Answer
The net magnetic force on wire 4 due to the currents in the other wires is $~~(-1.88\times 10^{-4}~N)~\hat{j}$
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.
The force on wire 4 due to the current in wire 5 is directed in the +y direction.
The force on wire 4 due to the currents in wire 1, wire 2, and wire 3 is directed in the -y direction.
We can find the net magnetic force on wire 4 due to the other wires:
$F = \frac{\mu_0~L~i^2}{2\pi~d}-\frac{\mu_0~L~i^2}{2\pi~d}-\frac{\mu_0~L~i^2}{(2\pi)~(2d)}-\frac{\mu_0~L~i^2}{(2\pi)~(3d)}$
$F = -\frac{\mu_0~L~i^2}{(2\pi)~(2d)}-\frac{\mu_0~L~i^2}{(2\pi)~(3d)}$
$F = -\frac{3~\mu_0~L~i^2}{(2\pi)~(6d)}-\frac{2~\mu_0~L~i^2}{(2\pi)~(6d)}$
$F = -\frac{5~\mu_0~L~i^2}{(2\pi)~(6d)}$
$F = -\frac{(5)(4\pi\times 10^{-7}~H/m)(10.0~m)(3.00~A)^2}{(2\pi)(6)(0.0800~m)}$
$F = -1.88\times 10^{-4}~N$
The net magnetic force on wire 4 due to the currents in the other wires is $~~(-1.88\times 10^{-4}~N)~\hat{j}$
