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I Partial or total derivative in Faraday's law

  1. Sep 30, 2016 #1
    I just realized there's a little difference between the differential and integral forms of Faraday's law I didn't notice earlier. In the differential form, it is the partial time derivative that is written, while in integral form, it is simply the time derivative.


    Why is that ?
  2. jcsd
  3. Sep 30, 2016 #2


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    In the differential form, ##\vec E## and ##\vec B## are functions of position (e.g. x, y, z) and time t.

    In the integral form, the integral(s) integrate out the position dependence of ##\vec E## and ##\vec B##, so the left and right sides of the equation depend on time only.
  4. Sep 30, 2016 #3


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    Note, however, that the left-hand side (integral form) is not the general one. It's only valid if the surface and its boundary of integration are at rest. This is a source of endless confusion to students. It's always save to start from the local (differential) form of the laws, i.e., Maxwell's equations. For the correct integral form of Faraday's Law (in Heaviside-Lorentz units),
    $$\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_{\Sigma} \mathrm{d}^2 \vec{S} \cdot \vec{B} = -\mathcal{E}=-\int_{\partial \Sigma} \mathrm{d} \vec{r} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right)$$
    see the excellent Wikipedia article

  5. Oct 1, 2016 #4
    Thanks ! That seems kind of important, it sucks we don't hear about it when we first learn about Maxwell's equations.

    Anyway, I'm still trying to understand this subtle point. Is the following form correct also in the general case (I just apply Stoke's theorem) ?

    $$\oint \vec E \cdot d \vec l = \iint_S (\nabla \times \vec E) \cdot d \vec S= \iint_S -\frac {\partial \vec B} {\partial t} \cdot d \vec S$$

    EDIT: Maybe Stokes theorem only applies when the area integrated and its boundary do not move ?
  6. Oct 1, 2016 #5


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    This is of course also correct. It's just integrating the fundamental law, i.e., Maxwell's equation (Faraday's law of induction) over a surface ##S## and then applies Stokes's theorem. The tricky point is to correctly move the time derivative out of the integral. If the surface (and thus also its boundary) is not moving, it's trivial. You just take it out of the integral. If the boundary, however, is moving, you have to use the correct formula, which is proven in the Wikipedia article:

  7. Oct 4, 2016 #6
    Ok, I finally got it ! Thank you for the explanations and the wiki link, great indeed.

    It really is bad the integral form of the law we usually learn, ##\oint \vec E \cdot d \vec l = - \frac {d} {dt} \iint_S \vec B \cdot d \vec S## is not correct in general. I don't understand why it is taught. We should learn, in addition to the differential form, either the inside partial time derivative form ##\oint \vec E \cdot d \vec l = \iint_S -\frac {\partial \vec B} {\partial t} \cdot d \vec S## or the probably more useful outside time derivative = EMF form, ##\oint (\vec E + \vec v \times \vec B) \cdot d \vec l = - \frac {d} {dt} \iint_S \vec B \cdot d \vec S##, or both.
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