Three long wires (wire 1, wire 2,and wire 3) are coplanar and hang vertically. The distance betweenwire 1 and wire 2 is 16.0 cm. On theleft, wire 1 ca

Force between two current carrying wires( wire 1 and wire 2)
${\displaystyle F_{12}=\frac{K{i}_1{i}_2}{2\pi r}}$
as both the wires carrying currents in opposite directions thenature of the fore on 2nd due to 1st is repulsive. hence to nullify the net force acting on the second conductorthird conductor should carry current in upward direction.
Then again 2nd and 3rd conductors carrying currents inopposite directions. hence again 2nd capacitor will experiencerepusive force.
If both the forces are equal then net force acting on thesecond is zero.
${\displaystyle F_{23}=\frac{K{i}_2{i}_3}{2\pi x}}$
x is distance
given net force should be zero, hence ${\displaystyle F_{12}=F_{23}}$
${\displaystyle \frac{i_1{i}_2}{2\pi r}=\frac{i_2{i}_3}{2\pi x}}$
${\displaystyle i_1=1.50A;i_2=3.40A;r=0.16m}$
similarly calculate the net force on the third due to 1st and2nd and following the same procedure we get
${\displaystyle \frac{i_1{i}_3}{(16+x)\times {10}^{-12}}=\frac{i_2{i}_3}{x}}$
so we have two unknowns and two equations try to calculatex and ${\displaystyle i_3}$; the directions of current ${\displaystyle i_3}$ is upwards as mentioned above