IB DP Physics: Electromagnetic induction: flux, Faraday's and Lenz's laws and generators (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions practise sub-topic D.4, Induction, from the IB Diploma Programme Physics guide, first assessment 2025. D.4 is additional higher level (HL only), so every question here is HL only. If you need the theory first, use the study guide, the revision notes, the IB Physics hub and the printable checklist. Use g = 9.81 N kg⁻¹.

Questions

1. A circular coil of radius 4.0 cm lies in a uniform magnetic field of 0.25 T. The plane of the coil makes an angle of 40° with the field direction. Calculate the magnetic flux through one turn. [3]

2. A coil has 250 turns. The flux through each turn falls uniformly from 5.0 × 10⁻³ Wb to 1.0 × 10⁻³ Wb in 20 ms.

(a) Calculate the magnitude of the induced emf. [2] (b) Explain what the minus sign in Faraday’s law tells you about the induced current. [2]

3. A straight rod of length 0.28 m is dragged at a steady 2.2 m s⁻¹ along horizontal conducting rails in a vertical uniform field of 0.38 T. The rod and rails have a combined resistance of 0.35 Ω.

(a) Calculate the emf induced across the rod. [2] (b) Calculate the horizontal pull needed to hold the speed steady. [3]

4. The south pole of a bar magnet is moved towards the end of a coil that is connected to a resistor. State and explain the polarity induced in the end of the coil nearest the magnet, and explain how energy conservation requires this. [4]

5. A magnet hanging from a spring oscillates vertically in and out of a coil connected to a resistor.

(a) Describe how the induced emf varies during one oscillation. [3] (b) Explain why the amplitude of the oscillation decreases with time. [2]

6. A rectangular coil of 150 turns measures 6.0 cm by 9.0 cm. It rotates at 25 Hz about an axis perpendicular to a uniform field of 0.12 T. Timing starts when the coil’s plane is perpendicular to the field.

(a) Calculate the peak emf. [3] (b) Calculate the emf at t = 3.0 ms. [2] (c) The frequency rises to 40 Hz. State the new peak emf and period. [2]

7. A generator’s emf has a peak of 8.4 V and a period of 16 ms. Its rotation speed is reduced to 75% of the original. Determine the original frequency, the new peak emf and the new period. [4]

8. A coil of 80 turns has a magnetic flux through it that changes as follows. From 0 to 0.20 s it rises uniformly from 0 to 6.0 × 10⁻³ Wb. From 0.20 s to 0.50 s it stays constant. From 0.50 s to 0.80 s it falls uniformly to zero. Using ε = −N ΔΦ/Δt, find the emf in each of the three intervals and describe the shape of the emf–time graph. [6]

9. A coil with an iron core is in series with a lamp, a battery and a switch. Explain why the lamp brightens gradually when the switch is closed, and why a spark can appear when it is opened. [4]

10. A wind-driven generator has a coil of 500 turns and area 0.020 m² rotating in a uniform field of 0.080 T at 6.0 revolutions per second.

(a) Calculate the peak emf. [3] (b) The wind speed rises so that the coil turns at twice the rate. State the new peak emf. [1] (c) Without changing the rotation rate, suggest two changes that would give the same new peak emf as in (b). [2] (d) State the orientation of the coil at an instant when the emf is zero, and explain why. [3]

11. A horizontal rod of mass 0.050 kg and length 0.50 m rests across two horizontal rails in a vertical uniform field of 0.40 T. The total resistance of the circuit is 0.80 Ω. A light string runs over a pulley to a hanging mass of 0.050 kg, and the rod is released. Friction is negligible.

(a) Show that the magnetic force on the rod is B²L²v / R when its speed is v. [2] (b) Find the terminal speed of the rod. Assume the rails and string are long enough. [2] (c) Show that, at terminal speed, the rate at which the hanging mass loses gravitational potential energy equals the rate of energy dissipation in the circuit. [3] (d) State and explain how the terminal speed changes if the resistance is doubled. [2]

Answers

1. The angle between the field and the normal is 90° − 40° = 50° [1]. A = π × 0.040² = 5.03 × 10⁻³ m² [1]. Φ = BA cos θ = 0.25 × 5.03 × 10⁻³ × cos 50° = 8.1 × 10⁻⁴ Wb [1]. [3] Examiner insight: using 40° directly is the standard error; write the angle to the normal explicitly so the step can be credited.

2. (a) ΔΦ = 4.0 × 10⁻³ Wb [1]. |ε| = 250 × 4.0 × 10⁻³ / 0.020 = 50 V [1]. (b) The induced current’s effect opposes the change that causes it [1]. Here the flux is decreasing, so the current produces a field in the same direction as the existing flux, to oppose the fall [1]. [4] Examiner insight: saying the induced field opposes “the field” loses the point; refer to the change in flux and say whether it is rising or falling.

3. (a) ε = BvL [1] = 0.38 × 2.2 × 0.28 = 0.23 V (0.234 V) [1]. (b) I = ε/R = 0.2341 / 0.35 = 0.669 A [1]. F = BIL = 0.38 × 0.669 × 0.28 [1] = 0.071 N [1]. [5] Examiner insight: keep unrounded values between parts; rounding the emf to 0.23 V first shifts the final force.

