A horizontal rod of mass 10g and length 10cm is placed on a smooth plate inclined at an angle of 600 with the horizontal and 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, then the value of B for which the rod remains stationary on the inclined plane is : (Given g=10m/s2)
1.73T
11.73T
1T
0.5T
A
0.5T
B
1.73T
C
1T
D
11.73T
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Solution
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Rod remains stationary only when component of magnetic force = Component of weight Bilcos60=mgsin60∘ B×1.73×10−1×12=10×10−3×10×√32 B=1Tesla
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