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 ${30}^o$ with the horizontal. The rod is not allowed to slide down by a 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. $7.14$A
• C. $11.32$A
• D. $5.98$A

Answer 1

Answer: (C)

$B=0.25T$

$\frac{m}{l}=0.5\frac{kg}{m}$

$\theta ={30}^o$

$F=Bil$

$F\cos 30$ balances $mg\sin 30$

$\therefore \left(Bil\right)cos{30}^o=mg\sin 30$

$\Rightarrow i=\frac{m}{l}\frac{g}{B}\frac{\sin 30}{\cos 30}=\frac{0.5\times 9.8}{0.25\times 866}\times \frac{1}{2}=11.32A$