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NEET PHYSICSEasy

The current in an inductor of self-inductance 4 H4 \text{ H} changes from 4 A4 \text{ A} to 2 A2 \text{ A} in 1 s1 \text{ s}. The emf induced in the coil is:

A

-2 V

B

2 V

C

-4 V

D

8 V

Step-by-Step Solution

The induced electromotive force (emf) E\mathcal{E} in an inductor is given by Faraday's law of induction: E=−LdIdt\mathcal{E} = -L \frac{dI}{dt}.

  1. Identify Given Values: Self-inductance (LL) = 4 H4 \text{ H} Initial Current (IiI_i) = 4 A4 \text{ A} Final Current (IfI_f) = 2 A2 \text{ A} Time interval (Δt\Delta t) = 1 s1 \text{ s}

  2. Calculate Rate of Change of Current: dIdt=If−IiΔt=2−41=−2 A/s\frac{dI}{dt} = \frac{I_f - I_i}{\Delta t} = \frac{2 - 4}{1} = -2 \text{ A/s}

  3. Calculate Induced EMF: E=−(4 H)×(−2 A/s)\mathcal{E} = - (4 \text{ H}) \times (-2 \text{ A/s}) E=8 V\mathcal{E} = 8 \text{ V}.

The positive sign indicates the direction opposes the change in current (Lenz's Law), but the magnitude is 8 V.

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