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Spin-Induced E Fields and Angular Momentum

G.R.Dixon, 5/13/2007

Let us define an uncharged current loop to be a non-spinning circle of negative line charge superimposed on an equal, spinning circle of positive line charge. The loop has a dipolar B field and no E field.

If the loop translates in its plane then there is a polarization of the positive charge; the loop acquires a dipolar electric field (in accordance with the general field transformations).

If we now model a disc-shaped permanent magnet as an array of such microscopic current loops, then each loop becomes electrically polarized when the magnet spins. Collectively the electric dipoles engender an electric field with radial components. This (conservative) field exists both within and outside the magnet. However, if the magnet is conducting (metallic) then conduction electrons flow such as to drive the E field to zero, both within and outside of the magnet. If the magnet is non-conducting (e.g. ceramic) then the E field will have observable effects when the magnet spins.

Now assuming B is not changed by a magnetís rotation, the rotational kinetic energy of the metallic magnet is

, (1)

where Imech is the mechanical momentum of inertia. However, in the ceramic magnet case the spin-induced electric field has energy, and the rotational kinetic energy is

, (2)

where Eelec is the electric field energy. This implies that the spinning ceramic magnet has more angular momentum than its metallic counterpart, all other things being equal. And of course it would have greater angular momentum than an unmagnetized disc.

This suggests that the best flywheel would be a spinning, ceramic, disc-shaped magnet. Such a magnet would theoretically require a greater energy investment to attain a given w, but that energy would be recouped when w is returned to zero.