= Solution
Parametrize the loop by $r'=R\hat e_r(\phi)$ and $d\ell'=R\hat e_\phi(\phi)d\phi$. At the observation point $r=z\hat e_z$, the <Biot-Savart law> for a <line current> gives
$$
dB=\frac{\mu_0I}{4\pi}
\frac{d\ell'\times(r-r')}{(R^2+z^2)^{3/2}}.
$$
The components perpendicular to the axis cancel by <rotational symmetry>, while the axial component of the <cross product> is $R^2d\phi$. Therefore
$$
\boxed{B(z)=\frac{\mu_0IR^2}{2(R^2+z^2)^{3/2}}\,\hat e_z}.
$$
The direction is fixed by the <right-hand rule> and reverses when the <electric current> reverses. At the centre,
$$
\boxed{B(0)=\frac{\mu_0I}{2R}\,\hat e_z}.
$$
This is the <on-axis magnetic field of a circular current loop>.
Back to article page
