JEE Main 2015 Physics, Electrostatics — question figure Two long currents carrying thin wires, both with current I, are held by insulating threads of…

Two wires will repel each other due to magnetic force, then the magnetic force per unit length is, $\frac{\text{d}\text{f}}{\text{d}\text{l}}=\frac{{\mu }_{0}{\text{I}}^{2}}{2\pi (2\text{L}\text{sin}\theta )}=\frac{{\mu }_{0}{\text{I}}^{2}}{4\pi \text{L}\text{sin}\theta }$dldf​=2π(2Lsinθ)μ0​I2​=4πLsinθμ0​I2​. And mass per unit length of each wire$=\frac{\text{d}\text{m}}{\text{d}\text{l}}=\lambda$=dldm​=λ. So, the magnetic force on the total length $L$L of the wire is ${\text{f}}_{\text{m}}=\frac{{\mu }_{0}{\text{I}}^{2}L}{4\pi \text{L}\text{sin}\theta }$fm​=4πLsinθμ0​I2L​, and weight $=\lambda L\text{g}$=λLg. 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$\Rightarrow \frac{\text{T}\text{sin}\theta }{\text{T}\text{cos}\theta }=\frac{{\text{f}}_{\text{m}}}{\text{mg}}\Rightarrow {\text{f}}_{\text{m}}=\lambda l'gtan\theta$⇒TcosθTsinθ​=mgfm​​⇒fm​=λl′gtanθ

$$\Rightarrow \frac{{\mu }_{0}{\text{I}}^{2}}{4\pi \text{L}sin\theta }L=\lambda L\text{g}\frac{\text{sin}\theta }{\text{cos}\theta }$$⇒4πLsinθμ0​I2​L=λLgcosθsinθ​

$\Rightarrow {\text{I}}^{2}=\frac{\lambda \text{g}\pi \text{L}}{{\mu }_{0}\text{cos}\theta }4{\text{sin}}^{2}\theta$⇒I2=μ0​cosθλgπL​4sin2θ$\text{⇒}\text{I}=2\text{sin}\theta \sqrt{\frac{\lambda \pi \text{gL}}{{\mu }_{0}\text{cos}\theta }}$⇒I=2sinθμ0​cosθλπgL​​