4. The flux through the coil is increasing [1]. The induced current opposes this increase, so the nearest end becomes a south pole [1] and repels the magnet [1]. The opposite polarity would pull the magnet in and give electrical energy plus kinetic energy for free, breaking energy conservation; instead, work done against the repulsion supplies the electrical energy [1]. [4] Examiner insight: link the polarity to a force on the magnet and then to work done; naming the polarity alone gains little.

5. (a) The emf alternates in sign as the magnet moves up then down [1]. The emf is largest in size as the magnet passes equilibrium, where its speed is greatest [1], and zero at the top and bottom, where it is momentarily at rest [1]. (b) The induced current heats the resistor [1]. That energy comes from the oscillating system, so the amplitude falls [1]. [5] Examiner insight: tie the size of the emf to the rate of change of flux, which follows the speed, not to the size of the flux.

6. (a) A = 0.060 × 0.090 = 5.4 × 10⁻³ m² [1]. ω = 2π × 25 = 157 rad s⁻¹ [1]. ε_max = NBAω = 150 × 0.12 × 5.4 × 10⁻³ × 157 = 15 V (15.3 V) [1]. (b) ε = ε_max sin ωt, with ωt = 157 × 0.0030 = 0.471 rad [1]. ε = 15.27 × sin 0.471 = 6.9 V [1]. (c) Peak emf = 15.27 × 40/25 = 24 V (24.4 V) [1]; period = 1/40 = 0.025 s [1]. [7] Examiner insight: work in radians for part (b), and show the substitution so that your method can be seen.

7. Original f = 1/0.016 = 62.5 Hz [1]. Peak emf is proportional to f, so the new peak = 0.75 × 8.4 = 6.3 V [1]. New f = 46.9 Hz [1], so new period = 0.016 / 0.75 = 21 ms (21.3 ms) [1]. [4] Examiner insight: the period goes up when the frequency goes down; say which quantity is multiplied by 0.75 and which is divided by it.

8. 0 to 0.20 s: ε = −80 × (6.0 × 10⁻³ / 0.20) = −2.4 V [1][1]. 0.20 to 0.50 s: ΔΦ = 0, so ε = 0 [1]. 0.50 to 0.80 s: ε = −80 × (−6.0 × 10⁻³ / 0.30) = +1.6 V [1][1]. The graph has three flat sections: negative, zero, positive [1]. [6] Examiner insight: the two non-zero emfs have opposite signs and different sizes because the ramps differ in duration; show each gradient separately.

9. When the switch is closed, the current grows and so does the flux through the coil [1]. A changing flux induces an emf in the same coil, which by Lenz’s law opposes the rise in current, so the current and the lamp brightness grow gradually [1]. When the switch is opened, the current falls rapidly and the flux collapses [1]. The induced emf now opposes the fall and can become large across the gap, causing a spark [1]. [4] Examiner insight: describe the self-induced emf as opposing the change in current, not the current itself.

10. (a) ω = 2π × 6.0 = 37.7 rad s⁻¹ [1]. ε_max = NBAω = 500 × 0.080 × 0.020 × 37.7 [1] = 30 V (30.2 V) [1]. (b) Peak emf is proportional to f, so it doubles to 60 V (60.3 V) [1]. (c) Any two from: double the number of turns, double the coil area, double the field strength [1][1]. (d) The coil’s plane is perpendicular to the field, with the normal along B [1]. The flux is then at its maximum [1], but momentarily not changing, so the rate of change of flux, and hence the emf, is zero [1]. [9] Examiner insight: “maximum flux therefore maximum emf” is a common wrong link; the emf follows the gradient of the flux.

11. (a) ε = BvL and I = ε/R = BvL/R [1]. F = BIL = B²L²v/R [1]. (b) At terminal speed the magnetic force equals the pull of the hanging mass, mg = 0.050 × 9.81 = 0.4905 N [1]. v = mgR / (B²L²) = 0.4905 × 0.80 / (0.16 × 0.25) = 9.8 m s⁻¹ [1]. (c) Rate of loss of potential energy = mgv = 0.4905 × 9.81 = 4.81 W [1]. ε = 0.40 × 9.81 × 0.50 = 1.962 V [1], so power in the circuit = ε²/R = 1.962² / 0.80 = 4.81 W, which is equal [1]. (d) The terminal speed doubles [1], because the current for a given speed is halved, so twice the speed is needed to give the same magnetic force [1]. [9] Examiner insight: in “show that” parts every step must be visible; the terminal speed also needs the statement that the net force is zero.

Where marks are usually lost

• Measuring θ from the surface of the loop instead of from its normal.
• Forgetting N in the emf of a coil.
• Describing Lenz’s law as opposing the field instead of opposing the change in flux.
• Substituting f where ω = 2πf is needed.
• Writing that emf is maximum when flux is maximum for a rotating coil.
• Not linking induced current to a force and to work done.

Next steps

• Review the revision notes and the study guide.
• Return to the IB Physics hub and the printable checklist.
• Try the free 10-minute diagnostics to find your next weak spot.
• Book a free trial class.

Official syllabus

These questions follow sub-topic D.4 (Induction) of the IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. D.4 is additional higher level content only.