Physics · Electrodynamics

Magnetic Field of a Long Coil

The magnetic field inside a long coil is uniform and proportional to the current and the turn density N/l.

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Formula

LaTeX: B = \mu_0 \cdot \frac{N}{l} \cdot I
B in T (tesla) · μ₀ = 1.257×10⁻⁶ V·s/(A·m) · N dimensionless · l in m · I in A

Variables & units – Magnetic Field of a Long Coil

SymbolMeaningUnit
BMagnetic flux density inside the coilT (Tesla)
μ₀Magnetic field constant (1.257×10⁻⁶)V·s/(A·m)
NNumber of turnsdimensionless
lLength of the coilm
ICurrentA

Derivation & background – Magnetic Field of a Long Coil

The formula follows from Ampere law for a long, densely wound cylindrical coil (solenoid) whose length is large compared with its diameter. Inside, the field is uniform and position-independent, outside almost zero. Not the absolute number of turns counts but the turn density n = N/l. An iron core amplifies the field by the permeability μ_r (a few hundred to a thousand for iron): B = μ_r·μ₀·(N/l)·I, the principle of the electromagnet.

Exam blueprint

Validity range

Applies inside long, densely wound coils (length much greater than diameter) without a core. An iron core multiplies by μ_r; at the ends the field drops to about half.

Derivation steps

Ampere law applied to a rectangular loop enclosing the winding.

  1. 1Line integral: only the interior segment of length l contributes, so B·l = μ₀·I_enclosed.
  2. 2The loop encloses N turns carrying I: B·l = μ₀·N·I, hence B = μ₀·(N/l)·I.

Rearrangements

Current

Sizing an electromagnet for a target flux density.

Number of turns

More turns on the same length increase the field linearly.

Task variant

What current produces B = 10 mT in a coil (N = 1000, l = 0.5 m)?

I = B·l/(μ₀·N) = 0.01 × 0.5/(1.257×10⁻⁶ × 1000) ≈ 4.0 A.

How many turns does a 10 cm coil need for 2 mT at I = 0.8 A?

N = B·l/(μ₀·I) = 0.002 × 0.1/(1.257×10⁻⁶ × 0.8) ≈ 199, so about 200 turns.

Common mistakes

Applying the formula to short, thick coils.

It holds only for l >> diameter; short coils have a noticeably weaker, non-uniform field.

Confusing turn count N and turn density n = N/l.

Tables often list n; then B = μ₀·n·I without dividing by l again.

Inserting the coil length in cm.

Convert l to metres, otherwise B is off by a factor of 100.

Forgetting the iron core or double-counting μ_r.

With a core B = μ_r·μ₀·(N/l)·I; apply μ_r exactly once.

Exam context

  • Common as the field source in combined tasks: Lorentz force on particles in a coil field, e/m determination with Helmholtz coils and induction experiments.

These mistakes cost points in real exams. The set drills them until they stick.

Formula cluster

Magnetic fields

The field source for Lorentz force and induction experiments.

Worked example

Coil with N = 500 turns on l = 25 cm at I = 2 A: B = 1.257×10⁻⁶ × (500/0.25) × 2 ≈ 5×10⁻³ T = 5 mT, roughly a hundred times the Earth magnetic field.

Applications

Electromagnets and relays, MRI coils, Helmholtz coils in the lab, loudspeakers, particle deflection (e/m determination)

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Which formula describes Magnetic Field of a Long Coil?

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How do you rearrange B = μ₀·(N/l)·I for Current?

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Which common mistake happens with Magnetic Field of a Long Coil?

