Physics · Magnetism

Magnetic field of a long straight wire

Q: How do I check units and scale of a wire’s field?

Permeability has unit $\mathrm{T\,m/A}$; multiplying by amperes and dividing by metres leaves tesla. Convert milliamperes and centimetres before substitution. Two centimetres is $0.020\,\mathrm m$, not two metres. The existing result $5.0\cdot10^{-5}\,\mathrm T$ is fifty microtesla, not fifty millitesla. The vacuum permeability used is a rounded value of a constant with measurement uncertainty. As an additional check, retain the current-to-distance ratio: the same ratio gives the same field magnitude in the same model. Units alone cannot detect using surface distance instead of axis distance or overlooking a return conductor.

The external field magnitude of a long straight current-carrying wire decreases inversely with shortest axis distance; the field vector circulates tangentially around the axis.

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Formula

Calculate

B = \frac{μ _{0} ∣ I ∣}{2 π r}

B in T; μ₀ in N/A²; I in A; r in m.

Variables & units

Variables & units
Symbol	Meaning	Unit
$B$	Magnitude of magnetic flux density at the external measurement point	T
$μ₀$	Vacuum magnetic constant	N/A²
$I$	Current in the wire	A
$r$	Shortest perpendicular distance to the conductor axis, not its surface	m

LaTeX

B=\frac{\mu_0|I|}{2\pi r}

QUANTA STUDYquanta-study.de

Long round conductor with uniform current density: normalized field rises linearly inside to one at radius a, then falls inversely outside. At half radius, field is half the surface value; on axis it is zero. — Original interior/exterior-model diagram following OpenStax §12.5. B(a) is surface field magnitude. For I = 10.0 A and a = 2.00 mm, B(a) ≈ 1.00 mT. External inverse-distance form does not apply inside; determine circulation separately with the right hand.

Magnetic field of a straight wire calculator

Find the magnetic flux density around a long straight wire from current and distance, in air or vacuum.

Calculation model

Find

Current I (magnitude)

Current in the wire as a magnitude, e.g. 10 A in a single long wire.

Perpendicular wire-axis distance r

Distance from the point to the centre of the wire, e.g. 2 cm; the point lies outside the wire.

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

Working · rearrange for B, substitute, evaluate

1. Rearrange for B
$B=\frac{\mu_0|I|}{2\pi r}$
2. Substitute values (in SI units)
- r = 2 cm = 0.02 m
$\frac{0.00000125664\cdot 10}{2\cdot 3.14159\cdot 0.02}=0.0001$
3. Result
B = 100 µT

μ₀≈1.25663706127·10⁻⁶ T·m/A, measured CODATA 2022 value; it is no longer exactly 4π·10⁻⁷ in the revised SI.

Result: Magnetic flux density B

B = 100 µT

With |I| = 10 A and r = 2 cm, B = 100 µT.

Quick tips

Estimate: μ₀/(2π) is 2 · 10⁻⁷ T·m/A, so B in µT is 0.2 times I in A divided by r in m: 10 A at 2 cm gives 100 µT.
Twice the distance, half the field: B falls off as 1/r, not as 1/r² like the field of a point charge.

Assumptions and limits

Steady current magnitude in a very long straight thin wire, external point in vacuum or approximately air; r is shortest distance from the wire axis outside the conductor, with ends far away. This mode returns only magnitude; determine direction separately from current direction and the right-hand rule.

Derivation & background

Magnetostatic Ampère’s law and cylindrical symmetry give $B=\frac{\mu_0|I|}{2\pi r}$ . The right-hand rule determines circulation for conventional current. r is axis distance; inside the wire use enclosed current. A finite segment and its return path need additional geometric conditions. Vacuum permeability has measurement uncertainty in the present SI.

Exam blueprint

Validity range

External field of a very long straight conductor with steady current in vacuum or a suitable air model. r is outside the cross-section; check endpoints, return paths, material and time dependence separately.

Derivation steps

Cylindrical symmetry simplifies the magnetostatic Ampère integral; determine field direction separately.

1Specify a very long straight steady current and an external point in a vacuum/air model.
2Measure r as shortest distance to the axis, not to the surface.
3Cylindrical symmetry gives tangential field of constant magnitude on a concentric circle.
4Choose traversal consistently with current and use $\oint\vec B\cdot\mathrm d\vec\ell=\mu_0I_{\mathrm{enc}}$ .
5The integral becomes $B2\pi r$ ; outside, the full current is enclosed.
6Solving gives magnitude $B=\frac{\mu_0|I|}{2\pi r}$ .
7Check right-hand circulation, inverse-distance dependence and possible return-wire fields.

