OK. Let me pick your brains. I'm working on the age-old question does
a magnetic field rotate with the magnet if you spin it. People have
established that if you put a loop near a Faraday generator and spin
the magnet you get no voltage. Some have taken this as proof that the
magnetic field does not rotate with the magnet. But others have shown
that the voltage induced in a loop this way is zero because the
voltage in the front side of the loop is exactly canceled by the back
side of the loop. Hence one person has even formulated a "law" that
says you cannot ever determine if the field rotates IF you only use
CLOSED loops!

So here's my idea. We make a magnet so it spins on it's axis. On the
face of the magnet and perpendicular to it we build a loop as
follows: one side (bottom) is a wire along spinning axis. This is a
flux line so there is no induction in this leg. Another side is wire
that is deployed along another flux line so there is no induction
there. The side nearest the magnet is vertical along it's face and
represents where the major voltage will be induced IF the field
rotates. The back side of the coil is longer, but in a weaker field
and represents where the canceling voltage will be induced. The two
wires to the meter leave together so as not to form a loop.

OK. Here's the trick. We now enclose the backside leg in a super-
conducting tube so that no fields can get to that leg to cancel the
voltage. Hence if we see any induction in the loop it proves the field
rotates!

Now here are my doubts and need for knowledge. How does a
superconducting tube work? Does it REALLY keep fields OUT of the
inside of the tube or does the incident magnetic field on the outside
of the tube create currents which create a secondary field which
EXACTLY CANCELS the original field. The point being that the field
inside the tube is not really zero but in truth two equal and opposite
fields which may or may not have relative motion between them. If the
latter, then one would need to know that relative motion to determine
the meaning of the results of the experiments. For example, if the
outside leg were enclosed in an IRON tube, I believe we'd find that
the field of the magnet would move with the magnet, while the
canceling field of the shield would be attached to those Iron atoms
and not be moving. The net result of using an IRON shield would be no
net voltage whether or not the field rotates.

So does this superconducting scheme have a prayer or am I a crank?

Benj
(who notes that answering the rotating magnet question is a "good
first step" toward finding that precise arrangement of magnets that
will produce "free energy"!)

Autymn D. C. wrote:
one star

Thanks for the reply Autymn Womyn, such as it is.
I take it this means that you also have no idea how a super-conducting
tube shields a magnetic field!

The question of the attachment of magnetic fields to reference frames
is not such a simple one, and apparently nobody has ever successfully
performed the electrostatic experiment to test the rotating magnet
question.

Thanks for thinking of me! And thanks for replying in Muttish!

Uncle Al wrote:
Benj wrote:

OK. Let me pick your brains. I'm working on the age-old question does
a magnetic field rotate with the magnet if you spin it. People have
established that if you put a loop near a Faraday generator and spin
the magnet you get no voltage. Some have taken this as proof that the
magnetic field does not rotate with the magnet. But others have shown
that the voltage induced in a loop this way is zero because the
voltage in the front side of the loop is exactly canceled by the back
side of the loop. Hence one person has even formulated a "law" that
says you cannot ever determine if the field rotates IF you only use
CLOSED loops!
[snip crap]

The Benj troll continues to pick up the USENET cranks apparently in
reverse order of general crankosity!

Given an axially symmetric dipole magnetic field, what does "rotate"
mean when the symmetry axis is the rotation axis? Are you implying
that lines of force are physical entities that can flow through a coil
and entrain charge?

I'm not implying "lines of force" are real, I'm implying that qVxB
forces are real! Namely that a relative motion between a magnetic
field and a charge (or conductor) produces measurable forces or
currents. Hence if the magnetic field is rotating with the magnet,
(People who believe this are called "bristle theorists" since they see
a magnet and its field like a rotating hairbursh) there should be an
induced emf in a conductor up the face of the rotating magnet. The
apparent problem, however, is that if you use a closed loop there is
an equal and opposite emf induced in the back side of the loop and the
output is zero whether or not the field actually rotates and the
experiment is indeterminant.

Benj
(who notes that answering the rotating magnet question is a "good
first step" toward finding that precise arrangement of magnets that
will produce "free energy"!)

Idiot.

Humor-impaired! :-)

1) Time is homogeneous.
2) Noether's theorem
3) Mass-energy is locally conserved.

yes, yes.

Uncle Al doesn't see any asterisks referring to strobes, magnets,
ceramic supercons, liquid nitrogen fog, or a salvaged manual
transmission from a 1970 Honda Civic (the classic infinite energy
source after removal of its ****anium restraining shim).

WHOA! I never heard about the ****anium restraining shim before!
Could that be my reason for failure up to now?

Benj

Autymn D. C. wrote:
On May 6, 10:29�am, Benj wrote:
Autymn D. C. wrote:
one star

Thanks for the reply Autymn Womyn, such as it is.
I take it this means that you also have no idea how a super-conducting
tube shields a magnetic field!

The star was for your possessive illiteratia again. Read my posts for
Archimedes Plutonium for how a superconductor works.

