The question might sound a bit strange at first so allow me to give some context. My prof has given a derivation of the magnetic flux where he used both $\Delta$ and $d$ for elementary line and surface elements. I am confused by the simultaneous usage of both infinite and finite elements, and their meaning compared to each other. In my own version of the proof I wanted to use only infinitesimals. I will post the derivation according to my prof first, afterwards I'll give my own version (which was intended to be completely analogous), where I tried to use only infinitesimals.

The general idea is to start from the work done by the Lorentz force, define the magnetic flux and link it back to the emf.

My prof's derivation: $$ \Delta W = \bar{F}.\Delta\bar{l} = q(\bar{v} \times \bar{B}).\Delta \bar{l} = q(\Delta \bar{l} \times \bar{v}).\bar{B} \quad \text{(permutation rule)} \\ \bar{v} = \frac{d\bar{s}}{dt} \Rightarrow \Delta W = \frac{q}{dt} (\Delta \bar{l} \times d\bar{s}).\bar{B} \\ \Delta \bar{l} \times d\bar{s} = d\Delta \bar{S} \Rightarrow q \frac{d}{dt}(\Delta \bar{S}.\bar{B}) \\ d\phi := \Delta \bar{S}.\bar{B} \Rightarrow \Delta W = q \frac{d\Delta\phi}{dt} \\ W = \varepsilon = -\frac{d\phi}{dt} \\ $$ The last part somehow follows from integration, the - sign will be explained later. I understand most parts, save for the mixed usage of $\Delta$ and $d$, and the last integration of $\Delta \phi$ (?). Maybe it's something stupid that I'm missing but this derivation confuses me.

My version: $$ dW = \bar{F}.d\bar{l} = q(\bar{v} \times \bar{B}).d \bar{l} = q(d \bar{l} \times \bar{v}).\bar{B} \\ \bar{v} = \frac{d\bar{s}}{dt} \Rightarrow d W = \frac{q}{dt} (d \bar{l} \times d\bar{s}).\bar{B} \\ d \bar{l} \times d\bar{s} = d\bar{S} ^{(*)} \Rightarrow \frac{q}{dt}(d \bar{S}.\bar{B}) \\ d\phi := d\bar{S}.\bar{B} \Rightarrow dW = q \frac{d\phi}{dt} \\ \text{Error} $$

That's how far I can get. I can't somehow integrate $dW$ and keep the $\frac{d\phi}{dt}$. The source of this problem seems to come from $(*)$. My prof somehow summons an extra d, where in my version it is missing and I don't know what a mathematical correct way would be of fixing this.

I wasted a few hours on trying to find a solution but to no avail. I don't immediately see any mistake on my part. I haven't had any differential topology yet, so it could very well be that my understanding of differentials etc. is wrong. Any suggestions or tips on whether how to fix it or if my version is even correct are welcome.

I know of different derivations that use Faraday's law and Maxwell's 3rd equation, but I want to understand this one first.

Edit The velocity in the Lorentz force is the velocity of the conductor. I thought of the charges as being quasi-stationary in the conductor so the only velocity they have is the conductor's velocity. (If that makes any sense)

Edit 2 I should have done this sooner but here is a short explanation of the used terms: $\Delta W$ is the difference in work and the finite version of $dW$. $\Delta \bar{l}$ is the difference between two lengths of the conducting wire and the finite version of $d\bar{l}$. $d\bar{s}$ is an infinitesimal displacement of the conducting wire due to the Lorentz force. $\bar{B}$ is the magnetic induction vector, perpendicular on $\bar{l}$ and $\bar{s}$. $\Delta \phi$ is the finite difference in the magnetic flux and $d\phi$ is the infinitesimal difference. $d\bar{S}$ is the an infinitesimal surface vector, $\Delta \bar{S}$ would be the finite limit of it. No idea what is meant by $d\Delta \bar{S}$, infinitesimal difference in of a finite difference of the surface? $\varepsilon$ is the emf. q is a charge. I hope this clears things up a bit.

I was not sure whether to post the question at physics or at math. My apologies if I made the wrong choice.

  • $\begingroup$ So there is another force exerted on the particle then? Otherwise $\Delta l \parallel v$ and $\Delta W=0$ from the start. $\endgroup$ – user154997 Aug 15 '17 at 23:55
  • $\begingroup$ @LucJ.Bourhis No, the velocity in the Lorentz force is the velocity of the moving conductor. I somehow thought of it as the charges in the conductor being quasi-stationary, so the only velocity they have is the velocity from the conductor. $d\bar{s}$ would then be an infinitesimal displacement of the conductor, which in this situation is not parallel to the conductor. I will edit my question to reflect this additional detail. $\endgroup$ – space Aug 16 '17 at 0:25
  • $\begingroup$ What is $\Delta l$ then? $\endgroup$ – user154997 Aug 16 '17 at 7:44
  • $\begingroup$ @LucJ.Bourhis $\Delta \bar{l}$ is the difference of two lengths of the conducting wire. I believe it is the finite version of $d\bar{l}$. $\endgroup$ – space Aug 16 '17 at 13:08
  • $\begingroup$ At the other end of your prof demonstration, $q$ disappears: typo? More annoying, how can a work be equal to an e.m.f.? $\endgroup$ – user154997 Aug 16 '17 at 13:47

