UPDATE September 23, 2024: Electromagnetism is not "linear" if we take into account the field energy. If electric fields E₁ = -E₂, then the energy density of the sum field E₁ + E₂ is zero, but each individual field has a non-zero energy density. Keeping this in mind resolves many apparent paradoxes.
If an electron e- is close to a conducting object, like a metal wire, that causes a backreaction in the distribution of electrons in the metal. This will significantly affect the inertia which the electron e- feels. In the "cylinder of electrons" in the diagram below we ignored the backreaction. The strong electric field
Eelectr + Ee-
is "diluted" over the entire length of the conducting metal wire. It does not add to the inertia of the electron.
It looks like that Maxwell's equations describe the electromagnetism of charged metal objects. The rules are different for charged electric insulators!
----
The Feynman Lectures on Physics (1964) treats the case where a test charge q moves parallel to to a current-carrying wire. A more difficult case is when q directly approaches the wire.
A wire as two overlapping cylinders of charge: a simple explanation of the magnetic force on an electron e- approaching the wire
Let us model a long wire as two uniformly charged cylinders. The protons form a positively charged cylinder which is static in the laboratory frame. The electrons form another cylinder which moves very slowly relative to the laboratory frame: the speed is of the order of micrometers per second.
wire with a current
●●●●●●●●●● protons
••••••••••••••••• ---> v electrons e-
●● ^ ●●●●● protons
| | W
| | ^ u
| | |
v • e-
W energy packet
^ y
|
------> x
Let us use this simple model:
1. As the electron e- approaches the wire, it receives an energy "packet" W from its Coulomb attraction to the protons. The packet W has a zero momentum in the x direction.
2. The electron must give the energy packet W back to the electrons in the wire, because there is a Coulomb repulsion. This time W must comove with the electrons in the wire, to the right at the velocity v. The packet must contain a momentum
p = W / c² * v
to the right.
3. The electron is left with a momentum p to the left. This is the "magnetic" force which accelerates the electron to the left.
Let us next analyze this in detail, trying to make sense of the field energy and momentum.
A uniformly negatively charged static cylinder and an approaching electron e-
=============== - cylinder of electrons
^ u W = F Δy
|
-v <--- • e-
|
v F Coulomb's force
^ ^
| |
Eelectr Ee-
^ y
|
------> x
The cylinder consists of the electrons inside a wire. As the electron e- approaches the cylinder, it loses a kinetic energy W. There is no change in the x momentum of e-.
Coulomb's force describes the interaction between charges: we ignore the momentum in the electric field. Since e- loses kinetic energy, its inertia is diminished. The electron accelerates to the left. In the frame of the protons in the wire, this acceleration in interpreted as the magnetic force.
If we take into account the field, then the situation looks different. The extra energy W is now in the combined field
Eelectr + Eq,
which is moving to the left. The momentum to the left of the electron e- and of the extra field energy W has grown. The interaction pushed this combined system to the left.
But now we face a dilemma: if the system acquired a momentum to the left, then the cylinder must have acquired a momentum to the right. The Lorentz force says that the cylinder did not acquire any momentum to the right. Is the Lorentz force formula incorrect, after all?
Conclusions
We are able to derive the magnetic force with a very simple model where we only take into account Coulomb's force, and ignore the electric field momentum.
But if we include the electric field momentum, then the Lorentz force formula seems to be broken. We will investigate this more in the next blog post.
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