To solve this problem, we need to calculate the force per unit length on the middle wire B due to the other two wires A and C.$\\$The force between two parallel current-carrying wires is given by:$\\ F = \frac{\mu_0 I_1 I_2}{2\pi d} \\$where $I_1$ and $I_2$ are the currents, and $d$ is the distance between the wires.$\\$In this case, all wires carry the same current $I. \\$The force per unit length on wire B due to wire A is:$\\ F_{BA} = \frac{\mu_0 I^2}{2\pi d} \\$The force per unit length on wire B due to wire C is:$\\ F_{BC} = \frac{\mu_0 I^2}{2\pi d} \\$Since the wires form a right triangle, the forces $F_{BA}$ and $F_{BC}$ are perpendicular.$\\$The net force per unit length on wire B is the vector sum of $F_{BA}$ and $F_{BC}. \\$Using the Pythagorean theorem:$\\ F_{net} = \sqrt{F_{BA}^2 + F_{BC}^2} \\$Substitute the values:$\\ F_{net} = \sqrt{\left(\frac{\mu_0 I^2}{2\pi d}\right)^2 + \left(\frac{\mu_0 I^2}{2\pi d}\right)^2} \\ F_{net} = \sqrt{2 \left(\frac{\mu_0 I^2}{2\pi d}\right)^2} \\ F_{net} = \frac{\mu_0 I^2}{2\pi d} \sqrt{2} \\ F_{net} = \frac{\sqrt{2} \mu_0 I^2}{2\pi d} \\$Therefore, the magnitude of the force per unit length on the middle wire B is $\frac{\mu_0 I^2}{\sqrt{2} \pi d}. \\$This corresponds to Option 3.