A horizontal rod of mass 0.01kg and length 10 cm is placed on a frictionless plane inclined at an angle 60° with the horizontal and with the length of rod parallel to the edge of the inclined plane. A uniform magnetic field is applied vertically downwards. If the current through the rod is 1.73 A, then the value of magnetic field induction B for which the rod remains stationary 'on the inclined plane is
A)1 T
B)3 T
C)2.5 T
D)4 T
Correct Answer:
A
Solution :
Here two forces are acting on the rod simultaneously. From the hypothetical free body diagram \[{{E}_{0}}.\] \[{{E}_{0}}.\] \[\Delta {{I}_{C}}=1mA={{10}^{-3}}A\] B = 1T