Question 130 of 311

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 $\lambda$ per unit length then the value of $I$ is:
($g=$ gravitational acceleration)

A$\sqrt{\dfrac{\pi \lambda gL}{{\mu }_{0}}tan\theta }$B$sin\theta \sqrt{\dfrac{\pi \lambda gL}{{\mu }_{0}cos\theta }}$C$2sin\theta \sqrt{\dfrac{\pi \lambda gL}{{\mu }_{0}cos\theta }}$D$2\sqrt{\dfrac{\pi gL}{{\mu }_{0}}tan\theta }$

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Answer: (C) $2sin\theta \sqrt{\dfrac{\pi \lambda gL}{{\mu }_{0}cos\theta }}$

Two wires will repel each other due to magnetic force, then the magnetic force per unit length is, $\dfrac{\text{d}\text{f}}{\text{d}\text{l}}=\dfrac{{\mu }_{0}{\text{I}}^{2}}{2\pi \left(2\text{L}\text{sin}\theta \right)}=\dfrac{{\mu }_{0}{\text{I}}^{2}}{4\pi \text{L}\text{sin}\theta }$. And mass per unit length of each wire$=\dfrac{\text{d}\text{m}}{\text{d}\text{l}}=\lambda$. So, the magnetic force on the total length $L$ of the wire is ${\text{f}}_{\text{m}}=\dfrac{{\mu }_{0}{\text{I}}^{2}L}{4\pi \text{L}\text{sin}\theta }$, and weight $=\lambda L\text{g}$. By equilibrium of wire, \text{T}\text{sin}\theta ={\text{f}}_{m}\text{&}\text{T}\text{cos}\theta \text{=} \text{W}\text{=}\lambda l'g$\Rightarrow \dfrac{\text{T}\text{sin}\theta }{\text{T}\text{cos}\theta }=\dfrac{{\text{f}}_{\text{m}}}{\text{mg}}\Rightarrow {\text{f}}_{\text{m}}=\lambda l'gtan\theta$
$\Rightarrow \dfrac{{\mu }_{0}{\text{I}}^{2}}{4\pi \text{L}sin\theta }L=\lambda L\text{g}\dfrac{\text{sin}\theta }{\text{cos}\theta }$

$\Rightarrow {\text{I}}^{2}=\dfrac{\lambda \text{g}\pi \text{L}}{{\mu }_{0}\text{cos}\theta }4{\text{sin}}^{2}\theta$$\text{⇒}\text{I}=2\text{sin}\theta \sqrt{\dfrac{\lambda \pi \text{gL}}{{\mu }_{0}\text{cos}\theta }}$