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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
Explanation
Balance $mg\sin\theta$ with the component $IlB\cos\theta$ of the horizontal magnetic force.
Detailed Solution
For equilibrium along the incline: $mg\sin30^\circ = IlB\cos30^\circ$
$I = \frac{mg}{lB}\tan30^\circ = \frac{0.5\times9.8}{0.25\times\sqrt{3}}$
$= 11.32$ A