Find the magnetic field $B$ at the point $P$ in figure.

MAgnetic field at $P$ due to $SM$ and $OQ$ is zero. Due to $QR$ and $RS$ are equal and outwards. Due to $MN$ and $NO$ are equal and inwards.

Due to $QR$ and $RS$, it will be outwards

$$\begin{gathered} B_1=2\left[\frac{{\mu }_0}{4\mathrm{\pi }}\frac{I}{2a}\left(\sin 0°+\sin 45°\right)\right] \\ =\frac{{\mu }_{0}I}{4\sqrt{2}\mathrm{\pi a}} \end{gathered}$$

Due to $MN$ and $NO$ inwards and net direction will be inwards.

$$\begin{gathered} B_2=2\left[\frac{{\mu }_0}{4\mathrm{\pi }}\frac{I}{a}\left(\sin 0°+\sin 45°\right)\right] \\ =\frac{{\mu }_{0}I}{2\sqrt{2}\mathrm{\pi a}} \\ \therefore B_{net}=B_2-B_1=\frac{{\mu }_{0 }I}{4\sqrt{2}\mathrm{\pi a}} \end{gathered}$$