Question

A rod $\text{OPQ}$ is rotating with angular speed $\omega$ about point $O$ as shown in the figure. A uniform inward magnetic field $B$ exists perpendicular to the plane of the rod. Potential difference between points $P$ and $Q$ of the rod is -

• A. $\frac{B\omega l^2}{2}$
• B. $B\omega l^2$
• C. Zero
• D. $\frac{B\omega l^2}{8}$

Answer 1

The correct option is C Zero

When a rod of length $l$ is rotated with angular velocity $\omega$ in a perpendicular magnetic field $B$, then an emf is induced across its ends and its value is given by,

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


From the figure, the potential difference between points $P$ and $O$ is,

$E_P-E_O=\frac{B\omega l^2}{2}\Rightarrow E_P=\frac{B\omega l^2}{2}+E_O$

The effective length between points $O$ and $Q$ is $l$.

So, the potential difference between points $Q$ and $O$ is,

$E_Q-E_O=\frac{B\omega l^2}{2}\Rightarrow E_Q=\frac{B\omega l^2}{2}+E_O$

Now, the potential difference between points $P$ and $Q$ is,

$E_P-E_Q=(\frac{B\omega l^2}{2}+E_O)-(\frac{B\omega l^2}{2}+E_O)=0$

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