Q. A conducting rod of length $2l$ is rotating with constant angular speed $\omega$ about its perpendicular bisector. A uniform magnetic field $\overrightarrow{B}$ exists parallel to the axis of rotation. The e.m.f. induced between two ends of the rod is
A) $B\omega l^{2}$
B) $\frac{1}{2}B\omega l^{2}$
C) $\frac{1}{8}B\omega l^{2}$
D) Zero
Explanation
(d) Potential difference between
$O$ and $A$ is $V_{0} - V_{A} = \frac{1}{2}Bl^{2}\omega$
$O$ and $B$ is $V_{0} - V_{B} = \frac{1}{2}Bl^{2}\omega$
So $V_{A} - V_{B} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.