 Solution

For an observation point in the wire's plane, both $d\mathbf{s}$ and $\mathbf{r}-\mathbf{s}$ lie in that plane. Their cross product is normal to it, and the Biot-Savart law integrates only such normal vectors. Thus the field is perpendicular to the plane at every point off the wire.

At the centre, all four sides contribute in the same normal direction by the right-hand rule. A single side is distance $a$ from the centre, so its contribution isMultiplying by four giveswhere $\mathbf{n}$ is the oriented normal fixed by the current.