Oh Aut! My GOD NO!!! I have to finally be forced to read all the
Archimedes Plutonium superconducting spew AND your responses to it?
YOW! Oh the humanity! I sort of figured it might be best if first got
a little "accepted" superconducting knowledge first....

The question of the attachment of magnetic fields to reference frames
is not such a simple one, and apparently nobody has ever successfully
performed the electrostatic experiment to test the rotating magnet
question.

Magn�ts are not Fermi-Dirac condensva; each domain must break down at
the atom which fluctvently impels as a cog. With the wire's
h�steresis, the inducted currend would be speed-dependent.

I don't think this is quite right, Aut. I mean it might be speed-
dependent, but the fact that a superconductor can support a magnet
shows that the response extends down to DC! (or very close to it).

Currently I am thinking that as the magnetic field moves over a
superconductor it induces an equal and opposite field from
superconducting circulations in the material. This means that the
canceling field is attached to the superconducting material and is NOT
moving along with the inducing field. Therefore, in my little
experiment, the so-called shielded leg would experience the moving
field and a stationary canceling field. This is no different than
trying to shield the wire with iron (which does not work either).

Hence it appears my idea is totally bogus, but I had hopes that
someone here might have had at least a passing acquaintance with
magnetic reactions and how super conductors shield a field. I have
heard the statement (probably on TV so that means it's probably
wrong!) that a super conducting material "forces the flux lines out of
the material" which would mean that inside a superconducting tube
there would only be ONE field, namely that of the canceling
induction. But I think it would take some experimental data to really
settle this.

I suspect my idea is totally bogus. Oh well, not the first time!

On May 5, 7:57 pm, Benj wrote:
OK. Let me pick your brains. I'm working on the age-old question does
a magnetic field rotate with the magnet if you spin it.

If we ever build a Beanstalk (tethered mass in geosynchronous
equatorial orbit) will the stalk carry a current from "cutting" the
Earth's field as it orbits?


Mark L. Fergerson

On May 5, 10:57*pm, Benj wrote:
OK. Let me pick your brains. *I'm working on the age-old question does
a magnetic field rotate with the magnet if you spin it.

Assuming you are spinning the magnet around an axis of cylindrical
symmetry: no. There is no concept of a rotating geometrically
invariant field. Which doesn't mean the field as described in normal
terms will not change. That's a different question -- and you are
free to think of the new field as a "rotating magnetic field" if you
like -- nobody can stop you. ;-)

Much ado about nothing.

...

If we ever build a Beanstalk (tethered mass in geosynchronous
equatorial orbit) will the stalk carry a current from "cutting" the
Earth's field as it orbits?

If conductive, perhaps -- remember B fields, like motion, are expressed
in vectors, so the cross product can be zero even if both terms are not
(they only need be parallel).

"Kevin G. Rhoads" wrote in message
...
| If we ever build a Beanstalk (tethered mass in geosynchronous
| equatorial orbit) will the stalk carry a current from "cutting" the
| Earth's field as it orbits?
|
| If conductive, perhaps -- remember B fields, like motion, are expressed
| in vectors, so the cross product can be zero even if both terms are not
| (they only need be parallel).


No, the Earth's magnetic field moves with the Earth - the magnetic pole
doesn't travel in a circle every day, yet it is offset from the Earth's
axis.
Therefore it rotates with the Earth.
Nobody will build a "beanstalk" anyway, the engineering and safety
requirements are too excessive and the uses too few. 23,500 miles of
steel cable (or even titanium or carbon fibre) won't be lightweight.

On May 12, 8:43 am, "Kevin G. Rhoads" wrote:
If we ever build a Beanstalk (tethered mass in geosynchronous
equatorial orbit) will the stalk carry a current from "cutting" the
Earth's field as it orbits?

If conductive, perhaps -- remember B fields, like motion, are expressed
in vectors, so the cross product can be zero even if both terms are not
(they only need be parallel).

It was a "trick" question. The Earth's field is generated somewhat
differently from that of an ordinary magnet; the angular velocity of
the conductive fluid within the Earth that generates its field is not
directly tied to the angular velocity of the surface (note that the
spin and magnetic poles of the Earth are not co-located and their
relationship changes over time).

So yes, a (conductive) Beanstalk will definitely carry a current.

Benj may feel cheated; I suggest he research the Homopolar Generator
and variants.


Mark L. Fergerson

| If we ever build a Beanstalk (tethered mass in geosynchronous
| equatorial orbit) will the stalk carry a current from "cutting" the
| Earth's field as it orbits?
|
| If conductive, perhaps -- remember B fields, like motion, are expressed
| in vectors, so the cross product can be zero even if both terms are not
| (they only need be parallel).

No, the Earth's magnetic field moves with the Earth - the magnetic pole
doesn't travel in a circle every day, yet it is offset from the Earth's
axis.

The Earth's magnetic field does not move with the Earth out to arbitrary
distances. In the ionosphere the field gets shaped by the interaction
of the solar wind with the internal generator of the Earth (among other
lesser influences). Your analysis is oversimplified to the point of
being flat out wrong. With all due respect: Please, try again.