Q.
A square loop of side $a$ is rotating about its diagonal with angular velocity $\omega$ in a perpendicular magnetic field $\overrightarrow{B}$. It has 10 turns. The $e.m.f$. induced is
A) $B_{a}^{2}\sin{\omega t}$
B) $B_{a}^{2}\cos{\omega t}$
C) $5\sqrt{2}\ Ba^{2}$
D) $10\ Ba^{2}\omega\sin{\omega t}$
Explanation
(d) $\phi = NBA\cos{\theta = 10\ Ba^{2}}\cos{\omega t}$
$e = - \frac{d\phi}{dt} = - \frac{d}{dt}\left( 10\ Ba^{2}\cos{\omega t} \right) = 10\ Ba^{2}\omega\sin{\omega t}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.