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 ${}^{\prime }{\theta }^{\prime }$ with the vertical. If wires have mass $\lambda$ per unit length then, the value of $I$ is :
( $g=$ gravitational acceleration)

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

Answer 1

Answer: (A)

forces per unit length are taken
$\tan \theta =\frac{\frac{{\mu }_0{i}^2}{2\pi d}}{{\lambda }_{g.}}$
$i^2=\frac{\lambda .g.\sin \theta }{{\mu }_0.\cos \theta }(2\pi )d[d=2L\sin \theta ]$
$i=2\sin \theta \sqrt{\frac{\lambda g\pi L}{{\mu }_0\cos \theta }}$