Field in a finite solenoid, with the medium

What field does the drive coil put in the tube, and how steep is its fringe?

The formula

H = \frac{N \cdot I}{\sqrt{ℓ^{2} + D^{2}}}, B = μ_{0} \cdot (1 + χ) \cdot H

LaTeX

H = \frac{N \cdot I}{\sqrt{ℓ^{2} + D^{2}}}, \qquad B = \mu_0 \cdot \left(1 + χ\right) \cdot H

Work it out

N Turns

Total turns on the drive coil, all layers.

I Current

The current actually in the coil — the current-rise page's figure, not the supply's voltage over the resistance.

ℓ Coil length

The winding's length along the tube.

D Coil mean diameter

Twice the mean winding radius — the former plus the winding depth.

χ Medium susceptibility

The medium's volume susceptibility, M ÷ H — from the medium-susceptibility page. Zero for an empty tube; about 0.15 for 5 % soft iron.

or see a worked example with typical numbers

The on-axis field of a finite solenoid at its centre is (N I/ℓ) times the cosine of the half-angle its end subtends, ℓ/√(ℓ² + D²). For a coil much longer than it is wide this is the familiar N I/ℓ; for one as long as it is wide it is 71 % of that.
At the end of the coil, on the axis, only the turns on one side contribute: H_end = (N I/2ℓ)·ℓ/√(ℓ² + (D/2)²), which tends to half the centre field for a long coil.
The medium is a long column along the axis, so its own poles are far away and M = χH, with B = μ₀(1 + χ)H.
The fringe gradient is estimated as the centre field over the coil diameter — the field falls from about H to about nothing over roughly a diameter beyond the end. It is the order of magnitude the body-force page needs, and no better than that.

A thin winding: all the turns at one mean diameter. A winding whose depth is a good fraction of its diameter has its outer turns further from the axis and contributes a little less than this; the error is a few per cent for the coils in the photographs.
The field is quoted on the axis. Off the axis inside a short coil it is higher near the winding and lower at the centre plane's edges — the medium across the bore does not all see the same H, and the figure here is the one a probe on the axis would read.
The medium is linear in H, M = χH, with no demagnetising correction — right for a long column of slurry in a tube along the axis, and wrong for a short plug of it. It is also wrong once the medium is near saturation, which the warning checks for.
Free-space μ₀ outside the medium. The tube wall, if it is a plastic, does nothing; a steel tube is a different problem.
Nothing here is Meyer's. The estate's Fig. 26A draws field lines round the coils; the field's strength on the axis is the sum of the turns' Biot–Savart contributions and has been since 1820.