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Q.

A metallic rod of mass per unit length 0.5 kgm–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

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a

7.14 A

b

5.98 A

c

14.76 A

d

11.32 A

answer is D.

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Detailed Solution

Mass per unit length of a metallic rod is ml = 0.5 kg m−1.Let I be the current flowing. For equilibrium,mgsin⁡30∘=IlBcos⁡30∘ ⇒I=mglBtan⁡30∘=0.5×9.80.25×3=11.32A
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