# +m.B or -m.B?

In The Feynman Lectures on Physics vol 2, RPF showed that the “true energy” of a magnetic dipole(current loop) in a constant external B field is actually $+\vec{m}.\vec{B}$ , not  $-\vec{m}.\vec{B}$ . Basically the reason as Feynman said is that we need to take into account the extra energy to keep the current in the loop constant. If we included the battery’s work to keep the current in the loop constant when performing the virtual work pulling a current loop from infinity to the final position, the total energy will be $+\vec{m}.\vec{B}$ .
But if we use  $+\vec{m}.\vec{B}$  instead of  $-\vec{m}.\vec{B}$ , we will get contraditory results in some cases. For example, we know that a magnetic  dipole tends to point along the field lines, but $+\vec{m}.\vec{B}$ predicted that it tends to point opposite to the field lines.

How should I explain this?

Let’s try to derive it by calculating the total magnetic field energy.

$U=\int \frac{1}{2\mu_0}B^2 dV$

$U=\int \frac{1}{2\mu_0}(\vec{B_{ext}}+\vec{B_{dipole}})^2 dV$

where

Consider a “real” dipole, a magnetized sphere of radius R, instead of an ideal one. A real dipole has a pure the magnetic field outside the sphere:

$B_{dipole}=\frac{\mu_0}{4\pi r^3}[3(\vec{m}.\hat{r})\hat{r}-\vec{m}]$  for r>0

and

inside the sphere, we assume the field is due to A spinning spherical shell of uniform charge with a magnetic dipole m. The magnetic field is:

$B_{dipole}=\frac{\mu_o\vec{m}}{2\pi R^3}$ for r->0

expanding the things inside the bracket

$U=\int \frac{1}{2\mu_0}(B_{ext}^2+B_{dipole}^2+\vec{2B_{ext}}.\vec{B_{dipole}}) dV$

The first and second term are independent of orientation, we are not interested in them, we can ignore them

$U=\int \frac{1}{2\mu_0}(2\vec{B_{ext}}.\vec{B_{dipole}}) dV$

$U=\int_{0}^{R} \frac{1}{2\mu_0}(\vec{2B_{ext}}.(\frac{\mu_o\vec{m}}{2\pi R^3})) dV+\int_{R}^{\infty} \frac{1}{2\mu_0}(2\vec{B_{ext}}.(\frac{\mu_0}{4\pi r^3}[3(\vec{m}.\hat{r})\hat{r}-\vec{m}])) dV$

The second integrand is an odd function, it involves

$\int_{0}^{\pi} sin\theta(3cos^2\theta-1)d\theta=0$

thus the only surviving term is from the first term

$U=\frac{2}{3}\vec{m}.\vec{B}$

Obviously I am just calculating something wrong,

But remember carefully…

that in the derivation of the energy of a magnetic field we get

$U=\frac{1}{2\mu_0}[\int B^2 dV-\int (\vec{A}\times\vec{B}).\vec{dA}]$

We can always extend this out to infinity, and if our currents are finite the last term vanishes. However if we have a constant magnetic field this is not the case, so we have to include the “surface” term

now we are left with

$U_{surface}=-\int (\vec{A_{ext}}\times\vec{B_{dipole}}).\vec{dA}$

$U_{surface}=-\int [\frac{\mu_0}{4\pi r^3}(3(\vec{m}.\vec{r})\hat{r}-\vec{m})\times(\frac{1}{2}\vec{B_{ext}}\times\vec{r})].\vec{dA}$

$U_{surface}=\int\frac{\mu_0}{4\pi r^3}(-\vec{m}\times ((\frac{1}{2}\vec{B_{ext}}\times\vec{r})).\vec{dA})$

$U_{surface}=\int\frac{\mu_0}{8\pi r^3}(\hat{r}(\vec{m}.\vec{B_{ext}})-\vec{B_{ext}}(\vec{m}.\hat{r})).\vec{dA})$

The calculations are pretty messy, there is nothing interesting to talk about, so I’ll just put the final result

$U_{surface}=\frac{1}{3}\vec{m}.\vec{B_{ext}}$

thus the total field energy is

$U=+\vec{m}.\vec{B_{ext}}$

as what Feynman said….. I think this is so cool.

on the other hand for electric dipole we will get

$U=-\frac{1}{3}\vec{p}.\vec{E_{ext}}$ for the energy density term

and

$U=-\frac{2}{3}\vec{p}.\vec{E_{ext}}$ for the surface term

so the total energy is $U=-\vec{p}.\vec{E_{ext}}$

I never thought that electricity and magnetism are very different. The differences between electric dipoles and magnetic dipoles are:

1. The “inner field” points from positive charge to negative charge in electric dipoles, while it points in the opposite direction for magnetic dipole.

2. Scalar potential $V$ behave in a different way with vector potential $\vec{A}$

But why does magnetic dipoles(elementary particles) doesn’t use the total field energy as the interaction energy?

Because only $-\vec{m}.\vec{B}$ can be converted into mechanical energy…

But the remaining $+2\vec{m}.\vec{B}$ also changes with the orientation! where does this energy goes?

It seems that the energy goes into some kind of mysterious energy that keeps elementary particles’ spin constant. If we leave it that way, the energy will not be conserved. To “protect” the interpretation of electromagnetic field energy and the tendency of potential energy to be minimal, we need to add a “mysterious energy” to the energy term.  But I think is not necessary, actually the field energy is derived by calculating the mechanical work to put the system together. In this case we don’t need to put an extra work to keep the magnitude of dipole moment constant but the field energy still takes that into account. So maybe the energy stored in the field is not always true in general. It does not work in this case. If we stick to the definition of potential energy $F=-grad(U)$, we wouldn’t have this trouble, we don’t need to care about the existance of the “mysterious energy”.