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Is it possible to "extract" energy from a magnet, making it lose its magnetism? Or, to put in another way, is magnetism a form of energy? (I am not talking about potential energy in a magnetic field). Since matter is equivalent to energy, is the property of magnetism equivalent as well?

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Perhaps, the following paper is relevant - "Extracting Energy from an External Magnetic Field" http://arxiv.org/abs/1208.1702

I have only quickly perused it, but it appears to provide a way to extract energy from a uniform, uni-directional magnetic field by surrounding it with a rotating cylindrical magnetic insulator.

Possibly, this approach could extract energy from a portion of a magnet's field which is small enough to be approximately uniform.

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You cannot extract energy from magnets. You can just disrupt the orientation of the molecules(dipoles) in the magnet(by heating or so..). The answer to the question is same as the answer to the question, "Can you remove charge of an electron??" or "Can i extract energy from the charge of an electron?". Its impossible. You can just disrupt the molecules or decompose the electron(to quarks).

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Magnetism (as we know it in magnets) is nothing more than the alignment of a great number of atoms in similar directions so they end up producing a noticeable magnetic field. In a way you're taking advantage of the inherent kinetic energy associated with electrons orbiting a nucleus. The question would then be could we take advantage of that energy - which we really can't without disrupting the atom itself. The energy we get from atoms (fusion/fission) is from the bonds in the nucleus. The orbital speed of the electrons are defined by their orbital (energy level), and to modify it disrupts the atom's makeup.

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Heating a magnetized bar demagnetizes it, it shows it is in a lower energy state than the ambient temperature . It can only gain energy and be destroyed not give energy. –  anna v Sep 4 '12 at 3:11
    
@annav It could still be in a metastable state, with heating required to take it over the barrier, but generally it will depend on the specifics of the thermodynamics. –  Emilio Pisanty Sep 4 '12 at 13:25
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