Dipolar magnetic field pressure

Question

It is relatively easy to derive the potential energy stored into the magnetic field of an uniformly magnetized sphere of radius $R$ and total magnetic moment $\mu$ : \begin{equation}\tag{1} U_{\text{magn}} = \frac{\mu_0 \, \mu^2}{4 \pi R^3}. \end{equation}

The field is uniform inside the sphere, and dipolar on the exterior side. It is exerting a pressure that I want to calculate. In thermodynamics, the pressure can be defined as the partial derivative of the "internal" energy relative to a change of volume : \begin{equation}\tag{2} P = -\, \frac{\partial U}{\partial V}. \end{equation} Since $V = 4 \pi R^3 / 3$, it is tempting to derive directly (1) (assuming $\mu = \text{constant}$), to get this relation : \begin{align} P_{\text{magn} \, 1} = -\, \frac{\partial \,}{\partial V} \Big( \frac{\mu_0 \, \mu^2}{3 V} \Big) &= \frac{\mu_0 \, \mu^2}{3 V^2} \\[12pt] &\equiv \frac{U_{\text{magn}}}{V}. \tag{3} \end{align} This recalls the stiff equation of state $p = \rho$.

I don't think that $\mu$ could be considered as an independant variable (changing the sphere radius may have an effect on the total dipolar moment, unless there's some kind of constraint). The polar field $B_{\text{pole}}$ is related to $\mu$ by this : \begin{equation}\tag{4} \mu(R, B_{\text{pole}}) = \frac{2 \pi B_{\text{pole}}^2 \, R^3}{\mu_0}. \end{equation} If I assume that $B_{\text{pole}}$ is the independant variable (it may be some constraint), then substituting (4) into (1) and doing the derivative gives this relation instead (a negative pressure = tension !) : \begin{equation}\tag{5} P_{\text{magn} \, 2} = -\, \frac{\partial \,}{\partial V} \Big( \frac{3 B_{\text{pole}}^2}{4 \mu_0} \, V \Big) = -\, \frac{U_{\text{magn}}}{V}. \end{equation} This recalls the cosmological constant relation of state : $p = -\, \rho$.

There's a third possibility (are there others ?). I can consider the magnetic flux $\Phi = B_{\text{inside}} \, \pi R^2$ as the independant variable (flux of the sphere's internal field passing through its own equator) : \begin{equation}\tag{6} \Phi = \frac{\mu_0 \, \mu}{2 R}. \end{equation} In this case, the pressure would be \begin{equation}\tag{7} P_{\text{magn} \, 3} = -\, \frac{\partial \,}{\partial V} \Big( \frac{\Phi^2}{\pi \mu_0 \, R} \Big) = \frac{U_{\text{magn}}}{3 V}, \end{equation} which recalls the electromagnetic equation of state from relativistic physics : $p = \frac{1}{3} \, \rho$.

I suspect that $P_{\text{magn} \, 3}$ should be the proper pressure of the magnetic field. But how to justify this ?

Take note that $U_{\text{magn}}/V$ is NOT the field energy density, since it's varying from one place to another (the field outside the sphere is not uniform, since it's dipolar). So I'm not sure how to interpret the $P$ above correctly, since it's a constant (i.e not depending on position).

So the question is the following :

What is the total magnetic pressure felt by an outside agent that changes a bit the volume of a magnetized sphere ? I expect this : \begin{equation}\tag{8} P_{\text{magn} \, 3} = \frac{U_{\text{magn}}}{3 V} = \frac{\mu_0 \, \mu^2}{(4 \pi R^3)^2} \equiv \frac{B_{\text{int}}^2}{4 \mu_0} \equiv \frac{B_{\text{pole}}^2}{4 \mu_0}. \end{equation}

Answer 1

The field inside the sphere is

$$\label{1}\tag{1} \mathbf B_{in} = \frac 2 3 \mu_0 \mathbf M$$

where $\mathbf M$ is the magnetization vector. The magnetic dipole moment is

$$\mathbf m = V \mathbf M$$

where $V = 4 \pi R^3 / 3$ is the volume of the sphere. I use the notation $\mathbf m$ for the dipole moment because $\mu$ is usually used for the magnetic permeability.

The magnetic pressure inside the sphere is defined as the total energy density inside the sphere:

$$\label{2} \tag{2}P_{in} = u_{in} = \frac{U_{in}}{V_s} =\frac{B_{in}^2}{2 \mu_0} = \frac{2\mu_0 M^2}{9} = \frac{2 \mu_0 m^2}{9 V^2}$$

The energy density $B^2/2 \mu_0$ takes into account the work done on both bound and free currents when establishing the field (see my answer to this question).

If you want the magnetic pressure outside the sphere, you just have to calculate the energy density outside the sphere, $B_{out}^2 / 2 \mu_0$.

The outside field is (assuming $\mathbf M$ is in the $z$ direction)

$$\label{3}\tag{3} \mathbf B_{out} = \frac{m}{r^3} [2 \cos \theta \mathbf \ {\hat r} + \sin \theta \mathbf \ {\hat \theta}]$$

from which

$$\label{4}\tag{4} P_{out} (r, \theta) = \frac{B_{out}^2}{2 \mu_0} = \frac{\mu_0 m^2}{32 \pi^2 r^6} [3 \cos^2 \theta +1] = \frac{\mu_0 M^2}{9} \left( \frac R r \right)^6 \left( \frac{3 \cos^2 \theta +1 } 2 \right)$$

By setting $r=R$, we can get the value of $P_{out}$ on the surface of the sphere:

$$\label{5}\tag{5} P_{out}(R, \theta) = \frac{\mu_0 M^2}{9} \left( \frac{3 \cos^2 \theta +1 } 2 \right)$$

You can notice that

$$P_{out}(R,n\pi) = P_{in} \ \ \ \ n \in \mathbb Z$$

The average pressure on the sphere surface is

$$\label{6}\tag{6} \frac 1 {4 \pi} \int_0^{2 \pi} d \phi \int_0^{\pi} P_{out}(R,\theta) \sin \theta d \theta = \frac{\mu_0 M^2}{9} = \frac{P_{in}}2$$

which corresponds to your (8).

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As requested: the energy density inside the sphere relative to the bound currents only is

$$u_{in,b} = \frac 1 2 \mathbf B \cdot \mathbf M = \frac{\mu_0 M^2} 3 $$

and the relative energy is

$$U_{in,b} = \frac{\mu_0 M^2} 3 V = \frac{\mu_0 m^2}{3 V}$$

which corresponds to your (1).

Answer 2

Ok, the solution to my own query is very simple. Why is $P_3$ the proper answer ? It's because of equilibrium thermodynamics, which defines the pressure as the variation of energy by a variation of volume (equation (2)) : \begin{equation}\tag{2} P \equiv -\, \frac{\partial U}{\partial V}. \end{equation} To correctly apply this formula, we need to identify clearly the independant variables. But to get an equilibrium situation, with reversible processes, we need to fix the magnetic flux, or else there will be some electromagnetic induction (by Faraday's law), and production of electromagnetic radiation and heat. This process is irreversible, and (2) wont be usefull !

I think this problem is highly pedagogical, and shows some of the unity of nature !