35
$\begingroup$

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?)

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – Criggie Jun 12 '18 at 3:45
  • $\begingroup$ 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. $\endgroup$ – JDługosz Jun 12 '18 at 8:53
23
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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>) $\endgroup$ – Hot Licks Jun 11 '18 at 22:51
  • 3
    $\begingroup$ 1) 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? $\endgroup$ – Ján Lalinský Jun 11 '18 at 23:44
  • $\begingroup$ 2) any scientific reference for the effects of 'anchoring' $\endgroup$ – lalala Jun 12 '18 at 12:46
  • $\begingroup$ @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 $\endgroup$ – th_phys Jun 12 '18 at 13:54
  • $\begingroup$ 1) 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 en.wikipedia.org/wiki/Ferromagnetism#Exchange_interaction $\endgroup$ – Ján Lalinský Jun 12 '18 at 18:17
8
$\begingroup$

This is a corrected version of my deleted answer.

Magnetic material can be magnetized, i.e. all the tiny magnetic domains (in a ferromagnet for example) can become "permanently" oriented by using a strong external magnetic field. One then has a permanent magnet that has potential energy stored in the orientation, an ordered structure . Similar to a crystal structure , which is stable until external effects supply energy to destroy it .

During magnetization of a material, there is what is called a magnetocaloric effect, which shows that the process of creating a magnetization is exothermic.

The increasing external magnetic field (+H) causes the magnetic dipoles of the atoms to align, thereby decreasing the material's magnetic entropy and heat capacity. Since overall energy is not lost (yet) and therefore total entropy is not reduced (according to thermodynamic laws), the net result is that the substance is heated (T + ΔTad).

This article summarizes how one can demagnetize a magnet:

Demagnetize a Magnet by Heating or Hammering

If you heat a magnet past the temperature called the Curie point, the energy will free the magnetic dipoles from their ordered orientation

Heating and hammering can happen when the magnet is used to mechanically move objects, or in a dynamo, due to friction and induced magnetic effects in surrounding materials

......

Self Demagnetization

Over time, most magnets naturally lose strength as long range ordering is reduced. Some magnets don't last very long, while natural demagnetization is an extremely slow process for others.

This is due to thermodynamic processes as at any temperature in a solid there are vibrations and rotations which generate black body radiation and loss of orientation.

......

Apply AC Current To Demagnetize a Magnet

One way to make a magnet is by applying an electrical field (electromagnet), so it makes sense you can use alternating current to remove magnetism, too.

This reverses the process that magnetized the material.

Perpetual motion machines which use permanent magnets all can only work until the effects of demagnetization from friction, heating, and stray magnetic fields (if an attempt is made to extract work) , will demagnetize the permanent magnets.

$\endgroup$
  • 2
    $\begingroup$ I have never seen a machine with claimed perpetual motion where the correct explanation is that the energy comes from demagnetizing a permanent magnet. The magnetic energy is about a million times less than typical chemical energies, a few joule per litre or about a joule per kilogram. It cannot be larger than the magnetocrystalline anisotropy energy, which is micro-eV per atom. $\endgroup$ – Pieter Jun 12 '18 at 7:11
  • $\begingroup$ @Pieter my previous answer was wrong, if you notice in this I attribute the demagnetization in perpetual motion machines to the reasons listed above. They could be in perpetual motion if there were no friction etc, and any attempt to extract work increases friction etc. $\endgroup$ – anna v Jun 12 '18 at 7:35
  • $\begingroup$ But demagnetization has nothing to do with machines coming to a halt. And my point is a bit more general, something that the free-energy crowd does not seem to realize, that there is almost no energy to be had from the magnetization. $\endgroup$ – Pieter Jun 12 '18 at 8:08
  • $\begingroup$ @Pieter right, because the magnetization process is exothermic, it releases energy when magnetized and needs energy to become demagnetized, cannot give energy with demagnetization. That was my mistaken answer and I corrected it here and gave links $\endgroup$ – anna v Jun 12 '18 at 8:23

protected by Qmechanic Jun 12 '18 at 5:47

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.