A metallic rod of linear density is 0.25 kg `m^(-1)` is lylong horizontally on a smooth inclined plane which makes an angle of `45^(@)` with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of strength 0.25 T is acting on it in the vertical direction. Calculates the electric current flowing in the rod to keep it stationary .

Explanation

The linear density of the rod I mass per unti length of the rod is 0.25 kg `m^(-1)`
`rArr (m)/(l) = 0.25 kg m^(-1)` .
Let l be the current flowing in the matallic rod. The direction of electron current is into the paper . The direction of magnetic force IB/is given by Fleming.s left hand rule
For equilibrum ,
mg sin `45^(@) - IB/ cos 45^(@)`
`rArr 1 = (l)/(B) (m)/(l) g tan 45^(@) = (0.25 kg m^(-1))/(0.25 l ) = 1 xx 9.8 m s^(-2)`
`rArr` I = 9.8 A
so, we need to supply current of `9 xx` A to keep the metallic rod stationary.