question_answer

           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