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

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

8 marks

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

This question in 9702/41 Oct/Nov 2017

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

8 marks

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

This question in 9702/42 Feb/March 2018

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

8 marks

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

This question in 9702/41 May/June 2018

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

8 marks

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

This question in 9702/43 May/June 2018

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

8 marks

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

This question in 9702/42 Oct/Nov 2018

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

8 marks

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

This question in 9702/42 Feb/March 2019

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

9 marks

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

This question in 9702/41 Oct/Nov 2019

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

9 marks

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

This question in 9702/43 Oct/Nov 2019

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

8 marks

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

This question in 9702/42 Feb/March 2020

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

13 marks

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

This question in 9702/41 May/June 2020

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

13 marks

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

This question in 9702/43 May/June 2020

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

7 marks

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

This question in 9702/42 Oct/Nov 2020

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

6 marks

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

This question in 9702/41 Oct/Nov 2021

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

5 marks

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

This question in 9702/42 Oct/Nov 2021

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

6 marks

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

This question in 9702/43 Oct/Nov 2021

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

10 marks

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

This question in 9702/42 May/June 2022

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

10 marks

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

This question in 9702/42 Oct/Nov 2022

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

10 marks

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

This question in 9702/41 May/June 2023

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

10 marks

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

This question in 9702/43 May/June 2023

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

10 marks

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

This question in 9702/44 May/June 2025