A2 · Paper 4 practice · Magnetic fields

Fields, forces, and induction.

A full set of ten original structured questions in the style of Paper 4, covering the whole of Magnetic fields: field lines, the force on a conductor and on a moving charge, the Hall effect, fields due to currents, and electromagnetic induction. Several questions chain one idea into the next within the topic. Each is linked to its lessons; attempt them all, then reveal the worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level Paper 4 style. They are not from past papers. They test the same concepts and skills the syllabus rewards. Use F = BIL sinθ, F = BQv sinθ, r = mv/BQ, Φ = BA and induced e.m.f. = rate of change of flux linkage where needed.
Keep these straight

Force, charge, flux.

01
Paper 4
[12 marks]

A straight wire passes at right angles through the uniform field between the poles of a magnet, as shown in Fig. 1.1. The flux density is 0.25 T and the length of wire in the field is 4.0 cm.

N S wire, current into page
Fig. 1.1

(a) State two rules that magnetic field lines always obey. [2]

(b) Define magnetic flux density. [2]

(c) The wire carries a current of 3.0 A. Calculate the force on the wire. [2]

(d) State the direction of this force, and the rule you used. [2]

(e) The wire is now turned so that the current makes an angle of 30° with the field. Calculate the new force. [2]

(f) State the angle between the current and the field for which the force is a maximum. [1]

(g) State the value of the force when the current is parallel to the field. [1]

  • (a)Any two: they run from north to south outside a magnet; they never cross; they are closer where the field is stronger ✓✓
  • (b)The force per unit current per unit length on a conductor at right angles to the field ✓✓
  • (c)F = BIL = 0.25 × 3.0 × 0.040 ✓ = 0.030 N ✓
  • (d)The force is vertical (up or down depending on the current sense) ✓; from Fleming's left-hand rule ✓
  • (e)F = BIL sinθ = 0.25 × 3.0 × 0.040 × sin30° ✓ = 0.015 N ✓
  • (f)90° (current perpendicular to the field) ✓
  • (g)Zero ✓
02
Paper 4
[11 marks]

A horizontal wire of length 5.0 cm rests on a top-pan balance and lies at right angles to a horizontal magnetic field of flux density 0.18 T. When a current of 4.5 A passes through the wire, the magnetic force on it acts vertically.

(a) Calculate the magnetic force on the wire. [2]

(b) Calculate the change in the balance reading, in grams, that this force produces. [2]

(c) Define the tesla. [2]

(d) Express the tesla in SI base units. [2]

(e) State and explain the effect on the force of doubling the current. [2]

(f) State what reading change would be seen if the wire were rotated to lie along the field. [1]

  • (a)F = BIL = 0.18 × 4.5 × 0.050 ✓ = 0.041 N ✓
  • (b)Δm = F/g = 0.0405 / 9.81 ✓ = 4.1 × 10⁻³ kg = 4.1 g ✓
  • (c)One tesla is the flux density giving a force of 1 N on each metre of a wire carrying 1 A at right angles to the field ✓✓
  • (d)From B = F/(IL): T = N A⁻¹ m⁻¹ = kg s⁻² A⁻¹ ✓✓
  • (e)The force doubles ✓, because F is proportional to I ✓
  • (f)No change: the force becomes zero (current parallel to the field) ✓
03
Paper 4
[12 marks]

An electron travelling at 3.0 × 10⁶ m s⁻¹ enters a uniform magnetic field of flux density 2.0 mT at right angles to the field. (electron mass 9.11 × 10⁻³¹ kg, charge 1.6 × 10⁻¹⁹ C)

B out of page v F
Fig. 3.1

(a) Explain why the electron follows a circular path at constant speed. [2]

(b) Calculate the magnitude of the magnetic force on the electron. [2]

(c) Show that the radius of the path is given by r = mv/BQ, and calculate r. [3]

(d) Calculate the period of the circular motion. [2]

(e) State and explain the effect on the radius of doubling the speed. [2]

(f) State the effect on the period of doubling the speed. [1]

  • (a)The magnetic force is always perpendicular to the velocity ✓, so it does no work and the speed is constant, while the constant-magnitude perpendicular force gives circular motion ✓
  • (b)F = BQv = 2.0 × 10⁻³ × 1.6 × 10⁻¹⁹ × 3.0 × 10⁶ ✓ = 9.6 × 10⁻¹⁶ N ✓
  • (c)The magnetic force provides the centripetal force: BQv = mv²/r ✓, so r = mv/BQ ✓. r = (9.11 × 10⁻³¹ × 3.0 × 10⁶)/(2.0 × 10⁻³ × 1.6 × 10⁻¹⁹) = 8.5 × 10⁻³ m (8.5 mm) ✓
  • (d)T = 2πm/BQ = 2π × 9.11 × 10⁻³¹ / (2.0 × 10⁻³ × 1.6 × 10⁻¹⁹) ✓ = 1.8 × 10⁻⁸ s ✓
  • (e)The radius doubles ✓, since r = mv/BQ is proportional to v ✓
  • (f)No change: the period is independent of speed ✓
04
Paper 4
[10 marks]

A slice of a current-carrying semiconductor is placed in a magnetic field perpendicular to its largest face. A steady Hall voltage appears across the slice.

