Q 12-06-063JEE MainJEE Main 2023 (29 Jan, Shift 1)Easy
A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field $B=0.8\ \text{T}$. When released the radius of the loop starts shrinking at a constant rate of $2\ \text{cm s}^{-1}$. The induced emf in the loop at an instant when the radius of the loop is $10\ \text{cm}$ will be ______ mV.
Numerical value type. Enter your answer.
Answer: 10
$\phi=B\pi r^2$, so
$$|\varepsilon|=B\cdot2\pi r\frac{dr}{dt}=0.8\times2\pi\times0.1\times0.02=1.0\times10^{-2}\ \text{V}\approx10\ \text{mV}$$
Solution by Sreeraj P, M.Sc Physics