A conducting ring of radius 1 m is placed in a uniform magnetic field B of 0.01 T oscillating with frequency 100 Hz with its plane at right angle to B. What will be the induced electric field?

From Faraday's law of electromagnetic induction, the induced emf is equal to negative rate of change of magnetic flux.

That is $\mathrm{e}=-\frac{\mathrm{\Delta \varphi }}{\mathrm{\Delta t}}$

Flux induced $=2\mathrm{BA} \mathrm{cos\theta }$

where B is magnetic field, A is area.

$\text{ Given, }\mathrm{\theta }=0^{\circ }, \mathrm{\Delta t}=\frac{1}{100} \mathrm{s}$

$\mathrm{\Delta \varphi }=2\times 0.01\times \mathrm{\pi }\times (1{)}^2\times 200\times \cos 0^{\circ }$

$\therefore e=\frac{-2\times 0.01\times \pi \times (1{)}^2\times 200}{100}=-4\pi \text{ volt }$

Circumference of a circle of radius $\mathrm{r}\text{ is }2 \mathrm{\pi r}\text{. }$

$\therefore$   Induced electric field E is

$\mathrm{E}=\frac{\left|\mathrm{e}\right|}{2\mathrm{\pi r}}=\frac{4\mathrm{\pi }}{2\mathrm{\pi r}}=\frac{2}{1}=2\mathrm{V}/\mathrm{m}$