Two straight long conductors $AOB$ and $COD$ are perpendicular to each other and carry currents $i_1$ and $i_2$. The magnitude of the magnetic induction at a point $P$ at a distance $a$ from the point $O$ in a direction perpendicular to the plane $ACBD$ is

Point $P$ lies on the line through $O$ perpendicular to the plane of the two wires, so its perpendicular distance from each long conductor is $a$. For a long straight conductor, the magnetic field magnitude at distance $a$ is $$B=\frac{\mu_0 i}{2\pi a}.$$ Hence the fields at $P$ due to the two conductors are $$B_1=\frac{\mu_0 i_1}{2\pi a}$$ and $$B_2=\frac{\mu_0 i_2}{2\pi a}.$$ Using the right-hand rule, these two magnetic fields at $P$ are mutually perpendicular because the conductors themselves are perpendicular and $P$ is on the common perpendicular through $O$. Therefore the resultant magnitude is $$B=\sqrt{B_1^2+B_2^2}.$$ Substituting, $$B=\frac{\mu_0}{2\pi a}\sqrt{i_1^2+i_2^2}.$$ This matches option $\mathrm{(C)}$.