Two current carrying identical coils are kept as shown in figure. The magnetic field at centre ' $O$ ' is ( $N$ and $R$ represent the number of turns and radius of each coil respectively, $\mu_0=$ permeability of free space)

The magnetic field at the center $O$ results from the vector sum of fields produced by two perpendicular coils.

Each coil contributes $B = \frac{\mu_0 N I}{2R}$ at its center. With coils perpendicular, their fields are orthogonal.

The net magnitude is $B_{\text{net}} = \sqrt{\left(\frac{\mu_0 N I}{2R}\right)^2 + \left(\frac{\mu_0 N I}{2R}\right)^2} = \sqrt{2} \cdot \frac{\mu_0 N I}{2R}$.

Simplifying yields $B_{\text{net}} = \frac{\mu_0 N I}{\sqrt{2} R}$.

Final answer: $\frac{\mu_0 N I}{\sqrt{2} R}$