Ampere's Law

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Ampere's Law is stated in words as:

magnetomotive force around any closed loop is equivalent to the total current flowing through the loop area.

This means

\oint_{\partial S}\textbf{H}\cdot d\textbf{l} = \int_S \textbf{J}\cdot d\textbf{S},

where \textbf{H}\, and \textbf{J}\, are the magnetic field intensity and current density vectors, respectively.

James Maxwell realised that Ampere's Law, as written above, is incomplete. In order to yield correct results for current continuity, it must read

\oint_{\partial S}\textbf{H}\cdot d\textbf{l} = \int_S \textbf{J}\cdot d\textbf{S} + \frac{d}{dt}\int_s\textbf{D}\cdot d\textbf{S}

otherwise, charge cannot be properly conserved. A simple illustration of this is a capacitor connected to a voltage source whose output varies with time (See figure).

Illustration of a parallel plate capacitor connected to a time-varying voltage source.
Enlarge
Illustration of a parallel plate capacitor connected to a time-varying voltage source.

At some point in time, current flows into the upper plate and out of the bottom plate. In this case, a positive charge builds up on the upper plate and an exactly opposing negative charge appears on the bottom plate. The field between the plates increases as time passes. This time-rate-of-change in the field between the plates can be thought of as a "displacement current" which must exactly equal the "real" current flowing in the wires leading to the capacitor plates.

Relying on Stoke's Theorem, we can readily derive the differential form of Ampere's Law, which is

\nabla\times\textbf{H}=\textbf{J}+\frac{d\textbf{D}}{dt}.

This expression is recognisable as one of Maxwell's Equations, with the exception of the use of the total time derivative. This is because we want to highlight the fact that in problems where the field coordinate system moves with respect to the observer with velocity \textbf{v}\,, we apply an argument similar to that used for Faraday's Law to get

\frac{d\textbf{D}}{dt} = \frac{\partial\textbf{D}}{\partial t} + \nabla\times(\textbf{D}\times\textbf{v})+\textbf{v}(\nabla \cdot\textbf{D}),

And therefore,

\oint_{\partial S}\textbf{H}\cdot d\textbf{l} = \int_S \textbf{J}\cdot d\textbf{S} + \int_s\frac{\partial \textbf{D}}{\partial t}\cdot d\textbf{S} + \oint_{\partial S} (\textbf{D}\times\textbf{v})\cdot d\textbf{l} + \int_S(\nabla\cdot\textbf{D})\textbf{v}\cdot d\textbf{S}.

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