A coil of self-inductance 0.25 H carries a current of 4 A. The current is then reduced at a steady rate for 0.05 s. The average self-induced emf during this time is 15 V. The current in the coil at the end of this time is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(1\ \mathrm{A}\)
When the current in a coil changes, an emf is induced in the same coil. Its size is \( L\,|dI/dt| \). For a steady change, \( dI/dt \) is the change in current divided by the time taken. We find this change first. The current falls, so we subtract the change from the starting current.
Given:
\( L=0.25\ \mathrm{H} \)
\( I_1=4\ \mathrm{A} \)
\( \Delta t=0.05\ \mathrm{s} \)
\( \varepsilon=15\ \mathrm{V} \)
\( \begin{aligned} \Delta I &= \frac{\varepsilon\,\Delta t}{L} \\ &= \frac{15\times0.05}{0.25} \\ &= \frac{0.75}{0.25} \\ &= 3\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} I_2 &= I_1-\Delta I \\ &= 4-3 \\ &= 1\ \mathrm{A} \end{aligned} \)
NCERT (2026-27):
Self-inductance – “… When the current is varied, the flux linked with the coil also changes and an emf is induced in the coil. … the induced emf is given by”
\( \displaystyle \varepsilon=-\frac{d(N\Phi_B)}{dt} \)
\( \displaystyle \varepsilon=-L\frac{dI}{dt} \)\( \qquad (6.14) \)