1) A horizontal rod of mass 0.01 kg and length 10 cm is placed on a friction less plane inclined at an angle 60° with trhe horizontal and with the length of rod parallel to the edge of the inclined plane. A uniform magnetic fiewld 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 D) 4 T Answer: Option AExplanation: Here two forces acting on the rod simultaneously From FBD, mg sing 60 = Bil cos 60° $B=\frac{mg}{il}\tan 60^{0}=\frac{0.01\times10}{173\times0.1}\times\sqrt{3}=1T$
