Magnetic body force on the medium, two ways

The same calculation run twice, side by side. Change anything on either side and the difference is shown output by output. This page is a link: both sets of numbers are in the address bar.

Side A open alone

M Medium magnetisation

kA/m

The magnetisation of the medium as a whole, per unit volume of slurry — for a 5 % micron-iron slurry in a 50 kA/m field, about 7.5 kA/m. Negative for a diamagnetic medium.

G Field gradient

A/m²

How fast H changes along the tube at the coil's end, ∇H. A 50 kA/m field falling off over 25 mm is 2 × 10⁶ A/m².

ℓg Length of the gradient region

How far along the tube the gradient extends — roughly the coil's radius at either end.

Side B open alone

M Medium magnetisation

kA/m

The magnetisation of the medium as a whole, per unit volume of slurry — for a 5 % micron-iron slurry in a 50 kA/m field, about 7.5 kA/m. Negative for a diamagnetic medium.

G Field gradient

A/m²

How fast H changes along the tube at the coil's end, ∇H. A 50 kA/m field falling off over 25 mm is 2 × 10⁶ A/m².

ℓg Length of the gradient region

How far along the tube the gradient extends — roughly the coil's radius at either end.

Swap sides

What changed

The two sides are the same. Change something on either.

Output	A	B	B against A
fm Body force	18850 N/m³	18850 N/m³	same
Δp Pressure across the fringe	471.2 Pa	471.2 Pa	same

Side A, with its numbers

18850 N/m³ = μ_{0} 7.5 kA/m 2000000 A/m², 471.2 Pa = 18850 N/m³ 25 mm

LaTeX

18850\,\mathrm{N/m³} = \mu_0 \cdot 7.5\,\mathrm{kA/m} \cdot 2000000\,\mathrm{A/m²}, \qquad 471.2\,\mathrm{Pa} = 18850\,\mathrm{N/m³} \cdot 25\,\mathrm{mm}

M itself depends on H inside the gradient (M = χ·H), so this is the force at the stated magnetisation, not a self-consistent one: across the fringe M falls with H and the true mean force is lower.
Across 2.5 cm of fringe this comes to 471.2 Pa — 48.14 mm of water column. Set that against the loop's pressure drop: a coil end pumps only if it exceeds the loss round the loop.

Side B, with its numbers

18850 N/m³ = μ_{0} 7.5 kA/m 2000000 A/m², 471.2 Pa = 18850 N/m³ 25 mm

LaTeX

18850\,\mathrm{N/m³} = \mu_0 \cdot 7.5\,\mathrm{kA/m} \cdot 2000000\,\mathrm{A/m²}, \qquad 471.2\,\mathrm{Pa} = 18850\,\mathrm{N/m³} \cdot 25\,\mathrm{mm}

M itself depends on H inside the gradient (M = χ·H), so this is the force at the stated magnetisation, not a self-consistent one: across the fringe M falls with H and the true mean force is lower.
Across 2.5 cm of fringe this comes to 471.2 Pa — 48.14 mm of water column. Set that against the loop's pressure drop: a coil end pumps only if it exceeds the loss round the loop.