Q. 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
A) 14.76 A
B) 5.98 A
C) 7.14 A
D) 11.32 A
Explanation
[Solution] The horizontal magnetic force balances gravity along the incline: $BIl\cos 30^\circ = mg\sin 30^\circ$, so $I = \frac{(m/l)g\tan 30^\circ}{B} = \frac{0.5 \times 9.8 \times 0.577}{0.25} = 11.32$ A. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.