Two long parallel wires carry currents of equal magnitude but in opposite directions. These wires are suspended from rod PQ by four chords of same length L as shown in Fig The mass per unit length of the wires is `lambda`. Determine the value of `theta` assuming it to be small.

Explanation

The forece per unit length between current carrying parallel
wires is `(dF)/(dL)=(mu_0I_1I_2)/(2pid)`.
If two wires carry current in
opposite directions, the magnetic force is repulsive, due to
which the parallel wires in Fig.
have moved out so that
equilibrium is reached.
The figure shows free body
diagram of each wire. In equilibrium, `SigmaF_y=0, 2T cos theta=(lambdaL_0)g......(i)`
`SigmaF_z=0, 2T sin theta=F_B......(ii)`
Now dividing Eq.(ii) by (i), we get `tan theta=(F_B)/(lambdaL_0g).....(iii)`
The magnetic force `F_B=((dF)/(dL))xxL_0=(mu_0I^2)/(4pi(sin theta)L) L_0......(iv)`
For small `theta`, `tan theta~=sin theta~= theta`
On substituting Eq. (iv) in (iii), we get `theta=Isqrt((mu_0)/(4pilambdagL))`