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If \(C\) is a closed curve enclosing a surface \(S\), then the magnetic field intensity \(\mathbf{H}\), the current density \(\mathbf{J}\), and the electric flux density \(\mathbf{D}\) are related by

1. \(\iint_S \mathbf{H}\cdot d\mathbf{s}=\oint_C \left(\mathbf{J}+\dfrac{\partial \mathbf{D}}{\partial t}\right)\cdot d\mathbf{l}\)
2. \(\oint_C \mathbf{H}\cdot d\mathbf{l}=\iint_S \left(\mathbf{J}+\dfrac{\partial \mathbf{D}}{\partial t}\right)\cdot d\mathbf{s}\)
3. \(\iint_S \mathbf{H}\cdot d\mathbf{s}=\iint_S \left(\mathbf{J}+\dfrac{\partial \mathbf{D}}{\partial t}\right)\cdot d\mathbf{s}\)
4. \(\oint_C \mathbf{H}\cdot d\mathbf{l}=\oint_C \left(\mathbf{J}+\dfrac{\partial \mathbf{D}}{\partial t}\right)\cdot d\mathbf{l}\)

Correct answer: \(\oint_C \mathbf{H}\cdot d\mathbf{l}=\iint_S \left(\mathbf{J}+\dfrac{\partial \mathbf{D}}{\partial t}\right)\cdot d\mathbf{s}\)

Solution

Ampere's circuital law with Maxwell's correction states that the line integral of \(\mathbf{H}\) around a closed curve equals the surface integral of conduction current density plus displacement current density. This is exactly the Ampere-Maxwell law in integral form.

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