Calculators/Physics/Magnetic field of a coil

Magnetic field of a coil

B = μ₀·n·I · field inside a solenoid

Magnetic field of a coil — illustration
Flux density B
6.283 mT
= 6.283e-3 T
Turn density n = N/L
2,500
turns/m

Inside a long coil the field is uniform: B = μ₀·n·I, where n = N/L (turns per metre) and μ₀ = 4π·10⁻⁷ T·m/A. It does not depend on the coil radius — more turns or a larger current gives a stronger field (the principle of the electromagnet).

Inside a long current-carrying coil (solenoid) a uniform magnetic field is created:

B=μ0 n IB=\mu_0\,n\,I

where n=NLn=\dfrac{N}{L} is the turn density (number of turns per metre), II the current and μ0=4π⋅10−7 T⋅mA\mu_0=4\pi\cdot10^{-7}\ \tfrac{\text{T·m}}{\text{A}} the permeability of free space.

Interestingly, the field does not depend on the radius of the coil — only on the turn density and the current. To strengthen the field we increase the number of turns or the current, or insert a ferromagnetic core (the principle of the electromagnet).

Examples

N=500, L=0.2 m, I=2 A
→ n = 2500 turns/m, B = μ₀·2500·2 ≈ 6.28 mT
Doubling the current
→ the field B doubles

Frequently asked questions

Why doesn't the coil's field depend on its radius?+

Because for a long solenoid what matters is the turn density (turns per metre), not their size. The field inside is uniform: B = μ₀·n·I, regardless of the diameter.

How do you strengthen an electromagnet?+

Increase the number of turns or the current, or insert a core of ferromagnetic material (such as iron), which multiplies the field many times over.

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