Consider I₁ and I₂ are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L₁ = self inductance of coil 1, M₁₂ = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be :

Solution

Physics sir · 4 steps

Coil 1 feels two flux changes: its own current and coil 2's current. Add both, each with a minus sign.

1. Flux through coil 1. Coil 1 is linked by flux from its own current and from the current in coil 2. So Φ₁ = L₁ I₁ + M₁₂ I₂.
2. Apply Faraday's law. The induced emf is e₁ = -dΦ₁/dt. The minus sign is Lenz's law, and it acts on the whole flux, so it acts on both terms.

   e₁ = -L₁dI₁/dt - M₁₂dI₂/dt

3. Match each term. The self term pairs L₁ with coil 1's own current I₁. The mutual term pairs M₁₂ with the other coil's current I₂.
4. So. The induced emf in coil 1 is e₁ = -L₁dI₁/dt - M₁₂dI₂/dt, which is option (d).

The idea to keep. Write the total flux through a coil first, self part plus mutual part, then put one minus sign in front of its rate of change. Self inductance goes with the coil's own current, mutual with the other coil's.

Answer from the NTA final answer key. Solution by PrepWiser.