JEE Main 2023 — Electromagnetic Induction Question with Solution

From the graph, it can be seen that the value of the magnetic field at $t=2 \mathrm{s}$ is $4 \mathrm{mT}$, whereas the magnetic field at $t=4 \mathrm{s}$ is $8 \mathrm{mT}$.

The formula to calculate the magnitude of the induced emf is given by

$\begin{array}{rcl}{\epsilon }_{\text{ind }}& =& \frac{d\varphi }{dt}\\ & =& \frac{d}{dt}\left(BA\right)\\ & =& A\frac{dB}{dt} ...\left(1\right)\end{array}$

$\frac{dB}{dt}$ is the slope of the given graph.

Hence, the value of the induced emf can be calculated as follows:

$\begin{array}{rcl}{\epsilon }_{\text{ind }}& =& 4 {\mathrm{m}}^2\times \left(\frac{8 \mathrm{mT}-4 \mathrm{mT}}{4 \mathrm{s}-2 \mathrm{s}}\right)\\ & =& 8 \mathrm{mV}\end{array}$