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

5 (a) An electron is accelerated from rest in a vacuum through a potential difference of 1.2 ×104V. Show that the final speed of the electron is 6.5 ×107m s–1. [2] (b) The accelerated electron now enters a region of uniform magnetic field acting into the plane of the paper, as illustrated in Fig. 5.1. magnetic field into plane of paper + + + path of + + + electron + + + Fig. 5.1 (i) Describe the path of the electron as it passes through, and beyond, the region of the magnetic field. You may draw on Fig. 5.1 if you wish. path within field: … … path beyond field: … … [3] Use (ii) State and explain the effect on the magnitude of the deflection of the electron in the magnetic field if, separately, 1. the potential difference accelerating the electron is reduced, … … … [2] 2. the magnetic field strength is increased. … … … [2]

9 marks

Mark scheme: 5 (a) ½mv2 = qV ……(or some verbal explanation) …..……………..…… B1 ½ × 9.11 × 10-31 × v2 = 1.6 × 10-19 × 1.2 × 104 ………………………… B1 v = 6.49 × 107 m s–1 ………….………………………………………… A0 [2] (b) (i) within field: circular arc ………………..………………..…………….. B1 in ‘downward’ direction ……………..………………….. B1 beyond field: straight, with no ‘kink’ on leaving field ………………… B1 [3] (ii) 1. v is smaller …………………………………………………………………. M1 deflection is larger ………………………………………………………… A1 [2] 2. (magnetic) force is larger ………………………………………………… M1 deflection is larger ……………………………………………………….. A1 [2]

This question in 9702/41 Oct/Nov 2005

8 A proton is moving with constant velocity v. It enters a uniform magnetic field that is normal to the initial direction of motion of the proton, as shown in Fig. 8.1. proton velocity v uniform magnetic field into plane of paper Fig. 8.1 A uniform electric field is applied in the same region as the magnetic field so that the proton passes undeviated through the fields. (a) On Fig. 8.1, draw an arrow labelled E to show the direction of the electric field. [1] (b) The proton is replaced by other particles. The electric and magnetic fields remain unchanged. State and explain the deviation, if any, of the following particles in the region of the fields. (i) an α-particle with initial velocity v … … … … [3] (ii) an electron with initial velocity 2v … … … … [3]

7 marks

Mark scheme: 8 (a) arrow labelled E pointing down the page B1 [1] (b) (i) Bqv = qE M1 forces are independent of mass and charge ‘cancels’ M1 so no deviation A1 [3] (ii) magnetic force > electric force M1 so deflects M1 ‘downwards’ A1 [3]

This question in 9702/41 May/June 2006

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

7 Negatively-charged particles are moving through a vacuum in a parallel beam. The particles For have speed v. Examiner’s The particles enter a region of uniform magnetic field of flux density 930 μT. Initially, the Use particles are travelling at right-angles to the magnetic field. The path of a single particle is shown in Fig. 7.1. negatively-charged arc of radius 7.9 cm particles, speed v uniform magnetic field, flux density 930 μT Fig. 7.1 The negatively-charged particles follow a curved path of radius 7.9 cm in the magnetic field. A uniform electric field is then applied in the same region as the magnetic field. For an electric field strength of 12 kV m–1, the particles are undeviated as they pass through the region of the fields. (a) On Fig. 7.1, mark with an arrow the direction of the electric field. [1] (b) Calculate, for the negatively-charged particles, (i) the speed v, v = … m s–1 [3] charge (ii) the ratio . mass ratio = … C kg–1 [3]

7 marks

Mark scheme: 7 (a) arrow pointing up the page B1 [1] (b) (i) Eq = Bqv C1 v = (12 × 103) / (930 × 10–6) C1 = 1.3 × 107 m s–1 A1 [3] (ii) Bqv = mv 2 / r C1 q/m = (1.3 × 107) / (7.9 × 10–2 × 930 × 10–6) C1 = 1.8 × 1011 C kg–1 A1 [3]

This question in 9702/41 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

7 Electrons are moving through a vacuum in a narrow beam. The electrons have speed v. For The electrons enter a region of uniform magnetic field of flux density B. Initially, the electrons Examiner’s are travelling at a right-angle to the magnetic field. Use The path of a single electron is shown in Fig. 7.1. region of magnetic field flux density B electron speed v Fig. 7.1 The electrons follow a curved path in the magnetic field. A uniform electric field of field strength E is now applied in the same region as the magnetic field. The electrons pass undeviated through the region of the two fields. Gravitational effects may be neglected. (a) Derive a relation between v, E and B for the electrons not to be deflected. Explain your working. … … … … … [3] (b) An α-particle has speed v and approaches the region of the two fields along the same path as the electron. Describe and explain the path of the α-particle as it passes through the region of the two fields. … … … [2]

