24. A metallic rod of mass per unit length $0.5 kg m^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^\circ$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction $0.25 T$ is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is (2018)
Solution:
For the rod to remain stationary on the incline, the forces along the plane must balance: $m g \sin\theta = I l B \cos\theta$.
Rearranging gives $I = \left(\frac{m}{l}\right) \frac{g \tan\theta}{B}$.
With $\frac{m}{l} = 0.5 kg m^{-1}$, $g = 9.8 m s^{-2}$, $\theta = 30^\circ$, and $B = 0.25 T$: $I = \frac{0.5 \times 9.8 \times \tan(30^\circ)}{0.25} = \frac{4.9}{0.25 \sqrt{3}} \approx 11.32 A$.
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