Two long current carrying thin, wires, both with current I, are held by insultating of length L and are in equilibrium as shown in the figure, with threads as shown in the figure, with threads making an angle `theta` with the vertial. If the wires have mass `lambda` per unit lenght then the value of I is (g =gavitational acclearation)

Options

A. `sin theta sqrt((pi lambda gL)/(mu_(0) cos theta))`
B. `2sin theta sqrt((pi lambda gL)/(mu_(0) cos theta))`
C. `sqrt((pi gL)/(mu_(0)) cos theta)`
D. ` sqrt((pi lambda gL)/(mu_(0)) tan theta)`

Explanation

The current I is flowing in opposite direction in the two wires.

Therefore, for repulsive force pern unit length of the wires,
`F_(mu_(0))/(4pi)*(2l^(2))/(2x)=(mu_(0))/(4pi)=(l^(2))/(L sin theta)[ x L sin theta]`
Weight of the wire per unit lenght`=lambda g`.
If T is the tension in the string, in equilibrium condition,
`T sin theta F and T cos theta= lambda g`
i.e., `tan theta=(F)/(lambdag)=(mu_(0)I^(2))/(4pi lambda g L sin theta)`
or `I=sqrt((4pi lambda gL sin^(2) theta)/(mu_(0) cos theta))=2 sin theta sqrt((pi lambda gL)/( mu_(0) cos theta)) [ :. sin theta tan theta=(sin^(2) theta)/(cos theta)]`
The option B is correct.