Rearrangements

Current magnitude from field magnitude

|I|=\frac{2\pi rB}{\mu_0}

External point in the long-wire model; infer current direction from field circulation.

Task variant

A long straight wire carries 4.00 A. Find its external field magnitude 4.00 cm from the axis in the air model.

Approach: $B=\frac{\mu_0|I|}{2\pi r}$ ; with $r=0.0400\,\mathrm m$ , $B\approx20.0\,\mathrm{\mu T}$ .

Common mistakes

Using the diameter instead of shortest distance r.

r is measured from the wire axis to the field point.

Exam context

Ampère’s law, the right-hand rule and superposition of wire fields.

Formula cluster

Magnetic fields from currents

Wire geometry and symmetry determine which field relation applies.

ΦB = BA cos θ

Magnetic flux through a surface

B = μ₀·(N/l)·I

Magnetic Field of a Long Coil

F = q(v × B)

Lorentz Force

F = ke·|q₁q₂|/r²

Coulomb's Law

Worked example

For $I=5.0\,\mathrm A$ and $r=0.020\,\mathrm m$ , $B=\frac{\mu_0|I|}{2\pi r}\approx5.0\cdot10^{-5}\,\mathrm T$ . Doubling distance halves field magnitude; direction separately follows current circulation.

Applications

Magnetic-field measurements, wire arrangements, Ampère’s law and estimating fields near current-carrying wires.

How do you calculate with Magnetic field of a long straight wire?

Here is how to work through a typical Magnetic field of a long straight wire (B = μ₀|I|/(2πr)) task step by step:

1
Task
A long straight wire carries 4.00 A. Find its external field magnitude 4.00 cm from the axis in the air model.
Solution path
Approach: $B=\frac{\mu_0|I|}{2\pi r}$ ; with $r=0.0400\,\mathrm m$ , $B\approx20.0\,\mathrm{\mu T}$ .

Understand and apply the equation

A straight current-carrying wire creates a circulating field. Outside a long wire its magnitude decreases with distance from the axis, not the surface. Symmetry explains the simple relation; finite length, a return wire and the wire interior expose its limits.

A compass beside a current-carrying wire

Current in a straight wire can turn a nearby compass. Magnetic field circles the wire axis rather than pointing radially towards it. Imagine upward conventional current: with your right thumb along current, curled fingers indicate field circulation. The long-wire model calculates this wire’s contribution; other parts of the circuit can add further fields.[1][2]

Prior knowledge: B is magnetic flux density in teslas. Current direction means conventional current.
Measure r from the axis. The exterior formula applies outside the wire cross-section, not down to its centre.
For current along +z and an observation point on +x, field points along +y. Reversing current reverses that direction.

[1][2]

How do you calculate field magnitude and direction outside a long wire?

B=\frac{\mu_0|I|}{2\pi r}

Magnitude describes the external field of an ideally long straight wire; r is the shortest axis distance.[1]

The field vector is tangent to a circle around the wire axis, rather than pointing radially away from the wire. Conventional current fixes circulation: right thumb along current, curled fingers along field. For current along +z, field at a point on +x points along +y. Reversing current reverses the vector without changing its magnitude at equal absolute current.[1]

Why a circle simplifies the integral

Ampère’s law first describes closed circulation of field. Multiplying B by circle circumference requires cylindrical symmetry: field is tangential and constant in magnitude on a coaxial circle. A single finite segment lacks this complete symmetry. Its contribution needs an end correction or Biot–Savart; total field also needs the rest of the circuit. Maxwell’s equations in integral form contain this magnetostatic circulation as a limiting case; time-varying electric flux requires the additional displacement-current term.[2][1][4]

\begin{aligned}\oint_C\vec B\cdot\mathrm d\vec\ell&=\mu_0 I_{\mathrm{enc}}\\B(2\pi r)&=\mu_0|I|\\B&=\frac{\mu_0|I|}{2\pi r}\end{aligned}

Simplification requires cylindrical symmetry of current distribution, not merely a circular integration path.[2]

