The current in a \( 10\,\mathrm{mH} \) coil increases uniformly from zero to one ampere in 0.01 s. What is the magnitude of the induced emf?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1\,\mathrm{V} \)
A changing current in a coil induces an emf in the same coil by self-induction, \( \varepsilon=-L\,dI/dt \). The current rises uniformly, so \( dI/dt \) is the total change in current divided by the time taken.
\( L=10\,\mathrm{mH}=10\times10^{-3}\,\mathrm{H} \)
\( \Delta I=1\,\mathrm{A} \)
\( \Delta t=0.01\,\mathrm{s} \)
The magnitude of the induced emf is:
\( \begin{aligned} |\varepsilon| &= L\,\frac{\Delta I}{\Delta t} \\ &= 10\times10^{-3}\,\mathrm{H}\times\frac{1\,\mathrm{A}}{0.01\,\mathrm{s}} \\ &= 1\,\mathrm{V} \end{aligned} \)