I saw a demo in an ElectroMagnetics course that illustrated this. The
TA took an aluminium tube, about a meter and a half long and 3/4" ID.
First he dropped a cylindrical plug made of wood down the thing and it
slid right down like we would expect. Next he took a similiar sized
plug with a permanent magnet embedded inside and dropped it down the
tube. It took about ten seconds to traverse the length and drop out
the bottom.
The changing magnetic field, changing as the magnet falls down, induces
a current that circulates through the tubing. The current, which varies
at each point as the magnet falls induces a secondary magnetic field that
opposes the original one. All this serves to buoy the magnet as it
slides down the tube.
One interesting point made by this illustration is the induced magnetic
field must oppose the original one. If otherwise, the magnet would
shoot out the bottom at an increasing rate of speed (ala Cannon!) and
defy every principle of conservation of energy written by man.
ok..jef
On Mon, 24 Apr 2000, Wagner Lipnharski wrote:
{Quote hidden}> Hi Paul, this is what I think, I could be wrong.
>
> The old theory says that changes in magnetic field generate current at
> any electric conductor inserted in that field. I believe it is reversed;
> "When crossing magnetic field lines, electrons in an electric conductor
> will be pushed to one side of that conductor".
>
> The stored energy in a steady magnetic field can be considered a result
> of energy transformation, so the current you see at the coil after
> removing power is in real back induced current. If this is not true,
> than a loaded secondary coil stealing energy from the magnetic field
> (previously generated by the primary coil) would be creating energy,
> what is not possible.
>
> A good example is feeding an auto-transformer bigger coil with 1A, the
> same 1A will be crossing the smaller coil. Removing power and if the
> reverse diode is installed only at the smaller coil, you will have a
> current through the diode higher than 1A, so I think it is not just a
> decay current, it is an "induced" one.
>
> Switch
> -----
> +V o----o o-------.
> |
> S
> S
> S
> S
> S
> ^ S bigger
> | S induced
> | o----------. current
> | S | |
> load S --- |
> current S A v
> S |
> | |
> o----------'
> |
> Gnd
>
>
> I use to think that when a magnetic field is collapsing, the power used
> to generate the field would be reapplied (with some losses) over the
> coils. A changing magnetic field will push electrons away from their
> atoms toward one side of the conductor. If there is a path it will be
> keep pushing away (current), lower the load impedance higher the
> current. It doesn't matter if there is just one or a thousand of wire
> turns, with load it will try to consume all the stored energy from the
> collapsing field.
>
> Without load, electrons will be just accumulating (voltage) more and
> more, creating a high voltage at the extremes of the coil, then, after
> the field collapsed completely electrons return back to their original
> atoms (well, something like that), wasting energy as heat. So the
> collapsing field energy will be consumed in any way, with or without a
> load.
>
> The "induced current" here is the same as moving a permanent magnet in
> the middle of a coil.
>
> Wagner
>
>
> "Paul B. Webster VK2BZC" wrote:
> >
> [snip]
> > That is woolly. There is no "back induced current", just one that
> > decays.
> > Cheers,
> > Paul B.
>
Jeffrey D. Spears
University of Michigan
College of Engineering
``Double-E, can't spell gEEk without it!''
-Captain Gerald M. Bloomfield II, USMC
(my brother)
