A current carrying conductor of mass m, length I 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=50 gm,. `g=10m//s^(2),l=1/2m,i=1A` and B=1 T and k=50 N/m

Explanation

The forces acting on the rod are .mg. downwards, `F_(mag)=ilB` downwards and `F_("spring")=2kx` upwards
Under the action of these forces the rod is in equilibrium. Then, `F_("net") = 0`
or mg+ilB=2kx
or `x=(mg+ilB)/(2k)`
`=((1/20)(10)+(1)(1/2)(1))/(2xx50)`
`=1/100m`