8 × 300 mm solenoid
On-axis magnetic field of an air-core solenoid 8 mm across and 300 mm long, solved from its geometry alone. 4.19 µT at the centre per amp-turn; the field holds within 1% of that over 261.2 mm of the axis and falls to 0.500 of it at the coil mouth.
Centre field
4.19 µT / A·turn
B = µ0·N·I / sqrt(L² + D²), exactly linear in turns × current
µ0·n·I overstates by
0.0%
the infinite-solenoid shortcut predicts 4.19 µT per amp-turn here
Within 1% of centre
261.2 mm
87.1% of the winding length
Closed form usable from
78 turns
below this a discrete winding is more than 1.0% from the current-sheet value
Field along the axis
Measured from the centre of the winding, per amp-turn. Positive and negative z are symmetric, so only one side is listed. The bar is the field as a fraction of the centre value.
| z from centre | Field per A·turn | Fraction of centre | |
|---|---|---|---|
| 0.0 mmcentre | 4.19 µT | 1.000 | |
| 37.5 mm | 4.19 µT | 1.000 | |
| 75.0 mm | 4.19 µT | 1.000 | |
| 112.5 mm | 4.18 µT | 0.997 | |
| 150.0 mmcoil mouth | 2.09 µT | 0.500 | |
| 225.0 mm | 0.00 µT | 0.001 | |
| 300.0 mm | 0.00 µT | 0.000 | |
| 450.0 mm | 0.00 µT | 0.000 |
What it takes to reach a field
The centre field is exactly linear in the product of turns and current, so this is a column rather than a second set of pages: 500 turns at 2 A and 100 turns at 10 A are the same row. Wire gauge, resistance and the heat all that current makes are not on this page — the geometry is.
100 amp-turns
418.73 µT
500 amp-turns
2.09 mT
1,000 amp-turns
4.19 mT
5,000 amp-turns
20.94 mT
Questions this coil answers
What is the magnetic field inside a 8 × 300 mm solenoid?
4.19 µT at the centre of the winding, per amp-turn, solved as B = µ0·N·I / sqrt(L² + D²). The field scales exactly with the product of turns and current, so 4.19 mT at 1,000 amp-turns — 500 turns at 2 A, or 100 turns at 10 A, give the same number. This is an air core: no ferromagnetic material is assumed anywhere on this page.
How wrong is µ0·n·I for a 8 × 300 mm coil?
It overstates the centre field by 0.0%. The infinite-solenoid shortcut predicts 4.19 µT per amp-turn against the finite coil's 4.19 µT, and the whole difference is the factor 1/sqrt(1 + (D/L)²) that the infinite form drops. This coil is long at an aspect ratio of 37.50 (length ÷ diameter), and the shortcut gets worse the shorter and fatter the coil is.
How uniform is the field along the axis of a 8 × 300 mm solenoid?
The on-axis field stays within 1% of its centre value over 261.2 mm — 87.1% of the 300 mm winding — and within 5% over 283.5 mm. At the mouth of the coil it has fallen to 0.500 of the centre value. A long solenoid tends to exactly 0.5 there; a short one keeps much more, because both ends of a short winding are close to the point being measured.
How many turns does a 8 × 300 mm coil need before the closed-form field is trustworthy?
78 turns. Below that the winding is a stack of separated rings rather than a sheet of current, and the closed form is out by more than 1.0%. At that count the discrete sum lands 1.0% from it, and it stays inside the band out to 234 turns. This is checked against an independent Biot–Savart sum over the individual turns, not against the same formula rearranged.
Does another coil size give the same field as a 8 × 300 mm one?
Swapping the two dimensions on this coil lands outside the published grid, so there is no mirror page for it here — but the identity holds for every pair that is on the grid.
One dimension different
Step one axis at a time. The centre field moves with the diagonal of the two dimensions, so a change in the larger one moves it more.
Method and limits. The on-axis field is the current-sheet solenoid solved in closed form: B(z) = (µ0·n·I/2)·[(L/2 - z)/sqrt(R² + (L/2 - z)²) + (L/2 + z)/sqrt(R² + (L/2 + z)²)], which at the centre reduces to µ0·N·I / sqrt(L² + D²). The model is a uniform cylinder of azimuthal current, so it assumes an air core (µr = 1 — a ferromagnetic core multiplies the field by an effective permeability that depends on the core's own shape and is not computable from the winding), a single layer of closely spaced turns, and no end plates or return path. Only the axis is solved: off-axis the field is lower near the mouth and the radial component is not zero. The turn-count floor is measured against an independent Biot–Savart sum over discrete turns rather than against a rearrangement of the same formula. Nothing on this page is fetched, interpolated or recalled; it is solved from the two dimensions in the URL and µ0.