Force on a moving charge — Paper 4 · A Level Physics 9702

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

7 An electron having charge –q and mass m is accelerated from rest in a vacuum through a potential difference V. The electron then enters a region of uniform magnetic field of magnetic flux density B, as shown in Fig. 7.1. uniform magnetic field into plane of paper path of electron Fig. 7.1 The direction of the uniform magnetic field is into the plane of the paper. The velocity of the electron as it enters the magnetic field is normal to the magnetic field. The radius of the circular path of the electron in the magnetic field is r. (a) Explain why the path of the electron in the magnetic field is the arc of a circle. … … … … [3] (b) Show that the magnitude p of the momentum of the electron as it enters the magnetic field is given by p = (2mqV ). [2] (c) The potential difference V is 120 V. The radius r of the circular arc is 7.4 cm. Determine the magnitude B of the magnetic flux density. B = … T [3] (d) The potential difference V in (c) is increased. The magnetic flux density B remains unchanged. By reference to the momentum of the electron, explain the effect of this increase on the radius r of the path of the electron in the magnetic field. … … … [2] [Total: 10]

10 marks

Mark scheme: 7(a) (magnetic) force (always) normal to velocity/direction of motion M1 (magnitude of magnetic) force constant or speed is constant/kinetic energy is constant M1 so provides the centripetal force A1 7(b) increase in KE = loss in PE or ½ mv 2 = qV M1 p = mv with algebra leading to p = √(2mqV) A1 7(c) Bqv = mv2 / r mv = Bqr or p = Bqr C1 (2 × 9.11 × 10–31 × 1.60 × 10–19 × 120)1/2 = B × 1.60 × 10–19 × 0.074 C1 B = 5.0 × 10–4 T A1 7(d) greater momentum M1 (p = Bqr and) so r increased A1

This question in 9702/41 May/June 2017

9 A Hall probe is placed near to one end of a current-carrying solenoid, as shown in Fig. 9.1. X solenoid Hall probe Y Fig. 9.1 The probe is rotated about the axis XY and is then held in a position so that the Hall voltage is maximum. (a) Explain why (i) a Hall probe is made from a thin slice of material, … … … [2] (ii) in order for consistent measurements of magnetic flux density to be made, the current in the probe must be constant. … … [1] (b) The probe is now rotated through an angle of 360° about the axis XY. At angle θ = 0, the Hall voltage VH has maximum value VMAX. On Fig. 9.2, sketch the variation with angle θ of the Hall voltage VH for one complete revolution of the probe about axis XY. + V MAX V H 0 0 90 180 270 360 θ/ ° – Fig. 9.2 [3] [Total: 6]

6 marks

Mark scheme: 9(a)(i) Hall voltage depends on thickness of slice C1 thinner slice, larger Hall voltage A1 9(a)(ii) Hall voltage depends on current in slice B1 9(b) sinusoidal wave, one cycle B1 at θ = 0 and at θ = 360°, VH = VMAX B1 at θ = 180°, VH = –VMAX B1

This question in 9702/42 May/June 2017

7 An electron having charge –q and mass m is accelerated from rest in a vacuum through a potential difference V. The electron then enters a region of uniform magnetic field of magnetic flux density B, as shown in Fig. 7.1. uniform magnetic field into plane of paper path of electron Fig. 7.1 The direction of the uniform magnetic field is into the plane of the paper. The velocity of the electron as it enters the magnetic field is normal to the magnetic field. The radius of the circular path of the electron in the magnetic field is r. (a) Explain why the path of the electron in the magnetic field is the arc of a circle. … … … … [3] (b) Show that the magnitude p of the momentum of the electron as it enters the magnetic field is given by p = (2mqV ). [2] (c) The potential difference V is 120 V. The radius r of the circular arc is 7.4 cm. Determine the magnitude B of the magnetic flux density. B = … T [3] (d) The potential difference V in (c) is increased. The magnetic flux density B remains unchanged. By reference to the momentum of the electron, explain the effect of this increase on the radius r of the path of the electron in the magnetic field. … … … [2] [Total: 10]

