Question

A horizontal wire $AB$ which is free to move in a vertical plane and carries a steady current is in equilibrium at a height $d$ over another parallel long wire $CD$, which is fixed in a figure. Show that when $AB$ is slightly depressed and released, it executes simple harmonic motion. Find the period of oscillation.

Answer 1

Before disturbance [ref. image 2]

When it is slightly depressed [ref. image 3]

Restoring force

$|F_R|=\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi (d-x)}-\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi l}$

$=\frac{{\mu }_0{I}_1{I}_{2}l(d-(d-x\left)\right)}{2\pi d(d-x)}\simeq \frac{{\mu }_0{I}_1{I}_{2}lx}{2\pi d^2}$

$\Rightarrow |F|\propto x$

$\Rightarrow$ Particle executes SHM

The solution has been clearly explained in the image.

Some point to note are:

The force on two parallel conducting wires is repulsive and is given by $\mu l\frac{I_1{I}_2}{2\pi d}$

Restoring force is only provided by magnetic force, therefore only these are taken into account.