A rod of length $l$ rotates with a uniform angular velocity $\omega$ about an axis passing through its middle point but normal to its length in a uniform magnetic field of induction$B$ with its direction parallel to the axis of rotation. The induced emf between the two ends of the rod is

Length of the rod between the axis of rotation and one end of the $rod=\frac{l}{2}$

Area swept out in one rotation $=\pi {\left(\frac{l}{2}\right)}^2=\left(\frac{\pi l^2}{4}\right)$

$\because$ Frequency of revolution $=\frac{\omega }{2\pi }$

$\therefore$Area swept out per second $=\frac{\pi l^2}{4}\left(\frac{\omega }{2\pi }\right)=\frac{l^2\omega }{8}$

So rate of change of magnetic flux $=\frac{B{l}^2\omega }{8}$ i.e., Magnitude of induced emf $=\frac{B{l}^2\omega }{8}$

Magnitude of induced emf between the axis and the other end is also $\left(\frac{B{l}^2\omega }{8}\right)$. These two emfs have opposite polarity. Hence, the potential difference between the two ends of the rod is zero.