5 marks

Mark scheme: 7 (a) force due to E-field is equal and opposite to force due to B-field B1 Eq = Bqv B1 v = E/B B1 [3] (b) either charge and mass are not involved in the equation in (a) or FE and FB are both doubled or E, B and v do not change M1 so no deviation A1 [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

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

4 marks

Mark scheme: 5 (a) (i) Bqv(sinθ) or Bqv(cosθ) B1 [1] (ii) qE B1 [1] (b) FB must be opposite in direction to FE B1 so magnetic field into plane of paper B1 [2] GCE AS/A LEVEL – October/November 2011 9702 42

This question in 9702/42 Oct/Nov 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

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

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

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

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

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

6 (a) Explain the use of a uniform electric field and a uniform magnetic field for the selection of the velocity of a charged particle. You may draw a diagram if you wish. … … … … [3] (b) Ions, all of the same isotope, are travelling in a vacuum with a speed of 9.6 × 104 m s−1. The ions are incident normally on a uniform magnetic field of flux density 640 mT. The ions follow semicircular paths A and B before reaching a detector, as shown in Fig. 6.1. detector A B vacuum uniform magnetic field, flux density 640 mT Fig. 6.1 Data for the diameters of the paths are shown in Fig. 6.2. path diameter / cm A 6.2 B 12.4 Fig. 6.2 The ions in path B each have charge +1.6 × 10−19 C. (i) Determine the mass, in u, of the ions in path B. mass = … u [4] (ii) Suggest and explain quantitatively a reason for the difference in radii of the paths A and B of the ions. … … … … [3]

10 marks

Mark scheme: 6 (a) electric and magnetic fields normal to each other B1 either charged particle enters region normal to both fields or correct B direction w.r.t. E for zero deflection B1 for no deflection, v R E / B B1 [3] (no credit if magnetic field region clearly not overlapping with electric field region) GCE A LEVEL – May/June 2014 9702 42 (b) (i) m R Bqr / v C1 = = = R=(640 × 10–3 × 1.6 × 10–19 × 6.2 × 10–2) / (9.6 × 104) C1 = = = R=6.61 × 10–26 kg C1 = = = R=(6.61 × 10–26) / (1.66 × 10–27) u = = = R=40 u A1 [4] (ii) q / m ∝ 1 / r or m constant and q ∝ 1 / r B1 q / m for A is twice that for B B1 ions in path A have (same mass but) twice the charge (of ions in path B) B1 [3]

This question in 9702/42 May/June 2014

2 (a) On the axes of Fig. 2.1, sketch the variation with distance from a point mass of the gravitational field strength due to the mass. gravitational field strength 0 0 distance Fig. 2.1 [2] (b) On the axes of Fig. 2.2, sketch the variation with speed of the magnitude of the force on a charged particle moving at right-angles to a uniform magnetic field. force 0 0 speed Fig. 2.2 [2] (c) On the axes of Fig. 2.3, sketch the variation with time of the power dissipated in a resistor by a sinusoidal alternating current during two cycles of the current. power 0 0 time Fig. 2.3 [3]

7 marks

Mark scheme: 2 (a) smooth curve with decreasing gradient, not starting at x = 0 M1 end of line not at g = 0 or horizontal A1 [2] (b) straight line with positive gradient M1 line starts at origin A1 [2] (c) sinusoidal shape B1 only positive values and peak / trough height constant B1 4 ‘loops’ B1 [3] 5 4

This question in 9702/43 Oct/Nov 2014

7 (a) State what is meant by quantisation of charge. … … [1] (b) Charged parallel plates, as shown in Fig. 7.1, produce a uniform electric field between the plates. + beam of protons – Fig. 7.1 The electric field outside the region between the plates is zero. A uniform magnetic field is applied in the region between the plates so that a beam of protons passes undeviated between the plates. (i) State and explain the direction of the magnetic field between the plates. … … … [2] (ii) The magnetic flux density between the plates is now increased. On Fig. 7.1, sketch the path of the protons between the plates. [2]

5 marks

Mark scheme: 7 (a) either charge exists in discrete and equal quantities or multiples of elementary charge / e / 1.6 × 10–19 C B1 [1] (b) (i) force due to magnetic field must be upwards B1 B-field into the plane of the paper B1 [2] (ii) sketch showing: deflection consistent with force in (b)(i) B1 reasonable curve B1 [2]

This question in 9702/43 Oct/Nov 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

