Ampere's circuit law states that the line integral of the tangential component of H around a dosed path is the same as the net current $I_{enc}$ enclosed by the path.

In other words, the circulation of H equals $I_{enc}$; that is,

$\oint \mathbf{H} \cdot d \mathbf{l}=I_{\mathrm{cnc}}$ ...(1)

Ampere's law is similar to Gauss's law and it is easily applied to determine H when the current distribution is symmetrical. It should be noted that eq. (1) always holds whether the current distribution is symmetrical or not but we can only use the equation to determine H when symmetrical current distribution exists. Ampere's law is a special case of Biot-Savart's law; the former may be derived from the latter.

By applying Stoke's theorem to the left-hand side of eq. (1), we obtain

$I_{\mathrm{enc}}=\oint_{L} \mathbf{H} \cdot d \mathbf{l}=\int_{S}(\mathbf{\nabla} \times \mathbf{H}) \cdot d \mathbf{S}$ ...(2)

But

$I_{\mathrm{enc}}=\int_{S} \mathbf{J} \cdot d \mathbf{S}$ ...(3)

Comparing the surface integrals in eqs. (2) and (3) clearly reveals that

$\nabla \times \mathbf{H}=\mathbf{J}$ ...(4)

This is the third Maxwell's equation to be derived; it is essentially Ampere's law in differential (or point) form whereas eq. (1) is the integral form. From eq. (4), we should observe that $\nabla \times \mathbf{H}=\mathbf{J} \neq 0$; that is, magnetostatic field is not conservative.