60 × 15 mm solenoid
On-axis magnetic field of an air-core solenoid 60 mm across and 15 mm long, solved from its geometry alone. 20.32 µT at the centre per amp-turn; the field holds within 1% of that over 5.2 mm of the axis and falls to 0.922 of it at the coil mouth.
Centre field
20.32 µT / A·turn
B = µ0·N·I / sqrt(L² + D²), exactly linear in turns × current
µ0·n·I overstates by
312.3%
the infinite-solenoid shortcut predicts 83.78 µT per amp-turn here
Within 1% of centre
5.2 mm
34.8% of the winding length
Closed form usable from
2 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 | 20.32 µT | 1.000 | |
| 1.9 mm | 20.21 µT | 0.995 | |
| 3.8 mm | 19.90 µT | 0.980 | |
| 5.6 mm | 19.40 µT | 0.955 | |
| 7.5 mmcoil mouth | 18.73 µT | 0.922 | |
| 11.3 mm | 17.00 µT | 0.837 | |
| 15.0 mm | 14.97 µT | 0.737 | |
| 22.5 mm | 10.89 µT | 0.536 |
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
2.03 mT
500 amp-turns
10.16 mT
1,000 amp-turns
20.32 mT
5,000 amp-turns
101.59 mT
Questions this coil answers
What is the magnetic field inside a 60 × 15 mm solenoid?
20.32 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 20.32 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 60 × 15 mm coil?
It overstates the centre field by 312.3%. The infinite-solenoid shortcut predicts 83.78 µT per amp-turn against the finite coil's 20.32 µT, and the whole difference is the factor 1/sqrt(1 + (D/L)²) that the infinite form drops. This coil is short and wide at an aspect ratio of 0.25 (length ÷ diameter), and the shortcut gets worse the shorter and fatter the coil is.
How uniform is the field along the axis of a 60 × 15 mm solenoid?
The on-axis field stays within 1% of its centre value over 5.2 mm — 34.8% of the 15 mm winding — and within 5% over 11.9 mm. At the mouth of the coil it has fallen to 0.922 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 60 × 15 mm coil need before the closed-form field is trustworthy?
2 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 0.7% from it, and it stays inside the band out to 6 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 60 × 15 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.