# 20 × 300 mm solenoid On-axis magnetic field of an air-core solenoid 20 mm across and 300 mm long, solved from its geometry alone. 4.18 µT at the centre per amp-turn; the field holds within 1% of that over 209.6 mm of the axis and falls to 0.501 of it at the coil mouth. Canonical page: https://makerportal.ai/lab/solenoid/20x300mm Page title: 20 × 300 mm Solenoid Field — 4.18 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:** 20 × 300 mm — diameter × length, aspect ratio 15.00 - **Centre field:** 4.18 µT / A·turn — B = µ0·N·I / sqrt(L² + D²), exactly linear in turns × current - **µ0·n·I overstates by:** 0.2% — the infinite-solenoid shortcut predicts 4.19 µT per amp-turn here - **Field at the mouth:** 0.501 × centre — a long solenoid tends to 0.500; a short one keeps more - **Within 1% of centre:** 209.6 mm — 69.9% of the winding length - **Within 5% of centre:** 259.5 mm — 86.5% of the winding length - **Closed form usable from:** 32 turns — below this a discrete winding is more than 1.0% from the current-sheet value - **At 1,000 amp-turns:** 4.18 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 300 mm winding. Positive and negative z are symmetric. | z from centre | Field per A·turn | Fraction of centre | | |---|---|---|---| | 0.0 mm | 4.18 µT | 1.000 | centre | | 37.5 mm | 4.18 µT | 1.000 | | | 75.0 mm | 4.17 µT | 0.997 | | | 112.5 mm | 4.12 µT | 0.985 | | | 150.0 mm | 2.09 µT | 0.501 | coil mouth | | 225.0 mm | 0.02 µT | 0.004 | | | 300.0 mm | 0.00 µT | 0.001 | | | 450.0 mm | 0.00 µT | 0.000 | | ## 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 | 417.95 µT | | 500 | 2.09 mT | | 1,000 | 4.18 mT | | 5,000 | 20.90 mT | ## Questions this page answers ### What is the magnetic field inside a 20 × 300 mm solenoid? 4.18 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 4.18 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 20 × 300 mm coil? It overstates the centre field by 0.2%. The infinite-solenoid shortcut predicts 4.19 µT per amp-turn against the finite coil's 4.18 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 15.00 (length ÷ diameter), and the shortcut gets worse the shorter and fatter the coil is. ### How uniform is the field along the axis of a 20 × 300 mm solenoid? The on-axis field stays within 1% of its centre value over 209.6 mm — 69.9% of the 300 mm winding — and within 5% over 259.5 mm. At the mouth of the coil it has fallen to 0.501 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 20 × 300 mm coil need before the closed-form field is trustworthy? 32 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 0.8% from it, and it stays inside the band out to 96 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 20 × 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. ## 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/20x300mm. Free to quote and cite with attribution and a link to the canonical page.