Physics · Electrodynamics

Solenoid magnetic field: long air core and finite geometry

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

In the long air-core model, current follows from $I=\frac{B\ell}{\mu_0N}$ and turn count from $N=\frac{B\ell}{\mu_0I}$. For $B=10\,\mathrm{mT}$, $N=1000$ and $\ell=0.5\,\mathrm m$, the rounded value $\mu_0=1.257\cdot10^{-6}\,\mathrm{H/m}$ gives about $3.98\,\mathrm A$. Conversely, two millitesla at 0.8 amperes and length 0.1 metres require approximately 198.9 turns mathematically. Round up to a whole turn for a minimum target. First check that the long-interior approximation applies. More turns often increase resistance at unchanged wire cross-section; fixed supply voltage then changes current. A real design must therefore also account for geometry, temperature and heat removal.

The equation approximates interior field away from the ends of a long tightly wound air-core solenoid. Turn density and current matter, not turn count alone.

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Formula

Calculate

B = μ_{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

Variables & units
Symbol	Meaning	Unit
$B$	Magnetic flux density inside the coil	T
$μ₀$	Magnetic field constant $(1.257\times{}10^{-6})$	V·s/(A·m)
$N$	Number of turns	dimensionless
$l$	Length of the coil	m
$I$	Current	A

LaTeX

B = \mu_0 \cdot \frac{N}{l} \cdot I

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Finite axial field of a short air-core solenoid falls from 2.355 mT at centre to 1.400 mT at an end. Long approximation 3.016 mT is drawn as a horizontal comparison. — Thin uniformly wound air-core solenoid: N=300, length 0.100 m, radius 0.0400 m, current 0.800 A. Finite axial expression gives 2.355 mT at centre and 1.400 mT at z=±0.050 m. Long comparison value 3.016 mT is not the finite interior-field curve.

Solenoid magnetic field calculator

Find the magnetic flux density, current or number of turns of a long air-core coil, valid inside the coil away from its ends.

Find

Number of turns N

Number of turns as a whole number, e.g. 400; for a minimum field round a calculated N up to whole turns.

Winding length ℓ

Length of the winding, not of the wire, e.g. 20 cm; it should be much larger than the diameter.

Current I

Current through the coil wire as a magnitude, e.g. 2 A.

Filled with example values. Type your own and the result updates straight away.

Result: Flux density B inside the coil

B = 5.027 mT

With N = 400, ℓ = 20 cm and I = 2 A, B = 5.027 mT.

Working · rearrange for B, substitute, evaluate

1. Rearrange for B
$B=\mu_0\frac N\ell I$
2. Substitute values (in SI units)
- ℓ = 20 cm = 0.2 m
$\frac{0.00000125664\cdot 400\cdot 2}{0.2}=0.00502655$
3. Result
B = 5.027 mT

μ₀ ≈ 1.25663706127·10⁻⁶ H/m from CODATA 2022; it has measurement uncertainty. This is not a universal iron-core model.

Quick tips

Only the turn density N/ℓ matters: twice the turns on twice the length give the same field at the same current.
Estimate with μ₀ ≈ 1.26 × 10⁻⁶ T·m/A: 1000 turns per metre and 1 A give about 1.26 mT.

Assumptions and limits

Long tightly and uniformly wound air-core solenoid, central interior away from ends; steady or suitably quasi-static current. Length greatly exceeds diameter, which must be checked first. Current and field are magnitudes; no iron cores, fringe/exterior fields or thermal sizing.

Derivation & background

Ampère’s law yields nearly uniform interior field in the ideally long limit. A finite solenoid has edge and exterior fields; at the axial end of a very long solenoid the field is about half the central value. A real iron core is not a universal constant multiplier: magnetization, saturation, hysteresis, gaps and geometry must be considered. Vacuum permeability has measurement uncertainty in the present SI.

Exam blueprint

Validity range

Approximation for central interior of a long tightly wound air-core solenoid with steady or suitably quasi-static current. Length greatly exceeds diameter; edge and exterior fields are not universally zero. Iron cores need appropriate material and geometric assumptions.

Derivation steps

Field symmetry simplifies Ampère’s law in the ideally long air-core model.

1Specify a long uniformly tightly wound air-core solenoid and interior region away from the ends.
2Turn density is $n=\frac N\ell$ .
3Choose a rectangle with interior segment length a parallel to the field.
4In the ideally long limit, the exterior segment contributes zero; transverse segments are perpendicular to the field.
5The path encloses $na$ turns: $Ba=\mu_0naI$ .
6Cancel to obtain $B=\mu_0nI$ ; finite geometry introduces edge and axial corrections.

Rearrangements

Current

I = \frac{B \cdot l}{\mu_0 \cdot N}

Only for the central interior of a long air-core coil at a specified current magnitude; check finite geometry and magnetic cores separately.

Number of turns

N = \frac{B \cdot l}{\mu_0 \cdot I}

At fixed length and current magnitude, the long air-core approximation grows with N. At fixed supply voltage current can change with resistance.

Task variant

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

$I=\frac{B\ell}{\mu_0N}=\frac{(0.010\,\mathrm T)(0.50\,\mathrm m)}{(1.257\times10^{-6}\,\mathrm{H\,m^{-1}})1000}\approx3.98\,\mathrm A$ Long air-core solenoid, fringe field neglected.

