Q 12-06-050JEE MainJEE Main 2024 (5 Apr, Shift 2)Easy
The current in an inductor is given by $I = (3t + 8)$, where $t$ is in seconds. The magnitude of the induced emf produced in the inductor is $12\ \text{mV}$. The self-inductance of the inductor is ______ mH.
Numerical value type. Enter your answer.
Answer: 4
$\varepsilon = L\dfrac{dI}{dt} = 3L = 12\ \text{mV} \Rightarrow L = 4\ \text{mH}$.
Solution by Sreeraj P, M.Sc Physics