Q. A conducting rod $AC$ of length $4l$ is rotated about a point $O$ in a uniform magnetic field $\overrightarrow{B}$ directed into the paper. $AO = l$ and $OC = 3l$. Then
A) $V_{A} - V_{O} = \frac{B\omega l^{2}}{2}$
B) $V_{O} - V_{C} = \frac{7}{2}B\omega l^{2}$
C) $V_{A} - V_{C} = 4\ B\omega l^{2}$
D) $V_{C} - V_{O} = \frac{9}{2}B\omega l^{2}$
Explanation
(c) By using $e = \frac{1}{2}Bl^{2}\omega$
For part $AO;\ e_{OA} = e_{O} - e_{A} = \frac{1}{2}Bl^{2}\omega$
For part $OC;\ e_{OC} = e_{O} - e_{C} = \frac{1}{2}B(3l)^{2}\omega$
$\therefore e_{A} - e_{C} = 4\ Bl^{2}\omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.