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AS 91526 · 2024 paper · electromagnetic induction

Question 2(a) Achieved

Practises Inductors

Harry is experimenting with coils and magnets. Each part gives its own values.

An average EMF of 250 µV is induced in a coil when the current through it changes from 10.0 A to 6.00 A in 0.400 s.

Calculate the self-inductance of the coil.

The 2024 paper, page 4 ↗ Marking schedule, page 3 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

  1. 1 Which equation links an induced EMF to how fast the current changes?

    ε = −L ΔI/Δt. Only the size is wanted, so drop the minus sign.

  2. 2 What are ΔI and ε in base units?

    ΔI = 10.0 − 6.00 = 4.00 A, and ε = 250 µV = 250 × 10−6 V.

  3. 3 Rearrange for L and substitute.

    L = εΔt / ΔI = (250 × 10−6 × 0.400) / 4.00 = 2.50 × 10−5 H.
Answer 2.50 × 10−5 H (25.0 µH)

What the marker wanted

  • Achieved The self-inductance, 2.5 × 10−5 H.

In our words, from NZQA’s 2024 marking schedule.

✕

Common mistake

Using 250 V instead of 250 × 10−6 V, which gives 25.0 H. NZQA’s 2024 report singles out unit prefixes as badly handled that year.

How did it go?

Got it moves it further out. Review brings it back tomorrow.