10 marks

Mark scheme: 7(a) (magnetic) force (always) normal to velocity/direction of motion M1 (magnitude of magnetic) force constant or speed is constant/kinetic energy is constant M1 so provides the centripetal force A1 7(b) increase in KE = loss in PE or ½ mv 2 = qV M1 p = mv with algebra leading to p = √(2mqV) A1 7(c) Bqv = mv2 / r mv = Bqr or p = Bqr C1 (2 × 9.11 × 10–31 × 1.60 × 10–19 × 120)1/2 = B × 1.60 × 10–19 × 0.074 C1 B = 5.0 × 10–4 T A1 7(d) greater momentum M1 (p = Bqr and) so r increased A1

This question in 9702/43 May/June 2017

8 A thin slice of conducting material is placed normal to a uniform magnetic field, as shown in Fig. 8.1. magnetic field 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. The current I in the slice is normal to the faces CDQP and FERS. A potential difference, the Hall voltage VH, is developed across the slice. (a) (i) State the faces between which the Hall voltage VH is developed. … and … [1] (ii) Explain why a constant voltage VH is developed between the faces you have named in (i). … … … … … … … [4] (b) Two slices have similar dimensions. One slice is made of a metal and the other slice is made of a semiconductor material. For the same values of magnetic flux density and current, state which slice, if either, will give rise to the larger Hall voltage. Explain your reasoning. … … … … [2] [Total: 7]

7 marks

Mark scheme: 8(a)(i) DERQ and CFSP B1 8(a)(ii) charge carriers moving normal to (magnetic) field B1 charge carriers experience a force normal to I (and B) B1 charge build-up sets up electric field across the slice or build-up of charges results in a p.d. across the slice B1 charge stops building up/VH becomes constant when FB = FE B1 8(b) VH inversely proportional to n/number density of charge carriers B1 number density of charge carriers (n) lower in semiconductors so VH larger for semiconductor slice B1 or VH proportional to v/drift velocity (B1) (for same current) drift velocity (v) higher in semiconductors so VH larger for semiconductor slice (B1)

This question in 9702/42 Oct/Nov 2017

9 (a) State what is meant by a field of force. … … … [2] (b) Explain the use of a uniform magnetic field and a uniform electric field for the selection of the velocity of charged particles. You may draw a diagram if you wish. … … … … … … [4] (c) A beam of charged particles enters a region of uniform magnetic and electric fields, as illustrated in Fig. 9.1. region of uniform magnetic and electric fields path of particle mass m charge +q velocity v magnetic field into plane of paper Fig. 9.1 The direction of the magnetic field is into the plane of the paper. The velocity of the charged particles is normal to the magnetic field as the particles enter the field. A particle in the beam has mass m, charge +q and velocity v. The particle passes undeviated through the region of the two fields. On Fig. 9.1, sketch the path of a particle that has (i) mass m, charge +2q and velocity v (label this path Q), [1] (ii) mass m, charge +q and velocity slightly larger than v (label this path V). [2] [Total: 9]

9 marks

Mark scheme: 9(a) region (of space) B1 where an object/particle experiences a force B1 9(b) electric and magnetic fields normal to each other B1 velocity of particle normal to both fields B1 forces (on particle) due to fields are in opposite directions B1 forces are equal for particles with a particular speed/for a selected speed/for speed given by v = E(q) / B(q) B1 9(c)(i) path labelled Q shown undeviated B1 9(c)(ii) reasonable curve in field and no ‘kink’ on entering, labelled V B1 deviated ‘upwards’ B1

This question in 9702/42 Oct/Nov 2017

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 9702/43 Oct/Nov 2017

9 A thin slice of conducting material has its faces PQRS and VWXY normal to a uniform magnetic field of flux density B, as shown in Fig. 9.1. magnetic field flux density B Q R W X direction of motion of electrons P S V Y Fig. 9.1 Electrons enter the slice at right-angles to face SRXY. A potential difference, the Hall voltage VH, is developed between two faces of the slice. (a) (i) Use letters from Fig. 9.1 to name the two faces between which the Hall voltage is developed. … and … [1] (ii) State and explain which of the two faces named in (a)(i) is the more positive. … … [2] (b) The Hall voltage VH is given by the expression BI VH = ntq. (i) Use the letters in Fig. 9.1 to identify the distance t. … [1] (ii) State the meaning of the symbol n. … … [1] (iii) State and explain the effect, if any, on the polarity of the Hall voltage when negative charge carriers (electrons) are replaced with positive charge carriers, moving in the same direction towards the slice. … … … [2] [Total: 7]

