A long horizontal wire `AB` which is free to move in a vertical plane and carries a steady current of `20 A`,is in equilibrium at a height of `0.01 m` over another parallel infinitely long wire `CD`,which is fixed in a horizontal plane and carries a steady current of `30 A`.Show that when `AB` is slightly depressed, it executes simple harmonic motion.the period of oscillations is `1/N` seconds.then find the value of `N`.(Use `pi^(2)=10 & g=10 m//s^(2)`)

Explanation

At equilibrium
`mg=(mu_(0)i_(1)i_(2))/(2pid)`...(1)
For a short displacement `x`
`F_(mg)=(mu_(0)i_(1)i_(2))/(2pid(d+x))-mg`
`=(mgd)/(d+x)-mg =mg[d/(d+x)-1]`
`=mg[(d-d-x)/(d+x)]`
`F_(mg)=(-mgx)/(d+x)`
`a=-g/(d)x`For very small `x`
`T=2pisqrt(d/g)`
`T=2pisqrt(0.1/g)`
`=(2pixx0.1)/pi`
`T=0.2=1/5sec`.