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 wires have mass $\lambda$ per unit length 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= 2L$sin\theta$

$T=cos\theta =mg=\lambda lg$-----(1)

(In the vertical direction)

$Tsin\theta =F_B=\frac{{\mu }_{0}Il}{4\pi LSin\theta }$ ----(2)

(In horizontal direction)

From (1) and (2):

$\frac{TSin\theta }{Tcos\theta }=\frac{{\mu }_0{I}^{2}l}{4\pi LSin\theta }I=\sqrt{\frac{4\pi \lambda Lg{Sin}^2\theta }{{\mu }_{0}cos\theta }}=2Sin\theta \sqrt{\frac{\pi \lambda gl}{{\mu }_{0}Cos\theta }}$