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9702 · 20.5

Electromagnetic induction — practice questions

Practice and worked examples for 9702 Electromagnetic induction. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A coil with 500 turns has a magnetic flux of 2.0 mWb passing through it. If this flux is uniformly reduced to 0.5 mWb in 0.25 seconds, calculate the magnitude of the induced EMF.

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  1. Calculate the initial magnetic flux linkage: λ1=NΦ1=500×(2.0×10−3 Wb)=1.0 Wb\lambda_1 = N\Phi_1 = 500 \times (2.0 \times 10^{-3} \text{ Wb}) = 1.0 \text{ Wb}.
  2. Calculate the final magnetic flux linkage: λ2=NΦ2=500×(0.5×10−3 Wb)=0.25 Wb\lambda_2 = N\Phi_2 = 500 \times (0.5 \times 10^{-3} \text{ Wb}) = 0.25 \text{ Wb}.
  3. Determine the change in magnetic flux linkage: Δλ=λ2−λ1=0.25 Wb−1.0 Wb=−0.75 Wb\Delta \lambda = \lambda_2 - \lambda_1 = 0.25 \text{ Wb} - 1.0 \text{ Wb} = -0.75 \text{ Wb}.
  4. Identify the time taken for the change: Δt=0.25 s\Delta t = 0.25 \text{ s}.
  5. Calculate the magnitude of the induced EMF using Faraday's Law: E=∣ΔλΔt∣=∣−0.75 Wb0.25 s∣=3.0 V\mathcal{E} = |\frac{\Delta \lambda}{\Delta t}| = |\frac{-0.75 \text{ Wb}}{0.25 \text{ s}}| = 3.0 \text{ V}. The magnitude of the induced EMF is 3.0 V.

Worked example 2

A straight conductor of length 25 cm moves at a constant speed of 8.0 m/s at right angles to a uniform magnetic field of flux density 40 mT. Calculate the EMF induced across the ends of the conductor.

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  1. Identify the formula for motional EMF: E=BLv\mathcal{E} = BLv, where B, L, and v are mutually perpendicular.
  2. Convert all quantities to SI units:
    • Magnetic flux density, B = 40 mT = 40×10−3T40 \times 10^{-3} T
    • Length of conductor, L = 25 cm = 0.25 m
    • Speed of conductor, v = 8.0 m/s
  3. Substitute the values into the formula: E=(40×10−3 T)×(0.25 m)×(8.0 m/s)\mathcal{E} = (40 \times 10^{-3} \text{ T}) \times (0.25 \text{ m}) \times (8.0 \text{ m/s})
  4. Calculate the result: E=(0.040)×(0.25)×(8.0) V\mathcal{E} = (0.040) \times (0.25) \times (8.0) \text{ V} E=0.080 V\mathcal{E} = 0.080 \text{ V}
  5. The induced EMF across the ends of the conductor is 0.080 V or 80 mV.
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