In the magnetostatic model, magnitude is constant on each concentric circle and field is tangent to the path. Choose traversal consistently with current so enclosed current and field contribution count positively. Circumference then replaces the line integral. Ampère’s law is the physical starting point; cancelling factors only justifies its use for this symmetry. With several asymmetrically placed wires, the law may still apply although B can no longer be factored out of the integral.[2]

A finite segment gives less field

\begin{aligned}B_{\mathrm{seg}}&=\frac{\mu_0|I|}{4\pi r}\left(\frac{h_1}{\sqrt{r^2+h_1^2}}+\frac{h_2}{\sqrt{r^2+h_2^2}}\right)\\B_{\mathrm{mid}}&=\frac{\mu_0|I|}{2\pi r}\frac{h}{\sqrt{r^2+h^2}}\end{aligned}

The perpendicular foot lies between the ends; h₁ and h₂ are positive axial endpoint distances. At the midpoint segment length is 2h.[1]

Integrating the Biot–Savart contribution along the straight segment gives the end correction. If both endpoint distances greatly exceed r, both fractions approach one. Such a segment contributes to a complete circuit: steady current does not simply terminate in space. Supply and return conductors generate their own fields. The segment result therefore is not automatically the total measured field. Outside the region between endpoints, integrate using the actual limits.[1]

Which distance belongs in the formula and how do you rearrange it?

\begin{aligned}|I|&=\frac{2\pi rB}{\mu_0}\\r&=\frac{\mu_0|I|}{2\pi B}\\{}[\mu_0]&=\mathrm{T\,m\,A^{-1}}\end{aligned}

Both rearrangements require known model conditions and nonzero denominators.[1]

Distances in metres are measured from the axis, not as wire diameters. We use rounded vacuum permeability 1.25663706·10⁻⁶ H/m. In the present SI it has measurement uncertainty rather than being exactly fixed by the historical ampere definition. Input rounding and geometric deviations dominate these examples. The formula first describes this current’s field; Earth’s field or nearby conductors are added vectorially afterwards. Measuring only total field magnitude without direction does not always uniquely determine current.[3][1]

Example: long-wire result and segment check

Examining a 24 cm segment

Given: I=7.50 A along +z; field point 3.00 cm along +x from the midpoint of a straight 24.0 cm segment. Vacuum/air model; other circuit contributions are treated separately.
Find: Find the ideally long result and finite-segment contribution.

Approach: first calculate the long-wire model, then multiply by the endpoint correction.
$\begin{aligned}B_\infty&=\frac{\mu_0I}{2\pi r}\\B_{\mathrm{mid}}&=B_\infty\frac{h}{\sqrt{h^2+r^2}}\end{aligned}$
Substitution: half the segment length is 0.120 m.
$\begin{aligned}B_\infty&=\frac{(1.25663706\cdot10^{-6})(7.50)}{2\pi(0.0300)}\,\mathrm T\approx50.0\,\mathrm{\mu T}\\\frac{h}{\sqrt{h^2+r^2}}&=\frac{0.120}{\sqrt{0.120^2+0.0300^2}}\approx0.970143\\B_{\mathrm{mid}}&\approx48.5071\,\mathrm{\mu T}\end{aligned}$

Result: The segment contributes about 48.5 µT along +y; the long-wire comparison is 50.0 µT.

Check: Relative to the long result, segment field is 2.99% smaller. The difference vanishes as h approaches infinity. A return conductor can still increase or decrease total field.

Example: unequal parallel currents

A zero-field point between two axes

Given: Long wires at x=−0.0400 m and +0.0400 m carry 8.00 A and 3.00 A, both along +z. Points lie on the joining line outside the wires; other fields are neglected.
Find: Find midpoint field and the zero-field point between the wires.

Approach: between equal-direction currents, field vectors oppose. Equate their magnitudes at the zero point.
$\begin{aligned}B_y(0)&=\frac{\mu_0}{2\pi a}(I_1-I_2)\\\frac{I_1}{d_1}&=\frac{I_2}{2a-d_1}\\d_1&=\frac{2aI_1}{I_1+I_2}\end{aligned}$
Substitution: d₁ is measured from the left axis.
$\begin{aligned}B_y(0)&\approx25.0\,\mathrm{\mu T}\\d_1&=\frac{(0.0800)(8.00)}{11.00}\,\mathrm m\approx0.0581818\,\mathrm m\\x_0&=d_1-0.0400\,\mathrm m\approx0.0181818\,\mathrm m\end{aligned}$

Result: Midpoint field is 25.0 µT along +y. The zero point lies at x≈+18.2 mm.

