrons, speed 1.7 × 107 m s–1 X d Fig. 6.1 The path of the electrons remains in the plane of the page. (i) State the direction of the magnetic field. … … [1] (ii) Show that the magnitude of the force exerted on each electron by the magnetic field is 1.3 × 10–14 N. [2] (iii) On Fig. 6.1, draw an arrow to indicate the direction of the centripetal acceleration of the electron where it enters the magnetic field at point X. [1] (iv) Use the information in (b)(ii) to calculate the distance d between the path of the electrons entering the magnetic field and the path of the electrons leaving it. d = … m [3] (c) The electrons in (b) are replaced with positrons that are moving with speed 3.4 × 107 m s–1 along the same initial path as the electrons. The positrons enter the magnetic field at point X on Fig. 6.1. On Fig. 6.1, draw a line to show the path of the positrons through the magnetic field. [3] [Total: 12]

12 marks

Mark scheme: 6(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 6(b)(i) into the page B1 6(b)(ii) F = Bqv C1 = 4.8  10–3  1.6  10–19  1.7  107 = 1.3  10–14 N A1 6(b)(iii) arrow at point X pointing down the page B1 6(b)(iv) F = mv2 / r C1 1.3  10–14 = (9.11  10–31)  (1.7  107)2 / r C1 (r = 0.020 m) A1 d = 2r d = 0.040 m 6(c) path shows upwards deflection such that the curvature is always anticlockwise within the field B1 circular path with larger radius B1 line enters field at X and leaves field at distance 2d vertically from X B1

This question in 9702/43 Oct/Nov 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

7 (a) Define magnetic flux density. … … … [2] (b) A particle of mass m and charge +Q moves at speed v into a region where there is a uniform magnetic field, as shown in Fig. 7.1. path of region of particle magnetic field particle Y Z Fig. 7.1 The uniform magnetic field is into the page and has flux density B. The particle enters the region of the field at point Y. (i) State an expression, in terms of some or all of m, Q, B and v, for the magnetic force F that acts on the particle when it is at point Y. F = … [1] (ii) On Fig. 7.1, draw an arrow at point Y to indicate the direction of the force in (b)(i). [1] (iii) On Fig. 7.1, draw a line to show a possible path for the particle through the region of the magnetic field. [1] (c) (i) Explain how an electric field can be used with the magnetic field to ensure that the particle in (b) now passes through point Z. … … … … … [3] (ii) Derive an expression for v in terms of B and the electric field strength E. v = … [2] [Total: 10]

10 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)(i) F = BQv B1 7(b)(ii) arrow at Y pointing vertically upwards B1 7(b)(iii) upwards deflection showing circular path B1 7(c)(i) electric field applied vertically downwards (may be shown on a labelled diagram) B1 electric force on particle in opposite direction to magnetic force (may be shown on a labelled diagram) B1 particle undeflected when magnitudes of electric and magnetic forces are equal B1 7(c)(ii) EQ = BQv B1 v = E / B A1

This question in 9702/41 May/June 2025

2 (a) (i) State what is represented by a gravitational field line. … … … [2] (ii) The Earth may be considered as a uniform sphere, as shown in Fig. 2.1. Earth Fig. 2.1 On Fig. 2.1, draw field lines to represent the Earth’s gravitational field outside the Earth. [2] (b) The Earth’s magnetic field may be considered as being due to the Earth acting as a long solenoid, as shown in Fig. 2.2. axis of magnetic rotation pole solenoid Equator magnetic pole Fig. 2.2 The magnetic poles do not align with the geographic poles, which are on the axis of rotation. Fig. 2.3 is a copy of Fig. 2.2 without the labels but with two magnetic field lines shown. Fig. 2.3 (i) On Fig. 2.3, label the magnetic poles with the letters N and S to indicate which one is the magnetic N pole and which one is the magnetic S pole. [1] (ii) On Fig. 2.3, draw field lines to represent the Earth’s magnetic field outside the Earth. [2] (c) An observer moves around the surface of the Earth. (i) Use your answer in (a)(ii) to explain why the observed gravitational field of the Earth does not vary around the surface. … … … [2] (ii) With reference to your answer in (b)(ii), describe how the observed magnetic field of the Earth varies around the surface. … … … … … [3] [Total: 12]

12 marks

Mark scheme: 2(a)(i) direction of force B1 force acting on a (test) mass B1 2(a)(ii) at least four radial lines from the Earth’s surface, equally spaced around the surface B1 arrows indicating direction towards Earth B1 2(b)(i) top pole labelled S and bottom pole labelled N B1 2(b)(ii) solenoid field pattern at the poles: B1 at least two field lines either side of both poles, close to the poles, clustered closely together, leaving the surface approximately perpendicularly to the surface and curving away from the axis of the poles as their distance from the surface increases solenoid field pattern above the equator: B1 at least one field line either side of the Earth connecting two points on the surface that are on the same side of the poles, one north of the equator and one south of it, passing above the surface near the magnetic equator approximately parallel to the surface 2(c)(i) (around the surface) lines are evenly spaced B1 all lines perpendicular to surface B1 or pointing down towards surface (at all points around the surface) 2(c)(ii) Any three bulleted points from: B3 • strongest at the poles • weakest near the Equator Up to two points from: • perpendicular to surface at the poles • parallel to the surface near the Equator • angle to surface increases from Equator to poles

This question in 9702/42 May/June 2025

7 (a) Define magnetic flux density. … … … [2] (b) A particle of mass m and charge +Q moves at speed v into a region where there is a uniform magnetic field, as shown in Fig. 7.1. path of region of particle magnetic field particle Y Z Fig. 7.1 The uniform magnetic field is into the page and has flux density B. The particle enters the region of the field at point Y. (i) State an expression, in terms of some or all of m, Q, B and v, for the magnetic force F that acts on the particle when it is at point Y. F = … [1] (ii) On Fig. 7.1, draw an arrow at point Y to indicate the direction of the force in (b)(i). [1] (iii) On Fig. 7.1, draw a line to show a possible path for the particle through the region of the magnetic field. [1] (c) (i) Explain how an electric field can be used with the magnetic field to ensure that the particle in (b) now passes through point Z. … … … … … [3] (ii) Derive an expression for v in terms of B and the electric field strength E. v = … [2] [Total: 10]

10 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)(i) F = BQv B1 7(b)(ii) arrow at Y pointing vertically upwards B1 7(b)(iii) upwards deflection showing circular path B1 7(c)(i) electric field applied vertically downwards (may be shown on a labelled diagram) B1 electric force on particle in opposite direction to magnetic force (may be shown on a labelled diagram) B1 particle undeflected when magnitudes of electric and magnetic forces are equal B1 7(c)(ii) EQ = BQv B1 v = E / B A1

This question in 9702/43 May/June 2025