JEE Main 2023 — Electromagnetic Induction Question with Solution

The problem involves a conducting circular loop placed in a uniform magnetic field with its plane perpendicular to the field. The radius of the loop is expanding at a constant rate, and we are asked to find the magnitude of the induced emf in the loop at an instant when the radius of the loop is $$2\mathrm{cm}$$.

The magnetic flux through a circular loop of radius $$r$$ and area $$A=\pi r^2$$ placed in a uniform magnetic field $$B$$ perpendicular to the plane of the loop is given by:

$${\Phi }_B=BA=B\pi r^2$$

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

$$\mathcal{E}=-\frac{d{\Phi }_B}{dt}$$

In this case, the radius of the loop is expanding at a constant rate of $$10^{-3}m/s$$, which means that the rate of change of the area of the loop is:

$$\frac{dA}{dt}=\frac{d}{dt}(\pi r^2)=2\pi r\frac{dr}{dt}=2\pi (0.02\mathrm{m})(10^{-3}m/s)=4\times 10^{-5}{\mathrm{m}}^2/s$$

The magnetic flux through the loop is changing at this rate, and the induced emf in the loop is given by:

$$\mathcal{E}=\left|\frac{d{\Phi }_B}{dt}\right|=\left|\frac{dB}{dt}\frac{dA}{dt}\right|=\left|B\frac{dA}{dt}\right|=\left|0.4\mathrm{T}\times 4\times 10^{-5}{\mathrm{m}}^2/s\right|=16\pi \mu \mathrm{V}$$

Therefore, the magnitude of the induced emf in the loop at an instant when the radius of the loop is $$2\mathrm{cm}$$ is $$50.24$$ $$\simeq$$ 50 $\mu \mathrm{V}$.