Solenoid Magnetic Field Calculator

How it works

B = μ₀·μᵣ·N·I / L = μ₀·μᵣ·n·I · μ₀ = 4π × 10⁻⁷ T·m/A · H = n·I · Φ = B·π·r² · at either end of a long solenoid the field is about B/2

Inside a long, tightly wound solenoid the fields of all the turns add to a nearly uniform field along the axis, and Ampère’s law gives B = μ₀nI — it depends only on the turns per unit length and the current, not on the radius. A ferromagnetic core multiplies the field by its relative permeability until the core saturates. The formula is accurate for coils several times longer than their diameter; short coils have weaker, less uniform fields. The derivation is in OpenStax University Physics Vol. 2, section 12.7, with μ₀ from CODATA.

Worked example

A 25 cm air-core coil with 500 turns carries 2 A. n = 500 / 0.25 = 2,000 turns per metre, so H = 2,000 × 2 = 4,000 A/m and B = 4π × 10⁻⁷ × 4,000 = 5.027 mT (50.27 G). Through a 2 cm radius the flux is 5.027 mT × π × 0.02² = 6.317 µWb. The coil is 6.25 times longer than its diameter, so the long-solenoid formula is accurate.

Sources: OpenStax University Physics Vol. 2 — Solenoids and Toroids · HyperPhysics — Solenoid field

Frequently asked questions