QJEEM23Phc78b: JEE Main 2023 Physics - Class 12 JEEM23Phc78b Solution: A conducting circular loop is placed in a uniform magnetic…

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:

$\Phi_B = B A = B \pi r^2$

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 $10^{-3} \mathrm{~m/s}$ , which means that the rate of change of the area of the loop is:

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

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 $\mu \mathrm{V}$ .