OxfordAQA International AS and A-level Physics: Flux, flux linkage and electromagnetic induction (9630) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.10.3 (Magnetic flux and flux linkage) and 3.10.4 (Electromagnetic induction), of OxfordAQA International AS and A-level Physics (9630), Version 4.4, which the specification marks as International A-level only. Read the study guide first, or use the revision notes. Give numerical answers to a sensible number of significant figures.

Questions

1. A rectangular loop measures 0.15 m by 0.060 m. It is in a uniform field of 0.28 T, perpendicular to the plane of the loop. (a) Calculate the magnetic flux through the loop. [2] (b) The loop is replaced by a coil of 120 turns of the same size. Calculate the flux linkage. [1]

2. A rectangular coil of 60 turns measures 0.10 m by 0.20 m. It is in a field of 0.090 T. (a) The normal to the coil makes an angle of 50° with the field. Calculate the flux linkage. [3] (b) Calculate the angle between the normal and the field when the flux linkage is 0.054 Wb. [2]

3. Write down the two laws of electromagnetic induction named after Faraday and Lenz. [3]

4. A student connects a coil to a sensitive meter and pushes a bar magnet into the coil. (a) Explain why the meter deflects only while the magnet is moving. [2] (b) The magnet is then pulled out faster. Describe and explain two differences in what the meter shows. [2]

5. The flux linkage of a coil changes with time as follows.

Time / s	0	0.20	0.40	0.50
Flux linkage / mWb	0	30	30	0

The flux linkage changes uniformly between readings. (a) Calculate the emf from 0 to 0.20 s. [2] (b) State and explain the emf from 0.20 s to 0.40 s. [1] (c) Calculate the size of the emf from 0.40 s to 0.50 s and say how its direction compares with part (a). [2]

6. A metal rod of length 0.30 m slides at a constant 0.80 m s⁻¹ along two horizontal rails in a vertical field of 0.45 T. The rod and rails form a circuit of total resistance 0.15 Ω. (a) Calculate the emf induced in the rod. [2] (b) Calculate the current. [1] (c) Calculate the horizontal force needed to keep the rod moving at constant speed. [2] (d) Explain, using Lenz’s law, why this force is needed. [1]

7. A small cylindrical magnet is dropped down a long vertical aluminium tube. It takes much longer to emerge than a non-magnetic object of the same size. Explain why. [4]

8. A coil rotates uniformly in a uniform field. Explain why the induced emf is zero at the instant the flux linkage is at its maximum. [3]

9. A rectangular coil of 80 turns measures 0.050 m by 0.040 m. It rotates at 1200 revolutions per minute about an axis perpendicular to a uniform field of 0.25 T. At t = 0 the normal to the coil is along the field. (a) Calculate the angular speed of the coil. [2] (b) Calculate the peak emf. [2] (c) Calculate the emf at t = 3.0 ms. [2] (d) The coil is now rotated at twice the speed. State the new peak emf and period. [2]

10. A coil of 400 turns has area 2.5 × 10⁻³ m². It lies perpendicular to a uniform field of 0.12 T directed into the page. It is pulled completely out of the field in 0.15 s. The resistance of the circuit is 8.0 Ω. (a) Find the flux linkage at the start. [2] (b) Find the mean induced emf. [2] (c) Calculate the average current. [1] (d) State and explain the direction of the current, clockwise or anticlockwise as seen with the field into the page. [2] (e) Calculate the electrical energy transferred in the circuit during the 0.15 s. [1]

11. A coil of 150 turns and area 5.0 × 10⁻³ m² rotates uniformly with period 0.040 s in a field of 0.080 T. The normal is along the field at t = 0. (a) Calculate the angular speed. [1] (b) Calculate the peak emf. [2] (c) State the times in the first period when the emf is zero, and say what is true of the coil then. [2] (d) Calculate the first time when the emf equals half the peak value. [2]

12. Show that the weber is equivalent to the volt second. [3]

Answers

1. (a) A = 0.15 × 0.060 = 9.0 × 10⁻³ m² [1]; Φ = BA = 0.28 × 9.0 × 10⁻³ = 2.5 × 10⁻³ Wb [1] (b) NΦ = 120 × 2.52 × 10⁻³ = 0.30 Wb [1] [3] Examiner insight: Show the area in m² first; a slip on the area or a missing power of ten affects both parts.

2. (a) A = 0.10 × 0.20 = 0.020 m² [1]; BAN = 0.090 × 0.020 × 60 = 0.108 Wb [1]; NΦ = 0.108 × cos 50° = 0.069 Wb [1] (b) cos θ = 0.054 / 0.108 = 0.50 [1]; θ = 60° [1] [5] Examiner insight: The angle is from the normal, so cos is right. Using sin 50° (0.083 Wb) is the typical slip; sketch the normal.

3. Faraday: the size of the induced emf is directly proportional to how fast the flux linkage changes [1]. Lenz: the induced emf (and any current) has a direction that opposes [1] the change causing it [1]. [3] Examiner insight: “Opposes the change” is what gains credit; “opposes the flux” or “opposes the magnet” is a looser statement that may not.

4. (a) The flux linkage through the coil changes only while the magnet moves [1]; an emf is induced only when flux linkage changes, so with the magnet at rest the rate of change is zero and there is no deflection [1]. (b) The deflection is larger because the rate of change of flux linkage is greater [1]; it lasts for a shorter time, as the magnet takes less time to travel the same distance [1]. [4] Examiner insight: Link each observation to the rate of change of flux linkage.

