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Chain formation onset

Will the particles chain up in the field?

The formula
λ=EckT,Ec=μ0⁢mp24⁢π⁢d3
λ
Dipolar coupling constant
Ec
Contact energy of two touching particles, J
kT
Thermal energy, k_B × T in kelvin, J
mp
Particle moment Ms·V, in A·m²
d
Particle diameter, m
LaTeX 
λ = \frac{Ec}{kT}, \qquad Ec = \frac{\mu_0 \cdot mp^2}{4 \pi \cdot d^3}

Work it out

The magnetic solid. Its saturation magnetisation sets each particle's moment, taken as fully magnetised.

µm

The coupling goes with the cube of this: ten times the diameter, a thousand times the coupling. The default is a ten-nanometre ferrofluid grain, which is the only size range where the answer is in doubt.

°C

Thermal energy is what keeps the particles apart.

Compare with a variation

Result

λ Dipolar coupling constant 19.58

Contact energy over thermal energy. Below 1, no chains; 1 to 3, short chains in the field; above 3, chains that persist.

Ec Contact energy 7.923e-11 nJ

The energy holding two touching particles head to tail, μ₀m² ÷ 4πd³. Room-temperature k_BT is 4 × 10⁻²¹ J.

With your numbers
19.58=7.923e-11 nJ4.047e-12 nJ,7.923e-11 nJ=μ0⁢8.901e-1924⁢π⁢10 nm3
LaTeX 
19.58 = \frac{7.923e-11\,\mathrm{nJ}}{4.047e-12\,\mathrm{nJ}}, \qquad 7.923e-11\,\mathrm{nJ} = \frac{\mu_0 \cdot 8.901e-19^2}{4 \pi \cdot 10\,\mathrm{nm}^3}

Worth knowing

  • λ is 19.6, above 3: the dipole attraction is well beyond thermal energy and the particles chain along the field into structures that persist. The medium in the field is non-Newtonian — it has a yield stress and a viscosity that depends on the shear — and the magnetoviscous calculation's assumption of separate spheres has failed; its figure is not a floor but the wrong picture. Thermal energy is 4.047e-12 nJ against a contact energy of 7.923e-11 nJ.
  • This is a threshold, not a structure: λ says whether chains form, and nothing about how long they are or what the chained medium's yield stress is. The thresholds are from simulation and experiment on ideal spheres, and the archive holds this page at lower confidence than the rest of the group.

This result is a link — the address bar holds your numbers, so it can be pasted into a post and opened to the same answer.

What this looks like

Dipolar coupling constant against particle diameter Particle diameter swept from 0.005 µm to 0.015 µm with everything else held at your numbers. The dashed lines cross where you are.
Dipolar coupling constant against particle diameterDipolar coupling constant rises from 2.45 to 66.1 as particle diameter rises from 0.005 µm to 0.015 µm. At your particle diameter of 0.01 µm it is 19.6.0204060800.0050.00750.010.01250.0150.01 µm19.6Particle diameter (µm)Dipolar coupling constant
The formula behind the curve
λ=EckT,Ec=μ0⁢mp24⁢π⁢d3
λ
Dipolar coupling constant
Ec
Contact energy of two touching particles, J
kT
Thermal energy, k_B × T in kelvin, J
mp
Particle moment Ms·V, in A·m²
d
Particle diameter, m
LaTeX 
λ = \frac{Ec}{kT}, \qquad Ec = \frac{\mu_0 \cdot mp^2}{4 \pi \cdot d^3}
What moves the answer Each input moved 10% either way, with the others held still, and the effect on dipolar coupling constant.
What moves the answerDipolar coupling constant is most sensitive to Particle diameter, which moves it by about 33.1% for a 10% change. It is least sensitive to Temperature, at about 0.687%.Change in the answer when each input moves by 10%-40%-20%0%20%40%Particle diameter±33.1Temperature±0.687
The formula behind the curve
λ=EckT,Ec=μ0⁢mp24⁢π⁢d3
λ
Dipolar coupling constant
Ec
Contact energy of two touching particles, J
kT
Thermal energy, k_B × T in kelvin, J
mp
Particle moment Ms·V, in A·m²
d
Particle diameter, m
LaTeX 
λ = \frac{Ec}{kT}, \qquad Ec = \frac{\mu_0 \cdot mp^2}{4 \pi \cdot d^3}

Method

  1. Find the particle's moment: the saturation magnetisation times the sphere's volume, πd³ ÷ 6. The particle is taken as fully magnetised, which in any field worth applying it is.
  2. The energy of two such moments touching head to tail is μ₀m² ÷ 4πd³ — the dipole–dipole energy at a centre separation of one diameter.
  3. Divide by k_BT for λ. Because m² goes with d⁶ and the denominator with d³, λ goes with d³: doubling the diameter multiplies it by eight.
  4. Read the regime off the thresholds: below 1 the particles stay separate; between 1 and 3 they chain along the field and disperse when it is removed; above 3 the chains persist and the medium is a structured fluid.

Assumptions

  • Identical spheres, fully magnetised, touching — the contact separation is one diameter and any surfactant shell is ignored. A shell of thickness δ raises the separation to d + 2δ and lowers the coupling by (d ÷ (d + 2δ))³, which for a ten-nanometre grain with a two-nanometre coat is a factor of 2.7; a coated ferrofluid is more stable than the bare figure says.
  • The thresholds at λ ≈ 1 and 3 are from simulation and experiment on monodisperse dipolar hard spheres in the dilute limit. Real powders are polydisperse, and the largest particles chain first; the loading also matters — at a few per cent the chains are sparse, and at tens of per cent they span the tube — and this page does not take the loading as an input, because λ is a threshold, not a structure.
  • What the page does not say: chain length, yield stress, or the viscosity of the chained medium. It says only when the separate-spheres assumption of the magnetoviscous calculation has failed, and above λ of a few it has.
  • Nothing here is Meyer's. The estate material puts a magnetic slurry through a coil; the calculation exists because a slurry of micron iron in a field is not a liquid of separate particles at all — the coupling is in the millions — and the medium in the coil is a very different thing from the medium in the pump.