Chapter 29 Problem 73

Problem

A thin metal rod of length $l$ rotates with angular velocity $\omega$ about an axis through one end. The rotation axis is perpendicular to the rod and is parallel to a uniform magnetic field ${\vec {B}}$ . Determine the emf developed between the ends of the rod

Solution

$d{\mathcal {E}}=Bvdr=B\omega rdr$

${\mathcal {E}}=\int d{\mathcal {E}}=\int _{0}^{l}B\omega rdr={\frac {1}{2}}B\omega l^{2}$