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Scientific sources

Common notations & search queries

B=mu0*n*IB=μ0·N/l·IMagnetfeld Spule FormelSolenoid Feldsolenoid magnetic field formulaFlussdichte Spule berechnenlange Spule homogenes Feld

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Frequently asked questions about Magnetic Field of a Long Coil

How do you calculate the magnetic field of a coil?+

For a long cylindrical coil the interior field is B = μ₀·(N/l)·I with the field constant μ₀ = 1.257×10⁻⁶ V·s/(A·m). Insert the number of turns N, the coil length l in metres and the current I in amperes. Example: N = 500, l = 0.25 m, I = 2 A gives B = 1.257×10⁻⁶ × 2000 × 2 ≈ 5 mT. What matters is the turn density N/l, not the absolute number of turns: 500 turns on a short length act more strongly than on a long one. The result applies in the uniform interior region; with an iron core you additionally multiply by the permeability μ_r.

Why is the field inside a long coil uniform?+

Each individual turn produces a curved ring field. With many turns packed closely one behind the other, these contributions superpose: inside, the field parts of all turns add up to parallel, equally dense field lines along the coil axis, while the transverse components of neighbouring turns cancel each other. Outside, the magnetic return flux spreads over a huge volume, so the external field is almost zero. The approximation works as long as the length is clearly larger than the diameter and you stay away from the ends: there the field lines splay apart and the flux density falls to about half the interior value.

What does an iron core do in a coil?+

It boosts the field dramatically: B = μ_r·μ₀·(N/l)·I with the permeability μ_r, which for iron ranges from a few hundred to several thousand depending on the grade. Microscopically the coil field aligns the elementary magnets (Weiss domains) of the iron, whose fields add to the external one. This turns a 5 mT air coil into an electromagnet with effects of several tesla, the principle of relays, lifting magnets and transformer cores. Two limits matter: from about 1.5 to 2 T iron saturates, all domains are aligned and more current yields hardly more field, and μ_r is not a true constant but depends on the operating point. After switch-off some residual magnetism (remanence) also remains.

How do you rearrange the coil formula for current or turns?+

The formula is a pure product, so rearranging is division: I = B·l/(μ₀·N) and N = B·l/(μ₀·I). Example: B = 10 mT in a coil with N = 1000 and l = 0.5 m requires I = 0.01 × 0.5/(1.257×10⁻⁶ × 1000) ≈ 4 A. Conversely, 2 mT at I = 0.8 A and l = 0.1 m demands N = 0.002 × 0.1/(1.257×10⁻⁶ × 0.8) ≈ 200 turns. Practical note: more turns mean a longer wire and thus more ohmic resistance, so at a fixed voltage the current drops. Designing real electromagnets therefore balances turn count, wire cross-section and heat removal against each other.

How large are typical magnetic fields in comparison?+

A map of orders of magnitude helps to judge results. The Earth magnetic field is about 50 µT = 5×10⁻⁵ T. A school coil without a core typically reaches 1 to 10 mT, twenty to two hundred times that. Strong permanent magnets (neodymium) achieve 0.3 to 1.4 T at their surface. Medical MRI machines run superconducting coils at 1.5 to 7 T, research magnets reach several tens of tesla. For comparison downwards: the magnetic fields of the human heart are around 10⁻¹⁰ T. So if your coil calculation yields 50 T, a power of ten has almost certainly slipped, usually through cm instead of m or mA instead of A.

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Create a curated FSRS exam set for B = μ₀·(N/l)·I: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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How do you calculate with Magnetic Field of a Long Coil?

Here is how to work through a typical Magnetic Field of a Long Coil (B = μ₀·(N/l)·I) task step by step:

  1. 1

    Task

    What current produces B = 10 mT in a coil (N = 1000, l = 0.5 m)?

    Solution path

    I = B·l/(μ₀·N) = 0.01 × 0.5/(1.257×10⁻⁶ × 1000) ≈ 4.0 A.

  2. 2

    Task

    How many turns does a 10 cm coil need for 2 mT at I = 0.8 A?

    Solution path

    N = B·l/(μ₀·I) = 0.002 × 0.1/(1.257×10⁻⁶ × 0.8) ≈ 199, so about 200 turns.