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 $30^o$ 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.....$A$
NEET 2018, Medium
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For equilibrium,
$m g \sin 30^{\circ}=I l B \cos 30^{\circ}$
$\mathrm{I}=\frac{\mathrm{mg}}{\mathrm{lB}} \tan 30^{\circ}$
$=\frac{0.5 \times 9.8}{0.25 \times \sqrt{3}}=11.32 \,\mathrm{A}$
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