JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2024
- Hard
- Numerical
The current in an inductor is given by \(\mathrm{I}=(3 \mathrm{t}+8)\) where \(t\) is in second. The magnitude of induced emf produced in the inductor is \(12 \mathrm{mV}\). The selfinductance of the inductor _______ \(\mathrm{mH}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
\(\mathrm{I}=3 \mathrm{t}+8\)
\(\varepsilon=12 \mathrm{mV}\)
\(|\varepsilon|=\mathrm{L}\left|\frac{\mathrm{dI}}{\mathrm{dt}}\right|\)
\(12=\mathrm{L} \times 3\)
\(\mathrm{~L}=4 \mathrm{mH}\)
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