Question

A wooden stick of length $3 l$ is rotating about one end with constant angular velocity $\omega $ . A uniform magnetic field $B$ is present in space, perpendicular to the rotational plane. One third of its upper end is coated with copper. What will be the potential difference between the ends of the stick?

• A. $ \frac{9 B \omega l^{2}}{2}$
• B. $ \frac{4 B \omega l^{2}}{2}$
• C. $ \frac{5 B \omega l^{2}}{2}$
• D. $\frac{\text{Bω} l^{2}}{2}$

Answer 1

Answer: (C)

When the rod rotates, there will be an induced current in the rod. The given situation can be treated as if a rod ‘A’ of length $′ 3 l ′$ rotating in the clockwise direction, while an other say rod ‘B’ of length $′ 2 l ′$ rotating in the anti clockwise direction with same angular speed $ ′ \omega ′$ .
As we know that, $e = \frac{1}{2} \text{Bω} l^{2}$
For ‘A’ : For ‘B’
$e_{A} = \frac{1}{2} \text{Bω} \left(\right. 3 l \left.\right)^{2}$ & $e_{B} = \frac{1}{2} B \left(\right. - \omega \left.\right) \left(\right. 2 l \left.\right)^{2}$
Resultant induced emf will be:
$e = e_{A} + e_{B} = \frac{1}{2} \text{Bω} l^{2} \left(\right. 9 - 4 \left.\right)$ $e = \frac{5}{2} \text{Bω} l^{2}$