7 marks

Mark scheme: 9(a)(i) PSYV and QRXW B1 9(a)(ii) electron moving in magnetic field deflected towards face QRXW M1 so face PSYV is more positive A1 9(b)(i) PV or SY or RX or QW B1 9(b)(ii) number of charge carriers per unit volume B1 9(b)(iii) negative and positive charge (carriers) would deflect in opposite directions M1 so no change in polarity A1

This question in 9702/42 Feb/March 2018

8 (a) Explain how a uniform magnetic field and a uniform electric field may be used as a velocity selector for charged particles. … … … … [3] (b) Particles having mass m and charge +1.6 × 10–19 C pass through a velocity selector. They then enter a region of uniform magnetic field of magnetic flux density 94 mT with speed 3.4 × 104 m s–1, as shown in Fig. 8.1. path of charged particle velocity selector 15.0 cm uniform magnetic field into page Fig. 8.1 The direction of the uniform magnetic field is into the page and normal to the direction in which the particles are moving. The particles are moving in a vacuum in a circular arc of diameter 15.0 cm. Show that the mass of one of the particles is 20 u. [4] (c) On Fig. 8.1, sketch the path in the uniform magnetic field of a particle of mass 22 u having the same charge and speed as the particle in (b). [2] [Total: 9]

9 marks

Mark scheme: 8(a) electric and magnetic fields at right-angles to one another (may be shown on a clearly labelled diagram) B1 particle enters fields (with velocity) normal to the (two) fields (may be shown on a clearly labelled diagram) B1 no deviation for particles with selected velocity B1 8(b) magnetic force equals/is the centripetal force C1 Bqv = mv2 / r C1 M = Bqr / v = (94 × 10–3 × 1.6 × 10–19 × 0.075) / (3.4 × 104) M1 division by 1.66 × 10–27 shown, to give m = 20 u A1 8(c) sketch: semicircle clear (in same direction) B1 with larger radius B1

This question in 9702/42 May/June 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 9702/41 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 An electron is travelling in a vacuum at a speed of 3.4 × 107 m s–1. The electron enters a region of uniform magnetic field of flux density 3.2 mT, as illustrated in Fig. 8.1. region of uniform magnetic flux density 3.2 mT 30° electron speed 3.4 × 107 m s–1 Fig. 8.1 The initial direction of the electron is at an angle of 30° to the direction of the magnetic field. (a) When the electron enters the magnetic field, the component of its velocity vN normal to the direction of the magnetic field causes the electron to begin to follow a circular path. Calculate: (i) vN vN = … m s–1 [1] (ii) the radius of this circular path. radius = … m [3] (b) State the magnitude of the force, if any, on the electron in the magnetic field due to the component of its velocity along the direction of the field. … [1] (c) Use information from (a) and (b) to describe the resultant path of the electron in the magnetic field. … … [1] [Total: 6]

6 marks

Mark scheme: 8(a)(i) = 1.7 × 107 m s–1 A1 8(a)(ii) mv2 / r = Bqv or r = mv / Bq C1 r = (9.11 × 10–31 × 1.7 × 107) / (3.2 × 10–3 × 1.60 × 10–19) C1 = 0.030 m A1 8(b) zero B1 8(c) helix/coil B1

This question in 9702/42 May/June 2019

8 Electrons enter a rectangular slice PQRSEFGH of a semiconductor material at right-angles to face PQFE, as shown in Fig. 8.1. magnetic field flux density B S R H G P Q E F direction of incident electrons Fig. 8.1 A uniform magnetic field of flux density B is directed into the slice, at right-angles to face PQRS. (a) The electrons each have charge –q and drift speed v in the slice. State the magnitude and the direction of the force due to the magnetic field on each electron as it enters the slice. … … … [2] (b) The force on the electrons causes a voltage VH to be established across the semiconductor slice given by the expression BI VH = ntq where I is the current in the slice. (i) State the two faces between which the voltage VH is established. face … and face … [1] (ii) Use letters from Fig. 8.1 to identify the distance t. … [1] (c) Aluminium (2713 Al ) has a density of 2.7 g cm–3. Assume that there is one free electron available to carry charge per atom of aluminium. (i) Show that the number of charge carriers per unit volume in aluminium is 6.0 × 1028 m–3. [2] (ii) A sample of aluminium foil has a thickness of 0.090 mm. The current in the foil is 4.6 A. A uniform magnetic field of flux density 0.15 T acts at right-angles to the foil. Use the value in (i) to calculate the voltage VH that is generated. VH = … V [2] [Total: 8]

