Explaining Faraday's Law With Lorentz Transformations of $E$ and $B$ Fields

Question

I've searched around for this but haven't come across a totally satisfying explanation yet. I'm trying to build a relatively simple model of how the Lorentz Transformations of the $E$ and $B$ fields can explain Faraday's Law, but I'm stumbling on one particular example -- B1 in the figure below.

In these sketches, I have a conducting loop moving in some magnetic field. In A1 and B1, the field is assumed to be purely magnetic. In B1 and B2, the magnetic field is non-uniform. According to Faraday's Law, an emf and current should be induced in the loop in each case.

In A1, the loop velocity is perpendicular to $B$, so the induced current is due to the magnetic force acting on charges in the loop ($q \vec{v} \times \vec{B}$).

A2 is A1 boosted to the frame of the loop. Here, there is no magnetic force on the charge carriers, but there is an electric force due to the electric field given by Lorentz Transforming the magnetic field of A1. The electric force creates the current. So far, so good.

B1 is what I'd like to have explained.

B2 is B1 boosted to the frame of the loop. I think this is similarly explained by saying that there is an electric force due to the electric field given by Lorentz Transforming the magnetic field of B1. Though I'm not entirely sure about this, because it seems like that would also be true if the magnetic field was uniform (in which case there should not be an induced current).

The reason B1 is confusing to me is because, by assumption, there is no electric field, and there is also no force due to the magnetic field since the velocity of the charge carriers is parallel to B. So what is inducing the current here?

Answer 1

Just a small detail, Faraday's law is about inducing electromotive forces (emf's), not currents. To make the link between current and emf, you'll need to go beyond Faraday's law and make some assumptions on the impedance of the loop.

For $B_1$, yes your field: $$ B = \rho\phi e_\phi -ze_z $$ satisfies $\nabla\cdot B = 0$. However, it is only locally defined; it does not define globally a magnetic field, since $\phi$ is defined up to an integer multiple of $2\pi$. For your argument, all you need is a globally defined magnetic field with no radial component. The simplest way is to think in terms of magnetic vector potential $A$ with only a radial component: $$ \begin{align} B_\phi &= \partial_zA_\rho & B_z &= -\frac{1}{\rho}\partial_\phi A_\rho \end{align} $$ You can take for example: $$ A_\rho = \rho z\cos\phi $$ to get: $$ \begin{align} B_\phi &= \rho\cos\phi & B_z &= z\sin\phi \end{align} $$ which is a well defined magnetic field.

Your previous argument still works, if the velocity of the loop is along $z$, the magnetic force is radial so there is no magnetic emf. You assume that there is no electric field, so no electric emf. Thus no emf. Alternatively, you can check for the flux. The magnetic field is stationary so you just need to look at flux variations due to the motion. The easiest way to calculate it is by looking at the flux through the disk, which is: $$ \Phi = \int B_z \rho d\rho d\phi $$ and $\Phi=0$ thanks to the $\sin\phi$ factor, in particular, it is constant. Thus Faraday's law checks out. As you can see, even if the magnetic field is non uniform, $\nabla\cdot B$ forces it to have a non changing magnetic flux, which is why Faraday's law is valid. In general, you can prove that: $$ \dot \Phi = -\oint (v\times B)\cdot dx $$ where the variation of the flux is due only to the motion of the loop in the stationary magnetic field, assuming that $\nabla\cdot B = 0$. Since by construction you want the RHS to be zero, your LHS will also be zero.

Btw, without the assumption $\nabla\cdot B = 0$, the formula becomes gets an extra term: $$ \dot \Phi = -\oint (v\times B)\cdot dx+\int(\nabla\cdot B)v\cdot d^2x $$ You can check that this is the case in your naive example $B = ze_z$.

Hope this helps.