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 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. Findd the period of oscillations.

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

$F_m=\frac{{\mu }_0}{2\mathrm{\pi }}\frac{i_1{i}_2}{x}$ (upwards)     …….(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.,

 $$\begin{gathered} F_m=F_g \\ \Rightarrow \frac{{\mu }_0{i}_1{i}_2}{2\mathrm{\pi d}}=mg \\ \Rightarrow \frac{{\mu }_0{i}_1{i}_2}{2{\mathrm{\pi d}}^2}=\frac{mg}{d} \end{gathered}$$

When $AB$ is depressed therefore, $F_m$ will increase while $F_g$ remains the same.Change in magnetic force will become the net restoring force, let $AB$ is displaced by $dx$ downwards. 

Differentiating Eq. (i) w.r.t. $x$ we get

                                                  $d{F}_m=\frac{{\mu }_0}{2\pi }\frac{i_1{i}_2}{x^2}·dx$

i.e. restoring force, $F=d{F}_m\propto -dx$

Hence, the motion of the wire is simple harmonic.

From Eqs. (ii) and (iii), we can write 

                                           $d{F}_m= -\left(\frac{mg}{d}\right)·dx$

$\therefore$ Acceleration of wire, $a= -\left(\frac{g}{d}\right)·dx$

Hence, Period of oscillation 

$$\begin{gathered} T=2\mathrm{\pi }\sqrt{\frac{\left|\mathrm{displacement}\right|}{\left|\mathrm{acceleration}\right|}}=2\mathrm{\pi }\sqrt{\left|\frac{\mathrm{dx}}{\mathrm{a}}\right|} \\ \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{d}}{\mathrm{g}}}=2\mathrm{\pi }\sqrt{\frac{0.01}{9.8}} \\ \mathrm{T}=0.2 \mathrm{s} \end{gathered}$$