Question

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 ${30}^{\circ }$ 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. $14.76A$
• B. $5.98A$
• C. $7.14A$
• D. $11.3A$

Answer 1

The correct option is D $11.3A$

$ilB\cos \theta =mg\sin \theta$

$i=\frac{mg}{lB}\tan \theta =\left(\frac{m}{l}\right)\frac{g\tan \theta }{B}$

$i=0.5\times \frac{9.8}{0.25}\times \frac{1}{\sqrt{3}}=11.3A$

Hence,

option $\left(D\right)$ is correct answer.