91Ó°ÊÓ

we have to show that$W_{\text{tot.}}=HVdB$

Any change in the magnetic environment of a coil of wire will cause a voltage (emf) to be "generated" in the coil, according to Faraday's law. And the magnitude of the induced emf can be calculated as follows:

$\mathcal{E}=N\frac{d{\mathrm{Φ}}_B}{dt}$

where ${\mathrm{Φ}}_B=AB$is the magnetic flux, so:

$\mathcal{E}=NA\frac{dB}{dt}......\left(1\right)$

The induced emf multiplied by the current flowing through the coil equals the coil's power, so:

$$\begin{gathered} P=\mathcal{E}I \\ P=NAI\frac{dB}{dt}.....\left(2\right) \end{gathered}$$

Whether or not the specimen is present, the magnetic field H inside the coil is N I / L. Substitute the following into (2):

$P=HLA\frac{dB}{dt}$

The length of the cylinder multiplied by the cross sectional area of the cylinder equals the volume of the cylinder, so:

$P=HV\frac{dB}{dt}$

The integration of power with respect to time equals the work or energy, so:

$\begin{array}{r}{W}_{\text{tot.}}=∫Pdt=HV∫\frac{dB}{dt}dt\\ W_{\text{tot.}}=HVdB\end{array}$

Hence proved.

$H_m=U-{\mathrm{μ}}_{0}HM$

The magnetic field of B can be written in terms of H and M as:

$B={\mathrm{μ}}_0\left(H+\frac{M}{V}\right)$

We get: for an infinitesimal change in B

$dB={\mathrm{μ}}_0\left(dH+\frac{dM}{V}\right)$

substitute into the result of part (a) to get:

$$\begin{gathered} W_{\text{tot.}}={\mathrm{μ}}_{0}HV\left(dH+\frac{dM}{V}\right) \\ {W}_{\text{tot.}}={\mathrm{μ}}_{0}HVdH+{\mathrm{μ}}_{0}HdM \end{gathered}$$

$\text{we can write}HdH\text{as}d\left(\frac{1}{2}{H}^2\right)\text{, so:}$

$W_{\text{tot.}}=d\left(\frac{1}{2}{\mathrm{μ}}_{0}V{H}^2\right)+{\mathrm{μ}}_{0}HdM$

The first component indicates the change in vacuum field energy; the field with the specimen is the same as the field without it; the work is the work required to change the magnetization of the sample inside the solenoid without this term.

$W={\mathrm{μ}}_{0}HdM$

where $W=-PV$is substituted for the work from part (b) to obtain (notice that $H_m$is the enthalpy and the subset e is used to differentiate between the enthalpy and the magnetic field):

$H_m=U-{\mathrm{μ}}_{0}HM$

What is the thermodynamic identity for this system?

The total entropy change is equal to the change in the system's entropy including the change in the coil's entropy.

$dS=d{S}_U+d{S}_M$

where,

$d{S}_U=\frac{dU}{T}$

and,

$d{S}_M=-\frac{W}{T}=\frac{{\mathrm{μ}}_{0}HdM}{T}$

hence,

$dS=\frac{dU}{T}-\frac{{\mathrm{μ}}_{0}HdM}{T}$

Note that the work, entropy, is a change in the entropy $d{S}_M$, since as the magnetization of the system increases, the entropy decreases (the disorderness), and thus both sides are multiplied by T to get:

$dU=TdS+{\mathrm{μ}}_{0}HdM$

How would you define analogues of the enthalpy and Gibbs free energy for a magnetic system? and also Derive the thermodynamic identities for each of these quantities.

We have the following definition of enthalpy:

$H=U+PV=U-W$

We get: for an infinitesimal change in enthalpy:

$d{H}_m=dU-{\mathrm{μ}}_{0}HdM-{\mathrm{μ}}_{0}MdH$

substitute from part (c) with $dU$to get:

$$\begin{gathered} d{H}_m=TdS+{\mathrm{μ}}_{0}HdM-{\mathrm{μ}}_{0}HdM-{\mathrm{μ}}_{0}MdH \\ d{H}_m=TdS-{\mathrm{μ}}_{0}MdH \end{gathered}$$

The Helmholtz free energy is given as follows:

$F=U-TS$

for a small change we have:

$dF=dU-TdS-SdT$

substitute from part (c) to get:

role="math" localid="1648491943269" $$\begin{gathered} F=TdS+{\mathrm{μ}}_{0}HdM-TdS-SdT \\ dF={\mathrm{μ}}_{0}HdM-SdT......\left(3\right) \end{gathered}$$

The Gibbs free energy is given as follows:

$G=F-TS$

substitute with $S_M=W/T={\mathrm{μ}}_{0}HM/T$to get:

$G=F-{\mathrm{μ}}_{0}HM$

for a small change we have:

$dG=dF-{\mathrm{μ}}_{0}HdM-{\mathrm{μ}}_{0}MdH$

substitute from (3) to get:

$$\begin{gathered} G={\mathrm{μ}}_{0}HdM-SdT-{\mathrm{μ}}_{0}HdM-{\mathrm{μ}}_{0}MdH \\ dG=-SdT-{\mathrm{μ}}_{0}MdH \end{gathered}$$

we have to show that$W_{\text{tot.}}=HVdB$

Any change in the magnetic environment of a coil of wire will cause a voltage (emf) to be "generated" in the coil, according to Faraday's law. And the magnitude of the induced emf can be calculated as follows:

$\mathcal{E}=N\frac{d{\mathrm{Φ}}_B}{dt}$

Here, ${\mathrm{Φ}}_B=AB$is the magnetic flux, so:

role="math" localid="1648489677675" $\mathcal{E}=NA\frac{dB}{dt}.....\left(1\right)$

The induced emf multiplied by the current flowing through the coil equals the coil's power, so:

role="math" localid="1648489689715" $$\begin{gathered} P=\mathcal{E}I \\ P=NAI\frac{dB}{dt}......\left(2\right) \end{gathered}$$

Whether or not the specimen is present, the magnetic field H inside the coil is $NI/L$. Substitute the following into (2):