Physics · Electromagnetic Induction

JEE Main 2024 — 6 April, Shift 2 — Question 39

In a coil, the current changes form -2 A to +2 A in 0.2 s and induces an emf of 0.1 V . The selfinductance of the coil is :

Answer: A

$\quad(\text { Emf })_{\text {induced }}=-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$

In magnitude form,

$\left|\operatorname{Emf}_{\text {ind }}\right|=\left|(-) \mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}\right|$

$\Rightarrow 0.1=\frac{(\mathrm{L})[+2-(-2)]}{0.2}$

$\Rightarrow \quad \mathrm{L}=\frac{0.1 \times 0.2}{4}=5 \mathrm{mH}$

Answer key and solution verified before publishing.

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