# When work is performed solely by magnetism, is there an equivalent loss of energy from the magnetic field?

When two magnets are placed within appropriate proximity and released, the attractive force will perform work and bring them together. Work is performed overcoming friction. Can we measure a reduction in the total energy of the magnetic field?

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Note: $dW=\vec{F}\cdot d\vec{l}$ and $\vec{F}=q \frac{d\vec{l}}{dt}\times \vec{B}$ so $dW=(q \frac{d\vec{l}}{dt}\times \vec{B})\cdot d\vec{l}=q(d\vec{l}\times\frac{d\vec{l}}{dt})\cdot \vec{B}=0$ –  kηives May 15 '12 at 0:07
@kηives not for pure magnetic dipoles. See physics.stackexchange.com/questions/10565/… –  kleingordon May 15 '12 at 1:02

## 2 Answers

Yes, if you can measure the energy stored in the magnetic field, you should find that it is reduced when two magnetic dipoles come together. By way of explanation: the potential energy of a pair of aligned, end-to-end, identical magnetic dipoles can be found using the equation in this Wikipedia article, plugging in $\vec{m}_j = \vec{m}_k = m\hat{z}$, $\vec{e}_{jk} = \hat{z}$, and $\vec{r}_{jk} = r\hat{z}$ to get

$$U = -\frac{\mu_0}{4\pi r^3}\bigl(3 (m)(m) - m^2\bigr) = -\frac{\mu_0 m^2}{2\pi r^3}$$

Here $r$ is the distance between the dipoles and $m$ is their dipole moment. This energy is stored in the dipoles' mutual magnetic field. As they get closer, the energy drops.

It's worth noting that the work done to bring these dipoles together is actually performed by the electric force, for real magnetic dipoles which consist of current loops.

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This would mean that you could "use up" a permanent magnet if you kept doing the experiment over and over? The magnet could lose all magnetism to work performed? –  Ike May 15 '12 at 1:15
No, not for an ideal magnet. –  David Z May 15 '12 at 1:33
@David: The work is done by the force keeping the currents in the loops, which might be electric, or it might be Pauli exclusion. The major dipoles are also spinning electrons which are not really current loops, and in this case, the best interpretation is that the magnetic field is doing the work directly. B can do work on spinning charges, just not on moving point particles. –  Ron Maimon May 15 '12 at 6:13
No, the best interpretation is not that the magnetic field is doing the work directly - not unless the magnetic attraction between electron spins is the sole major contributor to magnetic attraction, and if that were the case, paramagnetism would be a lot stronger than it is. The electrical attraction between the displaced electrons and the nuclei is also important. I suppose one could say that direct magnetic force is partly involved, but it is not the sole agent. –  David Z May 15 '12 at 19:23

From the statements offered I can only assume that we don't actually KNOW the answer. If the magnetic field strength is reduced due to the work performed, either the magnet goes inert or the energy is replaced from somewhere else. My original question was more of a conservation of energy thing and the answered must be "it must be magic".

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No. Your question was "Can we measure a reduction in the total energy of the magnetic field?" and the answer is yes. There is some discussion about the work, but this does not change the answer. –  tmac May 16 '12 at 21:50
@Don it appears likely that you are the original poster of the question, but I note two different accounts. That state of affairs is largely discouraged on Stack Exchange. I can merge the two account if you desire. Just tell me which one you would like to be the master. –  dmckee May 16 '12 at 21:58