A metallic rod of mass per unit length $0.5 ext{ kg m}^{-1}$ is lying horizontally on a smooth inclined plan

Question

A metallic rod of mass per unit length $0.5 	ext{ kg m}^{-1}$ 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 $0.25 	ext{ T}$ is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

Accepted Answer

For equilibrium, $mg \sin 30^{\circ} = IB \cos 30^{\circ}$. Thus, $I = \frac{mg}{B} 	an 30^{\circ} = \frac{0.5 	imes 9.8}{0.25 	imes \sqrt{3}} = 11.32 	ext{ A}$.

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