Q. A copper rod of length $l$ is rotated about one end perpendicular to the magnetic field $B$ with constant angular velocity $\omega$. The induced e.m.f. between the two ends is
A) $1/2\ B\omega l^{2}$
B) $3/4\ B\omega l^{2}$
C) $B\omega l^{2}$
D) $2B\omega l^{2}$
Explanation
(a) If in time $t$, the rod turns by an angle $\theta$, the area generated by the rotation of rod will be $= \frac{1}{2}l \times l\theta = \frac{1}{2}l^{2}\theta$
So the flux linked with the area generated by the rotation of rod
$\phi = B\left( \frac{1}{2}l^{2}\theta \right)\cos 0 = \frac{1}{2}Bl^{2}\theta = \frac{1}{2}Bl^{2}\omega t$
And so $e = \frac{d\phi}{dt} = \frac{d}{dt}\left( \frac{1}{2}Bl^{2}\omega t \right) = \frac{1}{2}Bl^{2}\omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.