# Magnetic vector potential and dipoles

- The magnetic vector potential $\myv A$: $$\myv B=\myv \grad\times \myv A$$
- Multipole expansion of field due to currents.
- Magnetic dipole.
- Examples (including a very strange result for the electron).

### Vector potential

Revisiting the Helmholtz theorem, it tells us that a vector field that has a curl: $$\myv \grad \times \myv B = \mu_0 \myv J,$$

but no divergence... $$\myv \grad \cdot \myv B = 0,$$

may be expressed as the curl of a vector function, that we'll call the **magnetic vector potential** $\myv A$,
$$\myv B = \myv \grad \times \myv A.$$

Vector identity (11): $$\myv \grad \times \myv B=\myv \grad \times(\myv \grad \times \myv A) = \myv \grad (\myv \grad \cdot \myv A)-\grad^2\myv A=\mu_0 \myv J .$$

[With electric potentials, we can add a scalar to the potential, and the electric field $-\myv E = \myv \grad (V+k)=\myv \grad V$ remains unchanged.]

The curl of any gradient vanishes, $\myv \grad \times \myv \grad f=0$,

So we can add the gradient of any scalar function $\lambda(\myv r)$ to the magnetic potential, $ \myv \grad \times (\myv A + \myv \grad \lambda )=\myv \grad \times \myv A $. This property can be used to chose a scalar field $\lambda $ (choose the 'gauge') such that $$\myv \grad \cdot \myv A=0$$ (the "Coulomb gauge").

Then, Ampere's law -- $\myv\grad\times \myv B=-\mu_o \myv J$ -- can alternately be stated in terms of the potential... $$\grad^2\myv A = -\mu_0\myv J .$$

This is 3 Poisson equations (one for each Cartesian coordinate).

We'll use Helmholtz' constructed potential which is the one appropriate for the Coulomb gauge to find the potential by integrating the current density:

$$\myv A (\myv r) = \frac{\mu_0}{4 \pi} \int \frac{\myv J(\myv r')}{\rr} d \tau'$$

### Multipole expansion of $\myv A.$

In electrostatics, when we had some localized charge distribution, we figured out...

- a general potential $V(\myv r),$
- far from the charge distribution,
- an expansion of $1/\rr$ in powers of $1/r.$

We can do the same for a localized *current distribution*.

We already did the heavy lifting with the multipole charge
distribution to
show....
$$\begineq \frac{1}{\rr} &=& \frac{1}{r} \sum_n \frac{r'}{r}^n P_n(\cos \theta')\\
&=&\frac{1}{r} + \frac{\myv r \cdot \myv r'}{r^3}+\frac{1}{2}\left(\frac{3(\myv r \cdot \myv r')^2}{r^5}-\frac{r'^2}{r^3}\right)+....\endeq$$

Consider some loop of current (pictured). What is the magnetic potential far from the loop? Using our just discovered integral for the potential in terms of the current ($I\,d\myv l$) instead of the current density ($\myv J\,d\tau$)... $$\myv A (\myv r) = \frac{\mu_0}{4 \pi} \int \frac{ I(\myv r')}{\rr} d \myv l'$$

And subbing in the expression for $1/\rr$... $$\begineq \myv A(\myv r) &=& \frac{\mu_0I}{4\pi} \left[\frac{1}{r}\oint d\myv l' + \frac{1}{r^2}\oint r' \cos \theta' d\myv l'\right.\\ &&\ \ \ \ \left. +\frac{1}{r^3}\oint r'^2\left(\frac{3}{2}\cos^2\theta'-\frac{1}{2}\right)d\myv l'+...\right].\endeq$$

The first term is the *monopole* term, then the *dipole* term, then
the *quadrupole*, etc.

The monopole term vanishes (why??): $$\oint d\myv l' = 0.$$

So, usually the dominant term is...

### The magnetic dipole

$$\myv A_\text{dip} =\frac{\mu_0I}{4\pi r^2}\oint r'\cos\theta' d\myv l'.$$Using a result from problem 1.61 it can be shown that...$$\oint r'\cos\theta' d\myv l' = \oint (\uv{r}\cdot \myv r')d\myv l' = -\uv{r} \times \int d\myv a'.$$where $\int d\myv a'$ is a surface integral over the surface bounded by the closed path of the line integral.

