Q. A conducting circular loop is placed in a uniform magnetic field of induction B tesla with its plane normal to the field. Now, the radius of the loop starts shrinking at the rate $\left( \frac{dr}{dt} \right)$. Then, the induced emf at the instant when the radius is $r,$is
A) $\pi rB\left( \frac{dr}{dt} \right)$
B) $2\pi rB\left( \frac{dr}{dt} \right)$
C) $\pi r^{2}\left( \frac{dB}{dt} \right)$
D) $\left( \frac{\pi r^{2}}{2} \right)B\left( \frac{dr}{dt} \right)$
Explanation
(b) Induced emf is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = - \frac{d\phi}{dt}$
If the radius of loop is $r$ at a time $t$, then the instantaneous magnetic flux is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \phi = \pi r^{2}B$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = \ - \frac{d}{dt}(\pi r^{2}B)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = \ - \pi B\left( \frac{2r\ dr}{dt} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = \ - 2\pi\ Br\frac{dr}{dt}$
Numerically, $e = 2\pi Br\left( \frac{dr}{dt} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.