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?

Solution

$$B=0.25 T$$

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

$$\theta =30^o$$

$$F=Bil$$

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

$$\therefore (Bil) cos30^{o} =mg\sin 30$$

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