# What is magnetic flux? How is it related to Faraday's law?

For example, how do we explain Faraday's wheel using Faraday's law? There should be no emf induced since the flux is constant. Also, if we are given a magnetized material, and we somehow manage to change its magnetic field, would this induce an emf in the material? I've always thought that it is the change in the 'external' magnetic field which produces an e.m.f, but if this were the case then inductors wouldn't function at all.

So if we change the 'internal' and 'external 'magnetic field in our magnetic material, then would the emf be calculated by taking into account how the total flux changes, or just that due to the external field?

Magnetic flux and Faraday's law are defined purely mathematically. Because mathematical definitions typically deal with idealized concepts, subtle issues start to show up when we try to apply such definitions to more complicated physical situations. In particular, we usually need some physical principles to go with the mathematics; the physics tells us how to apply the mathematics.

Magnetic flux and Faraday's Law provide good examples of such subtleties and complications. I'll walk through the definitions carefully, pointing out the stumbling blocks, and then come back to Faraday's wheel. Most importantly, when talking about Faraday's wheel, there's one subtlety that requires additional input from physics, which is how to define a circuit.

We start with the mathematical definition of flux: $$\Phi_B = \iint_S \vec{B} \cdot d\vec{A}.$$ You probably know that $\vec{B}$ is the magnetic field vector, $d\vec{A}$ is the differential area element (the direction of which is normal to the surface with some continuous choice of either "outward" or "inward" direction), and $S$ is the surface over which the integral is performed. There are numerous idealizations built into this integral. Among them, we are assuming that we know exactly how to delimit this surface with exactly zero thickness and perfectly distinct boundaries.

This can raise subtle issues when we try to talk about real physical situations. For example, we might want to talk about the flux through a wire loop. But a physical wire has nonzero thickness. Should the surface of integration end at the center of the wire? At its outer edge? How do you specify the outer edge if the wire is not in a plane? You'll frequently hear people saying that Faraday's law only applies to "infinitely thin" wires. I would suggest that it's more correct to say that Faraday's law only applies naively to thin wires; it can be used in more general situations with careful application.

Next, we come to Faraday's Law, which relates the electromotive force $\mathcal{E}$ and the the boundary of your surface to the time-derivative of the magnetic flux as $$\mathcal{E} = \int_{\partial S} \vec{E} \cdot d\vec{l} = -\frac{\mathrm{d}\Phi_B} {\mathrm{d}t},$$ where $\partial S$ is the boundary of the surface we used to define the flux. There are some more subtleties here. First, and quite simply, you have to be very specific about the surface and the boundary that you are dealing with. Second, the derivative is the total derivative, rather than the partial derivative. This can be important when the surface of integration is changing in time, and when the magnetic field is changing in space and time.

That last part is a really subtle issue, even though it's about something so major as how we define a circuit. I'll quote Jackson to explain it: "The electric field $\vec{E}$ is the electric field at $d\vec{l}$ in the coordinate system or medium in which $d\vec{l}$ is at rest, since it is that field that causes current to flow if a circuit is actually present." [Classical Electrodynamics, third edition, section 5.15.] So for a circuit in which the path of electrons is changing relative to the conductor through which they're moving, you have to use a changing circuit path.

Now, applied to Faraday's wheel (the homopolar generator), the first point is that you have to specifically define the surface that you're talking about — and hence the boundary of that surface. Students frequently rush past this point, and take the disc itself as the surface. But this is wrong because the boundary of that surface is the edge of the wheel, so technically you would be calculating the electromotive force around the wheel. But you're measuring the potential between the center and the edge, so that surface can't work. Instead, you need to take a surface whose boundary passes between the center contact and the contact on the edge, then goes out from each of them, and over to some galvanometer or something. A standard choice for the surface is a square that it is perpendicular to the wheel, with one edge along the wheel's axis and the next edge along a radius of the wheel. Now the paradox comes up when you realize that $\vec{B}$ is actually parallel to this surface, so $\vec{B} \cdot d\vec{A}$ is zero everywhere and always. So it would be reasonable to think that \begin{align} \frac{d \Phi_B}{dt} &= \frac{d}{dt} \iint_S \vec{B} \cdot d\vec{A} \\ &= \iint_S \frac{d}{dt} \left( \vec{B} \cdot d\vec{A} \right) \\ &= \iint_S \frac{d}{dt} \left( 0 \right) \\ &= \iint_S 0 \\ &= 0, \end{align} which would imply $\mathcal{E} = 0$. But that's not what's measured. It turns out that the derivation I just gave is wrong.

