Question

A conducting rod of length $l$ is hinged at point O. It is free to rotate in a vertical plane. There exists a uniform magnetic field $\overrightarrow{B}$ in horizontal direction. The rod is released from the position shown. The potential difference between the two ends of the rod is proportional to:

• A. $l^{3/2}$
• B. $l^2$
• C. $\sin \theta$
• D. $(\sin \theta {)}^{1/2}$

Answer 1

Answer: (A), (D)

$(\sin \theta {)}^{1/2}$
$l^{3/2}$

$V=\frac{B}{2}{R}^2\omega$

where $\omega =\sqrt{\left(\frac{2}{I}\right)\left(mg\frac{l}{2}\sin \theta \right)}$

$\omega \alpha \sqrt{\frac{\sin \theta }{l}}$

$I=k{l}^2$ ($I$- moment of intertia)

$\therefore V\alpha l^2\times \sqrt{\frac{\sin \theta }{l}}$

$V\alpha l^{\frac{3}{2}}$

and $V\alpha \sqrt{\sin \theta }$