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 horizontal position. When the rod makes an angle $\theta$ with the initial position, the potential difference between two ends of the rod is proportional to -

When the rod is in horizontal direction, the potential difference across its ends is zero.

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

Answer 1

The correct option is D ${\sin }^{1/2}\theta$


Let the angular velocity of the rod is $\omega$ when it makes an angle of $\theta$ with initial position.

Change in height of center of mass of the rod,

$h=\frac{l}{2}\sin \theta$

According to work energy theorem,

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

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


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

The emf induced in the rod when it makes an angle of $\theta$ with initial position,

$E=\frac{B\omega l^2}{2}$

$=\frac{B{l}^2}{2}\times \sqrt{\frac{3g\sin \theta }{l}}$

$\Rightarrow E\propto l^{3/2}{\sin }^{1/2}\theta$

Hence, option $\left(D\right)$ is the correct answer.