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

$N=\frac{B\ell}{\mu_0I}=\frac{(0.0020\,\mathrm T)(0.10\,\mathrm m)}{(1.257\times10^{-6}\,\mathrm{H\,m^{-1}})(0.80\,\mathrm A)}\approx199$ Within the same long air-core approximation, about 200 turns are required.

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 $n=\frac N\ell$ .

Tables often list n; then $B=\mu_0nI$ 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.

Only an appropriate homogeneous linear material model permits $B=\mu_r\mu_0\frac N\ell I$ . Actual iron cores need a material characteristic, operating point and geometry; $\mu_r$ is not a universal constant enhancement factor.

Exam context

Common as the field source in combined tasks: Lorentz force on particles in a coil field, $\frac{e}{m}$ determination with Helmholtz coils and induction experiments.

Formula cluster

Magnetic fields

The field source for Lorentz force and induction experiments.

F = q(v × B)

Lorentz Force

U_ind = −L·(dI/dt)

Self-Induction of a Coil

Worked example

Long air-core coil: $N=500$ , ℓ=0.25 m, $I=2 A$ . Approach: $B=\mu_0\frac N\ell I$ . With $\mu_0\approx1.257\cdot10^{-6}\,\mathrm{H\,m^{-1}}$ , $B\approx5.03\,\mathrm{mT}$ . Compared with an explicitly assumed 50 µT this is about one hundred times larger; end regions and actual cores are outside this calculation.

Applications

Field measurements in long air-core solenoids, model calculations for particle deflection and induction. Electromagnets and Helmholtz arrangements need their specific field and material model.

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
Task
What current produces $B = 10 mT$ in a coil (N = 1000, l = 0.5 m)?
Solution path
$I=\frac{B\ell}{\mu_0N}=\frac{(0.010\,\mathrm T)(0.50\,\mathrm m)}{(1.257\times10^{-6}\,\mathrm{H\,m^{-1}})1000}\approx3.98\,\mathrm A$ Long air-core solenoid, fringe field neglected.
2
Task
How many turns does a 10 cm coil need for 2 mT at $I = 0.8 A$ ?
Solution path
$N=\frac{B\ell}{\mu_0I}=\frac{(0.0020\,\mathrm T)(0.10\,\mathrm m)}{(1.257\times10^{-6}\,\mathrm{H\,m^{-1}})(0.80\,\mathrm A)}\approx199$ Within the same long air-core approximation, about 200 turns are required.

Understand and apply the equation

Many turns produce an almost uniform field in the middle of a long air-core solenoid. The simple equation depends on turns per length; it is neither a universal coil formula nor an arbitrary iron-core multiplier. Comparing with a finite solenoid shows where the approximation works and where its ends matter.

Many turns reinforce a shared field component

A current loop produces magnetic field along its axis. In a tightly wound solenoid, many loop contributions superpose. In the middle of a long air-core coil they reinforce a shared axial direction and vary little over small position changes, giving a simple approximation. At an end, loops on one side are missing; field decreases and becomes less uniform.[1][3]

Prior knowledge: N counts turns; turn density is number per winding length. Winding length is not total wire length.
The long air-core formula describes B in teslas, not H in amperes per metre.
With a shared current orientation, interior field points along the axis. Use the right hand for direction.

[1][3]

How do you calculate the field inside a long air-core solenoid?

B_\infty=\mu_0nI=\mu_0\frac N\ell I

N turns are uniformly distributed over winding length ℓ; I is current magnitude here. The subscript denotes the ideally long model.[1]

The equation approximates the interior of a tightly wound solenoid without a ferromagnetic core whose length greatly exceeds its diameter. Uniform means that magnitude and direction change little within the selected measurement region. It does not mean a finite coil has the same field everywhere, especially at its ends or outside. Air permeability is treated as vacuum permeability at the precision used here.[1]

An integration path counts enclosed turns

\begin{aligned}\oint\mathbf B\cdot d\boldsymbol\ell&=\mu_0I_{\mathrm{enc}}\\{}I_{\mathrm{enc}}&=naI\\{}Ba&=\mu_0naI\\{}B&=\mu_0nI\end{aligned}

a is the length of the interior integration segment, not necessarily the entire solenoid length. It encloses na turns.[2]

In the ideally long limit choose a rectangle: one long side is inside parallel to the field, the other outside in negligible field. Short sides are perpendicular to the field and contribute zero. The integral reduces to an interior segment with constant B. For a finite coil, external and edge contributions are not exactly zero; choosing a convenient rectangle cannot remove them. The assumed symmetry justifies the simplification. Maxwell’s equations in integral form place this steady-current Ampère model and the additional displacement current in context.[2][6]

When is a solenoid long enough and what changes at an end?

