To solve this problem, we need to determine the magnetic field at point $P$ due to the two perpendicular wires.$\\$Given:$\\$• Two long conducting wires $AOB$ and $COD$ are perpendicular to each other.$\\$• They carry currents $I_1$ and $I_2$ respectively.$\\$• Point $P$ is at a distance $d$ from the common point $O$ along a direction perpendicular to the plane of the wires.$\\$The magnetic field due to a long straight wire at a perpendicular distance $d$ is given by:$\\ B = \frac{\mu_0 I}{2\pi d} \\$For wire $AOB$ carrying current $I_1: \\ B_1 = \frac{\mu_0 I_1}{2\pi d} \\$For wire $COD$ carrying current $I_2: \\ B_2 = \frac{\mu_0 I_2}{2\pi d} \\$Since the wires are perpendicular, the magnetic fields $B_1$ and $B_2$ at point $P$ are also perpendicular.$\\$The resultant magnetic field $B$ at point $P$ is given by the vector sum:$\\ B = \sqrt{B_1^2 + B_2^2} \\$Substitute the expressions for $B_1$ and $B_2: \\ B = \sqrt{\left(\frac{\mu_0 I_1}{2\pi d}\right)^2 + \left(\frac{\mu_0 I_2}{2\pi d}\right)^2} \\ B = \frac{\mu_0}{2\pi d} \sqrt{I_1^2 + I_2^2} \\$Thus, the magnetic field at point $P$ is $\frac{\mu_0}{2\pi d} \sqrt{I_1^2 + I_2^2}. \\$This corresponds to Option 4.