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 paralllel long wire `CD` which is fixed in a horizontal plane and carries a steady current of `30 A`, as shown in figure. Show that when `AB` is slightly
depressed it executes simple harmonic motion. Find the period of oscillations.

Explanation

Let m be the mass per unit length of wire AB. At a height x
above the wire CD, magnetic force per unit length on wire AB
will be given by
`F_m=(mu_0)/(2pi) (i_1i_2)/x (upward)…(i)`

Weight per unit length of wire AB is
`f_g=mg (downwards)`
Here, m=mass per unit length of wire AB
At x=d, wire is in equilibrium, i.e.,
`F_m=F_g, or (mu_0)/(2pi) (i_1i_2)/d=mg`
When AB is depressed, x decrease therefore, `F_m` will increase,
while `F_g` remains the same. Let AB be displaced by dx downwards.
Differentiating EQ. (i) w.r.t.x, we get
`(dF_m)/(dx)=-(mu_0)/(2pi) (i_1i_2)/(x^2)dx=(-mu_0)/(2pi) (i_1i_2)/(d^2)dx implies dF_m=-((mg)/d)dx...(ii)`
i.e, restoring force, `dF_mprop-dx`
Hence, the motion of wire is simple harmonic.
From Eqs (ii) and (iii), we can write
`dF_m=-((mg)/d).dx (x=d)`
`:.` Acceleration of wire `a=(dF_m)/m=-(g/d).dx`
Hence, period of oscillation
`T=2pisqrt(|(dx)/a|)=2pi=sqrt((|"displacement"|)/(|"acceleration"|))`
or `T=2pisqrt(d/g)=2pi sqrt(0.01/9.8) or T=0.2s`