Calculate the magnetic force between two dia/para-magnetic bodies in a weak uniform field

Question

I'm designing a Cavendish-type gravity attraction experiment, and want to put an order of magnitude (spherical cow in a vacuum) estimate on the force I can expect between test and field masses that are slightly magnetic, to see whether I have to allow for it in the earth's field.

I have had no prior experience of attempting to do quantitative calculations in magnetics, and my research today has been very frustrating. I've been to Kaye and Laby, wikipedia and followed many of the links from there, and some CRC excerpts, with liberal use of Google. One disconcerting claim in one of the references was that many references use inconsistent units, and I'm sure I've found a few misprints. So I'm nervous about my data even before trying to start calculations.

I'm intending to use aluminium test masses and lead field masses. My basic yardstick is two spherical masses 100 mm apart, 50 mm diameter, for which I can expect a gravitational attraction of about 1.3 nN. I'm not expecting high accuracy, I would be happy with +/- 10%, so I need to know whether their magnetic attraction in the earth's field is in the order of, or well below, 100 pN.

The figures I have at the moment are earth's field is about 50 $\mu$T at my location, lead has a $\chi_v$ of $-18\times 10^{-6}$, and aluminium $\chi_v$ of $+22\times 10^{-6}$. I understand the $\chi_v$ should be in dimensionless units, though Wikipedia gives them in cm$^3$/mol, and K&L calls them SI. I'm therefore hesitant to even start working out what attraction/repulsion I can expect between the two volumes with a small induced magnetisation in each.

Question - Can somebody with experience in this type of calculation, with the confidence that they understand the units, compute the attraction between a lead and aluminium sphere, each 50 mm diameter, 100 mm apart, in a 50 $\mu$T magnetic field? Or a cube of similar volume, whatever's easier, the shape is not too important for orders of magnitude.

I suspect I would be OK, as I've not heard any consideration for magnetic effects being expressed around discussions of Cavendish's or later experiments. However, I'm a nervous soul, and would appreciate an order of magnitude estimate.

I have a number of workarounds. I can use Helmholtz coils to either cancel earth's field, or to introduce a large field to see if it changes the measurements. Another possibility is to cast the field mass with lead in an aluminium tube, with volume ratios in their $\chi_v$ ratios, so that the net composite with cylindrical symmetry will have near zero $\chi_v$. I may do the field disturbance thing anyway. But I'd like to know how many orders of magnitude there are between gravitational and magnetic force in this configuration.

Answer 1

I'll do the calculation in SI units. I'll refer to the lead sphere as sphere 1 and the aluminum sphere as sphere 2. I'll drop the "v" subscript on the volume susceptibility.

Your value for the volume magnetic susceptibility of aluminum in SI,

$$\chi_2=+2.2\times 10^{-5},\tag1$$

agrees with Wikipedia. This article doesn't have the value for lead, but I'll assume that your value,

$$\chi_1=-1.8\times 10^{-5},\tag2$$

is correct for SI. These volume susceptibilities are dimensionless, but they do have different values in SI and CGS, because various electromagnetic formulas are different in the two systems of units.

For my final formula, I'll generalize your setup to the case where the two spheres have different radii. In your case, these are 25 mm so

$$R_1=R_2=2.5\times 10^{-2}\text{ m}.\tag3$$

It isn't completely clear what "100 mm apart" means. I'll assume that you mean that there is 100 mm of space between the surfaces of the two spheres. This means that there is 150 mm between their centers, so

$$R=1.5\times 10^{-1}\text{ m}.\tag4$$

You're assuming that the Earth's magnetic field is 50 microteslas. This is measuring the magnetic flux density (also called magnetic induction) $\mathbf B$:

$$B=5\times 10^{-5}\text{ T}=5\times 10^{-5}\text{ kg}\cdot\text{s}^{-2}\cdot\text{A}^{-1}.\tag5$$

The relationship between the the magnetic flux density $\mathbf B$ and the magnetic field strength $\mathbf H$ is

$$\mathbf B=\mu_0 \mathbf H\tag6$$

where

$$\mu_0=1.26\times 10^{-6}\text{ kg}\cdot\text{m}\cdot\text{s}^{-2}\cdot\text{A}^{-2}\tag7$$

is the magnetic constant, also known as the magnetic permeability of the vacuum.

Thus the Earth's magnetic field strength is

$$H=\frac{B}{\mu_0}=39.8\text{ A}\cdot\text{m}^{-1}.\tag8$$

The magnetization $\mathbf M$ of a material with volume magnetic susceptibility $\chi_v$ in a magnetic field with magnetic field strength $\mathbf H$ is given by

$$\mathbf M=\chi_v\mathbf H.\tag9$$

This means that the magnitude of the magnetization of the lead sphere is

$$M_1=\chi_1H=7.16\times 10^{-4}\text{ A}\cdot\text{m}^{-1}\tag{10}$$

and the magnetization of the aluminum sphere is

$$M_2=\chi_2H=8.75\times 10^{-4}\text{ A}\cdot\text{m}^{-1}.\tag{11}$$

They are magnetized in opposite directions because the susceptibilities have opposite signs.

The magnetic moment $\mathbf m$ of an object with uniform magnetization is the magnetizaton times the volume:

$$\mathbf m=\mathbf MV.\tag{12}$$

Thus the magnetic moments of the two spheres are

$$m_1=\frac43\pi R_1^3M_1=4.69\times 10^{-8}\text{ A}\cdot\text{ m}^2\tag{13}$$

and

$$m_2=\frac43\pi R_2^3M_2=5.73\times 10^{-8}\text{ A}\cdot\text{ m}^2\tag{14}.$$

The force between two uniformly magnetized spheres is the same as if they were point magnetic dipoles at their centers. The force between two point magnetic dipoles with dipole moments $\mathbf{m}_1$ and $\mathbf{m}_2$ is

$$\mathbf F=\frac{3\mu_0}{4\pi r^5}\left[ (\mathbf{m}_1\cdot\mathbf{r})\mathbf{m}_2+ (\mathbf{m}_2\cdot\mathbf{r})\mathbf{m}_1+ (\mathbf{m}_1\cdot\mathbf{m}_2)\mathbf{r}- \frac{5(\mathbf{m}_1\cdot\mathbf{r})(\mathbf{m}_2\cdot\mathbf{r})}{r^2}\mathbf{r} \right]\tag{15}$$

where $\mathbf r$ is the separation vector between the two dipoles. Your two spheres will have antiparallel dipole moments (along the Earth's field) but it isn't clear how their separation vector is oriented relative to the Earth's field. For simplicity I'll assume that the separation is parallel to the field. In this case the magnitude of the (repulsive) force is

$$F=\frac{\mu_0}{4\pi}\frac{6m_1m_2}{R^4}=3.18\times 10^{-18}\text{ kg}\cdot\text{m}\cdot\text{s}^{-2}=3.18\times 10^{-18}\text{ N},\tag{16}$$

only a few billionths of a nanonewton.

In symbols, you can combine the various formulas to find that the force is

$$F=\frac{8\pi}{3}\chi_1\chi_2\frac{B^2R_1^3R_2^3}{\mu_0R^4}\tag{17}$$

in terms of input parameters.

I showed all the intermediate numerical values using base SI units — kilograms, meters, seconds, and amperes — so that you can see how the units work out to produce newtons.