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_(mag)=ilB `downwards and spring `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) )/(2 xx 50) = 1/200 m = 0.5 cm`