The current in a solenoid changes from 0.0A to 10.0A in 0.1 sec. If an average emf of 2V is induced in the circuit, the value of self-inductance is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.02 H
The average induced emf in self-induction is \( \varepsilon = L\left|\frac{\Delta I}{\Delta t}\right| \), so the self-inductance is the emf divided by the rate of change of current.
\( \displaystyle \frac{\Delta I}{\Delta t} = \frac{10\,\mathrm{A}-0}{0.1\,\mathrm{s}} = 100\,\mathrm{A\,s^{-1}} \)
\( \displaystyle L = \frac{\varepsilon}{\Delta I/\Delta t} = \frac{2\,\mathrm{V}}{100\,\mathrm{A\,s^{-1}}} = 0.02\,\mathrm{H} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
