Q95 / 120 · VITEEE 2017 · Physics

The current in a coil of $\mathrm{L}=40 \mathrm{mH}$ is to be increased uniformly from 1 A to $11 \mathrm{~A}$ in 4 milli sec. The induced e.m.f. will be

Correct answer
A
$100 \mathrm{~V}$
B
$0.4 \mathrm{~V}$
C
$440 \mathrm{~V}$
D
$40 \mathrm{~V}$

Solution

$\mathrm{e}=\frac{\mathrm{LdI}}{\mathrm{dt}}=\frac{40 \times 10^{-3}(11-1)}{4 \times 10^{-3}}=100 \mathrm{~V}$

Hence, the correct answer is option: 'A'.

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