Two coaxial circular coils share the same axis. Coil 1 has $N_1=250$ turns, coil 2 has $N_2=180$ turns and their mutual inductance is $M=0.07\text{H}$. The current in coil 1 varies as $I_1(t)=3\sin (2\pi \times 60t)\text{A}$.\n(i) Find the maximum magnitude of the emf induced in coil 2.\n(ii) At the instant when $I_1$ is increasing through zero, state the direction of the induced current in coil 2 relative to the direction of the current that would be produced in coil 1 by the same instantaneous emf.

Step-by-step solution

1. The induced emf in coil 2 is given by Faraday’s law for mutual inductance: ${\epsilon }_2=-M\frac{d{I}_1}{dt}$.
2. Differentiate $I_1(t)$: $\frac{d{I}_1}{dt}=3(2\pi \times 60)\cos (2\pi \times 60t)=360\pi \cos (2\pi \times 60t){\text{A s}}^{-1}$.
3. Hence ${\epsilon }_2=-M360\pi \cos (2\pi \times 60t)$. The magnitude is $|{\epsilon }_2|=M360\pi |\cos (2\pi \times 60t)|$.
4. Maximum occurs when $|\cos |=1$, so ${\epsilon }_{2\text{max}}=M360\pi =0.07\times 360\pi =25.2\pi \text{V}\approx 79.2\text{V}$.
5. For part (ii), when $I_1$ is increasing through zero, $\cos (2\pi \times 60t)=1$, so $d{I}_1/dt$ is positive. The magnetic flux linking coil 2 is increasing in the same sense as the primary field. By Lenz’s law the induced emf in coil 2 opposes this increase, therefore the induced current in coil 2 flows in the opposite direction to the primary current that would be produced by the same emf.

Concept tested: Mutual inductance and Lenz’s law

Published 21 September 2026 · How this solution was prepared and checked