Question

A conducting rod rotates with a constant angular velocity ${}^{\prime }{\omega }^{\prime }$ about the axis which passes through point ${}^{\prime }{O}^{\prime }$ and perpendicular to its length. A uniform magnetic field ${}^{\prime }{B}^{\prime }$ exists parallel to the axis of the rotation. Then potential difference between the two ends of the rod is :

• A. $6B\omega {\ell }^2$
• B. $B\omega {\ell }^2$
• C. $10B\omega {\ell }^2$
• D. $zero$

Answer 1

Answer: (D)

$zero$

Let us consider a small element $dx$ at a distance $x$ from the center of the rod rotating with angular velocity $\omega$ about its perpendicular bisector. The $emf$ induced in the small element due to its motion is given by

$de=B\omega xdx$

$emf$ induced between the center of the rod and one of its side is

$\begin{array}{c}de={\int }_0^{1}B\omega xdx\\ \Rightarrow e=B\omega {\left[\frac{x^2}{2}\right]}_0^{l/2}\\ \Rightarrow e=\frac{1}{8}B\omega l^2\end{array}$

The $emf$ at both ends is the same.

So, the potential difference between the two side is zero.

Hence option $D$ is correct answer.