Faraday disc volatage doubling

Question

imagine two cases, in both cases we have two faraday discs spinning on the same axis in the same direction but let's imagine each disc has it's B field lines pointing in the opposite direction so that in one disc the current goes from center to rim and in the other the current goes from rim back to center, 
in one case we simply electrically connect the rims together with a stationary connection and take the centers and connect them to brush contacts, in the  other case we add a stationary brush that connects the outer rims of the discs and still have the center brushes, tell me in which case there will be higher voltage?

I think that theoretically since even without sliding contacts attached there is a charge difference on a conducting disc that is spinning inside a homogeneous B field, I think that if the B field directions and current directions add up in series the effect should be the same on both discs or is it?

thanks.

Philip Wood · Answer

With the brush contacts joined by a wire stationary in the lab frame of reference there will be no induced emf in the wire and so the emf between brushes contacting the two axles will be, as you suggest, the sum of the emfs induced in the discs.

Things will be different when the edges of the disc are joined by a wire that rotates with the discs. To understand why, you need to think about the magnetic fields that you say run in opposite directions through the co-axial discs. Suppose the field goes from left to right through the right hand disc and from right to left through the left hand disc. Because magnetic field lines are continuous closed loops, there has to be a field coming (essentially) radially inwards in the region between the two discs, and this radial field will be cut by the rotating wire linking the discs. The emf generated in this wire will tend to cancel the sum of the emfs generated in the discs themselves.

Faraday disc volatage doubling

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