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NEET Medium

A metallic rod of mass per unit length $0.5 \text{ kg m}^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction $0.25 \text{ T}$ is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

14.76 A

5.98 A

7.14 A

11.32 A

Step-by-Step Solution

For equilibrium, $mg \sin 30^{\circ} = IB \cos 30^{\circ}$ . Thus, $I = \frac{mg}{B} \tan 30^{\circ} = \frac{0.5 \times 9.8}{0.25 \times \sqrt{3}} = 11.32 \text{ A}$ .

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