“Long” is not a fixed number of centimetres. Winding-length-to-radius ratio and observation location matter. The finite expression sums axial fields of many ideal thin circular turns with uniform turn density. The same coil gives different central and end values. This expression applies on the axis; off-axis points require a different spatial model.[1][3]

B_z(z)=\frac{\mu_0nI}{2}\left[\frac{z+\ell/2}{\sqrt{R^2+(z+\ell/2)^2}}-\frac{z-\ell/2}{\sqrt{R^2+(z-\ell/2)^2}}\right]

Thin cylindrical winding of radius R, uniform turn density and centre z=0; this expression describes the axis, not the full three-dimensional field.[1]

\begin{aligned}B_{\mathrm{centre}}&=B_\infty\frac{\ell}{\sqrt{\ell^2+4R^2}}\\{}B_{\mathrm{end}}&=\frac{B_\infty}{2}\frac{\ell}{\sqrt{\ell^2+R^2}}\end{aligned}

For length much greater than R, the centre approaches the long model while the axial end field approaches half that value.[1]

The finite-axis expression sums fields of thin current loops along the winding. Its centre form gives a controllable approximation error without claiming that every coil is one-dimensional. It is particularly useful for tightly wound single-layer models; thick windings need further integration over radius. The real external field closes the field lines and is not negligible everywhere around a finite coil.[3][1]

View diagram: Centre and ends of a short solenoid

Centre and ends of a short solenoid

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Second example geometry: N=300, ℓ=10 cm, R=4 cm, I=0.8 A. Ends lie at z=±5 cm. The dashed line is the long-coil interior value, not an exterior-field prediction. The finite curve contains axial contributions from this same coil.[1][3]

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Winding current fixes axial direction

Looking at an end, counterclockwise conventional current makes the axial field point out of that end towards you. Reversing current makes it point into the solenoid. The gripping rule assigns curled fingers to winding current and the thumb to interior field. Electron drift in the wire opposes conventional current and must not be substituted without reversing orientation.[3]

More turns alter field strength at fixed current. At fixed supply voltage, however, longer wire can increase resistance and change current. “Twice the turns gives twice the field” therefore requires the same actual current, winding length and valid geometry. A Helmholtz arrangement combines two coordinated loop fields and cannot be calculated solely using the long cylindrical-solenoid equation.[1][3]

Turn density, constant status and design

\begin{aligned}[\mu_0]&=\mathrm{T\,m\,A^{-1}}=\mathrm{H\,m^{-1}}\\{}[n]&=\mathrm{m^{-1}}\\{}I&=\frac{B\ell}{\mu_0N}\\{}N&=\frac{B\ell}{\mu_0I}\end{aligned}

N is a dimensionless count. A design target for N ultimately needs an attainable integer number of turns.[1]

We use rounded calculation value μ₀≈1.25663706·10⁻⁶ H/m. The CODATA 2022 table assigns measurement uncertainty to vacuum permeability; in the present SI it is no longer exactly fixed by the old ampere definition. This uncertainty is much smaller than winding and current deviations in our model examples. Convert centimetres to metres before calculating length; turns per centimetre correspond to one hundred times as many turns per metre.[4]

Calculate interior field and finite correction

Long model solenoid

Given: N=720, ℓ=0.360 m, I=1.20 A, radius 10 mm; air core, tightly wound, measured at centre. Use μ₀=1.25663706·10⁻⁶ H/m.
Find: Calculate long-model interior field and check centre approximation error.

Approach: obtain turn density from winding length, then check using the finite centre expression.
$\begin{aligned}n&=\frac N\ell\\{}B_\infty&=\mu_0nI\\{}\frac{B_{\mathrm{centre}}}{B_\infty}&=\frac\ell{\sqrt{\ell^2+4R^2}}\end{aligned}$
Insert SI values without interchanging diameter and length.
$\begin{aligned}n&=\frac{720}{0.360\,\mathrm m}=2000\,\mathrm{m^{-1}}\\{}B_\infty&=(1.25663706\cdot10^{-6})(2000)(1.20)\,\mathrm T\\{}&=0.003015928944\,\mathrm T\\{}\frac{B_{\mathrm{centre}}}{B_\infty}&=\frac{0.360}{\sqrt{0.360^2+4(0.010)^2}}\approx0.998460\end{aligned}$

Result: The long model gives 3.016 mT; the finite centre value is about 3.011 mT.

Check: The finite model is 0.154% below the long value, relative to that value. Doubling current doubles both while the linear air-core model holds.

Short solenoid: centre differs from end

Given: Thin air-core winding: N=300, ℓ=0.100 m, R=0.0400 m, I=0.800 A. Same calculation value of μ₀.
Find: Calculate axial field at centre and at an end.

Approach: since length and diameter are comparable, use finite expressions and retain the long value only as comparison.
$\begin{aligned}B_\infty&=\mu_0\frac N\ell I\\{}B_c&=B_\infty\frac\ell{\sqrt{\ell^2+4R^2}}\\{}B_e&=\frac{B_\infty}{2}\frac\ell{\sqrt{\ell^2+R^2}}\end{aligned}$
Calculate both correction factors separately.
$\begin{aligned}B_\infty&=3.015928944\,\mathrm{mT}\\{}B_c&=3.015928944\frac{0.100}{\sqrt{0.100^2+4(0.0400)^2}}\,\mathrm{mT}\approx2.35504\,\mathrm{mT}\\{}B_e&=\frac{3.015928944}{2}\frac{0.100}{\sqrt{0.100^2+0.0400^2}}\,\mathrm{mT}\approx1.40011\,\mathrm{mT}\end{aligned}$

Result: Centre≈2.355 mT, end≈1.400 mT. The long model noticeably overestimates the centre.

Check: Centre factor is 0.7809 and end factor 0.4642. Here the end value is not exactly half the actual centre value.

