Two long currents carrying thin wires, both with current $I$, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle ' $\theta$ ' with the vertical. If wires have a mass $\mathrm{\lambda }$ per unit length then the value of $\mathrm{I}$ is:
($g=$ gravitational acceleration)

• A

  $\sqrt{\frac{\mathrm{\pi }\lambda gL}{{\mathrm{\mu }}_0}\mathrm{t}\mathrm{a}\mathrm{n}\theta }$
• B

  $\mathrm{s}\mathrm{i}\mathrm{n}\theta \sqrt{\frac{\mathrm{\pi }\mathrm{\lambda }gL}{{\mathrm{\mu }}_{0\mathrm{ }}\mathrm{c}\mathrm{o}\mathrm{s}\theta }}$
• Correct answer
  C

  $2\mathrm{s}\mathrm{i}\mathrm{n}\theta \sqrt{\frac{\mathrm{\pi }\mathrm{\lambda }gL}{{\mathrm{\mu }}_0\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{\theta }}}$
• D

  $2\sqrt{\frac{\mathrm{\pi }gL}{{\mathrm{\mu }}_0}\mathrm{t}\mathrm{a}\mathrm{n}\theta }$

Solution

Two wires will repel each other due to magnetic force, then the magnetic force per unit length is, $\frac{\text{d}\mathit{\text{f}}}{\text{d}\mathit{\text{l}}}=\frac{{\mu }_0{\mathit{\text{I}}}^2}{2\pi \left(2\mathit{\text{L}}\text{sin}\theta \right)}=\frac{{\mu }_0{\mathit{\text{I}}}^2}{4\pi \mathit{\text{L}}\text{sin}\theta }$. And mass per unit length of each wire$=\frac{\text{d}\mathit{\text{m}}}{\text{d}\mathit{\text{l}}}=\lambda$. So, the magnetic force on the total length $L$ of the wire is ${\mathit{\text{f}}}_{\text{m}}=\frac{{\mu }_0{\mathit{\text{I}}}^{2}L}{4\pi \mathit{\text{L}}\text{sin}\theta }$, and weight $=\lambda L\text{g}$. By equilibrium of wire,  $\mathit{\text{T}}\text{sin}\theta ={\mathit{\text{f}}}_m\text{\hspace{0.17em} \& }\mathit{\text{T}}\text{cos}\theta \text{=}\text{W\hspace{0.17em}}\text{= }\lambda l\text{'}g$$\Rightarrow \frac{\mathit{\text{T}}\text{sin}\theta }{\mathit{\text{T}}\text{cos}\theta }=\frac{{\mathit{\text{f}}}_{\text{m}}}{\mathit{\text{mg}}}\Rightarrow {\mathit{\text{f}}}_{\text{m}}=\lambda l\text{'}gtan\theta$
$\Rightarrow \frac{{\mu }_0{\text{I}}^2}{4\pi \text{L}sin\theta }L=\lambda L\text{g}\frac{\text{sin}\theta }{\text{cos}\theta }$

$\Rightarrow {\mathit{\text{I}}}^2=\frac{\lambda \text{g}\pi \text{L}}{{\mu }_0\text{cos}\theta }4{\text{sin}}^2\theta$$\text{⇒}\mathit{\text{I}}=2\text{sin}\theta \sqrt{\frac{\lambda \pi \text{gL}}{{\mu }_0\text{cos}\theta }}$

Hence, the correct answer is option: 'C'.