JEE Main 2023 — Electromagnetic Induction Question with Solution

The induced emf in the circuit is given by Faraday's law of electromagnetic induction, which is $\mathcal{E}=-d\varphi /dt$, where $\varphi$ is the magnetic flux through the circuit.

The magnetic flux through the circuit is proportional to the magnetic field through the core, so we can write $\varphi =NBA$, where $N$ is the number of turns in the loop, $B$ is the magnetic field through the core, and $A$ is the cross-sectional area of the core.

As the magnetic field changes from $1.5\mathrm{T}$ in one direction to $-1.5\mathrm{T}$ in the opposite direction, the change in magnetic flux is $\Delta \varphi =2NBA$.

The induced emf drives a current $I$ through the resistor in the circuit, and the current and the resistance are related by Ohm's law, which is $I=\mathcal{E}/R$. Substituting the expression for $\mathcal{E}$ into this equation, we get $I=-d\varphi /dtR$.

The charge $Q$ that flows through the circuit during the change in magnetic field is given by $Q=\int Idt$. Substituting the expression for $I$ into this equation and integrating with respect to time, we get $Q=-\Delta \varphi /R$, where $\Delta \varphi$ is the change in magnetic flux and $R$ is the resistance of the circuit.

Substituting the given values into this expression, we get:

$$Q=-\frac{2NBA}{R}=-\frac{2(100)(1.5)(24\times 10^{-4})}{12}=-0.06\mathrm{C}=-60\mathrm{mC}$$

Therefore, the charge flowing through a point in the circuit during the change of magnetic field is $60\mathrm{mC}$, which is the same as the provided answer.