As shown in the figure, a metallic rod of linear density $0.45 k g m^{- 1}$ is lying horizontally on a smooth incline plane which makes an angle of $45^{0}$ with horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15T magnetic field is acting on it in the vertical upward direction, will be ${U s e g = 10 m / s^{2}}$

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30A

15A

10A

answer is A.

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Detailed Solution

$m g \sin 45 ° = I L B \cos 45 ° ∴ I = (\frac{m}{L}) \frac{g}{B} = \frac{(0.45) (10)}{0.15} = 30 A$

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