Answer
$7.8\times10^{-2}N$
Work Step by Step
The currents are running in the same direction, so the wires attract each other.
Use equation 20–7 to calculate the magnitude of the attractive force.
$$F=\frac{\mu_oI_1I_2}{2\pi d}\mathcal{l}_2$$
$$=\frac{(4\pi \times10^{-7} T\cdot m/A)(25A)^2}{2\pi(4.0\times10^{-2}m) }(25m)$$
$$=7.8\times10^{-2}N$$