8 marks

Mark scheme: 8(a) magnitude: (force =) Bqv B1 direction: P→Q or E→F or S→R or H→G B1 8(b)(i) EHSP and FGRQ B1 8(b)(ii) PE or QF or RG or SH B1 8(c)(i) any one correct starting point from: • (mass of 1 atom =) 27 × 1.66 × 10–27 • (amount of substance per unit volume =) 2.7 / 27 • 27 g (of substance) contains 6.02 × 1023 atoms • (2.7 g mass contains) 0.1 mol • (1 cm3 volume contains) 0.1 mol • (1 m3 volume contains) 105 mol C1 n = (2.7 × 103) / (27 × 1.66 × 10–27) = 6.0 × 1028 or n = (2.7 / 27) × 106 × 6.02 × 1023 = 6.0 × 1028 A1 8(c)(ii) VH = (0.15 × 4.6) / (6.0 × 1028 × 0.090 × 10–3 × 1.60 × 10–19) C1 = 8.0 × 10–7 V A1

This question in 9702/42 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

9 (a) An electron is travelling at speed v in a straight line in a vacuum. It enters a uniform magnetic field of flux density 8.0 × 10–4 T. Initially, the electron is travelling at right angles to the magnetic field, as illustrated in Fig. 9.1. region of uniform magnetic field path of electron Fig. 9.1 The path of the electron in the magnetic field is an arc of a circle of radius 6.4 cm. (i) State and explain the direction of the magnetic field. … … … [2] (ii) Show that the speed v of the electron is 9.0 × 106 m s–1. [3] (b) A uniform electric field is now applied in the same region as the magnetic field. The electron passes undeviated through the region of the two fields, as illustrated in Fig. 9.2. region of uniform electric and magnetic fields path of electron Fig. 9.2 (i) On Fig. 9.2, mark with an arrow the direction of the uniform electric field. [1] (ii) Use data from (a) to calculate the magnitude of the electric field strength. field strength = … N C–1 [2] (c) The electron in (b) is now replaced by an α‑particle travelling at the same speed v along the same initial path as the electron. Describe and explain the shape of the path in the region of the magnetic and electric fields. … … … [2] [Total: 10]

10 marks

Mark scheme: 9(a)(i) force is downwards/down the page or current is (right) to left B1 by left-hand rule, field is into plane of paper B1 9(a)(ii) magnetic force provides the centripetal force C1 Bqv = mv2 / r C1 v = Bqr / m = (8.0 × 10–4 × 1.60 × 10–19 × 6.4 × 10–2) / (9.11 × 10–31) = 9.0 × 106 m s–1 A1 9(b)(i) arrow showing field direction down the page B1 9(b)(ii) Bqv = Eq or v = E / B C1 E = 9.0 × 106 × 8.0 × 10–4 = 7.2 × 103 N C–1 A1 9(c) straight line/undeviated B1 condition for no deflection depends only on v or condition for no deflection does not depend on m or q B1

This question in 9702/42 May/June 2020

8 A slice of a conducting material has its face QRLK normal to a uniform magnetic field of flux density B, as illustrated in Fig. 8.1. S R M L magnetic flux P density B Q J K direction of movement of electrons Fig. 8.1 Electrons enter the slice travelling perpendicular to face PQKJ. (a) For the free electrons moving in the slice: (i) state the direction of the force on an electron due to movement of the electron in the magnetic field … … [1] (ii) identify the faces, using the letters on Fig. 8.1, between which a potential difference is developed. face … and face … [1] (b) Explain why the potential difference in (a)(ii) reaches a maximum value. … … … [2] (c) The number of free electrons per unit volume in the slice of material is 1.3 × 1029 m–3. The thickness PQ of the slice is 0.10 mm. The magnetic flux density B is 4.6 × 10–3 T. Calculate the potential difference across the slice for a current of 6.3 × 10–4 A. potential difference = … V [2] (d) The slice in (c) is a metal. By reference to your answer in (c), suggest why Hall probes are usually made using semiconductors rather than metals. … … … [2] [Total: 8]