The (wrong) derivation above used the partial derivative, and ignored movement of the surface. You can resolve this paradox if you remember that the derivative in question is the total derivative, and that the surface might be moving. Now, Jackson told us that we need to use a path that's stationary with respect to the medium. So the surface we're integrating over is moving with respect to stationary magnetic field we've been told about, and we need to differentiate the stationary field with respect to the moving surface. The standard way of describing the total derivative in a material that's moving with velocity $\vec{v}$ at a given point is to use the "material derivative" (also commonly known as the "convective derivative"): $$\frac{d}{dt} = \frac{\partial}{\partial t} + \vec{v} \cdot \vec{\nabla}.$$ To be a little more specific, you use this material derivative to get the derivative with respect to moving coordinates when you're given a field in static coordinates. Now we get \begin{align} \frac{d \Phi_B}{dt} &= \frac{d}{dt} \iint_S \vec{B} \cdot d\vec{A} \\ &= \iint_S \left[ \left( \vec{v} \cdot \vec{\nabla} \right) \vec{B} \right] \cdot d\vec{A} \\ &= \iint_S \left[ \vec{\nabla} \times \left( \vec{B} \times \vec{v} \right) + \vec{v} (\vec{\nabla} \cdot \vec{B}) \right] \cdot d\vec{A} \\ &= \iint_S \left[ \vec{\nabla} \times \left( \vec{B} \times \vec{v} \right) \right] \cdot d\vec{A} \\ &= \int_{\partial S} \left( \vec{B} \times \vec{v} \right) \cdot d\vec{l} \end{align} which is nonzero. [To go from the second to third line, I used a standard vector calculus identity with assumptions about $\vec{v}$ for our case. To go from the third to fourth, I used the fact that $\vec{\nabla} \cdot \vec{B} = 0$. To get to the last line, I used the Stokes theoreom. Remember that $\partial S$ is the boundary of $S$.]

Alternatively, we could differentiate with respect to stationary coordinates, but note that the surface is changing in those coordinates. If we look at a little segment of the boundary $\Delta \vec{l}$, and suppose it's moving with velocity $\vec{v}$, then after time $\Delta t$ has passed, this has introduced a new amount of area into the surface given by $$\Delta \vec{A} = (\vec{v} \Delta t) \times \Delta \vec{l}.$$ This changes the flux integral by $$\Delta \Phi_B = \vec{B} \cdot \Delta \vec{A} = \vec{B} \cdot (\vec{v} \times \Delta \vec{l}) \Delta t,$$ which implies $$\frac{\Delta \Phi_B}{\Delta t} = \vec{B} \cdot (\vec{v} \times \Delta \vec{l}).$$ Now we can take the limits of small $\Delta t$ and small $\Delta \vec{l}$, then integrate up the contributions from all those little $\Delta \vec{l}$ segments around $\partial S$ to get the total derivative of the flux: \begin{align} \frac{d \Phi_B}{dt} &= \int_{\partial S} \vec{B} \cdot (\vec{v} \times d\vec{l}) \\ &= \int_{\partial S} \left( \vec{B} \times \vec{v} \right) \cdot d\vec{l}. \end{align} To get from the first to second line, I just used the usual scalar triple product. And this is the same result as the one we got using the material derivative — again, it's nonzero.

I'm not saying that any of this is obvious at all; it's certainly not obvious from the simple mathematical definitions I've given above. But that's all just mathematics. The key point to remember when applying it to physics is Jackson's point: You need to define your circuit with respect to the medium through which the electrons are actually moving.

There's really nothing that tells us the answers a priori — that's why humans had to do the experiments to figure out how physics works. There's a pretty good discussion of subtleties regarding Faraday's Law on Wikipedia's Faraday paradox page, but the better discussion of the wheel comes on the Faraday induction page. And of course, we should all read Jackson more carefully.

• I can't say I understood it all; for one, I'm unclear on what you mean by 'differentiating the area', but the biggest question I have is how do we choose the surface? Could we not have chosen a diameter, or a chord rather than a radii? Also, why does naively calling the flux the 'area swept' times the magnetic field in this case give the correct answer?
– John
May 3, 2017 at 12:04
• I've edited my answer to emphasize that the real physics content here is Jackson's point about choosing the surface with respect to the medium. So that explains why it has to move. As to why we choose a radius instead of a diameter or a chord: we certainly could choose a diameter or a chord, but then we'd be calculating the electromotive force around a loop that includes that diameter or chord. So when you're doing the experiment, you'd have put your galvanometer contacts at either end of that diameter or chord. You can certainly do that; it's just not the usual problem we talk about.
– Mike
May 3, 2017 at 12:08
• I see... so if we wanted to measure the emf 'around' the disk, how would we connect the galvanometer? And will there be any eddy currents formed in this case? I can see that the rim of the disk will be at a potential difference from the center, but the current can only flow if there is a closed loop, which is only possible for the radii we've connected the galvanometer to.
– John
May 3, 2017 at 12:12
• And what exactly does differentiating the area of integration mean? Isn't the area of our surface a constant? Its position changes, but the area remains unchanged, right?
– John
May 3, 2017 at 12:32
• Regarding EMF around the disk, that's a good question, and there are several ways you could approach it. First, you could just place your two contacts at opposite edges. But then the simplest model for the circuit would have it going straight across the disk. You could also place the contacts close together along the edge, and then see how it varies as you slowly move the contacts farther apart. In each case, I predict no EMF because the velocity will be an odd function of distance along the line joining the contacts, so contacts at the same radius will lead to an integral of 0.
– Mike
May 3, 2017 at 13:19

Faraday's disc is one of two cases Feynman cites (Feynman lectures in Physics Vol 2, 17.2) in which an emf is induced but with no change in flux linkage. These are cases where the circuit is not a clearly defined loop. These exceptions don't strike at the foundations of electromagnetism: the induced emf is still predicted correctly by using $\vec{F}=q\vec{v}\times \vec{B}$.

Regarding magnetic materials, it is indeed changes in the total field that induce emfs in the material – as in the case of eddy currents.

• Comments are not for extended discussion; this conversation has been moved to chat. May 4, 2017 at 12:45