Question

In the figure shown the wires $\text{AB}$ and $\text{PQ}$ carry constant currents $I_1$ and $I_2$ respectively. $\text{PQ}$ is of uniformly distributed mass $m$ and length $l$. $\text{AB}$ and $\text{PQ}$ are both horizontal and kept in the same vertical plane. The $\text{PQ}$ is in equilibrium at height $h$, If the wire $\text{PQ}$ is displaced vertically by small distance, then its time period of oscillation in terms of $h$ and $g$

• A. $2\pi \sqrt{\frac{g}{h}}$
• B. $2\pi \sqrt{\frac{2h}{g}}$
• C. $\pi \sqrt{\frac{h}{2g}}$
• D. $2\pi \sqrt{\frac{h}{g}}$

Answer 1

The correct option is D $2\pi \sqrt{\frac{h}{g}}$

At equilibrium,

Magnetic repusive force balances the weight.

$\frac{{\mu }_0{I}_1{I}_2}{2\pi h}l=mg\Rightarrow h=\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi mg}.....\left(1\right)$

Let the wire be dispaced downward by a distance $x$ $(x<<h)$

Magnetic force on it will increase, so it goes back towrds its equilibrium position. Hence, it performs oscillations.

$F_{res}=\frac{{\mu }_0{I}_1{I}_2}{2\pi (h-x)}l-mg$

From eq $\left(1\right)$ we can write,

$\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi }=mgh$

$\Rightarrow F_{res}=\frac{mgh}{h-x}-mg=\frac{mg(h-h+x)}{h-x}$

$F_{res}=\frac{mg}{h-x}x=\frac{mg}{h}x(\because x<<h)$

comparing with eqn of SHM we have,

${\omega }^2=\frac{mg}{h}$

Also, the time period is, $T=\frac{2\pi }{\omega }$

or, $T=2\pi \sqrt{\frac{m}{\left(\frac{mg}{h}\right)}}=2\pi \sqrt{\frac{h}{g}}$

Hence, option $\left(D\right)$ is the correct answer.