When a permanent magnet attracts some object, lets say a steel ball, energy is converted into for instance kinetic energy and heat when attraction happens, and they eventually collide. Does this imply that energy is drawn from the magnetic field and the magnet is depleted, making it weaker and weaker for each magnetic attraction performed?

(if the answer requires Quantum Mechanical explanations, please elaborate :))


3 Answers 3


No, there is no need for the permanent magnets to lose any internal energy or strength when they are used to do work. Sometimes they may weaken but they don't have to.

The energy needed to do the work is extracted from the energy stored in the magnetic field (mostly outside the magnets), $\int B^2/2$, and if the magnets are brought to their original locations, the energy is returned to the magnetic field again. The process may be completely reversible and in most cases, it is.

Imagine two (thin) puck-shaped magnets with North at the upper side and South at the lower side. If you place them on top of each other, the magnetic field in the vicinity of the pucks is almost the same as the magnetic field from one puck – the same strength, the same total energy.

However, these two pucks attract because if you want to separate them in the vertical direction, you are increasing the energy. In particular, if you separate them by a distance much greater than the radius of the puck's base, the total magnetic field around the magnets will look like two copies of a singlet magnet's field and the energy doubles.

If the magnets are close, the magnetic energy is $E$; if they're very far in the vertical direction, it's $2E$. You may consider this position-dependent energy to be a form of potential energy (although there are some issues with this interpretation in the magnetic case when you consider more general configurations: in particular, a potential-energy description becomes impossible if you also include electric charges), potential energy that is analogous to the gravitational one. Gravitational potential energy may be used to do work but it may be restored if you do work on it (think about a water dam where water can be pumped up or down). Nothing intrinsic has to change about the objects (water) and the same is true for the magnets.

Let me mention that for a small magnet with magnetic moment $\vec m$ in a larger external magnetic field, the potential energy is simply $$ U = -\vec m\cdot \vec B $$ Independently of the magnetic field at other points, the potential energy is given simply by $\cos\theta$ from the relative orientation of the magnetic moment and the external magnetic field (times the product of absolute values of both of these vectors).

  • $\begingroup$ That's what I have first thought about, but it does not stand when we compare this solution to the electromagnet's solution. Griffith shows the generator provides the extra work. Magnetic force does not work. $\endgroup$
    – Shaktyai
    Commented Sep 19, 2012 at 9:52
  • $\begingroup$ @Luboš Motl any insight on that comment? $\endgroup$ Commented Sep 19, 2012 at 10:22
  • $\begingroup$ It's been discussed in detail in a previous Physics SE question I can't find right now. It's a misconception that magnets never do work. Of course that they do. They only don't do any work when acting on a charged particle without a magnetic moment: the magnetic Lorentz force is orthogonal to the velocity so the speed doesn't change. But magnets do work on magnets etc. It's very transparent e.g. exactly in the case of a particle with a magnetic moment for which the potential energy was written above. $\endgroup$ Commented Sep 19, 2012 at 11:25
  • $\begingroup$ Update: here is the thread about that issue: physics.stackexchange.com/q/10565 $\endgroup$ Commented Sep 19, 2012 at 11:28
  • 1
    $\begingroup$ Thanks! Great explanation, and good discussion in the referenced thread :) $\endgroup$ Commented Sep 19, 2012 at 12:49

An easy way to see the answer is to switch the question from magnetic forces to gravitational forces.

When the Earth attracts a steel ball, energy is converted into for instance kinetic energy and heat. Does this imply that the Earth is depleted, making it weaker and weaker for each gravitational attraction performed?

Of course the Earth's gravity doesn't "run out" by attracting things. (It actually gets stronger!) You release energy by bringing two gravitating objects together, but that doesn't contradict energy conservation, because you need to spend the exact same amount of energy to pull them apart again. Same with magnets.


I had the same problem a couple of month ago. If instead of a permanent magnet one uses an electromagnet, Griffith shows that the generator provides the extra energy. So for a permanent magnet it is reasonable to assume the energy is not drawn from the field (if it was the case,the magnetic field would work). For a permanent magnet, the most plausible source of energy is magnetic domain reconfiguration. In other word, everytime you use a permanent magnet it looses some energy.

  • $\begingroup$ I do not agree with the suggestion that "every time you use a permanent magnet it loses some energy"; not least because a permanent magnet does not contain any energy in the first place. As an analogy, by 'using' the gravitational field of the Earth to do work, the Earth does not lose "gravitational energy" (rather the opposite, in fact) $\endgroup$
    – Martin CR
    Commented Jan 9, 2022 at 15:26

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