Answer: \(\oint\vec H\cdot d\vec l=I_{enc}\); correcting it with the displacement current gives \(\nabla\times\vec H=\vec J+\dfrac{\partial\vec D}{\partial t}\).
Ampere's circuital law: the line integral of the magnetic field intensity \(\vec H\) around any closed path equals the net current enclosed by that path:
\[\oint_C\vec H\cdot d\vec l=I_{enc}=\int_S\vec J\cdot d\vec S\]
Differential form. By Stokes' theorem \(\oint_C\vec H\cdot d\vec l=\int_S(\nabla\times\vec H)\cdot d\vec S\). So \(\int_S(\nabla\times\vec H)\cdot d\vec S=\int_S\vec J\cdot d\vec S\) for every surface, giving
\[\nabla\times\vec H=\vec J\qquad\ldots(1)\]
The difficulty. Take divergence of (1): \(\nabla\cdot(\nabla\times\vec H)=0\Rightarrow\nabla\cdot\vec J=0\). But the equation of continuity says \(\nabla\cdot\vec J=-\dfrac{\partial\rho}{\partial t}\), which is not zero when charge density changes with time (e.g. a charging capacitor). So (1) holds only for steady currents.
Maxwell's correction. Add a term \(\vec J_d\): \(\nabla\times\vec H=\vec J+\vec J_d\). Taking divergence: \(0=\nabla\cdot\vec J+\nabla\cdot\vec J_d\), so
\[\nabla\cdot\vec J_d=-\nabla\cdot\vec J=\frac{\partial\rho}{\partial t}\]
From Gauss's law (Maxwell's first equation) \(\rho=\nabla\cdot\vec D\), so \(\dfrac{\partial\rho}{\partial t}=\nabla\cdot\dfrac{\partial\vec D}{\partial t}\). Hence \(\vec J_d=\dfrac{\partial\vec D}{\partial t}\), the displacement current density.
Maxwell's fourth equation:
\[\boxed{\nabla\times\vec H=\vec J+\frac{\partial\vec D}{\partial t}}\]
Integral form: \(\oint\vec H\cdot d\vec l=\int_S\left(\vec J+\dfrac{\partial\vec D}{\partial t}\right)\cdot d\vec S\).
Meaning: a magnetic field is produced both by conduction current and by a time-varying electric field. In free space (\(\vec J=0\)) it becomes \(\nabla\times\vec H=\varepsilon_0\dfrac{\partial\vec E}{\partial t}\), which together with Faraday's law predicts electromagnetic waves.