JEE Main 2015 — Physics Electromagnetism

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​​