Q. A copper disc of radius 0.1 m is rotated about its centre with 20 $rev - s^{- 1}$ in a uniform magnetic field of 0.1 T with its plane perpendicular to the field. The emf induced across the radius of the disc is
A) $\frac{\pi}{20}V$
B) $\frac{\pi}{10}V$
C) $20\pi\ mV$
D) $10\pi\ mV$
Explanation
(c) From Faraday’s law of electromagnetic induction
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = - \frac{d\phi}{dt} = - BAN$
Given, $\ B = 0.1\ T,\ N = 20,\ A = \pi r^{2} = \pi(0.1)^{2}$
$\therefore\ \ \ \ \ e = - 0.1 \times 20 \times \pi(0.1)^{2} = 20\pi\ mV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.