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This question regards Ewing's molecular theory of magnetism.

I.) Ewing's molecular theory of magnetism describes every magnetic substance as being a collection of dipoles that are initially in a state where there is no specific and uniform orientation as a result of the application of a magnetic field and because of this the net magnetic field strength of the substance becomes zero. Now, let us place a magnet close to this object. The magnetic field passes through this object (assume the substance is ferromagnetic), and the dipoles orient themselves in the direction of this magnetic field. Now due to magnetic field induction, the substance has gained magnetic properties. However so infinitely small, let us assume a certain interval of time had passed for the dipoles to orient themselves. Let us record this time and keep it aside for future reference. Now let us melt down this exact same substance used in our experiment and once again pass the very same magnetic field through the molten substance. Again let us record the time taken for the dipoles to orient themselves. Let us compare the time intervals we have recorded. Now my question is this: since the substance was originally a solid, the electrostatic force of attraction between adjacent molecules would be high; so magnetic field strength would have to be significant and would take some time to force the dipoles into its direction. Then we remove this field and supply heat energy to this substance. The individual kinetic energies of the molecules slowly overcome the electrostatic force of attraction between the adjacent molecules and the substance changes into a liquid. Since the force of attraction is significantly lesser than it was as a solid, if we were to apply a similar magnetic field through this medium, would the time taken to orient the dipoles be different than when we passed the field through the solid?



I hope you can answer the question for me.

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  • $\begingroup$ Hello, Ram! You should ask two questions separately, one question --- one post. Split your post in two. Note, the 'edit' button will be useful. $\endgroup$
    – Yrogirg
    Commented Sep 8, 2012 at 6:21
  • $\begingroup$ Your confusion is answered by doing thermodynamics with the external magnetic field as a parameter. The energy for melting is different in the magnetic field, and the thermodynamic relations ensure there are no paradoxes of energy when you move around in the P,T,B space. $\endgroup$
    – Ron Maimon
    Commented Sep 8, 2012 at 16:03

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Short answer: no, or at least not for this reason.

since the substance was originally a solid, the electrostatic force of attraction between adjacent molecules would be high; so magnetic field strength would have to be significant and would take some time to force the dipoles into its direction

Here you are assuming that the magnetic dipoles, which are due to the orientation of electronic spins,[1] requires a reorientation of the molecule, as would often be the case for electric dipoles. This is unjustified, since a molecule in a given spatial orientation can present a magnetic moment in either "up" or "down" directions, no matter how you choose your up/down axis. Generally, no molecular reorientation is involved in spin reorientation.

A spin S=1/2, in the simplest of cases, can flip its orientation without overcoming any barrier and without any relation to the molecular shape. Even for a single molecule/ion/atom magnet, which does present magnetic memory below a certain temperature, flipping its magnetic moment is only indirectly related to atomic motions, via spin-orbit coupling.

However, yes, in general magnetic memory times decrease when more thermal energy is available, so whatever short or long the time for reversing the spin orientation in any solid substance may be, I would expect it to be shorter in its molten version.

[1] The orbital magnetic moment is also relevant, but for the purposes of this argument it behaves just as a spin.

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Yeah, the time taken will be different and it will be higher because the molecules have more kinetic energy and will move more randomly, also their intermolecular space is more so there will be more time taken.

I am not sure as I have studied this chapter first time and am giving answer as a student. Kindly correct me if I am wrong please.

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  • $\begingroup$ Please abstain from using that much informal language. $\endgroup$ Commented Jul 29 at 8:02

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