Themes > Science > Physics > Electromagnetism > Magnetostatics > Magnetic Field > Maxwell's equations in integral form

In freshman physics course one usually does not study Maxwell's equations in differential form but rather one studies them in integral form. Let us now prove that the Maxwell's equations as presented in equations  do in fact give the equations that one has already studied in freshman physics. To accomplish this we perform a volume integral over the first two equations and an area integral over the second two equations.

Integrating over volume ($\int d\tau $) on Gauss' law for ${\bf E}$  and using Gauss' divergence theorem as in $\int ({\bf \bigtriangledown . E} )d\tau = \oint {\bf E}.d{\bf A}$, and $q = \int \rho d\tau $ yields  
 \begin{displaymath} {\Phi }_E^{\prime } \equiv \oint {\bf E}.d{\bf A} = 4 \pi k q\end{displaymath} (125)
which is Gauss' law for the electric flux ${\Phi }_E^{\prime }$ over a closed surface area. The magnetic equation similarly becomes  
 \begin{displaymath} {\Phi }_B^{\prime } \equiv \oint {\bf B}.d{\bf A} =0\end{displaymath} (126)
where ${\Phi }_B^{\prime }$ is the magnetic flux over a closed surface area.

Integrating over area ($\int d{\bf A}$) on Faraday's law and using Stokes' curl theorem, as in $\int (\bigtriangledown \times {\bf E}). d {\bf A} = \oint {\bf E}.d {\bf l}$, yields  
 \begin{displaymath} \oint {\bf E}.d {\bf l} + g \frac{\partial {\Phi }_B}{\partial t} = 0\end{displaymath} (127)

where ${\Phi }_B \equiv \oint {\bf B}.d{\bf A}$ is the magnetic flux (not necessarily over a closed surface area). Finally integrating Ampère's law over an area and using $i \equiv \int {\bf j.A}$, yields  
 \begin{displaymath} \oint {\bf B}.d {\bf l} - \frac{1}{gc^2} \frac{\partial {\Phi }_E}{\partial t} = \frac{4\pi k}{gc^2} i\end{displaymath} (128)
where ${\Phi }_B \equiv \oint {\bf B}.d{\bf A}$ is the electric flux (not necessarily over a closed surface area).

This completes our derivation of Maxwell's equations in integral form.


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