# Flux of $B$ through a moving surface

I'm struggling to follow a derivation in Griffiths, where he proves that the emf generated by the motion of a conducting loop in a magnetic field is

$$\mathcal{E} = -\partial_t \Phi$$ where $$\Phi$$ is the flux of the magnetic field through that surface.

There's the following sketch, illustrating the circuit at times $$t$$ and $$t+dt$$

I'm quite confused by the switch from a surface integral, $$\iint\mathbf{B}\cdot d\mathbf{a}$$, to a circulation $$\oint \mathbf{B}\cdot (\mathbf{v}\times d\mathbf{l})$$. That "therefore" seems quite the shortcut, or am I missing a "well-known" formula from calculus? I would expect the flux to involve the partial derivatives of $$\mathbf{B}$$, and I don't see an obvious way to "see" that it all reduces to the circulation around the element of area.

• Did you try to apply stoke's theorem? Commented Sep 3, 2023 at 4:54
• Using the curl of what? I guess one could replace B by curl A, and get a circulation with that, but it's not clear to me how it helps get to Faraday's equation (in terms of B) Commented Sep 3, 2023 at 5:08
• > I would expect the flux to involve the partial derivatives of B Why? Flux is product of field and area, there are no partial derivatives. Commented Sep 3, 2023 at 23:00
• If you're worried that magnetic field changes when point of evaluation is displaced by $d\mathbf l$, this is negligible, because when area vector changes by $d\mathbf A$, change of flux due to nonuniform $\mathbf B$ is of the order of $d\mathbf l \cdot \nabla \mathbf B \cdot d\mathbf A$, a differential of second order, while the major part of change of flux is $\mathbf B \cdot d\mathbf A$, a differential of first order. Commented Sep 3, 2023 at 23:04
• @JánLalinský The convective derivative formula of Jackson's below suggests that the reason partial derivatives of B are not involved is because it's divergence-free. But I'm not used to thinking of moving regions of integration, so I guess I'm missing the interpretation in the static frame. It makes sense to me that something like the convective derivative should appear. Commented Sep 4, 2023 at 19:05

Your partial derivative is misleading since flux depend only on time and apart from the movement of the loop there is no explicit time dependence.

Griffiths is integrating the flux through the lateral ribbon. Indeed, the difference in flux through the loop at different times is equal to the lateral flux due to Maxwell’s equation: $$\nabla\cdot B=0$$

Next you just need to calculate the are element of the lateral flux. It is given by leading order in $$dt$$: $$d^2A=vdt \times dl$$ which is why the surface integral becomes a line integral. You then divide by $$dt$$ to get the derivative.

Mathematically, you are calculating the Lie derivative of a 2-form, for which there are general formulas.

Hope this helps.

• I can see how $v\times dl$ can be reinterpreted as a line integral, but why should it run around the parallelogram's boundary (as opposed to just one segment)? An why do we ignore the field's spatial variation? (it is claimed that the derivation applies equally to inhomogeneous fields). Commented Sep 3, 2023 at 3:46
• $vdt\times dl$ is the area of the parallelogram directed along its normal. You just add all the parallelogram contributions by integrating about the line. You are not really choosing a specific segment.
– LPZ
Commented Sep 3, 2023 at 3:56
• Inhomogeneities of the field would involve higher order terms in $dt$ so are negligible for the first derivative. The second derivative $\ddot \Phi$ would involve the first spatial derivative of $B$ for example and so on.
– LPZ
Commented Sep 3, 2023 at 3:58

Actually, Jackson's "it is easy to show" (sic) footnote is helpful here:

So basically, the derivation considers that:

• there's a time variation $$\partial_t B$$
• there's a spatial gradient, which, thanks to calculus, yields two terms: a curl, and a divergence
• the divergence of B is zero
• the curl lets us use Stokes' theorem and convert the flux into a circulation
• This answer links the two approaches: physics.stackexchange.com/a/330662/376580 (but I'm still not clear why one would add segment contributions around the surface element) Commented Sep 3, 2023 at 6:14

I'm still puzzled by Griffiths' derivation (and seemingly every other textbook I've looked at), but this article by Davison "A Simple Proof that the Lorentz Force, Law Implied Faraday's Law of Induction, when B is Time Independent" referred to on wikipedia makes more sense to me: rather than considering a circulation around the parallelogram as a closed loop (why should this be?), they write the work done more explicitly as an integral over time (from $$t_1$$ to $$t_2$$), and over a single line segment, following the trajectory of a charge. This way, no \oint appears, because it's clear that the flux is integrated along the path from A to B, not around some parallelogram.

Another source following the same approach is U. Krey and A. Owen's "Basic theoretical physics", which goes like this