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JEE Main 20245 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2024 · 5 April · Evening Shift · Physics · Electromagnetic Induction

The current in an inductor is given by I=(3 t+8) where t is in...

The current in an inductor is given by I=(3t+8)\mathrm{I}=(3 \mathrm{t}+8) where t\mathrm{t} is in second. The magnitude of induced emf produced in the inductor is 12mV12 \mathrm{mV}. The self-inductance of the inductor _____ mH\mathrm{mH}.

Correct answer

0

Step-by-step solution

I=3t+8ε=12mV∣ε∣=L∣dIdt∣12=L×3 L=4mH\begin{aligned} & \mathrm{I}=3 \mathrm{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}\end{aligned}
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