A conducting circular loop of radius \frac10\sqrt\pi cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is :

Electromagnetic Induction PYQ — JEE Main 2023 · 24 Jan · Shift 1 · Q54 with Solution

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Correct answer: A

Use Faraday’s law

For a single circular loop, the induced emf magnitude is

\mathcal{E} = \left|\frac{d\Phi}{dt}\right| = A\left|\frac{dB}{dt}\right|

since the loop is perpendicular to the magnetic field, so

\Phi = BA

and $\cos 0^\circ = 1$ .

Find the area of the circular loop

Given radius:

r = \frac{10}{\sqrt{\pi}}\text{ cm} = \frac{10}{\sqrt{\pi}}\times 10^{-2}\text{ m}

Area of the loop:

A = \pi r^2 = \pi\left(\frac{10}{\sqrt{\pi}}\times 10^{-2}\right)^2

A = \pi \cdot \frac{100}{\pi} \cdot 10^{-4} = 100\times 10^{-4} = 10^{-2}\text{ m}^2

So,

A = 0.01\text{ m}^2

Find the rate of change of magnetic field

The magnetic field decreases uniformly from $0.5\,\text{T}$ to $0$ in $0.5\,\text{s}$ .

Hence,

\left|\frac{dB}{dt}\right| = \frac{0.5 - 0}{0.5} = 1\,\text{T/s}

Because the decrease is at a steady rate, the induced emf is constant throughout the interval. Therefore at $t=0.25\,\text{s}$ also, the emf has the same value.

Calculate induced emf

\mathcal{E} = A\left|\frac{dB}{dt}\right| = 0.01 \times 1 = 0.01\text{ V}

0.01\text{ V} = 10\times 10^{-3}\text{ V} = 10\text{ mV}

Match with the options

\boxed{\mathcal{E} = 10\text{ mV}}

So the correct option is A.

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