question_answer

 A horizontal rod of mass 10 gm and length 10 cm is placed on a smooth plane inclined at an angle of \[60{}^\circ \] 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, then the value of B for which the rod remains stationary on the inclined plane is

A)            1.73 Tesla                              

B)            \[\frac{1}{1.73}\] Tesla

C)            1 Tesla                                    

D)            None of the above

Correct Answer: C

Solution :

                   The given situation can be drawn as follows \[F=ilB\] \[\Rightarrow mg\sin {{60}^{o}}=ilB\cos {{60}^{o}}\] \[\Rightarrow B=\frac{0.01\times 10\times \sqrt{3}}{0.1\times 1.73}=1\ T\]