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When the current changes from +2 A to -2 A in 0.05 s, an emf of 8 V is induced in a coil. The coefficient of self-induction of the coil is

1. 0.2 H
2. 0.4 H
3. 0.8 H
4. 0.1 H

Correct answer: 0.1 H

Solution

The current changes from +2 A to -2 A, so the total change is $\Delta I = 4$ A. Using $e = L\,\frac{\Delta I}{\Delta t}$, we get $8 = L\times \frac{4}{0.05} = 80L$. Hence $L = 0.1$ H.

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