Two long parallel wires carry currents of equal magnitude but in opposite directions. These wires are suspended from rod $PQ$ by four chords of same length $L$ as shown in the figure. The mass per unit length of the wire is $\lambda$. Determine the value of $\theta$ assuming it to be small.

The force per unit length between current carrying parallel wires is
$\frac{dF}{dL}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$
If two wires carry current in opposite directions the magnetic force is repulsive, due to which the parallel wires in $Fig. \left(a\right)$ have moved out so that equilibrium is reached. $Fig \left(b\right)$ shows free body diagram of each wire in equilibrium.

$\Sigma {\text{F}}_{\text{y}}=0,$   $2 \text{T}\cos \theta =\left(\lambda {\text{L}}_0\right)\text{g}$             ……….$\left(1\right)$

$\Sigma {\text{F}}_2=0,$   $2 \text{T}\sin \theta ={\text{F}}_{\text{B}}$                     ………..$\left(2\right)$

Now dividing equation $\left(2\right) by \left(1\right)$, we get

$\tan \theta =\frac{F_B}{{\lambda }_0{\lambda }_g}$                          ……….$\left(3\right)$

The magnetic force $F_B=\left(\frac{dF}{dL}\right)\times L_0=\frac{{\mu }_0{\text{I}}^2}{4\pi \sin \theta }{\text{L}}_0$           ……….$\left(4\right)$

$For small\theta , \tan \theta \simeq \sin \theta =\theta$

On substituting equation $\left(4\right) in \left(3\right)$, we get

$\theta =I\sqrt{\frac{{\mu }_0}{4\pi \lambda \mathrm{gL}}}.$