A current carrying conductor of mass `m`, length `1 ` carrying a current `i` hangs by two identical springs each of stiffness `k`. For an outward magnetic field `B` find the deformation of the springs. Put `m=50gm.g=10m//s^(2),l=1//2m,i=1A` and `B=1T` and `k=50N//m`

Explanation

The forces acting on the rod are `'mg'` downwards, `F_(mg)=ilB` downwards and `F_(spri ng)=2kx` upwards
Under the action of these forces the rod is in equilibrium. Then, `F_(n et)=0(or)`
`mg+ilB=2kx(or)x=(mg+ilB)/(2k)`
`=(((1)/(20))(10)+(1)((1)/(2))(1))/(2xx50)=(1)/(200)m=0.5cm`