How are the Lorentz force, Maxwell's third law and Faraday's law of induction clasically related?

Question

Faraday's law of induction can be used in any situation where the magnetic flux is changing through a closed conducting loop. While giving the correct answer, it seems to me that for the following scenarios different things are happening.

1) Let's consider a closed loop that is standing still in my reference frame. When changing the magnetic field through this loop from Maxwell 3, $\vec{\nabla} \times \vec{E} = -\frac{\partial \vec B}{\partial t}$, it is easy to find that Faraday's induction law is correct here using Stokes theorem. Physically, electric field lines are created inside the loop and push the electrons to move, this creates the current. The emf is then just the electric potential.

2) Let's place ourselves at rest relative to the magnet and change the area of the loop. $\frac{\partial \vec B}{\partial t}=0$ at any point in space so no $\vec{E}$ is created due to the field. This means that Maxwell 3 can't be really used in this frame. In this case the magnetic Lorentz force $F=q\vec{v} \times \vec{B}$ can be used to explain the movement of the electrons. For simple cases like these it seems to me that I have been succesful in deriving Faraday's induction law.

3) Finally I'd like to add that there is a thing called the disc of Faraday (http://en.wikipedia.org/wiki/Homopolar_generator) where no flux is changing but an emf is induced by the Lorentz force.

To conclude: I have a very basic understanding of special relativity and feel what is coming. I have to ask you to not go too deep into that for the following reason: I am studying for my classical electromagnetism final and need to have an efficient knowledge of the connections between laws from a classical point of view. Here come my two questions:

• Is everything I said about the described scenarios correct?
• What is the most efficient way to place this in my head in a classical way? Should I just consider Maxwell 3 and the Lorentz force being different phenomena that in problems where the flux is changing happen to lead to the same result, namely Faraday's induction?

Answer 1

I assume what is meant by Faraday's law of induction is what Griffiths refers to as the "universal flux rule", the statement of which can be found in this question. This covers both cases 1) and 2), even though in 1) it is justified by the third Maxwell equation¹ and in 2) by the Lorentz force law.

The universal flux rule is a consequence of the third Maxwell equation, the Lorentz force law, and Gauss's law for magnetism (the second Maxwell equation). To the extent that those three laws are fundamental, the universal flux rule is not.

I won't comment on whether the universal flux rule is intuitively true. But the real relationship is given by the derivation of the universal flux rule from the Maxwell equations and the Lorentz force law. You can derive it yourself, but it requires you to either:

1. know the form of the Leibniz integral rule for integration over an oriented surface in three dimensions
2. be able to derive #1 from the more general statement using differential geometry
3. be able to come up with an intuitive sort of argument involving infinitesimal deformations of the loop, like what is shown here.

If you look at the formula for (1), and set $\mathbf{F} = \mathbf{B}$, you see that \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{a} &= \iint_{\Sigma} \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a} + \iint_{\Sigma} \mathbf{v}(\nabla \cdot \mathbf{B}) \cdot \mathrm{d}\mathbf{a} - \int_{\partial \Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell \\ &= - \iint_{\Sigma} \nabla \times \mathbf{E} \cdot \mathrm{d}\mathbf{a} - \int_{\partial\Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell \\ &= -\int_{\partial \Sigma} \mathbf{E} + \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell \end{align*}

where we have used the third Maxwell equation, Gauss's law for magnetism, and the Kelvin--Stokes theorem. The final expression on the right hand side is of course the negative emf in the loop, and we recover the universal flux rule.

Observe that the first term, $\iint_\Sigma \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a}$, becomes the electric part of the emf, so if the loop is stationary and the magnetic field changes, then the resulting emf is entirely due to the induced electric field. In contrast, the third term, $-\int_{\partial\Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell$, becomes the magnetic part of the emf, so if the magnetic field is constant and the loop moves, then the resulting emf is entirely due to the Lorentz force. In general, when the magnetic field may change and the loop may also move simultaneously, the total emf is the sum of these two contributions.

If you are an undergrad taking a first course in electromagnetism, you should know the statement of the universal flux rule, and you should be able to justify it by working out specific cases using the third Maxwell equation, the Lorentz force law, or some combination thereof, but I can't imagine you would be asked for the proof of the general case from scratch, as given above.

