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 not suitable for this application, unless wedesign for the inside current the same kind of circular contactwhich is required for the outside current. To be blunt,we could even rule out entirely the unitswhich 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 m per meter of length. For a half-axle of length L=1.5", that's 0.02 m (That wouldbe 0.01 m if both sides of the axle are usedto carry current, but substtuting brass for copper increases the resistance bya 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 be the resistivity of copper. The electric field E inside the disk is radialand the current density j is proportional to it (that's Ohm's law: j = E). 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 / (2r e) = E
(Incidentally, byGauss's law,there's a static charge o I/ on the axle.) The voltage U is the integral of E dr from axle (r0 ) to rim (r1 ) :
U = E dr = [ I / (2 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 / ( 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
Such a thickness should provides good enough structural integrityand doesn't force too large a gap between the magnets,(which would reduce the field, the flux and the voltage). This corresponds to 22-gauge coppersheet (thickness 0.025" = 0.635 mm). Also usable are24-gauge (0.508 mm) or 20-gauge (0.8128 mm). 18-gauge copper is just about1 mm (0.04") which is probably too thick.
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 whichinteract with the copper lattice). However, the magnetic field linesare 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 sameeffect as if the magnets were stationary! If we do that, we can bypassall of the precision machining and the risky business of maintaininga small gap between two powerful magnets: Just sandwich the copper disk between the two magnets and spin the whole thingas a massive flywheel!