8 marks

Mark scheme: 8(a)(i) downwards B1 8(a)(ii) PQRS and JKLM B1 8(b) (as charge separates) an electric field is created (between opposite faces) B1 (maximum value is reached when) electric force (on electron) is equal and opposite to magnetic force (on electron) B1 8(c) VH = BI / ntq = (4.6 × 10–3 × 6.3 × 10–4) / (1.3 × 1029 × 0.10 × 10–3 × 1.60 × 10–19) C1 = 1.4 × 10–12 V A1 8(d) semiconductors have a (much) smaller value for n B1 VH for semiconductors is (much) larger so more easily measured B1

This question in 9702/41 Oct/Nov 2020

8 A slice of a conducting material has its face QRLK normal to a uniform magnetic field of flux density B, as illustrated in Fig. 8.1. S R M L magnetic flux P density B Q J K direction of movement of electrons Fig. 8.1 Electrons enter the slice travelling perpendicular to face PQKJ. (a) For the free electrons moving in the slice: (i) state the direction of the force on an electron due to movement of the electron in the magnetic field … … [1] (ii) identify the faces, using the letters on Fig. 8.1, between which a potential difference is developed. face … and face … [1] (b) Explain why the potential difference in (a)(ii) reaches a maximum value. … … … [2] (c) The number of free electrons per unit volume in the slice of material is 1.3 × 1029 m–3. The thickness PQ of the slice is 0.10 mm. The magnetic flux density B is 4.6 × 10–3 T. Calculate the potential difference across the slice for a current of 6.3 × 10–4 A. potential difference = … V [2] (d) The slice in (c) is a metal. By reference to your answer in (c), suggest why Hall probes are usually made using semiconductors rather than metals. … … … [2] [Total: 8]

8 marks

Mark scheme: 8(a)(i) downwards B1 8(a)(ii) PQRS and JKLM B1 8(b) (as charge separates) an electric field is created (between opposite faces) B1 (maximum value is reached when) electric force (on electron) is equal and opposite to magnetic force (on electron) B1 8(c) VH = BI / ntq = (4.6 × 10–3 × 6.3 × 10–4) / (1.3 × 1029 × 0.10 × 10–3 × 1.60 × 10–19) C1 = 1.4 × 10–12 V A1 8(d) semiconductors have a (much) smaller value for n B1 VH for semiconductors is (much) larger so more easily measured B1

This question in 9702/43 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 two situations in which a charged particle in a magnetic field does not experience a force. 1. … … 2. … … [2] (b) A loosely coiled metal spring is suspended from a fixed point, as shown in Fig. 9.1. fixed point spring small mass flexible lead Fig. 9.1 Electrical connections are made to the ends of the spring by means of a flexible lead. The length of the spring is measured before the switch is closed and then again after the switch is closed. When the switch is closed, a magnetic field is set up around each coil of the spring. By reference to these magnetic fields, explain why there is a change in length of the spring. State whether the spring extends or contracts. … … … … … … [4] (c) With the switch in (b) closed, the small mass on the free end of the spring is now made to oscillate vertically. Use the principles of electromagnetic induction to explain why small fluctuations in the current in the spring are found to occur. … … … … [3] [Total: 9]

9 marks

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

This question in 9702/42 May/June 2021

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

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

6 Fig. 6.1 shows a thin slice of semiconducting material used in a Hall probe. I Q R X Y P S W Z I Fig. 6.1 (not to scale) Current I passes through the slice in the direction shown. The slice is placed in a uniform magnetic field of flux density B, so that two of its faces are perpendicular to the magnetic field. A steady Hall voltage VH is developed between face PQXW and face SRYZ. (a) (i) Use the letters in Fig. 6.1 to identify the faces that are perpendicular to the magnetic field. … and … [1] (ii) Explain how the steady Hall voltage VH is developed between faces PQXW and SRYZ. … … … … … [3] (b) The magnitude of VH is given by the equation BI VH = ntq. (i) State the meaning of the symbols n, t and q. You may refer to the letters in Fig. 6.1. n: … t: … q: … [3] (ii) Suggest, with reference to the equation, why the slice of the material used in a Hall probe is thin. … … … [2] [Total: 9]

