A metallic rod of mass per unit length 0.5 kg m–1 is lying… | NEET 2018

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Force on Current-Carrying Conductors

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

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Answer: (d)

Vertical $B$ and horizontal rod give a horizontal force $F = I l B$ .
Along the smooth incline: $F\cos\theta = mg\sin\theta \Rightarrow I = \frac{(m/l)\,g\tan\theta}{B}$
$I = \frac{0.5\times9.8\times\tan30^\circ}{0.25} = \frac{0.5\times9.8\times0.577}{0.25}$
$I \approx 11.32\text{ A}$ , so the answer is (d).

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