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Can a magnet ever wear out or lose strength?

If you break a magnet it (seemingly) gets weaker, but what about from normal use?

Or even very heavy use, like placing 2 magnets facing each other, so that they detract from each other, does that strain cause it to wear quicker?

(Note, I'm not looking for a merely yes or no answer; If yes, what will cause it to wear out quicker or slower. If no, why?)

Nat
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Welz
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    Not an answer, but I recall being told as a kid to never drop magnets because that could decrease the magnetism. Does that inspire an answer from anyone else? – Criggie Jun 12 '18 at 03:45
  • Oh yes; before you were born (I guess) they really did, and classroom magnets were stored with a bar across it to help it keep longer. Consumer-priced “rare earth” magnets changed everything. – JDługosz Jun 12 '18 at 08:53

2 Answers2

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Yes, a magnet, as time passes, will lose part of his strength. There are two main reasons:

  1. Thermal energy: it causes the disorientation of the atomic magnetic momenta.
  2. If you have a bar magnet free in space it’s easy to see (using Ampère’s law) that there is inside it a magnetic field $H$ opposite to the magnetisation of the magnet. In order to avoid this phenomenon you should anchor it (that is to say “linking north with south pole“) with a ferromagnet.

This two phenomena will cause atomic magnetic momenta to disorient, and, in so doing, the magnetic strength of the magnet will decrease.

The demagnetisation happens even if you apply a sufficiently strong magnetic field opposite to the one generated by the magnet.

EDIT: The proof of the existence of a field $H$ inside the magnet is now reported: let’s take a bar magnet as shown in figure Magnet bar with magnetisation $M$ that produces magnetic field $B$

The Ampère’s law tells us that $$ \int_\gamma H ds =0\; . $$ Now let’s call $\gamma_1$ the piece of curve inside the magnet and $\gamma_2$ the piece outside with length respectively $L_1$ and $L_2$. The Ampère’s law becomes $$ \int_{\gamma_1}Hds +\int_{\gamma_2} Hds =0\; . $$ Let be $H_1$ the mean $H$ field inside the magnet and $H_2$ the one outside. The integral turns into $$ H_1L_1+H_2L_2=0 $$ From here we have $$ H_1=-\frac{L_2}{L_1}H_2=-\frac{L_2}{L_1}\frac{B}{\mu_0}\; $$ And here we have what was to be demonstrated.

th_phys
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    Some "horseshoe" magnets were very sensitive to having their magnetic fields interrupted. I'm recalling old (Model T era) generators where simply removing the U-shaped magnet from the generator and handling it a little (basically removing it's "keeper") would "kill" it. And other old magnets were sensitive to shock. (Not such a big deal with more modern "hard" magnets like alnico ones>) – Hot Licks Jun 11 '18 at 22:51
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  • should happen to a ferromagnet only if its temperature gets close or above Curie temperature, shouldn't it? Otherwise magnetised medium is a stable state. 2) Inside bar magnet, H field is indeed antiparallel to magnetisation, but why would one try to avoid that? What happens to medium magnetisation depends on many things, like initial magnetisation, composition and shape, not just direction of the H field, doesn't it?
    • – Ján Lalinský Jun 11 '18 at 23:44
  • any scientific reference for the effects of 'anchoring'
    • – lalala Jun 12 '18 at 12:46
  • @JánLalinský 1) the situation above the Curie temperature (CT) is drastic: the ferromagnet becomes a paramagnet. What happens at temperatures well below the CT is due to the statistical distribution of energy in atoms: there’s the possibility that some atoms are so energetic that the magnetic momenta will disorient from the main direction of magnetisation. 2) look at it from the point of view of energy: given a magnetic moment $\mu$ and a magnetic field $H$, the energy is U=-\mu\cdot H$. The condition above is $H$ antiparallel to $\mu$, this corresponds to a max of energy, so an unstable point – th_phys Jun 12 '18 at 13:54
  • yes there are fluctuations on the microscopic level, but below Curie temperature those do not destroy macroscopic magnetisation 2) why do you think given magnetic moment's energy is given by $H$ field? That is a macroscopic field, on the microscopic level the field is different and depends on the type of atoms involved and their mutual microscopic arrangement in space. In ferromagnets, magnetizing interaction is much stronger than demagnetising interaction, see https://en.wikipedia.org/wiki/Ferromagnetism#Exchange_interaction
    • – Ján Lalinský Jun 12 '18 at 18:17