The universal flux rule only applies to the case of an idealized wire, modelled as a continuous one-dimensional closed curve in which current is constrained to flow, that possibly undergoes a continuous deformation. It cannot be used for cases like the Faraday disc. In such cases you will need to go back to the first principles, that is, the third Maxwell equation and the Lorentz force law. There is no shortcut or generalization of the flux rule that you can apply. You should be able to do this on an exam.

¹ This equation is also often referred to as "Faraday's law" (which I try to avoid) or the "Maxwell--Faraday equation/law" (which I will also avoid here because of the potential to cause confusion).

Answer 2

Maxwell's differential equations apply to fields (or potentials) at all points in space and time subject to the boundary conditions of the problem to be solved. Let's examine the specific boundary conditions you have implied:

1) For the case of the changing magnetic field and static wire loop, you seem to have two boundary conditions. Boundary condition A is a definition of the first time derivative of the magnetic field at all points in space and time. Boundary condition B is most likely to be an equipotential (a constant) along the surface of the wire, and equivalently that the electric field is 0 inside the wire. This may not be exactly the boundary problem you want. Suppose, for example, you wish to solve Maxwell's equations for the fields for a real permanent magnet and a real wire loop. The hard part is that a real wire loop with real current would also create a magnetic field $B_L$, and the Boundary condition A is a sum of the change in the loop magnetic field and the permanent magnetic field. However, since you are only attempting to show Faraday's induction law, you appear to be able to get away without examining the details of the summed magnetic field.

2) For the case of the changing loop, your boundary conditions appear to be similar, but too restrictive. Your Boundary Condition A now becomes the statement that the magnetic field cannot change. By definition of boundary condition A, the loop may not create a changing magnetic field. The permanent magnetic field "source" also may not change or move, by the same logic. To then change the diameter of the wire without changing the magnetic field can therefore only happen if both the current in the wire and the magnetic field is 0. If the magnetic field "source" is not zero, then changing the diameter of the wire will change the current in the wire and alter the magnetic field of the wire, which violates condition A. If the current in the wire is not zero, then changing the diameter of the wire will also alter the magnetic field. You cannot derive anything useful from this, because the only solution to the Maxwell equations that meets the boundary conditions is where all fields are precisely 0.

The thing to remember is that the Maxwell equations apply to Electromagnetic fields in the 4 dimensions of space and time. Charge and current distributions cannot be directly used as boundary conditions. Instead, you must convert charge and current distributions into constraints on the fields and/or potentials. Only then can you use Maxwell's equations to find the fields. Finally, after you find the fields, you may need to convert back from field boundaries to charge and current distributions in order to get the answers you are looking for.

Answer 3

You can think of the Lorentz Force as an empirical fact that explains the current flow through moving wires (the moving wire gives a $\vec{v}$, the $\vec{B}$ field exists, hence there is a force $q\vec{v}\times\vec{B}$).

Using the Lorentz Force, we find that the EMF around many loops of moving wires in steady magnetic fields gives an EMF equal to the (negative of) the time rate of change of the magnetic flux in the loop.

This leads to a natural generalization (called Maxwell 3) that there should be an EMF in any loop should be equal to the (negative of) the time rate of change of the magnetic flux in the loop.

It is a totally separate empirical fact that this (Maxwell 3) holds. Though with relativity it isn't as totally separate as it appears.

Not to imply that any of the above is historically accurate, of course.

Now the Lorentz force stands on its own (unless you use relativity) and is necessary to see how the Faraday disk works. For static fields (but moving charges) it is all that is needed, and Maxwell 3 is not needed. For changing fields, then you need Maxwell 3, but without relativity a changing field and a moving wire in constant field are really different situations and unrelated. The Lorentz Force and electrostatics (e.g. $\vec{\nabla}\times \vec{E}=\vec{0}$) work fine for static fields, including a Faraday disk. And in those situations they are not wrong, because in those situations, Maxwell 3 agrees with $\vec{\nabla}\times \vec{E}=\vec{0}$.