Answer
The size of the current in wire 2 is $~~0.502~A$
Work Step by Step
We can write the general expression for the magnetic force on a current-carrying wire due to a parallel current-carrying wire:
$F = \frac{\mu_0~L~i_1~i_2}{2\pi~d}$
Note that currents in the same direction attract each other, and currents in the opposite direction repel each other.
When wire 3 is a distance of $4.00~cm$ from wire 2, the net force on wire 2 is zero. Therefore, the force on wire 2 due to the current in wire 1 and the current in wire 3 must be equal in magnitude and opposite in direction.
Since $i_1 = 3~i_3$, then $d = (3)(4.00~cm) = 12.0~cm$
As the graph approaches the asymptote, the force on wire 2 due to the current in wire 3 approaches 0.
We can find the current in wire 2:
$\frac{F}{L} = -\frac{\mu_0~i_1~i_2}{2\pi~d} = -0.627~\mu N/m$
$i_2 = \frac{(0.627~\mu N/m)(2\pi~d)}{\mu_0~i_1}$
$i_2 = \frac{(0.627~\mu N/m)(2\pi)~(0.120~m)}{(4\pi\times 10^{-7}~H/m)(0.750~A)}$
$i_2 = 0.502~A$
The size of the current in wire 2 is $~~0.502~A$
