Answer
The x component of the magnetic force per unit length on wire 2 due to wire 1 is $~~8.84\times 10^{-11}~N/m$
Work Step by Step
We can find the distance between the currents:
$d = \sqrt{(2.40~cm)^2+(5.00~cm)^2}$
$d = 5.546~cm$
Note that currents in opposite directions repel each other. We can find the angle $\theta$ below the +x axis that the force is exerted on wire 2:
$\theta = tan^{-1}~(\frac{2.40~cm}{5.00~cm})$
$\theta = 25.64^{\circ}$
We can find the x component of the magnetic force per unit length on wire 2 due to wire 1:
$F = \frac{\mu_0~L~i_1~i_2}{2\pi~d}~cos~\theta$
$\frac{F}{L} = \frac{\mu_0~i_1~i_2}{2\pi~d}~cos~\theta$
$\frac{F}{L} = \frac{(4\pi\times 10^{-7}~H/m)(4.00~mA)(6.80~mA)}{(2\pi)(0.05546~m)}~cos~25.64^{\circ}$
$\frac{F}{L} = 8.84\times 10^{-11}~N/m$
The x component of the magnetic force per unit length on wire 2 due to wire 1 is $~~8.84\times 10^{-11}~N/m$
