Solenoid Magnetic Field Calculator
Enter turns, length and current to compute the solenoid interior field B=μ₀nI and total flux. N=1000, L=0.1 m, I=1 A → B≈0.0126 T.
Input Data
Results
At a glance:Solenoid interior field: an ideal long solenoid carrying current produces a uniform interior field B=μ₀·n·I, where μ₀=4π×10⁻⁷ T·m/A is the vacuum permeability, n=N/L is the turn density (N total turns, L length), I the current. In an ideal long solenoid (L≫diameter) the interior field is uniform and axial, and the exterior field is approximately zero. Total flux Φ=B·A (A cross-section). Applications: electromagnets, inductors, relays, MRI main magnets. Example: N=1000, L=0.1 m, I=1 A → n=10000 turns/m, B=4π×10⁻⁷×10000×1≈0.01257 T≈12.6 mT (about 250× Earth's ~50 µT field). Superconducting MRI solenoids reach 1–3 T. The tool takes turns, length, current and cross-section and outputs field (T and mT), turn density and total flux.
Formula
Vacuum permeability: μ₀ = 4π×10⁻⁷ T·m/A
Turn density: n = N/L
Solenoid field: B = μ₀·n·I = μ₀·N·I/L
Total flux: Φ = B·A
$$\mu_0 = 4\pi \times 10^{-7} \, \mathrm{T\cdot m/A}$$$$n = \frac{N}{L}$$$$B = \mu_0 n I = \frac{\mu_0 N I}{L}$$How to Use
- Enter turns N, length L (m) and current I (A) to get field B.
- Cross-section A computes total flux Φ=B·A; A=0 skips flux.
- The formula is most accurate for an ideal long solenoid (L≫diameter); a short solenoid gives a slightly smaller real field.
Case Studies
Basic solenoid
N=1000 turns, L=0.1 m, I=1 A.
n = N/L = 10000 turns/m.
B = μ₀·n·I = 4π×10⁻⁷ × 10000 × 1 ≈ 0.01257 T ≈ 12.6 mT.
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.