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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° 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

14.76 A
5.98 A
7.14 A
11.32 A
Explanation:

For equilibrium,
mg sin30° = I/B cos 30°
$I=\dfrac{mg}{lB}tan 30°$

$\dfrac{0.5\times 9.8}{0.25 \times\sqrt{3}}= 11.32A$
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