[Phys-L] Re: The Meissner Effect Revisited

[Bernard Cleyet <anngeorg@PACBELL.NET>]

Here's Ron's reply to "our" request.  You'll note that it includes JD's #3

If I were still at UCSC, I'd try to make a dropping magnet apparatus for
our second sound Dewar.  "We" for a time did a demo. using a Pb watch
glass.  I tried Sn, which worked, but, of course, not for long.  Hi Q
water supplies use pure Sn pipe.

bc, who wonders about the price of hi. temp. super c.  pipes.  Are they
   type I super C.?

---------------

Bernard,

The quote of Goodstein is:† "If the magnet ever made it inside the tube,
then it would be in freefall." to which Feynman responded, "Of
course!††It slows down going in, gets an extra kick going out and winds
up just where it would have been."†â€

David was simply applying energy conservation.† Since any loss in
gravitational potential energy as the magnet falls within the magnet
cannot be dissipated by the superconductor - ie, the induced currents
dissipate no energy - it must go to kinetic energy.† But the magnet does
get slowed down on entering, since the induced current produces an
upward force on the magnet (due to Faraday's law) - but it doesn't come
to a stop if it once enters the tube.† And, as Feynman said (correctly,
incidentally!) the magnet would be rejected as it leaves the tube, since
there cannot be any loss of energy.† That is, the final speed must be
the same as if it had just been dropped the same distance.

My talk took a different approach - ie, I looked at the induced current,
magnetic field due to the induced current and the force that field
exerts on the falling magnet to explain the same thing.† (If energy
conservation explains something, there must also be an explanation based
on the forces acting.)† When a magnet is dropped through a copper tube,
it is Faraday's law that induces the eddy currents, which in turn
produce the field and magnetic force that retards the motion of the
falling magnet. †The loss of energy is due to the resistive heating
within the copper. †But in a superconductor, once the magnet is inside
the tube, it is the Meissner effect and not Faraday's law that is
responsible for the induced current.† The net magnetic force acting on
the magnet due to the induced current and corresponding field is zero -
leaving only the gravitational force to act on the falling magnet.† (All
of this assumes an ideal "conventional" type I superconductor so that
magnetic field does not penetrate the body of the superconductor beyond
the microscopic London penetration depth.)† So the net effect is that
the magnet is in freefall while inside the superconducting tube.† Just
what Goodstein said and Feynman agreed to (as did the rest of us!).

cut


Ron



====================
Ronald Brown
Professor of Physics
Cal Poly - San Luis Obispo
805-756-2439
http://www.calpoly.edu/~rbrown


Chuck Britton wrote:

> I  would really appreciate it.
>
> Any 'ah-HA' experience of Feynman's is worth hearing.
>
>
> At 7:59 PM -0700 7/9/05, Bernard Cleyet wrote:
>
> > Do you want me to ask him.  I suspect he remembers me.
> >
> > bc
> >
> > > >
> >
> > > > "Then a surprising answer was offered by Caltech's Goodstein - which
> > > > evoked an immediate and gleeful "Of course!" from Feynman (and I might
> > > > add, from the rest of us as well!). It was like being on hallowed ground
> > > > ..."
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