3
$\begingroup$

According to http://young.physics.ucsc.edu/112/magnetic.pdf, the free energy $F$ in the presence of a magnetic field is given by \begin{equation} dF = -SdT -MdB, \end{equation} where $M$ is the total magnetic dipole moment of the system and $B$ is the "external magnetic field". Consequently, the magnetic moment can be calculated as \begin{equation} M = \left( \frac{\partial F}{\partial B} \right)_T. \end{equation} I am slightly confused since the author afterwards states that there is a difference between $B$, $B_\text{ext}$ and $\mu_0 H$ and that "from now on" he will just use the symbol $B$. But he used that symbol already from the beginning and only called the field "external", which I find confusing.

So my question: To be 100% rigorous, should these two equations contain $B_\text{ext}$ (by which I understand the flux density of the field measured in vacuum, i.e. with the sample removed)? Are the equations still valid if there are any free currents (in contrast to bound magnetic moments) in the material?

$\endgroup$

2 Answers 2

0
$\begingroup$

I quote the paper you are referring to:

"You will recall from E&M classes that the magnetic field inside a magnetic material can differ from the externally applied field Bext and from an auxiliary field $H$ which is often introduced. In particular the relation between $B$, $H$, and $M$ is $B = μ_0(H +M)$, where $μ_0$ is a constant. In this course we shall only consider “weakly magnetic” materials where $M ≪ H ≃ B/μ_0$, and so the difference between $B$, $B_{ext}$ and $μ_0H$ will be negligible. From now on, we will just label the field in Eqs. (1) and (2) by $B$ and ignore any fine distinctions."

In other words when the susceptibility is small in the context of the reference it does not matter which one, but I note that according HEINE, BYV. "THE THERMODYNAMICS OF BODIES IN STATIC ELECTROMAGNETIC FIELDS" it is possible to transform the equations $$ F=\int_{all space} dv\int_0^{\textbf{B}} \textbf{H} \cdot d\textbf{B}$$ $$S=\int_{allspace}dv \int_{0,T=const}^{\textbf{B}} \left(\frac{\partial\textbf{M}}{\partial T}\right)_{\textbf{B}} \cdot d\textbf{B}$$ to $$S=\int_{V}dv \int_{0,T=const}^{\textbf{B}_{ext}} \left(\frac{\partial\textbf{M}}{\partial T}\right)_{\textbf{B}_{ext}} \cdot d\textbf{B}_{ext}$$and $$ F=-\frac{1}{2}\int_{V'} dv\int_0^{\textbf{B}_{ext}} \textbf{M}_{ext} \cdot d\textbf{B}_{ext}-\int_{V} dv\int_0^{\textbf{B}_{ext}} \textbf{M} \cdot d\textbf{B}_{ext}$$ and without permanent magnets we have the simpler $$ F=-\int_{V} dv\int_0^{\textbf{B}_{ext}} \textbf{M} \cdot d\textbf{B}_{ext}$$ where $V$ is the region of the body.

$\endgroup$
4
  • $\begingroup$ Is $\int_0^B H\cdot dB$ equal to $H\cdot B$? If not, why? $\endgroup$
    – Riemann
    Commented Nov 5, 2023 at 12:12
  • $\begingroup$ If H were independent of B then the integral would be HB but that cannot happen. What can happen is that they are proportional, $B=\mu H$ and $\mu=const$, as in vacuum where $\mu = \mu_0$, or in a non-magnetic (dielectric) material and then the integral is HB/2. In a paramagnetic material you can have $\mu \approx const$ for sufficiently low H intensity as long as $T=const$. $\endgroup$
    – hyportnex
    Commented Nov 5, 2023 at 12:32
  • $\begingroup$ I think my confusion comes from the space dependence. So I thought that it implicitly meant $\int_V dr \int_0^{B(r)} H(r) dB$. But that looks strange and simplifies to $\int_Vdr H(r)\cdot B(r)$. However if $H$ and $B$ are considered to be space-independent, then it would be $V\int_0^B H\cdot dB$. Which of the two is true? $\endgroup$
    – Riemann
    Commented Nov 5, 2023 at 12:37
  • 1
    $\begingroup$ You have misunderstood the meaning of this integral. Here the function is between the two fields as $H=H(B)$, yes they are both space dependent, say $B=B(r)$ and then there is the indirect function as $h(r)=H =H(B(r))$ but there is a direct constitutive relationship between the two fields. Changing one will change the other at the SMAE place by the material relationship between the two. The integral $\int HdB$ is at one fixed location, and the integral $\int dv\int HdB$ is all of those locations together, and when is space homogeneous then it is the latter integral in your question. $\endgroup$
    – hyportnex
    Commented Nov 5, 2023 at 12:41
0
$\begingroup$

Consider a magnetizable body inside a solenoid. There are about 5 fields which one might wish to consider.

  1. The field ${\bf B}_{\rm pt}$ at some point inside the body
  2. The field ${\bf B}_{\rm in}$ inside the body, averaged over a volume large compared to the atoms
  3. The field ${\bf B}_o$ outside the body
  4. The field ${\bf B}_I$ that would be there, at the center of the solenoid, if the body were removed while maintaining the current $I$ in the solenoid fixed
  5. The field ${\bf B}_\phi$ that would be there, at the center of the solenoid, if the body were removed while maintaining the magnetic flux $\phi$ in the solenoid fixed

There are also ${\bf H}$ fields related to all the above. Confusion abounds!

You can ignore ${\bf B}_{\rm pt}$. You can also ignore ${\bf B}_o$. This may surprise you, but we don't need it and it is best not to invoke it. So the fields we are interested in are ${\bf B}_{\rm in}$, ${\bf B}_I$ and ${\bf B}_\phi$. In practice some treatments refer to ${\bf B}_I$ as "the applied field" or ${\bf B}_{\rm ext}$ or whatever. And some treatments refer to ${\bf B}_\phi$ using those same words! So you need to check carefully which field is being invoked to play the role of "applied field" or "external field".

In the limit of small magnetization one has ${\bf B}_I = {\bf B}_\phi = {\bf B}_{\rm in}$ so in that case it does not matter and many introductory treatments just treat that case. When the magnetization is not small, however, one needs to be more careful.

In the expression $dF = - S dT - M dB$ you should understand $B$ to refer to $B_\phi$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.