So,
$$\myv A_\text{dip} =\frac{\mu_0}{4\pi}
\frac{\myv m \times \uv r}{r^2},$$
where the **magnetic dipole moment, $\myv m$**, is:
$$\myv m\equiv I\int d\myv a = I\myv a.$$
The integral, $\myv a$, is the 'vector area' of the loop. If the loop is flat the magnitude of the vector
area is the same as the area inside the loop, and it points (right hand rule)
perpendicular to the area.

In spherical coordinates, if the magnetic dipole is pointing along the $z$-axis, we can express the magnetic potential of the dipole as: $$\myv A_\text{dip} =\frac{\mu_0}{4\pi}\frac{m\sin\theta}{r^2} \uv \phi .$$

Then we figure the magnetic field using the expression for the curl in spherical coordinates: $$\begineq \myv B_\text{dip}(\myv r) &=& \myv \grad \times \myv A_\text{dip} \endeq$$

$$=\frac{\mu_0 m}{4 \pi r^3}\left(2\cos \theta \,\uv{r} +\sin\theta\, \uv \theta\right).$$

### Problem 5.37

*A phonograph record of radius $R$, with a uniform surface charge $\sigma$,
is rotating at constant angular velocity $\omega.$ Find its magnetic dipole
moment.*

For a ring of current, its magnetic dipole moment $dm$ is $$dm= dI\, a= dI\, \pi r^2 = \sigma v \,dr\, \pi r^2 = \sigma \omega r \,dr\, \pi r^2.$$

So, the total magnetic dipole moment is

$$m = \int dm = \int_0^R \sigma \omega \pi r^3 \,dr = \pi \sigma \omega R^4 / 4.$$

### The gyromagnetic ratio - Problem 5.58

*A thin uniform 'donut' with total charge $Q$ and mass $M$ rotates about its
axis. Find the ratio of its magnetic dipole moment to its angular momentum.
This is called its " gyromagnetic ratio".*
$$I=\lambda \cdot v = \frac{Q}{2\pi R}\cdot (R \omega )= \frac{Q\omega}{2\pi}.$$

$$\Rightarrow \myv m = I \myv a = \frac{Q \omega}{2\pi}\pi R^2\uv z=\frac{Q \omega R^2}{2} \uv z.$$

$$\myv L = MvR\uv{z} = M\omega R^2\uv{z}.$$

The ratio of dipole moment to angular momentum--the gyromagnetic factor--is sometimes called a "g-factor": $$g= m/L = \frac{Q \omega R^2}{2 M \omega R^2} = \frac{Q}{2M}.$$

*What is the gyromagnetic ratio for a sphere?*

A sphere can be decomposed into a "sum of donuts" at different radii.

- $g$ does not depend on the radius.
- As long as the sphere has a uniform charge- and mass-density, the ratio $Q/M$ (each is separately proportional to the volume of an individual donut) will be a constant for all donuts.

So, for a sphere (or, for that matter, any body of rotation), the ratio will be the same as for a single donut: $$g=\frac{Q}{2M}.$$

According to quantum mechanics, the electron has angular momentum $L_e = \frac{1}{2}\hbar $.

So, if $e$ and $m_e$ are the charge and mass of an electron, you'd kind of expect that the magnetic moment of the electron--call it $\mu_e$--would be $$\begineq \mu_e&=&\frac{-e}{2m_e} \cdot \frac{1}{2}\hbar = \frac{-e \hbar }{4 m_e}\\ &=& -\frac{(1.6\times10^{-19})(1.05 \times 10^{-34})}{4*9.11\times10^{-31}} \\ &=& -4.61 \times 10^{-24}\text{A m}^2.\endeq$$

But experimentally, it is found that... $$\mu_e = (2.0023193043622 \pm0.0000000000015) \frac{-e \hbar }{4 m_e}.$$

Hmmm...

- It doesn't work to picture the electron as a rotating solid. (Though as far as experiments can tell, it's terrifically round down as far as we can measure.)
- Dirac's relativistic theory of quantum mechanics comes up with precisely an integer value $\mu_e=2$.
- In Quantum Electro Dynamics (QED - elaborated by Feynmann, Schwinger, Tomonaga) charged particles interact with each other through an exchange of photons. If you allow an isolated electron to interact with the photons it would give off, there is a slight correction, which agrees exactly (to within experimental uncertainty) with the measured value.

The **Bohr magneton**, $\mu_B$ is this combination of constants:
$$\mu_b=\frac{e\hbar}{2m_e}.$$
If Dirac had been right, the magnitude of the electron's magnetic dipole would have been *exactly* $\mu_B$.