Home / iGCSE Physics (0625) 4.5.1 Electromagnetic induction Paper 3 -Exam Style Questions- New Syllabus

iGCSE Physics (0625) 4.5.1 Electromagnetic induction Paper 3 -Exam Style Questions- New Syllabus

Question

(a) Fig. 9.1 shows a magnet spinning steadily near to a coil. There is a reading on the voltmeter.
Explain why there is a reading on the voltmeter.
(b) A step-down transformer reduces the mains voltage of $220\text{ V}$ a.c. to $12\text{ V}$ a.c. The transformer has $5000$ turns on the primary coil. Calculate the number of turns on the secondary coil.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

Topic $4.5.1$ — Electromagnetic induction (Part $\mathrm{(a)}$)
Topic $4.5.6$ — The transformer (Part $\mathrm{(b)}$)

▶️ Answer/Explanation

(a)
For the correct answer:
Induced e.m.f. (in the coil) due to a changing magnetic field (in the coil).

As the magnet spins, its magnetic field lines continuously cut through the stationary wire coil. This continuous movement creates a changing magnetic field linking with the coil. According to the principles of electromagnetic induction, this changing magnetic field induces an electromotive force (e.m.f.) or voltage across the coil’s ends. The voltmeter is connected across these ends, so it detects this induced e.m.f. and displays a reading.

(b)
For the correct calculated value:
$270$

To find the number of turns on the secondary coil, we use the transformer equation $\frac{V_p}{V_s} = \frac{N_p}{N_s}$. We are given the primary voltage $V_p = 220\text{ V}$, secondary voltage $V_s = 12\text{ V}$, and primary turns $N_p = 5000$. Rearranging the equation to solve for $N_s$ yields $N_s = \frac{V_s \times N_p}{V_p}$. Substituting the values gives $N_s = \frac{12 \times 5000}{220} = 272.72$. Rounding this to two significant figures yields exactly $270$ turns.

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