MadSci Network: Physics

Re: How do you resolve the Faraday Disc paradox?

Date: Thu Sep 16 02:45:50 1999
Posted By: Yaxun Liu, Grad student, Electrical Engineering, National University of Singapore
Area of science: Physics
ID: 936896916.Ph

These are very good questions which can help us understand the interaction of electromagnetic field and matter.

First let's resolve the paradox of Faraday's disc.
In fig.1 and fig.2 the Faraday's disc serves as a generator. In fig.1 the magnet is attached to the disc. When the disc is rotating, no electromotive force (EMF) is generated across its radius (OC) since the disc has no relative movement to the magnet. However, since the magnetic force line is moving across the wire segments CDEFO, a EMF is generated along CDEFO and a net EMF is resulted in the loop OCDEFO, which causes a current. In other words, it is equivalent to fixing the disc and the magnet and rotating the wire frame around them.

In fig.2 the magnet is detached from the disc. If the disc is rotating, there is a EMF generated along OC but not along CDEFO, and therefore a net EMF in the loop. However, if the disc is stationary but the magnet is rotating, EMF's are generated along both OC and CDEFO, but they cancel each other and there is no current in the loop.

In fig.3 and fig.4 the Faraday's disc serves as a motor. In fig.3 the magnet is attached to the disc while in fig.4 it is detached from the disc. In both cases there is a loop of current in OCDEFO therefore a torque on the current. This torque tends to rotate both the disc and the wire frame around the magnet. However, in fig.3 the disc is attached to the magnet and the wire frame is constrained by supporting structures, therefore both of them will not rotate. In fig.4 the wire frame is constrained by supporting structures but the disc can rotate, so it just rotates under the torque.

Now the question: Does the current in the disc (acting either as generator or motor) react against the mechanical magnet or against the magnetic field?

Answer: Generally, the current reacts against electromagnetic fields. The current is actually charged particles which are moving. Moving charged particles generate electromagnetic fields and "feel" forces exerted by electromagnetic fields. Charged particles interact with each other through "disturbance" of electromagnetic fields, which propagates at the speed of light. Electromagnetic fields have momentum and kinetic energy distributed over the whole space. When a moving charged particle changes its momentum or kinetic energy by interaction with the electromagnetic fields, it causes a disturbance of the momentum or kinetic energy distribution of the electromagnetic fields surrounding it, and this disturbance propagates through the space and affects other charged particles.

Question: If against the magnetic field, how is this reaction force "absorbed"? Does the answer to the above explain how attaching the magnet to the rotating disc prevent motor action but not generator action?  To generate equal amounts of output power with and without the magnet attached to the disc, is equal mechanical input power required?

Answer: The "reaction force" exerted by a charged particle on the electromagnetic fields can be described by the exchanges of momentum and kinetic energy between them, as explained in my answer to the previous question. This does not explain how attaching the magnet to the rotating disc prevent motor action but not generator action since in both cases the interaction mechanism is the same. According to the energy conservation theorem (until now, there is still absolutely no concrete evidence to overthrow it :-), to generate equal amounts of output power with and without the magnet attached to the disc, equal mechanical input power is required, if the friction between the magnet and the air can be ignored.

Question: Why do physics books always discuss the force on a charged particle moving in a magnetic field, but never, that I have seen, the force of a charged moving particle on the field?  Is there a force on the field (invoking Newton's laws dosen't "explain" a YES answer)

Answer: A charged moving particle does exert a "force" on the field, however, this "force" can only be represented by the exchange of momentum and energy between the particle and the electromagnetic fields, therefore, its calculation is based on the definition of the momentum and energy of the electromagnetic fields, which is complex. That may be the reason why most physics books do not discuss the force of charged moving particle on a field.


[1] L. Pearce Williams, Michael Faraday, A Biography, Reprint, originally published: New York: Basis Books, 1965, ISBN 0-306-80299-6, pp.194-197, pp.202-204. This book has a brief description of the Faraday's disc.

[2] Marijan Ribaric, Luka Sustersic, Conservation Laws and Open Questions of Classical Electromagnetics, World Scientific Publishing, ISBN 981-02-0151-6

[3] nrg/n-mach.html. This is an interesting article about the "homopolar generator" (actually it's just Faraday's disc with attached magnet). It also gives some links to related topics. There is a flood of materials about the "free-energy" machine based on  Faraday's disc on the web, but their credibility is deeply doubted. From my personal point of view, there is no mystery in the Faraday's disc which can not be explained by the classical physics.

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