Check: The zero point is closer to the weaker current. Both individual fields there are approximately 27.5 µT and cancel vectorially. Adding positive magnitudes would miss this zero.

Inside the wire, only enclosed current counts

B(r)=\frac{\mu_0|I|r}{2\pi a^2}\quad(0\le r\le a)

This interior expression requires a long round wire of radius a with uniform current density.[2]

Inside, enclosed current grows with circle area. Field therefore grows proportionally to r and is zero on the axis; the external inverse-distance form cannot extend to r=0. Interior and exterior forms meet at r=a. At high frequency, skin effect changes current distribution, so uniform density is not an appropriate assumption. Magnetic material can also require a different field calculation. This interior relation is a specific additional model, not a general remedy for every thick wire.[2]

The same field magnitude inside and outside

Given: A round nonmagnetic wire of radius a=2.00 mm carries I=8.00 A uniformly over its cross-section along +z. Observe on +x at r=1.00 mm and r=4.00 mm.
Find: Find both fields and the surface value.

Approach: inside, only current within the enclosed cross-section counts. Outside, the entire current is enclosed.
$\begin{aligned}B_{\mathrm{in}}(r)&=\frac{\mu_0I r}{2\pi a^2}\\B_{\mathrm{out}}(r)&=\frac{\mu_0I}{2\pi r}\\B(a)&=\frac{\mu_0I}{2\pi a}\end{aligned}$
Substitute radii in metres; directions follow the same circulation.
$\begin{aligned}B_{\mathrm{in}}(0.001)&=\frac{(1.25663706\cdot10^{-6})(8.00)(0.001)}{2\pi(0.002)^2}\,\mathrm T\approx0.400\,\mathrm{mT}\\B_{\mathrm{out}}(0.004)&\approx0.400\,\mathrm{mT}\\B(0.002)&\approx0.800\,\mathrm{mT}\end{aligned}$

Result: Both observation points have about 0.400 mT along +y. The model’s maximum is at the surface, about 0.800 mT.

Check: At half-radius inside, one quarter of current is enclosed, giving half the surface field. Outside, twice the radius halves field. Both expressions agree at r=a.

B grows inside the wire and decreases outside

QUANTA STUDYquanta-study.de

For the third example at 8 A, magnitudes at 1 mm and 4 mm match despite different expressions. Do not extend the exterior formula into the interior. The model assumes a round nonmagnetic wire with uniform current density.[2][1]

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Five exercises: distance, direction and model choice

Calculate · Foundation

A long thin wire in air carries 12.0 A. Find field magnitude 5.00 cm from its axis; use μ₀ above.

Show hint

Convert the denominator distance to metres.

Learning goal: Calculate external field in SI.

Show solution and calculation

Model and approach: Observe outside the long wire, with r measured from its axis.
$B=\frac{\mu_0|I|}{2\pi r}$ Approach: use the external inverse-distance expression.
Result 48.0 µT.

Check: at 10.0 cm field would be 24.0 µT.

B=\frac{(1.25663706\cdot10^{-6})(12.0)}{2\pi(0.0500)}\,\mathrm T\approx48.0\,\mathrm{\mu T}

[1]

Rearrange · Application

The isolated field contribution of a long wire is 60.0 µT at axis distance 4.00 cm. What current magnitude is needed?

Show hint

Other fields have explicitly already been removed.

Learning goal: Infer current from an isolated field contribution.

Show solution and calculation

Model and approach: The supplied field is this wire’s isolated external contribution.
$|I|=\frac{2\pi rB}{\mu_0}$ Approach: solve for absolute current.
Result about 12.0 A.

Check: doubling distance at fixed target field requires twice the current. Current direction additionally requires field circulation.

|I|=\frac{2\pi(0.0400)(60.0\cdot10^{-6})}{1.25663706\cdot10^{-6}}\,\mathrm A\approx12.0\,\mathrm A

[1]

Interpret · Application

An external-field graph shows $B_0$ at radius $r_0$ and $\frac{B_0}{2}$ at $2r_0$ . What follows at $4r_0$ ? What changes under current reversal?

Show hint

The curve gives magnitude; circulation is additional information.

Learning goal: Separate radial decay from direction reversal.

