Question

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

• A. $\frac{B{l}^2\omega }{2}$
• B. zero
• C. $\left(\frac{B{l}^2\omega }{8}\right)$
• D. $2B{l}^2\omega$

Answer 1

Answer: (B)

Length of the rod between the axis of rotation and one end of the rod $=\frac{l}{2}$
Area swept out in one rotation $=\pi (\frac{l}{2}{)}^2=(\frac{\pi l^2}{4})$
Angular velocity $=\omega rad{s}^{-1}$
Frequency of revolution $=\frac{\omega }{2\pi }$
Area swept out per second $=\frac{\pi l^2}{4}\left(\frac{\omega }{2\pi }\right)=\frac{l^2\omega }{8}$
Magnetic induction = $B$
Rate of change of magnetic flux $=\left(\frac{B{l}^2\omega }{8}\right)$
Magnitude of induced emf $=\left(\frac{B{l}^2\omega }{8}\right)$
Magnitude of induced emf between the axis and the other end is also $\left(\frac{B{l}^2\omega }{8}\right)$. These two emf’s are in opposite directions. Hence, the potential difference between the two ends of the rod is zero