On a smooth plane inclined at `30^(@)` with the horizontal, a thin current carrying metallic rod is placed parallel to the horizontal ground. The plane is in a uniform magnetic field of `0.15T` alon the vertical direction. For what value of current can the rod remain stationary ? the mass per unit lenght of the rod is `0.30`kg/m.

Explanation

Along the inclined plane, component of the magnetic force acting on the conductor and the component of the weight of the conductor will bring equilibrium. So,
`Bil cos theta= m g sin theta " "...(1)`
Here, force on the conductor BlI act horizontallly toward right.
Now, from equation (1)
`I=(mg sin theta)/(Bl cos theta)=(mg)/(Bl) tan theta`
`:. (m)/(l)=0.30 kg//m, g=9.8 m//s^(2)`
`B=0.15T, theta=30^(@)`
`:. I=(0.30xx9.8)/(0.15) tan 30^(@)`
`=2xx9.8xx(1)/(sqrt(3))=11.3A`