\[\oint_C \mathbf{E} \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times \mathbf{E}) \cdot \mathbf{n} dS\]
for a vector field \[\mathbf{E}\]
with continuous derivatives.Let
\[\mathbf{F}\]
be the magnetic field strength \[\mathbf{H}\]
and use Maxwell's Law \[\mathbf{\nabla} \times \mathbf{H} = \mathbf{j} + \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\]
to get \[\oint_C \mathbf{H} \cdot d \mathbf{r} = \int \int_S (\mathbf{j} + \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}) \cdot \mathbf{n} dS\]
If the electric field is steady all time derivatives are zero, and we obtain Oersted's Law:
\[\oint_C \mathbf{H} \cdot d \mathbf{r} = \int \int_S \mathbf{j} \cdot \mathbf{n} dS\]