question_answer

Two long current carrying thin wires, both with current I, and 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:                                                       [JEE MAIN 2015]

(g = gravitational acceleration)

A) \[2\sqrt{\frac{\pi gL}{{{\mu }_{0}}}\tan \theta }\]

B) \[\sqrt{\frac{\pi \lambda gL}{{{\mu }_{0}}}\tan \theta }\]

C) \[\sin \theta \sqrt{\frac{\pi \lambda gL}{{{\mu }_{0}}\cos \theta }}\]

D) \[2\sin \theta \sqrt{\frac{\pi \lambda gL}{{{\mu }_{0}}\cos \theta }}\]

Correct Answer: D

Solution :

[d]

\[T\sin \theta =\frac{{{\mu }_{0}}{{I}^{2}}x}{4\pi L\sin \theta }\]

\[T\cos \theta =\lambda xg\]

\[T\cos \theta =\frac{{{\mu }_{0}}{{I}^{2}}}{4\pi L\sin \lambda g}\Rightarrow I=2\sin \theta \sqrt{\frac{\pi \lambda gL}{{{\mu }_{0}}\cos \theta }}\]