9 marks

Mark scheme: 6(a)(i) PQRS and WXYZ B1 6(a)(ii) force on charge carriers is perpendicular to both (magnetic) field and current B1 as charge carriers are deflected to one side, an electric field is set up B1 (steady VH when) electric and magnetic forces on charge carriers are equal (and opposite) B1 6(b)(i) n: number density of charge carriers B1 t: distance PW (or SZ or QX or RY) B1 q: charge on each charge carrier B1 6(b)(ii) VH inversely proportional to t B1 (so t needs to be small for) VH to be large enough to measure B1

This question in 9702/41 Oct/Nov 2022

6 Fig. 6.1 shows a thin slice of semiconducting material used in a Hall probe. I Q R X Y P S W Z I Fig. 6.1 (not to scale) Current I passes through the slice in the direction shown. The slice is placed in a uniform magnetic field of flux density B, so that two of its faces are perpendicular to the magnetic field. A steady Hall voltage VH is developed between face PQXW and face SRYZ. (a) (i) Use the letters in Fig. 6.1 to identify the faces that are perpendicular to the magnetic field. … and … [1] (ii) Explain how the steady Hall voltage VH is developed between faces PQXW and SRYZ. … … … … … [3] (b) The magnitude of VH is given by the equation BI VH = ntq. (i) State the meaning of the symbols n, t and q. You may refer to the letters in Fig. 6.1. n: … t: … q: … [3] (ii) Suggest, with reference to the equation, why the slice of the material used in a Hall probe is thin. … … … [2] [Total: 9]

9 marks

Mark scheme: 6(a)(i) PQRS and WXYZ B1 6(a)(ii) force on charge carriers is perpendicular to both (magnetic) field and current B1 as charge carriers are deflected to one side, an electric field is set up B1 (steady VH when) electric and magnetic forces on charge carriers are equal (and opposite) B1 6(b)(i) n: number density of charge carriers B1 t: distance PW (or SZ or QX or RY) B1 q: charge on each charge carrier B1 6(b)(ii) VH inversely proportional to t B1 (so t needs to be small for) VH to be large enough to measure B1

This question in 9702/43 Oct/Nov 2022

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

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

5 (a) Define electric field. … … … [2] (b) Fig. 5.1 shows two parallel conducting plates that are in a vacuum. The plates are separated by a distance of 6.7 cm and have a potential difference (p.d.) of 430 V between them. +430 V conducting plate electron, speed 6.7 cm 2.6 × 107 m s–1 conducting plate 0 V Fig. 5.1 (i) On Fig. 5.1, draw four field lines to represent the electric field between the plates. [2] (ii) Determine the strength E of the electric field between the plates. E = … N C–1 [2] (iii) An electron travels at a speed of 2.6 × 107 m s–1 towards the region between the plates, as shown in Fig. 5.1. On Fig. 5.1, draw the path of the electron as it moves between and beyond the plates. [2] (c) A uniform magnetic field is now applied in the region of the electric field in Fig. 5.1, so that the electron in (b)(iii) travels undeviated through the region. (i) Determine the direction of the uniform magnetic field. … [1] (ii) Explain, with reference to the forces exerted by the two fields on the electron, why the path of the electron is undeviated. … … … [2] (iii) Determine the flux density B of the uniform magnetic field. Give a unit with your answer. B = … unit … [2] [Total: 13]

13 marks

Mark scheme: 5(a) force per unit charge B1 force on positive charge B1 5(b)(i) four straight vertical parallel lines, approximately evenly spaced B1 arrows downwards B1 5(b)(ii) E = V / d C1 E = 430 / 0.067 = 6.4  103 N C–1 A1 5(b)(iii) smooth curve within plates and straight lines outside plates B1 direction of deflection shown as upwards B1 5(c)(i) into the page B1 5(c)(ii) forces are in opposite directions B1 (undeviated) when (magnitudes of) forces are equal B1 5(c)(iii) Eq = Bqv C1 B = E / v = (6.4  103) / (2.6  107) = 2.5  10–4 T A1

