= Solution
For a <field> point in the plane of the loop, both the directed line element $d\mathbf s$ and the displacement $\mathbf r-\mathbf s$ lie in that plane. Their <cross product> is normal to it, so every term of the <Biot-Savart law> is normal, and their integral is normal as well. This argument also allows the <field> to vanish at particular points.
Choose the loop in the $xy$-plane with centre at the origin and let positive current be counterclockwise as seen from $+z$. On its lower side, write $\mathbf s=(t,-a,0)$, $-a\leq t\leq a$, so
$$
d\mathbf s\times(-\mathbf s)=a\,dt\,\mathbf e_z.
$$
That side contributes
$$
\frac{\mu_0I}{4\pi}\int_{-a}^a\frac{a\,dt}{(t^2+a^2)^{3/2}}\,\mathbf e_z
=\frac{\mu_0I}{4\pi}\left[\frac{t}{a\sqrt{t^2+a^2}}\right]_{-a}^a\mathbf e_z
=\frac{\sqrt2\mu_0I}{4\pi a}\mathbf e_z.
$$
All four contributions have the same direction and size by <rotation>. Thus
$$
\boxed{\mathbf B(0)=\frac{\sqrt2\mu_0I}{\pi a}\,\mathbf n},
$$
where $\mathbf n$ follows the <right-hand rule> for the current.
Back to article page
