Question

A neutral metal rod AB of length $l$ rotates with a constant angular velocity $\omega$ about an axis passing through O and normal to its length as shown in the figure.The potential difference between A & B will be :
($e$ is the charge of an electron)

• A. $V_A-V_B=\frac{1}{3}\frac{m{\omega }^2{l}^2}{e}$
• B. $V_A-V_B=\frac{m{l}^2{\omega }^2}{e}$
• C. $V_B-V_A=\frac{1}{2}\frac{m{\omega }^2{l}^2}{e}$
• D. $V_B-V_A=\frac{1}{4}\frac{m{\omega }^2{l}^2}{e}$

Answer 1

Answer: (D)

$V_B-V_A=\frac{1}{4}\frac{m{\omega }^2{l}^2}{e}$

Work done in taking a point charge from A to B is same as the change in Kinetic Energy from A to B .
Velocity at A, $r\omega =\frac{\omega l}{4}$
Velocity at B, $r\omega =\frac{3\omega l}{4}$
$e(V_B-V_A)=\frac{1}{2}m{\omega }^2(\frac{9l^2}{16}-\frac{l^2}{16})$
Hence $V_B-V_A=\frac{m{\omega }^2{l}^2}{4e}$