5. (a) ε = Δ(NΦ)/Δt = 30 × 10⁻³ / 0.20 [1] = 0.15 V [1] (b) Zero, because the flux linkage is constant (zero gradient) [1]. (c) ε = 30 × 10⁻³ / 0.10 = 0.30 V [1]; the direction is opposite to part (a), as the flux linkage is decreasing rather than increasing [1] [5] Examiner insight: Convert mWb to Wb and give the reason for part (b) in terms of the gradient, not just “nothing is happening”.

6. (a) ε = BLv [1] = 0.45 × 0.30 × 0.80 = 0.11 V [1] (b) I = ε/R = 0.108 / 0.15 = 0.72 A [1] (c) F = BIL [1] = 0.45 × 0.72 × 0.30 = 0.097 N [1] (d) The induced current opposes the change in flux (the increase in area of the circuit), so the magnetic force on the rod opposes its motion; an applied force is needed to balance it [1] [6] Examiner insight: Use the unrounded emf in later steps; using 0.11 V gives 0.73 A and a slightly different force.

7. The moving magnet changes the flux through the tube wall [1]; this induces emfs and so eddy currents in the aluminium [1]; by Lenz’s law, the currents produce a field that opposes the motion, so the magnet is pushed back upwards [1]; the magnet’s energy is transferred to thermal energy in the tube, so it falls at a lower speed than a non-magnetic object [1] [4] Examiner insight: Give the cause (changing flux), the effect (eddy currents) and the consequence (opposing force) in that order.

8. The emf is the rate of change of flux linkage [1]. At the maximum of the NΦ–t graph the gradient is zero, so the emf is zero [1]. Also, the sides of the coil are moving parallel to the field, so no field lines are cut [1] [3] Examiner insight: Saying the emf is large because the flux is large is a common error. The emf depends on the gradient, not the value.

9. (a) f = 1200/60 = 20 Hz [1]; ω = 2πf = 126 rad s⁻¹ [1] (b) BAN = 0.25 × (0.050 × 0.040) × 80 = 0.040 Wb [1]; ε₀ = BANω = 0.040 × 125.7 = 5.0 V [1] (c) ωt = 125.7 × 3.0 × 10⁻³ = 0.377 rad [1]; ε = 5.03 × sin 0.377 = 1.9 V [1] (d) Peak emf = 10 V (ω doubles) [1]; period = 0.025 s (half of 0.050 s) [1] [8] Examiner insight: Convert revolutions per minute to hertz before using 2π, and use radians in sin ωt.

10. (a) NΦ = BAN = 0.12 × 2.5 × 10⁻³ × 400 [1] = 0.12 Wb [1] (b) Final flux linkage is zero, so Δ(NΦ) = 0.12 Wb [1]; ε = 0.12 / 0.15 = 0.80 V [1] (c) I = 0.80 / 8.0 = 0.10 A [1] (d) The flux linkage into the page is decreasing [1]; the induced current opposes this by producing a field into the page inside the coil, so the current is clockwise [1] (e) E = εIt = 0.80 × 0.10 × 0.15 = 0.012 J [1] [8] Examiner insight: For the direction, state whether the flux is rising or falling before giving the sense of the current; the sense alone, however correct, shows no reasoning.

11. (a) ω = 2π/T = 2π / 0.040 = 157 rad s⁻¹ [1] (b) ε₀ = BANω = 0.080 × 5.0 × 10⁻³ × 150 × 157.1 [1] = 9.4 V [1] (c) At t = 0, 0.020 s and 0.040 s [1]; at these times the coil plane is perpendicular to the field and the flux linkage is at its maximum (or a maximum in the opposite sense) [1] (d) sin ωt = 0.50, so ωt = π/6 = 0.5236 rad [1]; t = 0.5236 / 157.1 = 3.3 × 10⁻³ s [1] [7] Examiner insight: For (d), the exact angle π/6 (one twelfth of a period) gives a check on your answer.

12. B = F/(IL), so the tesla is N A⁻¹ m⁻¹, and Wb = T m² = N m A⁻¹ [1]; N m = J, so Wb = J A⁻¹ [1]; J = V A s, so Wb = V A s A⁻¹ = V s [1] [3] Examiner insight: On a “show that” question, write each step in the chain of units; the final result alone does not show the working.

Where marks are usually lost

• Putting the angle to the coil’s plane into cos θ instead of the angle to the normal.
• Forgetting the number of turns, or including it in the flux through one turn.
• Using a flux value, not a change in flux, to find an emf.
• Not converting mWb, cm² or minutes to base units.
• Quoting a current direction with no statement about whether the flux linkage is rising or falling.
• Using degrees for sin ωt when ωt is in radians.
• Explaining eddy-current damping without mentioning the changing flux and the opposing force.
• Rounding the emf early.

Next steps

• Use the revision notes.
• Re-read the study guide for missed questions.
• Visit the OxfordAQA A-level Physics hub and tick off the checklist.
• Try one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams from May/June 2020 onwards, published by OxfordAQA. Sections 3.10.3 Magnetic flux and flux linkage and 3.10.4 Electromagnetic induction (International A-level only).