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A2 · Practice questions · Electromagnetic induction

E.m.f. from a changing flux linkage.

Six original Cambridge-style questions on magnetic flux and flux linkage, Faraday's and Lenz's laws, rotating and moving conductors, and the charge that flows when a coil leaves a field.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Rate of change, not the flux itself.

01
Analysis
[3 marks]

Define magnetic flux and magnetic flux linkage, and state Faraday's law of electromagnetic induction.

  • Magnetic flux through an area is the product of the flux density and the area at right angles to the field, Φ = BA, measured in weber (1 Wb = 1 T m²). ✓
  • Magnetic flux linkage for a coil of N turns is the product of the flux through the coil and the number of turns, NΦ. ✓
  • Faraday's law: the magnitude of the induced e.m.f. is directly proportional to (equal to, in SI units) the rate of change of magnetic flux linkage. ✓
02
Calculation
[3 marks]

A flat coil of 350 turns and area 2.4 × 10⁻³ m² lies with its plane at right angles to a uniform magnetic field of flux density 0.15 T. The field is switched off and falls uniformly to zero in 60 ms. Calculate the change in flux through one turn, the average e.m.f. induced in the coil, and the average current if the coil has a resistance of 7.0 Ω.

  • ΔΦ = ΔB × A = 0.15 × 2.4 × 10⁻³ = 3.6 × 10⁻⁴ Wb. ✓
  • E = NΔΦ / Δt = 350 × 3.6 × 10⁻⁴ / 0.060 = 2.1 V. ✓
  • I = E / R = 2.1 / 7.0 = 0.30 A. ✓
03
Analysis
[3 marks]

A bar magnet is dropped, north pole downwards, through a horizontal coil whose ends are connected to a resistor. (a) State the magnetic polarity of the upper face of the coil while the magnet approaches it, and the law used. (b) Explain why the magnet's acceleration is less than g while it passes through the coil. (c) Explain how this illustrates the conservation of energy.

  • (a) By Lenz's law the induced current opposes the change, so the upper face becomes a north pole to repel the approaching north pole. ✓
  • (b) The repulsion (and then attraction from below as the magnet leaves) is an upward force on the magnet opposing its motion, so the resultant downward force is less than its weight. ✓
  • (c) The magnet loses more gravitational potential energy than it gains as kinetic energy; the difference is the electrical energy that is dissipated in the resistor, so energy is conserved rather than created from nothing. ✓
04
Calculation
[3 marks]

A rectangular coil of 500 turns and area 1.2 × 10⁻³ m² is placed in a uniform magnetic field of flux density 40 mT. Calculate the flux linkage when the plane of the coil is perpendicular to the field, and when the normal to the coil makes an angle of 60° with the field. The coil is then turned from the first position to a position where its plane is parallel to the field in 25 ms; calculate the average e.m.f. induced.

  • Perpendicular: NΦ = NBA = 500 × 0.040 × 1.2 × 10⁻³ = 2.4 × 10⁻² Wb. ✓
  • At 60°: NΦ = NBA cos 60° = 2.4 × 10⁻² × 0.50 = 1.2 × 10⁻² Wb. ✓
  • The flux linkage falls from 2.4 × 10⁻² Wb to zero, so E = 2.4 × 10⁻² / 0.025 = 0.96 V. ✓
05
Calculation
[2 marks]

An aircraft with a wingspan of 36 m flies horizontally at 240 m s⁻¹ where the vertical component of the Earth's magnetic field is 45 μT. Calculate the e.m.f. induced between the wing tips.

  • In 1 s the wing sweeps an area of 36 × 240 m², so the flux cut per second is B × L × v. ✓
  • E = BLv = 45 × 10⁻⁶ × 36 × 240 = 0.39 V. ✓
06
Calculation
[4 marks]

A coil of 120 turns and area 8.0 × 10⁻³ m² has a resistance of 2.4 Ω. It lies with its plane perpendicular to a uniform field of flux density 0.25 T and is pulled completely out of the field in 0.15 s.

(a) Calculate the initial flux linkage of the coil. (b) Calculate the average e.m.f. induced and the average current while the coil is being removed. (c) Calculate the total charge that flows round the coil, and state how this charge would change if the coil were removed in 0.30 s instead.

  • (a) NΦ = NBA = 120 × 0.25 × 8.0 × 10⁻³ = 0.24 Wb. ✓
  • (b) E = NΔΦ / Δt = 0.24 / 0.15 = 1.6 V. ✓
  • I = E / R = 1.6 / 2.4 = 0.67 A. ✓
  • (c) Q = IΔt = 0.667 × 0.15 = 0.10 C; since Q = NΔΦ / R it does not depend on how long the removal takes, so in 0.30 s the same 0.10 C flows (at half the current for twice the time). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.