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

7.14 A

5.98 A

11.32 A

14.76 A

Solution

11.32 A

For equilibrium,

$m g \sin 30^{o} = I / B \cos 30^{o} \Rightarrow I = \frac{m g}{l B} \tan 30^{o} = \frac{0.5 x 9.8}{0.25 x \sqrt{3}} = 11.32 A$

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