The electric current flowing through an inductor having self-inductance \( 10 \) mH, is given by the relation: \( I = (t^{3} + 10t^{2} - 20) \) A. The emf induced in the inductor at \( t = 5 \) s is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( -1.75\,\mathrm{V} \)
A changing current in an inductor induces a self-induced emf \( \varepsilon = -L\,dI/dt \) that opposes the change.
\( \displaystyle \frac{dI}{dt} = \frac{d}{dt}(t^{3} + 10t^{2} - 20) = 3t^{2} + 20t \)
\( \text{At } t = 5\,\mathrm{s}: \ dI/dt = 3(25) + 20(5) = 175\,\mathrm{A\,s^{-1}} \)
\( L = 10\,\mathrm{mH} = 10^{-2}\,\mathrm{H} \)
\( \varepsilon = -L\,dI/dt = -(10^{-2})(175) = -1.75\,\mathrm{V} \)
The constant term \( -20 \) does not affect the emf, since only the rate of change of current matters.