Question

Two long current 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 mass per unit length of wire is $\lambda$, then the value of $I$ is ($g$ = gravitational acceleration)

• A. $2\sqrt{\frac{\pi gL}{{\mu }_0}tan\theta }$
• B. $2sin\theta \sqrt{\frac{\pi \lambda gL}{{\mu }_{0}cos\theta }}$
• C. $sin\theta \sqrt{\frac{\pi \lambda gL}{{\mu }_{0}cos\theta }}$
• D. $\sqrt{\frac{\pi \lambda gL}{{\mu }_0}tan\theta }$

Answer 1

The correct option is B $2sin\theta \sqrt{\frac{\pi \lambda gL}{{\mu }_{0}cos\theta }}$


Consider a free body diagram wire.
When the wire is in equilibrium position, it will carry current in opposite direction.
$F_B=\frac{{\mu }_o{I}^{2}L}{2\pi d}$
Where $L$ is the length of each wire and $d$ is the separation between the wires.

From the diagram, we can get $d$ = $2Lsin\theta$
$Tcos\theta =mg=\lambda Lg$-----(1)
(In the vertical direction)
$Tsin\theta =F_B=\frac{{\mu }_0{I}^{2}L}{4\pi Lsin\theta }$ ----(2)
(In horizontal direction)

From (1) and (2):
$\frac{Tsin\theta }{Tcos\theta }=\frac{{\mu }_0{I}^{2}L}{4\pi Lsin\theta \lambda Lg}I=\sqrt{\frac{4\pi \lambda Lg{sin}^2\theta }{{\mu }_{0}cos\theta }}=2sin\theta \sqrt{\frac{\pi \lambda gL}{{\mu }_{0}cos\theta }}$