Question

A rod of length I 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)

Zero

Answer 2

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

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

Angular velocity $= \omega rads^{- 1}$

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 $= \frac{Bl^{2}\omega}{8}$

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 direction hence the potential difference between the two ends of the rod is zero.