# 125 × 40 mm solenoid On-axis magnetic field of an air-core solenoid 125 mm across and 40 mm long, solved from its geometry alone. 9.57 µT at the centre per amp-turn; the field holds within 1% of that over 11.3 mm of the axis and falls to 0.884 of it at the coil mouth. Canonical page: https://makerportal.ai/lab/solenoid/125x40mm Page title: 125 × 40 mm Solenoid Field — 9.57 µT per amp-turn This markdown document and the HTML page above are rendered from the same solved values at build time, by the same functions. Nothing here is written by a language model and nothing is fetched at request time. ## Key figures - **Geometry:** 125 × 40 mm — diameter × length, aspect ratio 0.32 - **Centre field:** 9.57 µT / A·turn — B = µ0·N·I / sqrt(L² + D²), exactly linear in turns × current - **µ0·n·I overstates by:** 228.1% — the infinite-solenoid shortcut predicts 31.42 µT per amp-turn here - **Field at the mouth:** 0.884 × centre — a long solenoid tends to 0.500; a short one keeps more - **Within 1% of centre:** 11.3 mm — 28.2% of the winding length - **Within 5% of centre:** 25.6 mm — 64.1% of the winding length - **Closed form usable from:** 3 turns — below this a discrete winding is more than 1.0% from the current-sheet value - **At 1,000 amp-turns:** 9.57 mT — 500 turns at 2 A, or 100 turns at 10 A — the field only sees the product ## Field along the axis Field on the axis, per amp-turn, measured from the centre of the 40 mm winding. Positive and negative z are symmetric. | z from centre | Field per A·turn | Fraction of centre | | |---|---|---|---| | 0.0 mm | 9.57 µT | 1.000 | centre | | 5.0 mm | 9.50 µT | 0.992 | | | 10.0 mm | 9.28 µT | 0.969 | | | 15.0 mm | 8.93 µT | 0.932 | | | 20.0 mm | 8.47 µT | 0.884 | coil mouth | | 30.0 mm | 7.33 µT | 0.766 | | | 40.0 mm | 6.09 µT | 0.636 | | | 60.0 mm | 3.91 µT | 0.408 | | ## Turns and current The centre field is exactly linear in the product of turns and current, so only that product appears here. | Amp-turns (N × I) | Centre field | |---|---| | 100 | 957.48 µT | | 500 | 4.79 mT | | 1,000 | 9.57 mT | | 5,000 | 47.87 mT | ## Questions this page answers ### What is the magnetic field inside a 125 × 40 mm solenoid? 9.57 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 9.57 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 125 × 40 mm coil? It overstates the centre field by 228.1%. The infinite-solenoid shortcut predicts 31.42 µT per amp-turn against the finite coil's 9.57 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 0.32 (length ÷ diameter), and the shortcut gets worse the shorter and fatter the coil is. ### How uniform is the field along the axis of a 125 × 40 mm solenoid? The on-axis field stays within 1% of its centre value over 11.3 mm — 28.2% of the 40 mm winding — and within 5% over 25.6 mm. At the mouth of the coil it has fallen to 0.884 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 125 × 40 mm coil need before the closed-form field is trustworthy? 3 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 0.5% from it, and it stays inside the band out to 9 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 125 × 40 mm one? Yes — a 40 × 125 mm coil has exactly the same centre field of 9.57 µT per amp-turn, because the closed form depends on the two dimensions only through sqrt(L² + D²), which is symmetric in them. Everything else differs: that coil holds 0.518 of its centre field at its own mouth against this one's 0.884. ## 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. ## Related tool [Magnetic Field Simulator](https://makerportal.ai/lab/magnetic-field-tracer) — Trace field lines through a loop, Helmholtz pair, magnetic bottle, solenoid or dipole with exact Biot–Savart and RK4. Free, runs in your browser. --- Source: MakerPortal — https://makerportal.ai/lab/solenoid/125x40mm. Free to quote and cite with attribution and a link to the canonical page.