Particle magnetisation and saturation

How far toward saturation does a magnetic particulate go in this field, at this temperature?

r = \coth(x) - \frac{1}{x}

Method

1. Look up the material's saturation magnetisation Mₛ — 1.7 × 10⁶ A/m for iron, 4.8 × 10⁵ for nickel, 1.4 × 10⁶ for cobalt.
2. Find the particle's volume as a sphere, π d³ ÷ 6, and multiply by Mₛ. That is the particle's magnetic moment: the whole particle acts as one moment, so a particle a thousand times the diameter has a moment a billion times larger.
3. Convert the field to tesla and the temperature to kelvin. Multiply the moment by the field for the magnetic energy; multiply Boltzmann's constant (1.38 × 10⁻²³ J/K) by the temperature for the thermal energy. Their ratio is the Langevin argument x.
4. The fraction of saturation is the Langevin function of x: coth(x) − 1/x. It is x/3 for small x (the field barely wins) and approaches 1 for large x (every moment aligned).
5. Multiply the fraction by Mₛ for the actual magnetisation, in amperes per metre.

Assumptions

• Each particle is a single magnetic domain carrying one moment that is free to rotate — the superparamagnetic picture. That holds for particles below a few tens of nanometres. Larger particles carry many domains, and their magnetisation curve is governed by domain-wall motion instead, which this does not model; for them, what the calculation gets right is the conclusion, that they are saturated, not the path there.
• Particles do not interact with each other. A dense slurry's particles feel each other's fields, which makes the medium magnetise more readily than the isolated-particle figure says.
• The saturation values are bulk, room-temperature figures. They fall with temperature and are lower for oxidised or impure particles.
• The medium's own magnetisation is reported per unit volume of particulate. A slurry that is 10 % iron by volume has one tenth of this magnetisation per unit volume of slurry.