Question

A metallic rod of mass per unit length $0.5kg{m}^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of ${30}^o$ with the horizontal. The rod is not allowed to slide down by a passing a current through it when a magnetic field of induction $0.25T$ is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

• A. $14.76A$
• B.
  $7.14A$
• C.
  $11.32A$
• D.
  $5.98A$

Answer 1

The correct option is C

$11.32A$

$B=0.25T$
$\frac{m}{l}=0.5\frac{kg}{m}$
$\theta ={30}^o$
$F=Bil$
$F\cos {30}^o=mg\sin {30}^o$
$\therefore \left(Bil\right)cos{30}^o=mg\sin {30}^o$
$\Rightarrow i=\frac{m}{l}\frac{g}{B}\frac{\sin {30}^o}{\cos {30}^o}$

$=\frac{0.5\times 9.8}{0.25\times 0.866}\times \frac{1}{2}=11.32A$