The average emf induced in a coil in which current changes from 2 A to 4 A in 0.05 sec is 8 V. The self-inductance of coil is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.2 H
The self-induced emf is proportional to the rate of change of current, \( \varepsilon = L\,\frac{\Delta I}{\Delta t} \) (magnitude).
\( \Delta I = 4\,\mathrm{A} - 2\,\mathrm{A} = 2\,\mathrm{A} \)
\( \displaystyle L = \frac{\varepsilon\,\Delta t}{\Delta I} = \frac{8\,\mathrm{V}\times0.05\,\mathrm{s}}{2\,\mathrm{A}} = 0.2\,\mathrm{H} \)
Note: The same question was also asked in CUET (UG) 2026 on 19-05-2026 (Slot 1).
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 – …
