Discussion
If the current in a coil of inductance 10 mH is increased from 1 A to 5 A in0.2 s, then the emf produced in it is
Answer: E
The emf is 0.2 V.
The induced emf has magnitude \(L\frac{\Delta I}{\Delta t}\).
\[ \begin{aligned} e &= 10 \times 10^{-3} \times \frac{5 - 1}{0.2} \\ &= 0.01 \times 20 = 0.2\ \text{V} \end{aligned} \]
Using the final current 5 A instead of the change 4 A gives 0.25 V, and dropping the milli prefix gives 200 V; neither is offered. The emf opposes the change in current (Lenz's law), so it acts against the increase while the current rises.
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