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Many 'proofs' of Faraday's law use the third of the Maxwell's equations $\nabla \times \bf {E} = -\dfrac {\partial B}{\partial \it t}$ to simplify the flux integral by first splitting it into two integrals. One for the time-dependent magnetic field and one for the change in flux when the surface is time dependent.

$$\dfrac {d\phi_{B}}{dt} \Bigg|_{t=t_{0}}= \int \int_{\Sigma (t_{0})} \dfrac {\partial \textbf {B}}{\partial t} \Bigg |_{t = t_{0}} \cdot d \textbf {a} + \dfrac {d}{dt} \int \int_{\Sigma(t)} \textbf B( t_{0}) \cdot d \bf a $$

Why is the second of these integrals even needed? $\nabla \cdot \bf B$ $ = 0$ so from vector calculus we already know that the flux across surfaces with the same boundary curve $\partial \Sigma$ is the same...

This also doesn't make sense intuitively - $\Sigma$ is not a physical surface so the flux should't depend on it's actual shape but only it's boundary (i.e. the (physical) conducting loop). I get the derivations but can't see why the SHAPE of the surface should effect the flux. So what's going on here? Or is the divergence of time-dependent magnetic fields not zero in some cases?

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  • $\begingroup$ I see from the comments below that you've managed to answer your own question. The best practice is probably to write your own answer to this question and then accept it — that way someone else searching in the future might be able to benefit from your expertise. $\endgroup$ Sep 26, 2023 at 11:26
  • $\begingroup$ Sure I'll do that but this particular question isn't about an actual concept or anything - it's just that I didn't read the proof correctly.. $\endgroup$ Sep 26, 2023 at 11:40
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    $\begingroup$ Look at it this way: if you made a mistake then it's completely plausible that someone out there will make the same mistake in the future — and writing up an answer could help them out. $\endgroup$ Sep 26, 2023 at 11:43
  • $\begingroup$ Yeah I felt the same..:) Just wasn't sure whether these kind of questions were encouraged or not. $\endgroup$ Sep 26, 2023 at 12:25

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My bad - some ambiguous language in the proof really made things confusing.

Most sources will use the term 'time-dependent surface' to mean that the surface is changing (i.e. changing shape and/or position) because the LOOP ITSELF is changing (i.e. moving through the field or changing shape). So the second of these integrals actually gives the MOTIONAL e.m.f.

Changing ONLY the surface while keeping it's boundary curve (i.e. the conducting loop) and the magnetic field the same won't effect anything - this follows directly from the fact that $\nabla \cdot \bf B$ $= 0$ which means that it can be written as $\bf B$ $=$ $\nabla \times \bf A$ and it's flux will be the same across any surface with the same boundary. (Apply Stokes' theorem to $\bf A$)

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The double integral is an integral over an open surface $S$, your $\Sigma (t)$, bounded by closed loop $L$ as shown below.

enter image description here

The dot product ${\bf B}\cdot \rm d{\bf a} $ evaluates the scalar magnetic flux passing through area $\rm d{\bf a} $.

Pulling $\frac {\partial}{\partial t}$ out of the integral means that it can be changed to $\frac{d}{dt}$.


enter image description here

The flux through the plane, surface area $S_1$, is the same as the flux through the hemisphere, surface area $S_2$.

The mathematics is here - Electric flux through hemisphere

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  • $\begingroup$ I get that, but I don't get WHY the flux should depend on the shape of the surface at all - as long as it has the same boundary, $\nabla \cdot \bf B$$ = 0$ implies that the flux only depends on the boundary curve. $\endgroup$ Sep 26, 2023 at 8:06
  • $\begingroup$ I have added another diagram to my answer. $\endgroup$
    – Farcher
    Sep 26, 2023 at 8:18
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I don't understand the question completely, but I'll say that the surface integral of divergence-less vector fields only depend on the integration boundaries involved.

In other words since $$\nabla\cdot \mathbf{B}=0,$$ we have in general $$\mathbf{B}=\nabla\times\mathbf{A},$$ and so by Stoke's theorem $$\int(\nabla\times\mathbf{A})\cdot d\mathbf{a}=\oint\mathbf{A}\cdot d\mathbf{l}$$ which means that the flux only depends on the boundary curve.

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  • $\begingroup$ Exactly - that's where my confusion is...if we already KNOW that the flux only depends on the boundary curves then why do all these proofs even the case when the surface itself is time-dependent? Surely it's obvious that the flux won't be effected ass long as the the shape of the loop loop is left the same. $\endgroup$ Sep 26, 2023 at 10:19
  • $\begingroup$ en.wikipedia.org/wiki/Faraday%27s_law_of_induction#Proof - Take this derivation from Wikipedia for example. $\endgroup$ Sep 26, 2023 at 10:23
  • $\begingroup$ I got it - I was just reading the proof wrong. By 'time-dependent' surface Wikipedia meant that the surface changes because the LOOP is moving. $\endgroup$ Sep 26, 2023 at 10:30
  • $\begingroup$ Should I delete this question? I mean it's mostly about my brain fade rather than a conceptual problem :) $\endgroup$ Sep 26, 2023 at 10:31

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