The current in an inductor is given by I=(3 t+8) where t is in second. The magnitude of induced emf produced in the inductor is 12 mV. The self-inductance of the inductor _________ mH.

Solution

Physics sir · 3 steps

The emf depends on how fast the current changes, so differentiate I and the constant 8 drops out.

1. Rate of change. The induced emf depends on dI/dt, not on I itself. Differentiating I = 3t + 8 gives dI/dt = 3 A s⁻¹, and this is the same at every instant.

   |ε| = L dI/dt

2. Solve for L. L = |ε|/(dI/dt) = (12 × 10⁻³ V)/(3 A s⁻¹) = 4 × 10⁻³ H.
3. So. The self-inductance is 4 mH, so the answer is 4.

The common slip: 36. Multiplying the emf by the rate instead of dividing gives 12 × 3 = 36, not 4. Rearrange |ε| = L dI/dt to L = |ε| ÷ dI/dt.

The idea to keep. An inductor reacts only to change. Differentiate the current first, then use |ε| = L dI/dt. Any constant term in I has no effect on the emf.

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