JEE Main 2024 — Electromagnetic Induction Question with Solution

The induced emf ($\epsilon$) in an inductor is given by Faraday's law of electromagnetic induction, which in its differential form for an inductor can be expressed as:

$$\epsilon =L\frac{dI}{dt}$$

where:

• $\epsilon$ is the induced emf in the inductor,
• $L$ is the inductance of the inductor,
• $\frac{dI}{dt}$ is the rate of change of current through the inductor.

Given that the current $I=(3t+8)$, where $t$ is in seconds, we can find the rate of change of current by differentiating $I$ with respect to $t$.

$$\frac{dI}{dt}=\frac{d}{dt}(3t+8)=3$$

The given magnitude of induced emf is $12\text{mV}=12\times 10^{-3}\text{V}$ (since $1\text{mV}=10^{-3}\text{V}$).

Now, plug these values into the formula to find $L$:

$$12\times 10^{-3}=L\cdot 3$$

Solving for $L$ gives:

$$L=\frac{12\times 10^{-3}}{3}=4\times 10^{-3}\text{H}=4\text{mH}$$

Therefore, the self-inductance of the inductor is 4 mH.