Q. 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
Explanation
(c) The given situation can be drawn as follows
$F = ilB \Rightarrow mg\sin{60{^\circ}} = ilB\cos{60{^\circ}}$
$\Rightarrow B = \frac{0.01 \times 10 \times \sqrt{3}}{0.1 \times 1.73} = 1T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.