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

We know that current-carrying wires carrying antiparallel currents repel each other. Due to this repulsion, they acquire the configuration as shown in the figure. The force between two parallel current-carrying wires is given by
`F=(mu_(0))/(4pi)(2i_(1)i_(2)L)/(d)`
The force is repulsive if the currents are in opposite direction. The free body diagram of the problem is as shown in Fig. 29-22b
Calculations : If `L_(0)` is the length of each wire, for its vertical and horizontal equilibrium we have, respectively,
`2Tcostheta=L_(0)lambdag`,
`2Tsintheta=F`
Therefore
`tantheta=(F)/(L_(0lambdag)`
But as,
`F=(mu_(0))/(4pi)[(2Ixxl)/(2Lsintheta)L_(0)]` (as `d=2Lsintheta`)
So,
`tantheta=(mu_(0))/(4pi)(I^(2))/(lambdagLsintheta)`
Since as for small `theta,tantheta=sintheta=theta`
`theta=Isqrt((mu_(0))/(4pilambdagL))`
This force is quite small in magnitude and this is the reason why we do not observe its effect in our daily life.