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 in figure. Potential difference between the two ends of the rod is proportional to

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

Answer 1

Answer: (A), (D)

The correct options are
A $l^{3/2}$
D ${\left(\sin \theta \right)}^{1/2}$

Since the gravitational potential decreases to increase the kinetic energy of the rod,

$mg\frac{l}{2}sin\theta =\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2$

$⟹\omega =\sqrt{\frac{3g}{l}sin\theta }$

Since, the magnetic field and the direction of velocity of rods is perpendicular.

Total potential difference developed across the ends $={\int }_0^{x}B\left(x\omega \right)dx$ $=\frac{B{l}^2}{2}\sqrt{\frac{3g}{l}sin\theta }$ $\propto l^{3/2}\sqrt{sin\theta }$.