A metallic rod of mass per unit length is lying horizontally on a smooth inclined plane which makes an angle of with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction is acting on

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Solution

Q.
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 [2018]

1 7.14 A

2 5.98 A

3 14.76 A

4 11.32 A

Ans.
(4)

Mass per unit length of a metallic rod is $\frac{m}{l} = 0.5$ $kg$ $m^{- 1} .$

Let $I$ be the current flowing.

For equilibrium, $m g \sin 30 ° = I l B \cos 30 °$

$I = \frac{m g}{l B} \tan 30 ° = \frac{0.5 \times 9.8}{0.25 \times \sqrt{3}} = 11.32$ $A$

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