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 betwben the two ends of the rod is

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

Answer 1

Answer: (A)

Given $\to$ Arod of length $l$
angular velocity $\to \omega$
A magnetic field $B$ parallel to the axis of rotation.

We knaw that, $d\epsilon =Bdx\times \omega x$.
$d\epsilon =B\omega xdx$
Integrating both sides,
$\epsilon ={\int }_{-1{/}_2}^{1/2}B\omega xdx.$
$\epsilon =B\omega \underset{-1/2}{\overset{1/2}{\int }}xdx$
$\epsilon =B\omega {\left(\frac{x^2}{2}\right)}_{-1/2}^{4/2}.$
$\epsilon =\frac{B\omega \left[\frac{l^2}{2}-\frac{l^2}{4}\right]}{\epsilon =0}$
$\epsilon =0$
Hence potential difference between $2$ ends of the rod is zero.