Q. A circular metal plate of radius $R\ $is rotating with a uniform angular velocity$\ \omega\ $with its plane perpendicular to a uniform magnetic field B. Then the emf developed between the centre and the rim of the plate is
A) $\pi\omega BR^{2}$
B) $\omega BR^{2}$
C) $\pi\omega BR^{2}/2$
D) $\omega BR^{2}/2$
Explanation
(d) From Faraday’s law of electromagnetic induction, the emf induced between center and rim is equal to rate of change of magnetic flux.
[image:6_image63.emf]
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = - \frac{d\phi}{dt}$
Where, $d\phi = B\ dA,$where $B$ is magnetic field and $dA$ the area.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = - \frac{B\int_{0}^{R}{dA}}{T}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = - \frac{B \times \pi R^{2}}{T}$
Also, $\omega = \frac{2\pi}{T},$ where $T$ is periodic time,
$\ \ \ \ \ \ \ \ \ \ \ \ e = - \frac{B\pi R^{2}}{2\pi/\omega}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ - \frac{BR^{2}\omega}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.