Question

A rod of length $l$ rotates with a uniform angular velocity $\omega$ about its perpendicular bisector. A uniform magnetic field $B$ exists parallel to the axis of rotation. The potential difference between the two ends of the rod is

• A. $\text{Zero}$
• B. $\frac{1}{2}Bl{\omega }^2$
• C. $Bl{\omega }^2$
• D. $2Bl{\omega }^2$

Answer 1

The correct option is A $\text{Zero}$

Consider an element at a distance $x$ from the centre of the rod, as shown in the figure.

Emf induced across the element due to rotation,

$E=Bvl=B\left(\omega x\right)dx[\because v=\omega x]$

Now, the total emf developed across the rod or potential difference between the ends,

$E=V_Q-V_P={\int }_{-l/2}^{l/2}B\omega xdx$

$=B\omega {\int }_{-l/2}^{l/2}xdx=B\omega {\left[\frac{x^2}{2}\right]}_{-l/2}^{l/2}=\frac{B\omega }{2}[\frac{l^2}{4}-\frac{l^2}{4}]$

$\therefore V_Q-V_P=0$

Hence, $\left(\text{A}\right)$ is the correct answer.