This question in 9702/41 May/June 2024

5 (a) Define electric field. … … … [2] (b) Fig. 5.1 shows two parallel conducting plates that are in a vacuum. The plates are separated by a distance of 6.7 cm and have a potential difference (p.d.) of 430 V between them. +430 V conducting plate electron, speed 6.7 cm 2.6 × 107 m s–1 conducting plate 0 V Fig. 5.1 (i) On Fig. 5.1, draw four field lines to represent the electric field between the plates. [2] (ii) Determine the strength E of the electric field between the plates. E = … N C–1 [2] (iii) An electron travels at a speed of 2.6 × 107 m s–1 towards the region between the plates, as shown in Fig. 5.1. On Fig. 5.1, draw the path of the electron as it moves between and beyond the plates. [2] (c) A uniform magnetic field is now applied in the region of the electric field in Fig. 5.1, so that the electron in (b)(iii) travels undeviated through the region. (i) Determine the direction of the uniform magnetic field. … [1] (ii) Explain, with reference to the forces exerted by the two fields on the electron, why the path of the electron is undeviated. … … … [2] (iii) Determine the flux density B of the uniform magnetic field. Give a unit with your answer. B = … unit … [2] [Total: 13]

13 marks

Mark scheme: 5(a) force per unit charge B1 force on positive charge B1 5(b)(i) four straight vertical parallel lines, approximately evenly spaced B1 arrows downwards B1 5(b)(ii) E = V / d C1 E = 430 / 0.067 = 6.4  103 N C–1 A1 5(b)(iii) smooth curve within plates and straight lines outside plates B1 direction of deflection shown as upwards B1 5(c)(i) into the page B1 5(c)(ii) forces are in opposite directions B1 (undeviated) when (magnitudes of) forces are equal B1 5(c)(iii) Eq = Bqv C1 B = E / v = (6.4  103) / (2.6  107) = 2.5  10–4 T A1

This question in 9702/43 May/June 2024

6 An electric field and a magnetic field are used to form a velocity selector. Charged particles, called ions, pass into a region of uniform electric and magnetic fields that is between parallel plates, as shown in Fig. 6.1. plate + + + + + + path of ions region of electric – – – – – – and magnetic fields plate Fig. 6.1 (a) The potential difference (p.d.) between the plates of the velocity selector is V. The separation of the plates is d and the magnetic flux density is B. Show that the speed u of ions that pass undeviated through the velocity selector is given by V u = . Bd [2] (b) Positive ions with kinetic energy 4.1 × 10–17 J and mass 3.2 × 10–27 kg pass undeviated through the velocity selector when V is equal to 980 V and d is equal to 3.6 × 10–2 m. Determine B. B = … T [3] (c) A proton passes undeviated through the velocity selector. An alpha particle enters the velocity selector at the same speed as the proton. State how the expression in (a) predicts that the alpha particle also passes undeviated through the velocity selector. … … [1] (d) By reference to Fig. 6.1 and to the forces acting on a positive ion, determine the direction of the magnetic field. Explain your reasoning. … … … … … … [3] (e) The positive ions in (b) enter the velocity selector with greater kinetic energy. On Fig. 6.1, sketch the path of these ions. [2] [Total: 11]

11 marks

Mark scheme: 6(a) FB = FE B1 either: Bqu = qE and E = V / d leading to u = V / Bd B1 or: Bqu = qV / d leading to u = V / Bd 6(b) EK = ½mu2 C1 u = √[(2  4.1  10–17) / (3.2  10–27)] C1 = 1.6  105 m s–1 B = 980 / (3.6  10–2  1.6  105) A1 = 0.17 T 6(c) expression is independent of mass and charge A1 6(d) either: electric force is downwards so magnetic force is upwards B1 or: no resultant force so magnetic force is upwards (positive ions so) current is from left to right B1 from (Fleming’s) left-hand rule, magnetic field is into the page B1 6(e) curved path inside plates with consistent direction of curvature and with no discontinuity at entry or in curvature B1 direction of deflection is upwards B1

This question in 9702/42 Feb/March 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 9702/41 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