Question

A metallic rod of length $l$ is placed normal to the magnetic field $B$ and revolved in a circular path about one of the ends with angular frequency $\omega$. The potential difference across the ends will be :

• A. $\frac{1}{2} B^{2} l \omega$
• B. $\frac{1}{2} B \omega l^{2}$
• C. $\frac{1}{8} B \omega l^{2}$
• D. $B \omega l^{2}$

Answer 1

The Correct Option is B

Suppose a conducting rod of length $ l $ rotates with a constant angular speed $\omega$ about a point at one end.
A uniform magnetic field $\vec{B}$ is directed perpendicular to the plane of rotation as shown in figure. Consider a segment of rod of length drat a distance r from O. This segment has a velocity $v=r\omega.$

The induced emf in this segment is
$d e=B v d r=B(r \omega) d r$
Summing the emfs induced across all segments, which are in series, gives the total emf across the rod.
$\therefore e=\int\limits_{0}^{l} d e=\int\limits_{0}^{l} B r \omega d r=\frac{B \omega l^{2}}{2}$
$\therefore e=\frac{B \omega l^{2}}{2}$
From right hand rule we can see that P is at higher potential than O.
Thus, $V_{P}-V_{O}=\frac{B \omega l^{2}}{2}$