A coil has an inductance of 0.02 H and the current is increasing at the rate of 200 A/s. The emf induced in it, is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -4\,\mathrm{V} \)
A changing current in a coil induces a self-induced emf, and the negative sign is Lenz's law: the induced emf opposes the increase of current.
\( \displaystyle \varepsilon = -L\,\frac{dI}{dt} \)
\( \displaystyle \varepsilon = -0.02\,\mathrm{H} \times 200\,\mathrm{A\,s^{-1}} = -4\,\mathrm{V} \)
The induced emf is negative because the current is increasing.
Note: The question does not say whether the sign or only the magnitude of the induced emf is wanted. The answer given here is the signed value, −4 V, which shows that the emf opposes the rise of current. If only the magnitude is meant, Option (3), +4 V, also fits. When a negative and a positive value of the same size both appear among the options, choose the signed value, because it also shows the direction of the induced emf.