Physics · Electromagnetic Induction

JEE Main 2024 — 5 April, Shift 2 — Question 59

The current in an inductor is given by $I=(3 t+8)$ where $t$ is in second. The magnitude of induced emf produced in the inductor is 12 mV . The selfinductance of the inductor $\qquad$ mH .

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

$I=3 t+8$ $\varepsilon=12 \mathrm{mV}$

$|\varepsilon|=\mathrm{L}\left|\frac{\mathrm{dI}}{\mathrm{dt}}\right|$

$12=\mathrm{L} \times 3$

$\mathrm{L}=4 \mathrm{mH}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 2
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Self-Inductance and Mutual Inductance and Energy Density