An accuracy requirement becomes a minimum length

Given: An ideal thin uniformly wound air-core solenoid has radius R=1.50 cm. The infinite approximation may overestimate central axial field by at most 1.00%, relative to the finite value. For a check use N=600, I=0.500 A and ℓ=22.0 cm.
Find: Find minimum length and check the chosen coil.

Approach: use the central finite-length factor and solve the relative overestimate for length.
$\begin{aligned}\frac{B_\infty}{B(0)}-1&=\sqrt{1+\frac{4R^2}{\ell^2}}-1\le\delta\\\ell&\ge\frac{2R}{\sqrt{(1+\delta)^2-1}}\end{aligned}$
Substitute δ=0.0100, then calculate the finite field of the 22 cm coil.
$\begin{aligned}\ell_{\min}&=\frac{0.0300}{\sqrt{1.0100^2-1}}\,\mathrm m\approx0.2116\,\mathrm m\\B_\infty&=\mu_0\frac{600}{0.220}(0.500)\approx1.71360\,\mathrm{mT}\\B(0)&=\frac{B_\infty}{\sqrt{1+\frac{4(0.0150)^2}{0.220^2}}}\approx1.69788\,\mathrm{mT}\end{aligned}$

Result: Minimum length is about 21.16 cm. The chosen 22.0 cm length meets the requirement at the centre.

Check: The overestimate is about 0.9255% relative to finite central field. This does not establish 1% accuracy at the end, where a different spatial correction applies.

Five different checks of the solenoid model

Calculate · Foundation

A long air-core solenoid has N=450, ℓ=0.300 m and I=0.600 A. Use μ₀=1.25663706·10⁻⁶ H/m. Find central interior field.

Show hint

Do not multiply only turn count by current.

Learning goal: Calculate turn density and field in SI.

Show solution and calculation

Model and approach: Central region of a long air-core coil, away from its ends.
$\begin{aligned}n&=\frac N\ell\\B&=\mu_0nI\end{aligned}$ Approach: N and ℓ give $n=1500\,\mathrm{m^{-1}}$ .
Field is 1.131 mT.

Check: doubling length at fixed turns halves field while the long model remains valid.

B=(1.25663706\cdot10^{-6})\frac{450}{0.300}(0.600)\,\mathrm T=1.130973354\,\mathrm{mT}

[1]

Rearrange · Application

In the long air-core model, N=800 and ℓ=0.400 m should produce 2.00 mT. What current is required? Same μ₀ value.

Show hint

Convert millitesla to tesla first.

Learning goal: Design current for a specified field.

Show solution and calculation

Model and approach: Solve for current under the same long air-core assumptions.
$I=\frac{B\ell}{\mu_0N}$ Approach: rearrange the field equation for I.
I≈0.795775 A.

Check: turn density 2000 m⁻¹ with this current reproduces 2.00 mT.

I=\frac{(0.00200)(0.400)}{(1.25663706\cdot10^{-6})(800)}\,\mathrm A\approx0.795775\,\mathrm A

[1]

Interpret · Application

You look at the right end of an air-core solenoid. Conventional current is counterclockwise in this view. Does interior field point right towards you or left away? What changes on reversing current?

Show hint

Keep the stated end view fixed.

Learning goal: Connect winding-current arrows with field direction.

Show solution and calculation

Approach: apply the gripping rule to conventional winding current.
Field points right towards you; reversing current points it left.

Check: equal absolute current leaves field magnitude unchanged.

\mathbf B(-I)=-\mathbf B(I)

[3]

Error analysis · Application

For a thin air-core solenoid with $\ell=2R$ a calculation claims centre field is exactly μ₀nI because all turns look alike. Find the correct factor and relative overestimate measured against the finite value.

Show hint

Substitute $\ell=2R$ into the centre expression.

Learning goal: Correct an invalid finite-coil field assumption.

Show solution and calculation

Approach: finite geometry gives factor $\frac{1}{\sqrt2}$ .
Relative to the finite value, the long value is too large by $\sqrt2-1\approx0.414\quad(41.4\,\%)$ .

Check: similar-looking loops do not establish infinite translational symmetry.

\begin{aligned}\frac{B_c}{B_\infty}&=\frac{2R}{\sqrt{4R^2+4R^2}}=\frac1{\sqrt2}\\{}\frac{B_\infty-B_c}{B_c}&=\sqrt2-1\approx0.4142\end{aligned}

[1]

Transfer · Challenge

Two long air-core solenoids have the same length and radius. Coil B has twice as many turns and twice the wire length at equal wire cross-section; in this simplified model its resistance is doubled. Both reach steady state at the same constant voltage. Compare their fields.

Show hint

Equal voltage does not imply equal current here.

Learning goal: Account for assumptions when changing turns at fixed voltage.

Show solution and calculation

Approach: Ohm’s law gives half the current in B.
Twice the turn density times half the current gives the same interior field.

Check: this requires the stated resistance scaling and equal conditions; heating and extra layers are not automatically included.

