Question

A conducting rod of length $l$ is hinged at point $O$. It is free to rotate in a vertical plane. 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)

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

An emf induced in a conductor of length $l$, rotating with $\omega$ in a magnetic field $B$ is given by,

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



When the rod is rotated by an angle $\theta$, its centre of mass is lowered by,

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

Using, energy conservation Principle

Initial Total Energy = Final Total Energy

$\Rightarrow \frac{1}{2}I{\omega }^2=mgh$

$\Rightarrow \frac{1}{2\times 3}m{l}^2{\omega }^2=mg\frac{l}{2}\sin \theta$

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

$\therefore E=\frac{B\omega l^2}{2}=\frac{B\sqrt{\frac{3\sin \theta }{l}}.l^2}{2}$

$\Rightarrow E=\frac{\sqrt{3}}{2}B{l}^{3/2}(\sin \theta {)}^{1/2}$

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

Hence, options $\left(A\right)$ and $\left(D\right)$ are correct.