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Q.

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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a

30A

b

15A

c

10A

d

3A

answer is A.

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Detailed Solution

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

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As shown in the figure, a metallic rod of linear density 0.45kg m−1 is lying horizontally on a smooth incline plane which makes an angle of 450 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 {Use g=10m/s2}