http://mysite.du.edu/~jcalvert/tech/faraday.htmBrowse the rest of the site BTW -- great education!
The resolution to the riddle is that the magnetic field is a product of rotational symmetry. Rotating the magnet
necessarily, cannot have any effect, because it's already spinning!
Using a horseshoe magnet adds confusion, but the fields remain the same.
Where is the counter-torque generated? On the loop shorting across the disc: between the axis and the slip ring contact.
It still seems unbelievable, because one can mentally construct a situation where B field is penetrating the disc, and exists nowhere else (or at least, nowhere asymmetric and nonrotating). But this is a faulty construction! The laws of electromagnetism do not allow sharp cutoffs to fields, nor is a field allowed to decay over any distance, without also fringing away.
Thus, whether by distance or by shielding, the magnetic field must always be whole (it loops back on itself, because \$\nabla \cdot \vec{B} = 0\$ ). And anywhere within that field that you might place a return wire, it must necessarily be exposed to the same enclosed flux. You can equally well ask, what should happen if the disk is nonrotating and the contact is revolved around it? And again, the answer is the same, a voltage is created.
Tim