# Are there any geometric restrictions on the line and surface elements in the Maxwell-Faraday equation?

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?

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