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

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 (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 9702/42 Oct/Nov 2015

9 A particle of charge +q and mass m is travelling with a constant speed of 1.6 × 105 m s–1 in a vacuum. The particle enters a uniform magnetic field of flux density 9.7 × 10–2 T, as shown in Fig. 9.1. uniform magnetic field out of the plane of particle the paper charge +q flux density 9.7 × 10–2 T mass m speed 1.6 × 105 m s–1 path of particle Fig. 9.1 The magnetic field direction is perpendicular to the initial velocity of the particle and perpendicular to, and out of, the plane of the paper. A uniform electric field is applied in the same region as the magnetic field so that the particle passes undeviated through the fields. (a) State and explain the direction of the electric field. … … … [2] (b) Calculate the magnitude of the electric field strength. Explain your working. electric field strength = … V m–1 [3] (c) The electric field is now removed so that the positively-charged particle follows a curved path in the magnetic field. This path is an arc of a circle of radius 4.0 cm. q Calculate, for the particle, the ratio m. ratio = … C kg–1 [3] (d) The particle has a charge of 3e where e is the elementary charge. (i) Use your answer in (c) to determine the mass, in u, of the particle. mass = … u [2] (ii) The particle is the nucleus of an atom. State the number of protons and the number of neutrons in this nucleus. number of protons = … number of neutrons = … [1] [Total: 11]

11 marks

Mark scheme: 9 (a) direction of force due to electric field opposite to force due to magnetic field B1 electric field is up the page B1 [2] (b) force due to electric field = force due to magnetic field or Eq = Bqv B1 E = Bv C1 = 9.7 × 10–2 × 1.6 × 105 = 1.6 (1.55) × 104 V m–1 A1 [3] (c) q / m = v / Br C1 = 1.6 × 105 / (9.7 ×10–2 × 4.0 × 10–2) C1 = 4.1 (4.12) × 107 C kg–1 A1 [3] (d) (i) m = (3 × 1.60 × 10–19) / (4.12 × 107) C1 m = 1.16 × 10–26 / 1.66 × 10–27 = 7(.0) u (allow 7.1 u) A1 [2] (ii) 3 protons, 4 neutrons A1 [1]

This question in 9702/42 Feb/March 2016

9 A thin rectangular slice of aluminium has sides of length 65 mm, 50 mm and 0.10 mm, as shown in Fig. 9.1. direction of magnetic field Z Y 0.10 mm 50 mm X current 3.8 A Q R P S 65 mm Fig. 9.1 (not to scale) Some of the corners of the slice are labelled. A current I of 3.8 A is normal to face RSXY of the slice. In aluminium, the number of free electrons per unit volume is 6.0 × 1028 m−3. A uniform magnetic field of magnetic flux density B equal to 0.13 T is normal to face QRYZ of the aluminium slice in the direction from Q to P. A Hall voltage VH is developed across the slice and is given by the expression BI VH = . ntq (a) Use Fig. 9.1 to state the magnitude of the distance t. t = … mm [1] (b) Calculate the magnitude of the Hall voltage VH. VH = … V [2] [Total: 3]

3 marks

Mark scheme: 9 (a) 0.10 mm B1 [1] (b) VH = (0.13 × 3.8) / (6.0 × 1028 × 0.10 × 10–3 × 1.60 × 10–19) C1 = 5.1 × 10–7 V A1 [2]

This question in 9702/41 May/June 2016

9 A thin rectangular slice of aluminium has sides of length 65 mm, 50 mm and 0.10 mm, as shown in Fig. 9.1. direction of magnetic field Z Y 0.10 mm 50 mm X current 3.8 A Q R P S 65 mm Fig. 9.1 (not to scale) Some of the corners of the slice are labelled. A current I of 3.8 A is normal to face RSXY of the slice. In aluminium, the number of free electrons per unit volume is 6.0 × 1028 m−3. A uniform magnetic field of magnetic flux density B equal to 0.13 T is normal to face QRYZ of the aluminium slice in the direction from Q to P. A Hall voltage VH is developed across the slice and is given by the expression BI VH = . ntq (a) Use Fig. 9.1 to state the magnitude of the distance t. t = … mm [1] (b) Calculate the magnitude of the Hall voltage VH. VH = … V [2] [Total: 3]

3 marks

Mark scheme: 9 (a) 0.10 mm B1 [1] (b) VH = (0.13 × 3.8) / (6.0 × 1028 × 0.10 × 10–3 × 1.60 × 10–19) C1 = 5.1 × 10–7 V A1 [2]

This question in 9702/43 May/June 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

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