Physics - Electrostatics Question with Solution | TestHub

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 }}$