# Magnetic systems in thermodynamics

References: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.17;

F. Mandl, Statistical Physics, Second Edition, (John Wiley & Sons, 1988) – Section 1.4.

Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Section 5.3.

We can derive thermodynamic identities for a magnetic system in which pressure and volume are constant. In such a system, work is done by an external electrical power source such as a battery, with the work resulting in changes to the magnetization and magnetic field present in the system, rather than changes in pressure or volume.

As an example, consider a solenoid with a (very long, so we can approximate the field produced as equivalent to that produced by an infinite solenoid) length ${L}$ and ${n}$ turns per unit length. Suppose that the interior of the solenoid is filled with a cylinder of magnetic material with a magnetization (magnetic dipole moment per unit volume) of ${\mathbf{M}}$. By applying Ampère’s law to a loop with one edge inside and the opposite edge outside the solenoid, and using the fact that the magnetic field outside a solenoid is zero (see Griffiths referenced above, example 5.9), we have

$\displaystyle \int\left(\frac{1}{\mu_{0}}\mathbf{B}-\mathbf{M}\right)\cdot d\boldsymbol{\ell}=I_{f} \ \ \ \ \ (1)$

where ${I_{f}}$ is the free current enclosed by the loop. By choosing the loop edge to have a unit length we have ${I_{f}=nI}$, where ${I}$ is the current in the wire. The integrand is defined to be the auxiliary field ${\mathbf{H}}$:

$\displaystyle \mathbf{H}\equiv\frac{1}{\mu_{0}}\mathbf{B}-\mathbf{M} \ \ \ \ \ (2)$

Since both ${\mathbf{B}}$ and ${\mathbf{M}}$ are parallel to the axis of the solenoid, the integrand is non-zero only on the edge of the loop inside the solenoid and we have

$\displaystyle \mathcal{H}=nI \ \ \ \ \ (3)$

I’m using ${\mathcal{H}}$ to represent magnetic field to distinguish it from ${H}$, which represents enthalpy.

This is true for a steady current, but now suppose we vary the current to change the magnetic field strength ${\mathbf{B}}$. According to Faraday’s law, a change in the magnetic flux ${\Phi}$ through a loop causes a back-emf ${\mathcal{E}}$ to be produced which opposes the change in flux, according to

$\displaystyle \mathcal{E}=-\frac{d\Phi}{dt} \ \ \ \ \ (4)$

The work done by the battery is therefore the work required to overcome this back-emf and thus keep the current flowing at its new value. The total magnetic flux through the solenoid is the flux through a single turn (which is ${BA}$, where ${A}$ is the cross-sectional area of the solenoid) multiplied by the total number of turns, which is ${nL}$. Therefore

$\displaystyle \frac{d\Phi}{dt}=nAL\frac{dB}{dt} \ \ \ \ \ (5)$

The power (work per unit time) generated by the battery is the voltage times the current, which is ${-\mathcal{E}I}$, (negative because the battery’s voltage is opposite to the back-emf) so the work done in time ${dt}$ is

 $\displaystyle dW$ $\displaystyle =$ $\displaystyle -\mathcal{E}I\;dt\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(nAL\frac{dB}{dt}\right)\left(\frac{\mathcal{H}}{n}\right)dt\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle AL\mathcal{H}\;dB\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle V\mathcal{H}\;dB \ \ \ \ \ (9)$

where ${V=AL}$ is the volume of the solenoid.

From 2, we have

 $\displaystyle dB$ $\displaystyle =$ $\displaystyle \mu_{0}\left(d\mathcal{H}+dM\right)\ \ \ \ \ (10)$ $\displaystyle dW_{tot}$ $\displaystyle =$ $\displaystyle \mu_{0}V\left(\mathcal{H}\;d\mathcal{H}+\mathcal{H}\;dM\right) \ \ \ \ \ (11)$

The first term can be written as

$\displaystyle \mu_{0}V\mathcal{H}\;d\mathcal{H}=\frac{\mu_{0}V}{2}d\left(\mathcal{H}^{2}\right) \ \ \ \ \ (12)$

In a vacuum, the energy density in a magnetic field is

$\displaystyle E_{m}=\frac{1}{2\mu_{0}}B^{2} \ \ \ \ \ (13)$

and ${B=\mu_{0}\mathcal{H}}$ so

$\displaystyle E_{m}=\frac{\mu_{0}}{2}\mathcal{H}^{2} \ \ \ \ \ (14)$

Thus, 12 is the change in the magnetic field energy in the entire solenoid, assuming there is a vacuum inside the solenoid. [Actually, in a linear magnetic material, ${B=\mu\mathcal{H}}$, where ${\mu}$ is the permeability of the material, and it is only in a vacuum that ${\mu=\mu_{0}}$. Thus, 14 is the magnetic energy density in a vacuum.]

