A metallic rod of mass per unit length 0.5 kg m-1 is lying horizontally on a smooth inclined plane which makes an angle of 30 with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is
- 7.14 A
- 5.98 A
- 14.76 A
- 11.32 A
Answer
(4) 11.32 A
Solution
The rod is horizontal and B is vertical, so the magnetic force F = IlB is horizontal.
For equilibrium, components along the incline balance:
IlB 30 = mg 30
I = (m/l) g 30 B
= 0.5 9.8 1 30.25
= 11.32 A
Answer: (4)