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
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a
1.73 tesla
b
(1/1.73) tesla
c
1 tesla
d
0.5 tesla
answer is C.
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Detailed Solution
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