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Physics LibreTexts

15.3: Adiabatic Demagnetization

  • Page ID
    7307
  • We are now going to do the same argument for adiabatic demagnetization.

    We are going to calculate an expression for \((∂T/∂B)_S\). The expression will be positive, since \(T\) and \(B\) increase together. We shall consider the entropy as a function of temperature and magnetic field, and, with the variables

    Screen Shot 2019-07-08 at 12.11.09 PM.png

    we shall start with the cyclic relation

    \[ \left(\frac{\partial S}{\partial T}\right)_{B}\left(\frac{\partial T}{\partial B}\right)_{S}\left(\frac{\partial B}{\partial S}\right)_{T}=-1. \label{15.3.1}\]

    The middle term is the one we want. Let’s find expressions for the first and third partial derivatives in terms of things that we can measure.

    In a reversible process \(dS = dQ/T\), and, in a constant magnetic field, \(dQ = C_BdT\). Here I am taking S to mean the entropy per unit volume, and CB is the heat capacity per unit volume (i.e. the heat required to raise the temperature of unit volume by one degree) in a constant magnetic field.

    Thus we have \( \left(\frac{\partial S}{\partial T}\right)_{B}=\frac{C_{B}}{T}\).

    The Maxwell relation corresponding to \( \left(\frac{\partial S}{\partial P}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{P}\) is \(\left(\frac{\partial S}{\partial B}\right)_{T}=\left(\frac{\partial M}{\partial T}\right)_{B}\). Thus Equation \ref{15.3.1} becomes

    \[ \left(\frac{\partial T}{\partial B}\right)_{S}=-\frac{T}{C_{B}}\left(\frac{\partial M}{\partial T}\right)_{B}\].

    Now for a paramagnetic material, the magnetization, for a given field, is proportional to B and it falls off inversely as the temperature (that’s the equation of state). That is, M = aB/T. and therefore \( \left(\frac{\partial M}{\partial T}\right)_{B}=-\frac{a B}{T^{2}}=-\frac{M}{T}\). Equation 15.3.2 therefore becomes

    \[ \left(\frac{\partial T}{\partial B}\right)_{s}=\frac{M}{C_{B}}.\]

    You should check the dimensions of this equation.

    The cooling effect is particularly effective at low temperatures, when \(C_B\) is small.