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}

magnetic field of a sphere

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}

  • 2
    $\begingroup$ Just want to notify that I changed my answer according to what we discussed about the energy density. $\endgroup$
    – valerio
    Jul 17, 2017 at 12:27
  • $\begingroup$ I found the solution to my query. To be able to use equ. (2) for the equilibrium pressure (from thermodynamics), we do need to fix the magnetic flux. This is fundamental, or else there will be induction, and production of electromagnetic radiation and heat. This process is irreversible. $\endgroup$
    – Cham
    Jul 17, 2017 at 12:57

2 Answers 2


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).

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).

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Jul 16, 2017 at 15:06
  • 1
    $\begingroup$ There's a mistake in your B_out : $r^6$ should be $r^3$. Also, use tags to help the discussion (comments). $\endgroup$
    – Cham
    Jul 17, 2017 at 12:50
  • $\begingroup$ And I suggest that you add the calculation for $u_b$ too, to get the total "bound" energy, which gives the same as my (1). $\endgroup$
    – Cham
    Jul 17, 2017 at 12:52
  • $\begingroup$ I'm adding an answer to my question, since I have found the complete (simpe) solution. Oh and if you add the average pressure on the outside of the sphere, and substract the internal pressure, you'll get the expression (8). $\endgroup$
    – Cham
    Jul 17, 2017 at 12:58
  • $\begingroup$ @valerio92, to make your answer more consistent, so I could give you the 50points, please add the total bound energy to compare with my (1), and give the average pressure variation on the sphere (averaging your (4) on all $\vartheta$ and $\varphi$, and substract it from the internal pressure). Don't forget the extra $\sin{\vartheta}$ for the average on all surface ! You should get my (8). $\endgroup$
    – Cham
    Jul 17, 2017 at 13:19

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 !


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.