\frac{B_B}{B_A}=\frac{N_B}{N_A}\frac{I_B}{I_A}=2\cdot\frac12=1

[1]

Explore further: Real magnetic cores

A homogeneous linear magnetic material can be described by permeability under suitable geometric conditions. A real iron core has operating-point-dependent magnetization, often hysteresis, saturation and leakage flux. Simply multiplying an air-core result by an arbitrary large constant μᵣ therefore does not reliably predict an open electromagnet. Air gaps and the complete magnetic path may require a magnetic-circuit or field model.[5]

For changing current, the quasi-static field expression is appropriate only when propagation and displacement-current effects are small for the question. Resistive losses and acceptable wire temperature limit attainable current. A theoretical target value for B does not produce a ready-to-operate design. The long solenoid is a clear starting point for field measurements and particle deflection, not a replacement for checking actual construction. When current changes, self-induction describes the additional winding voltage in the appropriate inductance model.[2]

What to take away

Long air-core field depends on turns per length and current. More turns alone do not guarantee stronger field at fixed supply voltage.
Check length-to-radius ratio and observation location. Central field does not automatically apply at the end.
An iron core needs a material model; its effect is not a universal constant multiplier.

[1][5]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Refined question headings and added scientifically relevant paragraph links; retained existing derivations, examples, diagrams, exercises and sources.
October 1, 2026Distinguished long air-core model from finite axial fields; wrote Ampère reasoning, two original examples, five exercises and material limits.
October 1, 2026Added an accessible entry and prerequisites, explained model steps, and added a complete example and quantitative diagram; retained existing examples, five exercises and sources.

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Quanta. Magnetic Field of a Long Coil. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/magnetfeld-spule

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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 the central interior of a long tightly wound air-core solenoid, approximately $B=\mu_0\frac N\ell I$ . Winding length must greatly exceed diameter; directly at an end the long-interior approximation is inappropriate. N counts turns, while turn count divided by length gives turn density. The example with 500 turns, length 0.25 m and current 2 A gives about 5.03 mT using the usual rounded vacuum permeability. Convert centimetres and milliamperes before substitution. A real iron core cannot generally be represented by an arbitrary constant multiplier: geometry, air gaps, magnetization and saturation must fit the material model. The simple expression is a conditional model, not a universal formula for every coil.

Why is the field inside a long coil uniform?+

Many closely distributed turns superpose their loop fields. In the ideally infinite solenoid, symmetry gives a uniform axial interior field and Ampère’s law determines its magnitude. A real long solenoid approaches this result in its central interior, away from the ends. Uniformity therefore requires a specified accuracy and measurement region. A finite solenoid retains edge and external fields. On its axis, the end field of a very long solenoid approaches roughly half the central interior value. That factor is not generally exact for a short coil; check its finite-axis expression instead. A picture of similar-looking turns does not replace these model conditions.

What does an iron core do in a coil?+

An iron core changes the field through magnetization. A suitable homogeneous linear model can use relative permeability, but actual ferromagnetic cores often have operating-point- and history-dependent response. Air gaps, the return path and leakage flux additionally change field distribution. Consequently, an air-core value of 5 mT multiplied by an arbitrary large material number does not reliably predict several tesla. Saturation reduces further enhancement, while hysteresis can leave remanence after switch-off. A concrete design needs the actual material characteristic and complete geometry. This answer does not assert one universal saturation threshold for every iron grade. The useful distinction is between a specified linear material model and a real magnetic circuit.

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

In the long air-core model, current follows from $I=\frac{B\ell}{\mu_0N}$ and turn count from $N=\frac{B\ell}{\mu_0I}$ . For $B=10\,\mathrm{mT}$ , $N=1000$ and $\ell=0.5\,\mathrm m$ , the rounded value $\mu_0=1.257\cdot10^{-6}\,\mathrm{H/m}$ gives about $3.98\,\mathrm A$ . Conversely, two millitesla at 0.8 amperes and length 0.1 metres require approximately 198.9 turns mathematically. Round up to a whole turn for a minimum target. First check that the long-interior approximation applies. More turns often increase resistance at unchanged wire cross-section; fixed supply voltage then changes current. A real design must therefore also account for geometry, temperature and heat removal.

How large are typical magnetic fields in comparison?+

Field comparisons indicate orders of magnitude, not limits for every construction. Earth’s field is roughly several tens of microtesla and depends on location. A comparison value of 50 microtesla equals 0.050 millitesla. Our long model solenoid with 720 turns over 0.360 m at 1.20 A gives about 3.02 millitesla, approximately sixty times that comparison value. Much larger fields require different currents, windings, cooling or materials; the field equation alone does not verify these engineering conditions. An unusually large calculated value prompts an input and model check but does not mathematically prove a mistake. Entering centimetres instead of metres can, for example, introduce a factor of one hundred.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §12.6 Solenoids and Toroids

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Solenoid magnetic field: long air core and finite geometry

The equation approximates interior field away from the ends of a long tightly wound air-core solenoid. Turn density and current matter, not turn count alone.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

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Symbol

Meaning

Unit

B

Magnetic flux density inside the coil

μ₀

Magnetic field constant

(1.257\times{}10^{-6})

V·s/(A·m)

N

Number of turns

dimensionless

l

Length of the coil

I

Current

Derivation & background

Exam blueprint

Validity range

Derivation steps

Field symmetry simplifies Ampère’s law in the ideally long air-core model.

