ExamBro
ExamBro
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.
Verified Solution

Answer & Solution

Correct Answer

(B) \(4\)

Step-by-step Solution

Detailed explanation

\(\mathrm{I}=3 \mathrm{t}+8\)

\(\varepsilon=12 \mathrm{mV}\)

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

\(12=\mathrm{L} \times 3\)

\(\mathrm{~L}=4 \mathrm{mH}\)

Same subject
Explore more questions on app