(a) Explain how the Hall voltage is set up. [3]

(b) The Hall voltage is Vₐ = BI / (ntq). A slice has charge-carrier density n = 1.0 × 10²² m⁻³, thickness t = 0.50 mm, and carries a current of 20 mA. The Hall voltage is 2.4 mV. Calculate the flux density. [3]

(c) Explain why a semiconductor, rather than a metal, is used for a Hall probe. [2]

(d) State what a Hall probe is used to measure. [1]

(e) State the effect on the Hall voltage of doubling the flux density. [1]

  • (a)The moving charge carriers feel a magnetic force BQv and are pushed to one side ✓; charge builds up there, creating a transverse electric field (p.d.) ✓; this grows until its electric force balances the magnetic force, giving a steady Hall voltage ✓
  • (b)B = Vₐ n t q / I = (2.4 × 10⁻³ × 1.0 × 10²² × 0.50 × 10⁻³ × 1.6 × 10⁻¹⁹) / 0.020 ✓✓ = 0.096 T ✓
  • (c)A semiconductor has a much smaller carrier density n ✓, so for the same B and I the Hall voltage is much larger and easier to measure ✓
  • (d)Magnetic flux density ✓
  • (e)It doubles ✓
05
Paper 4
[11 marks]

Fig. 5.1 shows a velocity selector. Positive ions pass through a region of crossed electric and magnetic fields. The electric field is produced by parallel plates 2.0 cm apart with a potential difference of 600 V across them, and the magnetic field has flux density 0.050 T.

+ E B in v
Fig. 5.1

(a) Show that an ion travels straight through only if its speed is v = E/B. [2]

(b) Calculate the electric field strength between the plates, and hence the selected speed. [3]

(c) Explain what happens to ions that are travelling faster than the selected speed. [2]

(d) The selected ions, each of charge +1.6 × 10⁻¹⁹ C, then enter a magnetic field of flux density 0.20 T and move in a circle of radius 8.0 cm. Calculate the mass of an ion. [3]

(e) State the name of an instrument that uses this arrangement. [1]

  • (a)For a straight path the electric and magnetic forces balance: qE = qvB ✓, so v = E/B ✓
  • (b)E = V/d = 600 / 0.020 = 3.0 × 10⁴ V m⁻¹ ✓; v = E/B = 3.0 × 10⁴ / 0.050 ✓ = 6.0 × 10⁵ m s⁻¹ ✓
  • (c)A faster ion has a larger magnetic force (BQv) than electric force ✓, so it is deflected toward the magnetic-force side and does not pass straight through ✓
  • (d)r = mv/BQ, so m = BQr/v = (0.20 × 1.6 × 10⁻¹⁹ × 0.080) / (6.0 × 10⁵) ✓✓ = 4.3 × 10⁻²⁷ kg ✓
  • (e)A mass spectrometer ✓
06
Paper 4
[10 marks]

Two long straight parallel wires P and Q are a few centimetres apart and carry currents in the same direction.

(a) Describe the shape of the magnetic field around a single long straight wire, and how its direction is found. [2]

(b) State whether P and Q attract or repel, and explain why in terms of their fields. [3]

(c) State what happens to the force if the current in Q alone is reversed. [1]

(d) State what happens to the force if the currents in both P and Q are reversed. [1]

(e) Describe the magnetic field inside a long solenoid. [2]

(f) State the effect on this field of inserting a soft iron core. [1]

  • (a)Concentric circles centred on the wire ✓; direction from the right-hand grip rule (thumb along the current, fingers curl the field) ✓
  • (b)They attract ✓; between the wires the two circular fields point in opposite senses and partly cancel ✓, so each wire is pushed toward the weaker field, that is toward the other wire ✓
  • (c)The force becomes a repulsion ✓
  • (d)Unchanged: they still attract ✓
  • (e)Strong and almost uniform along the inside ✓, shown by parallel, equally spaced lines; like a bar magnet outside ✓
  • (f)It greatly increases the flux density ✓
07
Paper 4
[11 marks]

A flat coil of 200 turns, each of area 1.5 × 10⁻³ m², lies with its plane perpendicular to a magnetic field. The flux density increases steadily from 0 to 0.40 T in a time of 0.20 s.

(a) Define magnetic flux. [1]

(b) Calculate the change in magnetic flux linkage. [3]

(c) Calculate the e.m.f. induced in the coil. [2]

(d) State Faraday's law of electromagnetic induction. [2]

(e) State Lenz's law, and the conservation principle it expresses. [2]

(f) State the effect on the induced e.m.f. of doubling the number of turns. [1]

  • (a)Magnetic flux Φ = BA, the product of flux density and the perpendicular area ✓
  • (b)ΔΦ = BA = 0.40 × 1.5 × 10⁻³ = 6.0 × 10⁻⁴ Wb ✓; Δ(NΦ) = 200 × 6.0 × 10⁻⁴ ✓ = 0.12 Wb (weber-turns) ✓
  • (c)ε = Δ(NΦ)/Δt = 0.12 / 0.20 ✓ = 0.60 V ✓
  • (d)The induced e.m.f. is proportional to the rate of change of magnetic flux linkage ✓✓
  • (e)The induced current is in a direction so as to oppose the change producing it ✓; this is conservation of energy ✓
  • (f)It doubles ✓
08
Paper 4
[10 marks]

Fig. 8.1 shows how the magnetic flux linkage through a search coil varies with time as a magnet is moved near it.