1Specify a long uniformly tightly wound air-core solenoid and interior region away from the ends.
2Turn density is $n=\frac N\ell$ .
3Choose a rectangle with interior segment length a parallel to the field.
4In the ideally long limit, the exterior segment contributes zero; transverse segments are perpendicular to the field.
5The path encloses $na$ turns: $Ba=\mu_0naI$ .
6Cancel to obtain $B=\mu_0nI$ ; finite geometry introduces edge and axial corrections.

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:

Task

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

Solution path

$I=\frac{B\ell}{\mu_0N}=\frac{(0.010\,\mathrm T)(0.50\,\mathrm m)}{(1.257\times10^{-6}\,\mathrm{H\,m^{-1}})1000}\approx3.98\,\mathrm A$ Long air-core solenoid, fringe field neglected.

Task

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

Solution path

Understand and apply the equation

Many turns reinforce a shared field component

Prior knowledge: N counts turns; turn density is number per winding length. Winding length is not total wire length.
The long air-core formula describes B in teslas, not H in amperes per metre.
With a shared current orientation, interior field points along the axis. Use the right hand for direction.

[1][3]

How do you calculate the field inside a long air-core solenoid?

B_\infty=\mu_0nI=\mu_0\frac N\ell I

N turns are uniformly distributed over winding length ℓ; I is current magnitude here. The subscript denotes the ideally long model.[1]

An integration path counts enclosed turns

\begin{aligned}\oint\mathbf B\cdot d\boldsymbol\ell&=\mu_0I_{\mathrm{enc}}\\{}I_{\mathrm{enc}}&=naI\\{}Ba&=\mu_0naI\\{}B&=\mu_0nI\end{aligned}

a is the length of the interior integration segment, not necessarily the entire solenoid length. It encloses na turns.[2]

When is a solenoid long enough and what changes at an end?

B_z(z)=\frac{\mu_0nI}{2}\left[\frac{z+\ell/2}{\sqrt{R^2+(z+\ell/2)^2}}-\frac{z-\ell/2}{\sqrt{R^2+(z-\ell/2)^2}}\right]

Thin cylindrical winding of radius R, uniform turn density and centre z=0; this expression describes the axis, not the full three-dimensional field.[1]

\begin{aligned}B_{\mathrm{centre}}&=B_\infty\frac{\ell}{\sqrt{\ell^2+4R^2}}\\{}B_{\mathrm{end}}&=\frac{B_\infty}{2}\frac{\ell}{\sqrt{\ell^2+R^2}}\end{aligned}

For length much greater than R, the centre approaches the long model while the axial end field approaches half that value.[1]

View diagram: Centre and ends of a short solenoid

Centre and ends of a short solenoid

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Winding current fixes axial direction

Turn density, constant status and design

\begin{aligned}[\mu_0]&=\mathrm{T\,m\,A^{-1}}=\mathrm{H\,m^{-1}}\\{}[n]&=\mathrm{m^{-1}}\\{}I&=\frac{B\ell}{\mu_0N}\\{}N&=\frac{B\ell}{\mu_0I}\end{aligned}

N is a dimensionless count. A design target for N ultimately needs an attainable integer number of turns.[1]

Calculate interior field and finite correction

Long model solenoid

Given: N=720, ℓ=0.360 m, I=1.20 A, radius 10 mm; air core, tightly wound, measured at centre. Use μ₀=1.25663706·10⁻⁶ H/m.
Find: Calculate long-model interior field and check centre approximation error.

Approach: obtain turn density from winding length, then check using the finite centre expression.
$\begin{aligned}n&=\frac N\ell\\{}B_\infty&=\mu_0nI\\{}\frac{B_{\mathrm{centre}}}{B_\infty}&=\frac\ell{\sqrt{\ell^2+4R^2}}\end{aligned}$
Insert SI values without interchanging diameter and length.
$\begin{aligned}n&=\frac{720}{0.360\,\mathrm m}=2000\,\mathrm{m^{-1}}\\{}B_\infty&=(1.25663706\cdot10^{-6})(2000)(1.20)\,\mathrm T\\{}&=0.003015928944\,\mathrm T\\{}\frac{B_{\mathrm{centre}}}{B_\infty}&=\frac{0.360}{\sqrt{0.360^2+4(0.010)^2}}\approx0.998460\end{aligned}$

Result: The long model gives 3.016 mT; the finite centre value is about 3.011 mT.

Check: The finite model is 0.154% below the long value, relative to that value. Doubling current doubles both while the linear air-core model holds.

Short solenoid: centre differs from end

Given: Thin air-core winding: N=300, ℓ=0.100 m, R=0.0400 m, I=0.800 A. Same calculation value of μ₀.
Find: Calculate axial field at centre and at an end.

Approach: since length and diameter are comparable, use finite expressions and retain the long value only as comparison.
$\begin{aligned}B_\infty&=\mu_0\frac N\ell I\\{}B_c&=B_\infty\frac\ell{\sqrt{\ell^2+4R^2}}\\{}B_e&=\frac{B_\infty}{2}\frac\ell{\sqrt{\ell^2+R^2}}\end{aligned}$
Calculate both correction factors separately.
$\begin{aligned}B_\infty&=3.015928944\,\mathrm{mT}\\{}B_c&=3.015928944\frac{0.100}{\sqrt{0.100^2+4(0.0400)^2}}\,\mathrm{mT}\approx2.35504\,\mathrm{mT}\\{}B_e&=\frac{3.015928944}{2}\frac{0.100}{\sqrt{0.100^2+0.0400^2}}\,\mathrm{mT}\approx1.40011\,\mathrm{mT}\end{aligned}$

Result: Centre≈2.355 mT, end≈1.400 mT. The long model noticeably overestimates the centre.