The second term in 11 is the work required to change the magnetization of the sample inside the solenoid:

 $\displaystyle dW$ $\displaystyle =$ $\displaystyle \mu_{0}\mathcal{H}V\;dM\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mu_{0}\mathcal{H}\;d\mathcal{M} \ \ \ \ \ (16)$

where ${\mathcal{M}\equiv VM}$ is the total magnetization of the sample.

Assuming that only the energy and magnetization change, we can work out a thermodynamic identity for magnetic matter. By analogy with the earlier derivation, we can imagine a system in which the entropy changes by either the energy ${U}$ changing or the magnetization ${\mathcal{M}}$ changing. The entropy change due to an increase in energy at temperature ${T}$ is

$\displaystyle dS_{U}=\frac{dU}{T} \ \ \ \ \ (17)$

How does the entropy change when the magnetization ${\mathcal{M}}$ increases? A larger magnetization usually means that the dipoles in the sample are more ordered, so we’d expect an increase in ${\mathcal{M}}$ to cause a decrease in ${S}$. Thus we’d expect

$\displaystyle \left(\frac{\partial S}{\partial\mathcal{M}}\right)_{U}<0 \ \ \ \ \ (18)$

From 16, we’d like ${\left(\frac{\partial S}{\partial\mathcal{M}}\right)}$ to be something times ${\mathcal{H}}$. To find the ‘something’, we can look at the units. ${S}$ has units of ${\mbox{J K}^{-1}}$. From 2, ${\mathcal{M}}$ has units of a${\left(\mbox{volume}\right)\times B/\mu_{0}}$ which works out to ${\mbox{m}^{3}\left(\mbox{Tesla}\right)\left(\mbox{J m}^{-1}\mbox{Amp}^{-2}\right)^{-1}=\mbox{m}^{3}\left(\mbox{J Amp}^{-1}\mbox{m}^{-2}\right)\left(\mbox{J}^{-1}\mbox{m }\mbox{Amp}^{2}\right)=\mbox{m}^{2}\mbox{Amp}}$. The derivative ${\left(\frac{\partial S}{\partial\mathcal{M}}\right)}$ therefore has units of ${\mbox{J m}^{-2}\mbox{Amp}^{-1}\mbox{K}^{-1}}$. Multiplying this by ${T/\mu_{0}}$ gives a quantity with units of ${\mbox{Amp m}^{-1}}$ which are the units of ${\mathcal{H}}$, so we propose

$\displaystyle \frac{T}{\mu_{0}}\left(\frac{\partial S}{\partial\mathcal{M}}\right)_{U}=-\mathcal{H} \ \ \ \ \ (19)$

The total entropy change is therefore

 $\displaystyle dS$ $\displaystyle =$ $\displaystyle dS_{U}+dS_{\mathcal{M}}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{dU}{T}-\frac{\mu_{0}\mathcal{H}}{T}d\mathcal{M} \ \ \ \ \ (21)$

This gives a thermodynamic identity of

$\displaystyle dU=T\;dS+\mu_{0}\mathcal{H}\;d\mathcal{M} \ \ \ \ \ (22)$

As we might expect, the last term is just the work done on the system as given by 16.

By analogy with the definition of enthalpy for a pressure-volume system, the enthalpy here is

$\displaystyle H=U-\mu_{0}\mathcal{H}\mathcal{M} \ \ \ \ \ (23)$

This is the energy required to create a magnetic system from scratch, with an internal energy ${U}$ and magnetization ${\mathcal{M}}$ in a field ${\mathcal{H}}$. Presumably the second term is negative, since aligning the dipoles with the external field reduces the energy of the system.

The corresponding thermodynamic identity is

 $\displaystyle dH$ $\displaystyle =$ $\displaystyle dU-\mu_{0}\mathcal{H}\;d\mathcal{M}-\mu_{0}\mathcal{M}\;d\mathcal{H}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle T\;dS-\mu_{0}\mathcal{M}\;d\mathcal{H} \ \ \ \ \ (25)$

The Helmholtz free energy is still defined as

$\displaystyle F=U-TS \ \ \ \ \ (26)$

since there is no reference to pressure or volume. The thermodynamic potential is

 $\displaystyle dF$ $\displaystyle =$ $\displaystyle dU-T\;dS-S\;dT\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -S\;dT+\mu_{0}\mathcal{H}\;d\mathcal{M} \ \ \ \ \ (28)$

The Gibbs free energy is thus defined as

 $\displaystyle G$ $\displaystyle =$ $\displaystyle U-TS-\mu_{0}\mathcal{H}\mathcal{M}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle F-TS \ \ \ \ \ (30)$

Its thermodynamic identity is

 $\displaystyle dG$ $\displaystyle =$ $\displaystyle dF-\mu_{0}\mathcal{H}\;d\mathcal{M}-\mu_{0}\mathcal{M}\;d\mathcal{H}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -S\;dT-\mu_{0}\mathcal{M}\;d\mathcal{H} \ \ \ \ \ (32)$