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Questions  

A horizontal rod of mass 10 g and length 10 cm is placed on a smooth plane inclined at an angle  of  600 with the horizontal with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field of induction B is applied vertically downwards. If the current through the rod is 1.73 ampere, the value of B for which the rod remains stationary on the inclined plane is

a
1.73 tesla
b
(1/1.73) tesla
c
1 tesla
d
0.5 tesla

detailed solution

Correct option is C

force on the conductor due magnetic field IlB along horizontal direction and its component along the plane is ilB cosθ which when balances mgsinθ it will be in equilibriummgsinθ = ilBcosθ    B=mgiltanθ

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