The Maxwell-Faraday equation in integral form states $$\oint_{\partial \Sigma} \mathbf{E} \cdot \mathrm{d}\boldsymbol{\ell} = - \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{S}$$

Consider first the LHS. Notice how it has a $\mathrm{d}\boldsymbol{\ell}$ term. Since the LHS is a line integral, this term indicates that we are summing over the projection of the electric field along the tangent vector at this small infinitesimally small segment of the curve times the length of this segment. So there is some curve (i.e. a wire) used in that half.

Consider now the RHS. Notice how it has a $\mathrm{d}\mathbf{S}$ term in it. This indicates this surface integral sums over the projection of the magnetic field along the normal vector at a point on some surface S times this infinitesimal area. So a surface must be present.

So we have a line and a surface. What are the geometric and/or physical restrictions on this "curve" and this "surface" that must hold for Faraday's equation to be true?

For example, one restriction I would imagine could be the line must make up the surface, as with a coil.

Addition: $$ \oint_{\partial \Sigma} \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \iint_{\Sigma} \mathbf{J} \cdot \mathrm{d}\mathbf{S} + \mu_0 \varepsilon_0 \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{E} \cdot \mathrm{d}\mathbf{S}$$

Are the surface and line referenced in the Maxwell Faraday equation the same ones referenced in the Ampère's circuital law (with Maxwell's addition)? That is, do the same restrictions apply?

  • $\begingroup$ Basically yes, same as for Stokes' theorem: $\partial \Sigma$ closed boundary of smooth orientable surface $\Sigma$. $\endgroup$
    – udrv
    Commented Dec 27, 2015 at 8:57

1 Answer 1


Yes there is a relationship. The surface that is being integrated over on the RHS must be bounded by the closed loop that is being integrated around on the LHS. Other than that, there is no restriction. The same is true for Ampere's law in integral form.

So an interesting example would be to consider a circular, closed loop. We could sum the line integral of the E-field around this loop and say it was equal to (minus) the rate of change of magnetic flux found by integrating the magnetic field through a flat disc whose edge is defined by the loop. But, the RHS would be exactly the same if you evaluated the flux over any area that is defined by connection to the closed loop - for instance any "bag-like" shape where the neck of the bag is defined by the loop.

  • $\begingroup$ +1 does this restriction hold for the differential analogs of Maxwell's integral equations or is it specific to the integral ones? $\endgroup$ Commented Dec 27, 2015 at 9:57
  • $\begingroup$ Also, what about when you have a vacuum? $\endgroup$ Commented Dec 27, 2015 at 10:03
  • $\begingroup$ @StanShunpike Only the integral forms apply to geometric shapes. The differential forms apply at a point. Mawell's equations (when written in the correct way) apply in vacuum or otherwise. $\endgroup$
    – ProfRob
    Commented Dec 27, 2015 at 10:14

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