Re: [Phys-L] Falling magnet and eddy currents

[John Denker <jsd@av8n.com>]

On 9/18/22 10:09 AM, Antti Savinainen via Phys-l wrote:

> A student asked why there are no eddy currents in the long solenoid
> when the magnet is fully inside, since the situation looks quite
> similar to the aluminium tube case. Of course, one clear difference
> is that the long coil has a much larger resistance than the aluminium
> tube which would limit the effect of eddy currents on the falling
> magnet if there were an induced emf in the first place. However,
> there is a change in magnetic flux through an individual coil turn in
> the solenoid (quite similarly to the aluminium tube) when the magnet
> passes by, although there is no net change of magnetic flux through
> the solenoid as a whole.

	Sorry for the belated response. I've had a lot of
	demands on my time lately. I put in a 20-hour
	workday yesterday, as the election inspector in
	charge of a very busy polling place. And that's on
	top of a setup session late Monday. We had a whole
	bunch of new technology, i.e. computerized sign-in
	as opposed to the pen-and-paper methods used in
	previous years. We had to figure out a lot of stuff
	on the fly. That increased the excitement, but we
	got it done. Even the partisan poll watchers, who
	were not predisposed to be friendly toward us, were
	amazed at how smoothly things went.

Perhaps I am missing something, but I see the solenoid
question as relatively simple. The executive summary goes
like this:

 -- There *are* eddy voltages within the solenoid, if we
    look inside.

 -- The algebraic sum of the eddy voltages is zero. So if we
  don't look inside, i.e. if we treat the solenoid as a
  two-terminal black box, we see zero voltage across the
  terminals, which explains the lack of dissipation.

So, as so often happens with problematic questions, I would
argue that the question is based on a false assumption,
compounded by language that is open to misinterpretation
(focused on the word "in" in the opening sentence, depending
on whether it means "across the terminals" as opposed to
"within").

In more detail:

We agree that a long solenoid exhibits zero voltage across
its terminals. The converse is also important: A stubby
solenoid *does* exhibit voltage across its terminals when
you drop a magnet through it. I say "stubby" to refer to a
small length, since we are about to use "short" to mean
something else. ==> Here "short" and "shorted" are used in
the electrical sense, to refer to a short circuit.

Step 1: Model the long uniform tube as a stack of rings. In
the language of solenoids, each ring is a shorted turn. That
is, we have a stack of stubby solenoids, each with a short
circuit across its terminals. These shorted turns will have
eddy currents as well as eddy voltages. They will dissipate
like crazy. So this is a good model of the uniform tube.

  Tangential remark: If you have a practical inductor or
  transformer, if the insulation fails you get a shorted
  turn. This is considered a very bad thing.

Step 2: Replace the stack of shorted turns with a stack of
unshorted turns. Many of the turns in the vicinity of the
falling magnet will have nonzero voltages. The voltages are
still there, as in step 1, but the currents are now zero
because each turn is open circuited, so there is no
dissipation, so the magnet falls freely.

  Tangential remark: The *core* of a practical transformer,
  if it is made of conducting material, is structured as
  a stack of insulated lamellae, insulated to minimize the
  presence of shorted turns within the material. So the
  concept of a stack of unshorted turns has considerable
  real-world relevance.

Step 3: Hook up all the unshorted turns in series. So we now
have a normal solenoid. The series combination adds up to
zero voltage. This should be obvious by symmetry: The part
of the solenoid the neighborhood of the falling magnet has
mirror symmetry with respect to the middle of the magnet.
Meanwhile, the far-away parts have zero field, so the
distance to the end of the longish (but not infinitely long)
solenoid doesn't matter much. So the two-terminal voltage of
the longish solenoid is zero, as observed.

Step 4: Even if we short the terminals of the longish
solenoid, there is no dissipation, because there is no
current, by Ohm's law (zero voltage over some nonzero
resistance in the coil). [I'm assuming ordinary materials,
not superconductors.]

===

I think this series of models (shorted turns, unshorted
turns, series combination) accounts for all the
observations, in reasonably simple terms. If not, please
re-ask the question.