Question

A long horizontal wire $AB$, which is free to move in a vertical plane and carries a steady current of $20A$, is on equilibrium at a height of $0.01m$ over another parallel long wire $CD$ which is fixed in a horizontal plane and carries a current of $30A$ as shown in figure. Show that when $AB$ is slightly depressed, it executes simple harmonic motion. The period of oscillations is $\frac{x}{10}sec$. Find x.

Answer 1

Let $m$ be the mass per unit length of wire $AB$. At a height $x$ above the wire $CD$, the magnetic force per unit length on wire $AB$ will be given by:
$F_m=\frac{{\mu }_0}{2\pi }\frac{i_1{i}_2}{d}$ (upward) $......\left(i\right)$
Weight per unit length of wire $AB$ is, $F_g=mg$
Here, $m=$ mass per unit length of wire $AB$
At $x=d$, wire equilibrium, i.e,
$F_m=F_g$

$\frac{{\mu }_0}{2\pi }\frac{i_1{i}_2}{d}=mg$
When $AB$ is depressed, $x$ decreases.

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:
$\frac{d{F}_m}{dx}=\frac{{\mu }_0}{2\pi }\frac{i_1{i}_2}{x^2}dx=\frac{{-\mu }_0}{2\pi }\frac{i_1{i}_2}{d^2}dx\Rightarrow d{F}_m=-\left(\frac{mg}{d}\right)dx.....\left(ii\right)$
i.e, Restoring force, $d{F}_m\propto -dx$
Hence, the motion of wire is simple harmonic.
From Eqs. $\left(ii\right)$ and $\left(iii\right)$ we can write:
$d{F}_m=-\left(\frac{mg}{d}\right)dx$ $(x=d)$
Acceleration of wire, $a=\frac{d{F}_m}{m}=-\left(\frac{g}{d}\right)\cdot dx$
Hence, period of oscillation:
$T=2\pi \sqrt{\left|\frac{dx}{a}\right|}=2\pi \sqrt{\frac{\left|displacement\right|}{\left|acceleration\right|}}$
$T=2\pi \sqrt{\frac{d}{g}}=2\pi \sqrt{\frac{0.01}{9.8}}$
$T=0.2s$