The accumulated <point charge> $Q(t)$ produces
$$
\boxed{\mathbf E(r,t)
=\frac{Q(t)}{4\pi\epsilon_0r^2}\mathbf e_r}.
$$
Since $\dot Q=I$, the <displacement current> density is
$$
\boxed{\epsilon_0\frac{\partial\mathbf E}{\partial t}
=\frac{I}{4\pi r^2}\mathbf e_r}.
$$

Apply the integral <Ampère-Maxwell equation> to the boundary of a <spherical cap> of radius $r$ and polar angle $\theta<\pi/2$. By axial symmetry $\mathbf B=B_\phi\mathbf e_\phi$ is constant along the boundary, whose circumference is $2\pi r\sin\theta$. The displacement-current flux through the cap is
$$
\int_{\rm cap}\frac{I}{4\pi r^2}\,dA
=\frac{I}{4\pi r^2}\,2\pi r^2(1-\cos\theta)
=\frac I2(1-\cos\theta).
$$
Thus
$$
2\pi r\sin\theta\,B_\phi
=\frac{\mu_0I}{2}(1-\cos\theta).
$$
Using $(1-\cos\theta)/\sin\theta=\tan(\theta/2)$ gives
$$
\boxed{\mathbf B(r,\theta)
=\frac{\mu_0I}{4\pi r}
\tan\!\left(\frac\theta2\right)\mathbf e_\phi}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2 / 15D / b / Solution

{title2=$\epsilon_0\partial_tE$}
{wiki}

In vacuum, the displacement-current density is $\epsilon_0\partial E/\partial t$. Its contribution to magnetic circulation makes the <Ampère-Maxwell equation> compatible with changing electric charge.

= Ampère's law
{c}
{synonym}

Codex @codex  0

 Incoming links: Displacement current