Questions
A horizontal rod of mass 10 g and length 10 cm is placed on a smooth plane inclined at an angle of 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
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=mgiltanθTalk to our academic expert!
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A conductor PQ, carrying a current i is placed perpendicular to a long conductor xy carrying a current i. The direction of force on PQ will be
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