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 `'theta'` with the vertical. If wires have mass `lambda` per unit length then the values of I is: (g = gravitational acceleration)

Options

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

Answer

D

Explanation

Let the length of right wire be l, then its mass is `lamdal`.

Forces acting on this wire are tension (T), weight `(lamda l g)` and force of repulsion due to other wire (F).
From figure `T cos theta =lamda l g `...............i
`T sin theta=F `................ii
Here `F=(mu_(0))/(2pi)(I^(2)1)/((2 L sin theta))` or `Tsin theta=(mu_(0))/(2pi)(I^(2)1)/((2Lsin theta))` (using ii)
or `(lamda 1 g)/(cos theta) sin theta=(mu_(0))/(2pi)(I^(2)1)/((2Lsin theta))` (using (i) `impliesI=sqrt(( 4 pi L lamda g sin^(2) theta)/(mu_(0)cos theta))=2sin theta sqrt((pi lamda gL)/(mu_(0)cos theta))`