JEE Main 2026 — Electromagnetic Induction Question with Solution

The magnetic flux linked with the circular loop is given by $\Phi =BA\cos \theta$. Since the plane of the loop is perpendicular to the magnetic field, the angle between the area vector and the magnetic field is $\theta =0^{\circ }$.

$\Phi =B\cdot \pi r^2$

Given $r=20\text{ cm}=0.2\text{ m}$, the area of the loop is:

$A=\pi (0.2{)}^2=0.04\pi {\text{ m}}^2$

The induced emf in the loop is given by Faraday's law of induction:

$|E|=\frac{d\Phi }{dt}=A\frac{dB}{dt}$

Given the magnetic field $B=2t^2+2t+3$, we differentiate it with respect to time $t$:

$\frac{dB}{dt}=\frac{d}{dt}(2t^2+2t+3)=4t+2$

At $t=3\text{ s}$, the rate of change of the magnetic field is:

$\frac{dB}{dt}=4(3)+2=14\text{ T/s}$

Now, substituting the values into the emf equation:

$|E|=0.04\pi \times 14=0.56\pi \text{ V}$

The induced current $I$ in the loop is:

$I=\frac{|E|}{R}=\frac{0.56\pi }{2}=0.28\pi \text{ A}$

Using $\pi =\frac{22}{7}$, we get:

$I=0.28\times \frac{22}{7}=0.04\times 22=0.88\text{ A}$

We are given that the induced current is $\frac{\alpha }{50}\text{ A}$. Equating the two values:

$\frac{\alpha }{50}=0.88$

$\alpha =0.88\times 50=44$

Answer: $44$