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)

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