Q. A copper disc of radius $0.1\ m$ is rotated about its centre with 10 revolutions per second in a uniform magnetic field of $0.1\ tesla$ with its plane perpendicular to the field. The e.m.f. induced across the radius of disc is
A) $\frac{\pi}{10}\ V$
B) $\frac{2\pi}{10}\ V$
C) $\pi \times 10^{- 2}V$
D) $2\pi \times 10^{- 2}V$
Explanation
(c) $e = \frac{1}{2}B\omega r^{2} = \frac{1}{2} \times 0.1 \times 2\pi \times 10 \times (0.1)^{2} = \pi \times 10^{- 2}\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.