Here is one way I can make sense of it all. The following schematic describes the system I am going to work with:

enter image description here

There is a wire $\newcommand{\wire}{\mathcal{L}}\wire$ translated by an infinitesimal amount $\renewcommand{\vec}[1]{\boldsymbol{#1}}d\vec{L}$ (I will denote vectors with bold letters). The infinitesimal element of length of the wire is $d\vec{l}$. The dashed lines are fixed wires to close the circuit, through which flows an intensity $I=\lambda v$ where $\lambda$ is the density of moving charges per unit of length and $v$ is the speed of those charges. The infinitesimal work of the Lorentz force on the whole wire $\wire$ is then

$$dW = \int_\wire \lambda dl (\vec{v}\times \vec{B})\cdot d\vec{L}.$$

But $dl\,\vec{v}=v\,d\vec{l}$ as the charges are constrained to flow along the wire. Then using the formula for the current $I$,

$$dW = \int_\wire I (d\vec{l}\times \vec{B})\cdot d\vec{L}=\int_\wire I \vec{B}\cdot(d\vec{L}\times d\vec{l}).$$

But then $I$ is uniform along the wire, so we can take it out. Furthermore, we recognise in $d\vec{L}\times d\vec{l}$ the infinitesimal oriented area $d\vec{S}$ of the surface $\newcommand{\swiped}{\mathcal{S}}\delta\swiped$ delineated by the two positions of the wire and by the fixed wires represented by dashed lines. So

$$dW = I\iint_{\delta\swiped} \vec{B}\cdot d\vec{S}.$$

We recognise the flux $d\Phi$ of $\vec{B}$ through $\delta\swiped$. It is an infinitesimal flux because $\delta\swiped$ is infinitesimal in one direction. Thus we finally get

$$dW = I\,d\Phi.$$

But then we should also have

$$\frac{dW}{dt} = \mathcal{E}I$$

where $\mathcal{E}$ is the e.m.f., and therefore


  • $\begingroup$ Yes, I think we're going in the good direction. I understand your derivation, the only thing that's still confusing me is the following: as I see it, you say $d\bar{L} \times d\bar{l} = \delta S$. (I'm ok with that) What is the meaning of $d^2\bar{S}$ then? An even smaller infinitesimal area that when integrated gives $\delta S$? How does $d^2\bar{S}$ differ from $d\bar{S}$? What's the difference in the meaning of $\delta$ and $d$? I think those question form the bridge that I need to cross before I fully understand the derivation? My apologies if they are stupid questions. $\endgroup$ – space Aug 16 '17 at 15:56
  • $\begingroup$ $\delta\swiped$ is a surface, i.e. a geometrical object, a two-dimensional set of points, so I shied away from using the differential notation! But the area of $\delta\swiped$ would be a first-order differential, as is the flux of $\vec{B}$ through it. As for $d^2S$, it is indeed not usual a notation. I edited my answer to put the usual $dS$ indeed. $\endgroup$ – user154997 Aug 16 '17 at 16:00
  • $\begingroup$ So $\delta\swiped = \iint_{\delta\swiped}d\bar{S}$ simply means that the infinitesimal surface $\delta\swiped$ is written as the sum of infinitesimal surface vectors $d\bar{S}$? Looking at your derivation again, I'm struggling a bit with your last two equations. Why should we also have $\frac{dW}{dt} = \varepsilon I$? It's also not entirely clear to me how you go from $dW$ to $\varepsilon (= W)$ . My apologies if it seems I'm dragging this out, I just want to understand it fully. $\endgroup$ – space Aug 16 '17 at 16:52
  • $\begingroup$ Yes to your first: dS is double infinitesimal whereas $\delta S$ is only single infinitesimal. Maybe I should have done it all with a non-infinitesimal motion of the wire: would that be clearer? As for $dW/dt=\mathcal{E}I$, this is electrical circuit 101: power=intensity times voltage.. $\endgroup$ – user154997 Aug 16 '17 at 17:24
  • $\begingroup$ No I prefer it in the infinitesimal way as it feels more general to me. I haven't had any differential geometry courses yet so certain things were never really defined in a rigorous way. It all makes sense now. Yes, lame of me to forget the power equation. Am I correct to say then that $dW = Id\phi \Rightarrow \frac{dt}{dt}dW = Id\phi \Rightarrow \frac{dW}{dt} = I\frac{d\phi}{dt}$ but we also have $P = \frac{dW}{dt} = \varepsilon I$ and from this follows $\varepsilon = \frac{d\phi}{dt}$. Anyway, thanks for your time and effort. I understand the derivation now. $\endgroup$ – space Aug 16 '17 at 17:46

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