The current changes in a coil from 10 A to 2 A in 0.2 s. The induced emf produced is 50 V. The self inductance of the coil is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.25 H
The emf induced in a coil by a changing current is given by the law of self induction, \( \varepsilon = -L\,\Delta I/\Delta t \). The minus sign only shows that the emf opposes the change (Lenz's law), so the magnitudes are used to find L.
\( |\Delta I| = |2 - 10|\,\mathrm{A} = 8\,\mathrm{A} \)
\( \displaystyle \frac{|\Delta I|}{\Delta t} = \frac{8\,\mathrm{A}}{0.2\,\mathrm{s}} = 40\,\mathrm{A\,s^{-1}} \)
\( \displaystyle L = \frac{|\varepsilon|}{|\Delta I/\Delta t|} = \frac{50\,\mathrm{V}}{40\,\mathrm{A\,s^{-1}}} = 1.25\,\mathrm{H} \)
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 – …
