Polar plates must be used to reduce the magnetic losses from the center holes.
Ring magnets which have very large center holes (like the huge
NR025 from
Applied Magnets) are simply
not suitable for this application, unless we're willing
to design for the inside current the same kind of circular contact
which is required for the outside current.
To be blunt,
we could even rule out entirely the units
which have holes larger than
1" (ID) in the above table.
The highlighted RY046
(from K&J Magnetics ) is nice enough.
Two stacked pairs of these, would yield 70%
more voltage (up to 570 mV at 10,000 rpm).
With 3 pairs, we would obtain 700 mV (at a cost of $96.00).
If made out of pure copper, a quarter-inch rod will have a resistance of about
0.5 mW per meter of length.
For a half-axle of length L=1.5", that's
0.02 mW (That would
be 0.01 mW if both sides of the axle are used
to carry current, but substtuting brass for copper increases the resistance by
a factor of 4.)
It would be an overkill
to have a copper disk with an axle-to-rim resistance
much below that.
Let's estimate what this entails for the thickness (e) of the disk:
Consider the disk at rest.
Let s be the resistivity of copper.
The electric field E inside the disk is radial
and the current density j
is proportional to it
(that's Ohm's law: j = sE).
At a distance r
from the axis, the current density is equal to the total current I
divided by the lateral area of the relevant cylinder. Thus:
j = I / (2pr e)
= s E
(Incidentally, by Gauss's law, this means
that there's a static charge
eo I/s
on the axle.)
The voltage U is the integral of E dr
from axle (r0 ) to rim (r1 ) :
U = ò
E dr =
[ I / (2p s e) ] Log ( r1 / r0 )
This gives the resistance of the disk as R = U/I.
On the other hand, the resistance of the half-axle of length L is:
R = L / (p s r02 )
Using 2r0 = 0.25", 2r1 = 2"
and L = 1.5",
those two are equal when:
e = ( r02 / L )
Log ( r1 / r0 )
= 0.026 " = 0.66 mm
So, a copper disk with a thickness of 1 mm
will contribute less resistance than the axle itself.
A thicker disk would needlessly increase the gap between the magnets,
thereby reducing the field, the flux and the voltage...
Making it Simpler :
As mentioned above, the lines of current are rigidly attached to the copper disk
(as they are related to the trajectories of charged particles which
interact with the copper lattice). However, the magnetic field lines
are not similarly bound to the magnet.
If a magnet rotates around its axis of symmetry,
the magnetic field stays the same. Thus,
nothing is induced on the copper disk if the magnets spins...
So, if we let the magnets spin at the same rate as the copper disk, we obtain exactly the same
effect as if the magnets were stationary! If we do that, we can bypass
all of the precision machining and the risky business of maintaining
a small gap between two powerful magnets:
Just sandwich the copper disk between the two magnets and spin the whole thing
as a massive flywheel!