NΦ / Wb t / s 0.20 0.50 0.30 0.70
Fig. 8.1

(a) State how the induced e.m.f. is obtained from this graph. [1]

(b) Calculate the induced e.m.f. during the first 0.30 s. [3]

(c) State the induced e.m.f. between 0.30 s and 0.70 s, and explain your answer. [2]

(d) Compare the size and sign of the e.m.f. after 0.70 s with that in the first 0.30 s. [2]

(e) State the general effect on the induced e.m.f. of moving the magnet more quickly. [2]

  • (a)It is the gradient (rate of change) of the flux-linkage against time graph ✓
  • (b)ε = Δ(NΦ)/Δt = (0.50 − 0.20)/0.30 ✓✓ = 1.0 V ✓
  • (c)Zero ✓; the flux linkage is constant, so there is no rate of change and no e.m.f. ✓
  • (d)Same size (1.0 V from the equal but opposite gradient over 0.40 s gives 0.75 V) but opposite sign: the gradient is −(0.30)/0.40 = −0.75 V ✓, so it is 0.75 V in the opposite direction ✓
  • (e)The e.m.f. is larger ✓, because the flux linkage changes at a greater rate ✓
09
Paper 4
[13 marks]

A metal rod of length 0.25 m rests on two horizontal rails and is pulled along them at a steady 4.0 m s⁻¹. A uniform magnetic field of flux density 0.30 T is perpendicular to the plane of the rails, as in Fig. 9.1. The rails are joined by a resistor of resistance 0.50 Ω.

B into page R v
Fig. 9.1

(a) Show that the e.m.f. induced across the rod is given by ε = BLv, and calculate its value. [3]

(b) Calculate the current driven through the resistor. [2]

(c) The current in the rod lies in the magnetic field. Calculate the force on the rod due to this current. [2]

(d) State the direction of this force relative to the motion, and name the law that gives it. [2]

(e) Explain, using energy, why a force must be applied to keep the rod moving at constant speed. [2]

(f) State and explain the effect on this opposing force of doubling the speed. [2]

  • (a)In a time Δt the rod sweeps area ΔA = L v Δt, so ΔΦ = B L v Δt ✓; ε = ΔΦ/Δt = BLv ✓. ε = 0.30 × 0.25 × 4.0 = 0.30 V ✓
  • (b)I = ε/R = 0.30 / 0.50 ✓ = 0.60 A ✓
  • (c)F = BIL = 0.30 × 0.60 × 0.25 ✓ = 0.045 N ✓
  • (d)It opposes the motion (acts backwards) ✓; this follows from Lenz's law ✓
  • (e)Work done against this opposing force ✓ is transferred to electrical energy and then heat in the resistor; with no applied force the rod would slow down ✓
  • (f)It doubles ✓; the e.m.f. and current both double, and F = BIL is proportional to the current, hence to the speed ✓
10
Paper 4
[13 marks]

This question links three lessons of the topic. A long solenoid carries a current and a coil of 500 turns is wound around its middle.

(a) Describe the magnetic field inside the solenoid, and state the effect of placing a soft iron core inside it. [2]

(b) The current is switched off, and the flux linkage of the 500-turn coil falls from 0.080 Wb to zero in 0.040 s. Calculate the average e.m.f. induced. [3]

(c) State the direction of the induced current relative to the change, and the law used. [2]

(d) An electron now enters the solenoid's field of flux density 0.015 T at 2.0 × 10⁶ m s⁻¹, at right angles to the field. Calculate the radius of its circular path. [3]

(e) State and explain the effect on this radius of doubling the flux density. [2]

(f) State why the magnetic force on the electron does not change its speed. [1]

  • (a)Strong and almost uniform inside (parallel, equally spaced lines) ✓; a soft iron core greatly increases the flux density ✓
  • (b)ε = Δ(NΦ)/Δt. Here the quoted flux linkage falls from 0.080 to 0 Wb ✓; ε = 0.080 / 0.040 ✓ = 2.0 V ✓
  • (c)The induced current acts to oppose the fall in flux (it tries to maintain it) ✓; this is Lenz's law ✓
  • (d)r = mv/BQ = (9.11 × 10⁻³¹ × 2.0 × 10⁶)/(0.015 × 1.6 × 10⁻¹⁹) ✓✓ = 7.6 × 10⁻⁴ m (0.76 mm) ✓
  • (e)The radius halves ✓, since r = mv/BQ is inversely proportional to B ✓
  • (f)The force is always perpendicular to the velocity, so it does no work ✓

Mark this once you have attempted all ten questions and checked your working against the solutions. Revealing the solutions alone does not count.