The current in a coil falls from \( 10.0\,\mathrm{A} \) to \( 0.0\,\mathrm{A} \) in \( 0.1\,\mathrm{s} \). If an average emf of \( 100\,\mathrm{V} \) is induced in the coil, then the self-inductance of the coil is.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1\,\mathrm{H} \)
The emf induced by a changing current is given by the self-inductance of the coil, \( \varepsilon = L\,|\Delta I/\Delta t| \) (Faraday's law applied to the coil's own flux).
\( \displaystyle \left|\frac{\Delta I}{\Delta t}\right| = \frac{10.0 - 0.0}{0.1} = 100\,\mathrm{A\,s^{-1}} \)
\( \displaystyle L = \frac{\varepsilon}{|\Delta I/\Delta t|} = \frac{100}{100} = 1\,\mathrm{H} \)
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 – …
