question_answer

4 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 : (g = gravitational acceleration) [JEE Main Solved Paper-2015 ]

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 :

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