How to explain different voltage and symmetry?

Question

Imagine a magnet is falling along the axis of a uniform metal ring, there is an Eddy current in the ring due to changing magnetic flux and the direction of the current is determined by Lenz's law. Take two points on the ring on each end of the same diameter, let's call them A and B. Current flows from A to B it means A has higher voltage than B, then current flows from B to A, so B has higher voltage than A! I am missing something here; can you please help me understand this? What is voltage between A and B (suppose we have perfect symmetry).

Answer 1

There are two concepts of "voltage between two points $a,b$" in use.

One is dependent on the path chosen: it is based on the integral of total electric field, through the chosen path $\gamma$ joining the points $a,b$:

$$ \text{voltage_v1} = U_{\gamma} = \int_\gamma \mathbf E\cdot d\mathbf s. $$

This clearly depends on the choice of the path, not only on the points $a,b$. In your example, when two close points $a$, $b$ are assumed, then the integral will be zero when direct joining line is chosen, and will have non-zero value when the closed loop is chosen.

The other (and a lot less confusing) concept of voltage is that it is difference of electric potential $V$. Electric potential can be always defined based on the conservative part of the electric field, and the most usual way is to assume it is given by the Coulomb integral of charge distribution, irrespective of presence of induced field:

$$ V(\mathbf x) = \int K\frac{\rho(\mathbf x')}{|\mathbf x- \mathbf x'|}~d^3\mathbf x'. $$

This way, drop of potential when going from $a$ to $b$ is function of the points $a,b$ only (and the chosen positive direction - here, from $a$ to $b$).

$$ \text{voltage_v2} = U_{a\to b} = \text{potential drop when going from $a$ to $b$} = V(a) - V(b). $$

This concept of voltage is used not only in electrostatics, but also in the theory and practice of AC circuits. E.g. voltage on a perfect inductor is $LdI/dt$, which does not depend on the choice of path, only on the endpoints (usually chosen to be terminals of the inductor).

In your scenario with falling magnet, the magnet produces induced electric field, which causes induced EMF in the ring (or a pipe), which drives current there. But contribution of this induced field to conservative electric field is zero, so it changes nothing in the electric potential field and thus no change in voltages.

However, in addition to induced electric field, moving magnet may also bring in its own (very weak, proportional to velocity) conservative electric field. Even if the magnet is not charged in its own frame of reference, it may carry charge distribution in frames where it is moving. This kind of motion-induced charge density on the magnet is non-zero wherever magnetization current $\mathbf J_M$ has a component in line of the magnet velocity $\mathbf v$ (if the component is opposite to $\mathbf v$, then the charge density is negative). This can usually happen on parts of magnet's surface, where there is non-zero magnetization current.

For example, if magnet in shape of a short cylinder falls in the pipe sideways (poles facing the walls), then one half of the magnet's side mantle carries positive charge, and the opposite half carries negative charge, and this means there is conservative electric field due to this charge distribution. This means electric potential of air near one half side of the magnet will be different than potential of the air near the opposite half side.

However, as the pipe is conductive, there won't be much difference of potential in the pipe due to this - charge distribution on the pipe will adapt and create mirror charge that cancels magnet's conservative electric field in the pipe conductor.

Answer 2

Voltage is a difference in electric potential between two points. However, electric potential is only defined for electrostatic fields, i.e. fields due to separated positive and negative charge. In the case of a moving magnet, there is indeed an electric field induced in the nearby conductor, but it is not an electrostatic field.

Put another way, there are two ways to generate electric fields: separated electric charge, and changing magnetic fields. Only the first way brings about an electric potential.

Edit: I should add that in your example there is an emf generated in the ring. This example shows that emf is not always associated with voltage.