Why is iron ferromagnetic?

Thanks, Mat Baluch. A nice solution by the way. Yes I did consider the two state model.. (two equillibrium states, parallel and antiparallel to the magnetic field), which of course simplifies calcuations.

We can then apply the Boltzmann Probability Distribution equation for the two states:

$\frac { { n }_{ parallel } }{ { n }_{ antiparallel } } ={ e }^{ \frac { \Delta E }{ { k }_{ B }T } }$ ... (1)

where, $\Delta E = { E }_{ antiparallel }-{ E }_{ parallel }$

and, ${ E }_{ state }=-\overrightarrow { { \mu }_{ state } } \cdot \overrightarrow { B }$ .

We also know $N\ = { n }_{ parallel } + { n }_{ antiparallel }$ ... (2)

(1) and (2) give us both $n$ values.

Then simply, ${ \mu }_{ effective }=\quad { \mu }_{ atom}{ n }_{ parallel }-{ \mu }_{ atom }{ n }_{ antiparallel }$

Please refer to Mat's solution for calculation details.

Very nice problem, a perfect example of energy/entropy competition ! Let us see why.

Following the hypothesis of the problem, each iron atom can be represented as a magnetic dipole with:

$\mu_{atom} = \sqrt{4(4+2)} \times \mu_{B} = 4,54 \times 10^{-23} J.T^{-1}$

Each of these dipoles, under the magnetic field $B$ , has an energy:

$u = - \overrightarrow{\mu_{atom}}.\overrightarrow{B}$

In the lowest energy state, the magnetic moment of each atom would be aligned with the magnetic field. But for a non-zero temperature, there is a competition between energy and entropy, and the system at equilibrium is in a state that minimises its free energy function: $F = U - T.S$

The mathematical solving of this problem could be quite sophisticated, with as many degrees of freedom as all the $( \theta , \phi )$ orientations of magnetic moments in spherical coordinates. Rather, let us take the assumption that each atom can only be in two states, with their magnetic moment either "up" or "down" relative to the magnetic field.

A macro-state, which groups several micro-states for which the thermodynamic functions (energy, magnetic moment...) have the same value, is given by the number $p$ of atoms in the "up" state (and thus $N - p$ is the number of atoms in the "down" state). For such state, the energy is:

$U = (N-2p) \times \mu_{atom} \times B$

and entropy:

$S = k_{B} \times ln(W) = k_{B} \times ln\binom {N} {p}$

since for such marco-state, their are $\binom {N}{p}$ micro-states, corresponding to the choice of the $p$ atoms "up".

Hence:

$F = U - T.S = (N-2p) \mu_{atom} B - k_{B} T ln\binom {N} {p}$

To find the value of $p$ that minimises $F$ , we could use Sterling approximation (as there are many atoms...) and derive the result.

We find that $p$ that minimises $F$ is such that: $\boxed { \frac{p}{N} \approx 0.61 }$

Here is, for example, the behaviour of the free energy, for $N = 1000$ , as a function of $p$ . The free energy is minimum for $p \approx 610$ :

We can now conclude: under the magnetic field $B$ , $61 \%$ of atoms are aligned with this field (and $39 \%$ are in anti-aligned ). The resulting magnetic moment is:

$\mu = 0,61 N \mu_{atom} + 0,39 N ( - \mu_{atom} ) = 0,22 \times N \times \mu_{atom}$

With:

$N = \frac{\rho}{m_{Fe}} = 8,4895 \times 10^{28}$ iron atoms in a unit volume, we find:

$\boxed { \mu = 8,48 \times 10^{5} J.T^{-1} }$

There is a slight difference with the answer given by the author, possibly due to the simple two-state model.