Check: Centre factor is 0.7809 and end factor 0.4642. Here the end value is not exactly half the actual centre value.

An accuracy requirement becomes a minimum length

Given: An ideal thin uniformly wound air-core solenoid has radius R=1.50 cm. The infinite approximation may overestimate central axial field by at most 1.00%, relative to the finite value. For a check use N=600, I=0.500 A and ℓ=22.0 cm.
Find: Find minimum length and check the chosen coil.

Approach: use the central finite-length factor and solve the relative overestimate for length.
$\begin{aligned}\frac{B_\infty}{B(0)}-1&=\sqrt{1+\frac{4R^2}{\ell^2}}-1\le\delta\\\ell&\ge\frac{2R}{\sqrt{(1+\delta)^2-1}}\end{aligned}$
Substitute δ=0.0100, then calculate the finite field of the 22 cm coil.
$\begin{aligned}\ell_{\min}&=\frac{0.0300}{\sqrt{1.0100^2-1}}\,\mathrm m\approx0.2116\,\mathrm m\\B_\infty&=\mu_0\frac{600}{0.220}(0.500)\approx1.71360\,\mathrm{mT}\\B(0)&=\frac{B_\infty}{\sqrt{1+\frac{4(0.0150)^2}{0.220^2}}}\approx1.69788\,\mathrm{mT}\end{aligned}$

Result: Minimum length is about 21.16 cm. The chosen 22.0 cm length meets the requirement at the centre.

Check: The overestimate is about 0.9255% relative to finite central field. This does not establish 1% accuracy at the end, where a different spatial correction applies.

Five different checks of the solenoid model

Calculate · Foundation

A long air-core solenoid has N=450, ℓ=0.300 m and I=0.600 A. Use μ₀=1.25663706·10⁻⁶ H/m. Find central interior field.

Show hint

Do not multiply only turn count by current.

Learning goal: Calculate turn density and field in SI.

Show solution and calculation

Model and approach: Central region of a long air-core coil, away from its ends.
$\begin{aligned}n&=\frac N\ell\\B&=\mu_0nI\end{aligned}$ Approach: N and ℓ give $n=1500\,\mathrm{m^{-1}}$ .
Field is 1.131 mT.

Check: doubling length at fixed turns halves field while the long model remains valid.

B=(1.25663706\cdot10^{-6})\frac{450}{0.300}(0.600)\,\mathrm T=1.130973354\,\mathrm{mT}

[1]

Rearrange · Application

In the long air-core model, N=800 and ℓ=0.400 m should produce 2.00 mT. What current is required? Same μ₀ value.

Show hint

Convert millitesla to tesla first.

Learning goal: Design current for a specified field.

Show solution and calculation

Model and approach: Solve for current under the same long air-core assumptions.
$I=\frac{B\ell}{\mu_0N}$ Approach: rearrange the field equation for I.
I≈0.795775 A.

Check: turn density 2000 m⁻¹ with this current reproduces 2.00 mT.

I=\frac{(0.00200)(0.400)}{(1.25663706\cdot10^{-6})(800)}\,\mathrm A\approx0.795775\,\mathrm A

[1]

Interpret · Application

You look at the right end of an air-core solenoid. Conventional current is counterclockwise in this view. Does interior field point right towards you or left away? What changes on reversing current?

Show hint

Keep the stated end view fixed.

Learning goal: Connect winding-current arrows with field direction.

Show solution and calculation

Approach: apply the gripping rule to conventional winding current.
Field points right towards you; reversing current points it left.

Check: equal absolute current leaves field magnitude unchanged.

\mathbf B(-I)=-\mathbf B(I)

[3]

Error analysis · Application

For a thin air-core solenoid with $\ell=2R$ a calculation claims centre field is exactly μ₀nI because all turns look alike. Find the correct factor and relative overestimate measured against the finite value.

Show hint

Substitute $\ell=2R$ into the centre expression.

Learning goal: Correct an invalid finite-coil field assumption.

Show solution and calculation

Approach: finite geometry gives factor $\frac{1}{\sqrt2}$ .
Relative to the finite value, the long value is too large by $\sqrt2-1\approx0.414\quad(41.4\,\%)$ .

Check: similar-looking loops do not establish infinite translational symmetry.

\begin{aligned}\frac{B_c}{B_\infty}&=\frac{2R}{\sqrt{4R^2+4R^2}}=\frac1{\sqrt2}\\{}\frac{B_\infty-B_c}{B_c}&=\sqrt2-1\approx0.4142\end{aligned}

[1]

Transfer · Challenge

Two long air-core solenoids have the same length and radius. Coil B has twice as many turns and twice the wire length at equal wire cross-section; in this simplified model its resistance is doubled. Both reach steady state at the same constant voltage. Compare their fields.

Show hint

Equal voltage does not imply equal current here.

Learning goal: Account for assumptions when changing turns at fixed voltage.

