Question

A horizontal rod of mass $10g$ and length $10$ $cm$ is placed on a smooth plate inclined at an angle of ${60}^0$ with the horizontal and 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 :
(Given $g=10m/s^2$)

• A. $1.73$ $T$
  
• B. $\frac{1}{1.73}$ $T$
• C. $1$ $T$
  
• D. $0.5$ $T$

Answer 1

The correct option is C $1$ $T$

Rod remains stationary only when component of magnetic force $=$ Component of weight
$Bilcos60=mgsin{60}^{\circ }$
$B\times 1.73\times {10}^{-1}\times \frac{1}{2}=10\times {10}^{-3}\times 10\times \frac{\sqrt{3}}{2}$
$B=1Tesla$