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

Answer 1

The correct option is D $2\sin \theta \sqrt{\frac{\pi \lambda gL}{{\mu }_o\cos \theta }}$


Forces per unit length are considered
From figure, we get
$\tan \theta =\frac{\frac{{\mu }_o{i}^2}{2\pi d}}{\lambda .g}$
$\Rightarrow i^2=\frac{\lambda .g.\sin \theta }{{\mu }_o.\cos \theta }\left(2\pi \right)d\because [d=2L\sin \theta ]$

$\therefore i=2\sin \theta \sqrt{\frac{\lambda g\pi L}{{\mu }_o\cos \theta }}$