33
$\begingroup$

From my knowledge of magnetism, if a magnet is heated to a certain temperature, it loses its ability to generate a magnetic field. If this is indeed the case, then why does the Earth's core, which is at a whopping 6000 °C — as hot as the sun's surface, generate a strong magnetic field?

$\endgroup$

closed as off-topic by stafusa, Kyle Kanos, AccidentalFourierTransform, Jon Custer, Rory Alsop Dec 10 '17 at 23:11

  • This question does not appear to be about physics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

26
$\begingroup$

The core of the Earth isn't a giant bar magnet in the sense that the underlying principles are different. A bar magnet gets its magnetic field from ferromagnetism while Earth's magnetic field is due to the presence of electric currents in the core.

Since the temperature of the core is so hot, the metal atoms are unable to hold on to their electrons and hence are in the form of ions. These ions and electrons are in motion in the core which forms current loops. The individual currents produce magnetic fields which add up to form the magnetic field around the Earth.

$\endgroup$
  • 20
    $\begingroup$ It is just molten iron. No need to talk about "unable to hold on to their electrons" - that is true in any piece of iron. $\endgroup$ – Pieter Dec 9 '17 at 10:46
  • 11
    $\begingroup$ Yes, you don't need that bit about ionisation: metals don't have a band gap, so the electrons can be lifted into the conductance band at arbitrarily low temperature. It's possible to operate a dynamo working like Earth's at temperatures much lower than 6000 K, you just need to make sure the metal is liquid (e.g. Sodium 98°C). doi.org/10.1103/PhysRevLett.98.044502 $\endgroup$ – leftaroundabout Dec 9 '17 at 11:06
  • $\begingroup$ (Comment moved to separate answer) $\endgroup$ – leftaroundabout Dec 9 '17 at 11:41
  • $\begingroup$ I would not say it's molten iron, since the pressure is so freaking high in the inner core, it's mostly solid. $\endgroup$ – Ehouarn Perret Dec 10 '17 at 6:30
59
$\begingroup$

The crucial part is that earth's outer core is fluid, and that it's conductive. That the material happens to be iron which we know as ferromagnetic is actually rather unimportant, because the geomagnetic field is not created as a superposition of atomic spins like in a permanent magnet. Rather, it's generated via Ampère's law from macroscopic currents, like in an electromagnet made from a wound copper coil with a current going though it. (It is an electromagnet, really.)

The reason there are such currents is that the liquid is in convective motion, probably fuelled by thermal transport from the radioactive decay in the inner core. When a conducting liquid moves, it “pulls any magnetic field line with it”. Starting from a small background field, if the flow is complex and rapid enough, this tends to amplify over time.

This dynamo effect can be described theoretically, numerically, or with laboratory-scale experiments using liquid sodium (sodium being nonmagnetic, but a good conductor and easy to melt). It is not hindered by high temperatures (rather, the high temperatures are often necessary to ensure flow and/or conductance). And it takes place not only on Earth, but also on many other objects:

  • The Sun's dynamo uses the plasma (hydrogen/helium), i.e. the fluid is not a metal at all, nor a liquid, but an ionised gas. This is again driven by convection.
  • The gas giants Jupiter and Saturn have dynamos that apparently consist of hydrogen too, but because of the comparatively low temperatures but still immense pressure it's likely in a metallic state.
  • The ice giants Neptune and Uranus have unusually tilted and irregular magnetic fields. It is assumed that this is due to a dynamo not in the core region like for earth, but in a shell rather higher up in the planet's structure. It probably consists of a mixture of hot, pressurised, liquid water, ammonia and methane, which has enough dissolved ions to be a good conductor.
  • Rocky planets and moons often have earth-like dynamos, most notably Mercury and Ganymede.

Remanent magnetism, the kind that we know from permanent magnets and which only works below the Curie temperature, is only important on rocky planets which don't have a dynamo. The most prominent example is Mars, but this magnetic field is much weaker than the aforementioned dynamo-generated ones.

$\endgroup$
  • $\begingroup$ Where are the moving charges that original induced a magnetic field come from? It is supposed/accepted that this steam from ions in the mantle? $\endgroup$ – Alchimista Dec 9 '17 at 12:37
  • 1
    $\begingroup$ Using your words, what was the cause for a " small background field" ? $\endgroup$ – Alchimista Dec 9 '17 at 13:14
  • 3
    $\begingroup$ @Alchimista ah. Well, I don't think it's known what the original background field was. In doubt it could just be the Sun's. ...Which may seem to kick the stone down the road – but really it doesn't, because in a plasma, the mass difference between electrons and ions means magnetic fields can arise purely from kinetics, you just need sufficiently violent movement. Mostly though, the point is that you don't need to start out with any noteworthy field, because the positive feedback will blow it up exponentially up to a certain strength. $\endgroup$ – leftaroundabout Dec 9 '17 at 13:21
  • 4
    $\begingroup$ @Alchimista It is widely believed that almost any flow of a conducting liquid will give rise to self-excitation of a magnetic field provided that, first, the flow topology is not too simple (e.g. a flow in one direction or a purely rotational flow) and, second, the so-called magnetic Reynolds number is large enough. Here is a review article in quite simple language by the authors of the first "spin metal until magnetic field appears" experiment. $\endgroup$ – Džuris Dec 9 '17 at 16:57
  • 1
    $\begingroup$ @no_choice99 sure. It can't possibly be ferromagnetic above the Curie temperature (which is of course lower than the melting point), as the OP already stated. $\endgroup$ – leftaroundabout Dec 9 '17 at 21:20

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