Show solution and calculation

Approach: Ohm’s law gives half the current in B.
Twice the turn density times half the current gives the same interior field.

Check: this requires the stated resistance scaling and equal conditions; heating and extra layers are not automatically included.

\frac{B_B}{B_A}=\frac{N_B}{N_A}\frac{I_B}{I_A}=2\cdot\frac12=1

[1]

Explore further: Real magnetic cores

What to take away

Long air-core field depends on turns per length and current. More turns alone do not guarantee stronger field at fixed supply voltage.
Check length-to-radius ratio and observation location. Central field does not automatically apply at the end.
An iron core needs a material model; its effect is not a universal constant multiplier.

[1][5]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Refined question headings and added scientifically relevant paragraph links; retained existing derivations, examples, diagrams, exercises and sources.
October 1, 2026Distinguished long air-core model from finite axial fields; wrote Ampère reasoning, two original examples, five exercises and material limits.
October 1, 2026Added an accessible entry and prerequisites, explained model steps, and added a complete example and quantitative diagram; retained existing examples, five exercises and sources.

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Quanta. Magnetic Field of a Long Coil. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/magnetfeld-spule

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Frequently asked questions about Magnetic Field of a Long Coil

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

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Solenoid magnetic field: long air core and finite geometry

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Solenoid magnetic field calculator

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Derivation & background

Exam blueprint

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Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Magnetic fields

Worked example

Applications

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

Many turns reinforce a shared field component

How do you calculate the field inside a long air-core solenoid?

An integration path counts enclosed turns

When is a solenoid long enough and what changes at an end?

Centre and ends of a short solenoid

Winding current fixes axial direction

Turn density, constant status and design

Calculate interior field and finite correction

Long model solenoid

Short solenoid: centre differs from end

An accuracy requirement becomes a minimum length

Five different checks of the solenoid model

A long air-core solenoid has N=450, ℓ=0.300 m and I=0.600 A. Use μ0=1.25663706·10−6 H/m. Find central interior field.

In the long air-core model, N=800 and ℓ=0.400 m should produce 2.00 mT. What current is required? Same μ0 value.

You look at the right end of an air-core solenoid. Conventional current is counterclockwise in this view. Does interior field point right towards you or left away? What changes on reversing current?

For a thin air-core solenoid with ℓ=2R\ell=2Rℓ=2R a calculation claims centre field is exactly μ0nI because all turns look alike. Find the correct factor and relative overestimate measured against the finite value.

Two long air-core solenoids have the same length and radius. Coil B has twice as many turns and twice the wire length at equal wire cross-section; in this simplified model its resistance is doubled. Both reach steady state at the same constant voltage. Compare their fields.

Explore further: Real magnetic cores

What to take away

Sources for this explanation

Frequently asked questions about Magnetic Field of a Long Coil

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Magnetic Field of a Long Coil for exams

Solenoid magnetic field: long air core and finite geometry

Variables & units

Solenoid magnetic field calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Magnetic fields

Worked example

Applications

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

Many turns reinforce a shared field component

How do you calculate the field inside a long air-core solenoid?

An integration path counts enclosed turns

When is a solenoid long enough and what changes at an end?

Centre and ends of a short solenoid

Winding current fixes axial direction

Turn density, constant status and design

Calculate interior field and finite correction

Long model solenoid

Short solenoid: centre differs from end

An accuracy requirement becomes a minimum length

Five different checks of the solenoid model

A long air-core solenoid has N=450, ℓ=0.300 m and I=0.600 A. Use μ0=1.25663706·10−6 H/m. Find central interior field.

In the long air-core model, N=800 and ℓ=0.400 m should produce 2.00 mT. What current is required? Same μ0 value.

You look at the right end of an air-core solenoid. Conventional current is counterclockwise in this view. Does interior field point right towards you or left away? What changes on reversing current?

For a thin air-core solenoid with ℓ=2R\ell=2Rℓ=2R a calculation claims centre field is exactly μ0nI because all turns look alike. Find the correct factor and relative overestimate measured against the finite value.

Two long air-core solenoids have the same length and radius. Coil B has twice as many turns and twice the wire length at equal wire cross-section; in this simplified model its resistance is doubled. Both reach steady state at the same constant voltage. Compare their fields.

Explore further: Real magnetic cores

What to take away

Sources for this explanation

Frequently asked questions about Magnetic Field of a Long Coil

Subject references and further reading

Related formulas

A long air-core solenoid has N=450, ℓ=0.300 m and I=0.600 A. Use μ₀=1.25663706·10⁻⁶ H/m. Find central interior field.

In the long air-core model, N=800 and ℓ=0.400 m should produce 2.00 mT. What current is required? Same μ₀ value.

For a thin air-core solenoid with $\ell=2R$ a calculation claims centre field is exactly μ₀nI because all turns look alike. Find the correct factor and relative overestimate measured against the finite value.

A long air-core solenoid has N=450, ℓ=0.300 m and I=0.600 A. Use μ₀=1.25663706·10⁻⁶ H/m. Find central interior field.

In the long air-core model, N=800 and ℓ=0.400 m should produce 2.00 mT. What current is required? Same μ₀ value.

For a thin air-core solenoid with $\ell=2R$ a calculation claims centre field is exactly μ₀nI because all turns look alike. Find the correct factor and relative overestimate measured against the finite value.