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

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 $=\left(\frac{Bl^{2}\,\omega}{8}\right)$
Magnitude of induced emf $=\left(\frac{Bl^{2}\,\omega}{8}\right)$
Magnitude of induced emf between the axis and the other end is also $\left(\frac{Bl^{2}\,\omega}{8}\right)$. These two emf’s are in opposite directions. Hence, the potential difference between the two ends of the rod is zero