Show solution and calculation

Approach: external inverse-distance dependence gives $B(4r_0)=\frac{B_0}{4}$ .
Reversal preserves the whole magnitude curve but reverses the vector everywhere.

Check: the curve alone does not reveal current sign.

\frac{B(4r_0)}{B(r_0)}=\frac14

[1]

Error analysis · Application

A long round wire has radius 3.00 mm. The measurement point is 1.00 mm outside its surface. A calculation uses r=1.00 mm. How does this alter external field?

Show hint

Draw radius and additional distance end to end.

Learning goal: Use axis distance rather than surface distance.

Show solution and calculation

Approach: correct axis distance is 4.00 mm.
The wrong value overestimates field by factor four.

Check: only the correct distance describes the physical measurement location.

\frac{B_{\mathrm{wrong}}}{B_{\mathrm{right}}}=\frac{4.00\,\mathrm{mm}}{1.00\,\mathrm{mm}}=4

[1]

Transfer · Challenge

A long round wire of radius 2.00 mm carries 10.0 A with uniform current density. Find B at 1.00 mm from its axis and compare the incorrectly applied external formula.

Show hint

The integration circle encloses only one quarter of cross-sectional area.

Learning goal: Calculate enclosed current inside a wire.

Show solution and calculation

Approach: enclosed current is 2.50 A.
Interior field is 0.500 mT; external form using total current would give 2.00 mT.

Check: correct interior field falls to zero on the axis rather than diverging.

\begin{aligned}I_{\mathrm{enc}}&=I\frac{r^2}{a^2}=2.50\,\mathrm A\\B&=\frac{\mu_0 I r}{2\pi a^2}\approx0.500\,\mathrm{mT}\end{aligned}

[2]

What to take away

Field circles the wire. Determine direction from current and observation point, not a positive magnitude alone.
Distance is measured from the axis. A finite segment and a complete circuit are different calculation objects.
Inside, the formula depends on enclosed current. For uniform current density, B rises linearly to the surface.

[1][2]

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.

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Change history

October 2, 2026Highlighted field magnitude, direction and axis distance in question headings; added paragraph connections to Maxwell’s integral equations and Biot–Savart with primary-source attribution.
October 1, 2026Separated long external field, finite-segment correction and wire interior; wrote two original examples and five model/calculation exercises in DE/EN.
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 straight wire. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/magnetfeld-langer-gerader-leiter

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Frequently asked questions about Magnetic field of a long straight wire

What does the magnetic field of a long straight wire describe?+

The wire current creates a magnetic field circulating tangentially around the conductor axis. Outside an ideally long straight wire its magnitude is $B=\frac{\mu_0|I|}{2\pi r}$ . Distance is measured from the axis and appears in the denominator. Doubling absolute current therefore doubles field; doubling distance halves it. Direction additionally follows the right-hand rule for conventional current. Field does not point radially away from the wire. Current reversal preserves the magnitude curve while reversing every field vector. This relation first describes the contribution of this wire; fields from return conductors or other sources must be added with their directions.

How do I check a calculation for a long wire’s magnetic field?+

First check that the point lies outside the conductor and sufficiently far from its ends. Measure shortest axis distance and convert centimetres to metres. In the existing example, five amperes at two centimetres give approximately $50\,\mathrm{\mu T}$ . Doubling distance must reduce magnitude to about $25\,\mathrm{\mu T}$ at unchanged current and model. Independently verify direction with the right-hand rule. For a finite segment the long-wire value is only an approximation, and a complete circuit also contains return paths. An unexpected measurement can therefore result from additional field contributions even when the single-wire calculation is correct.

What assumptions apply to the long-wire model?+

The shortcut assumes a very long straight conductor with steady current and suitable cylindrical symmetry. It describes external field in vacuum or an appropriate air model. Near a wire end or return conductor, these simple symmetry conditions fail. A finite segment needs its actual endpoint distances. Inside a thick wire, only current enclosed by the integration circle belongs in Ampère’s law. With uniform current density, interior field grows proportionally to axis distance rather than diverging. Other current distributions, magnetic materials or rapid time dependence require additional assumptions. Extending the external inverse-distance expression all the way to the axis is therefore not a valid physical limit.

How do I check units and scale of a wire’s field?+

Permeability has unit $\mathrm{T\,m/A}$ ; multiplying by amperes and dividing by metres leaves tesla. Convert milliamperes and centimetres before substitution. Two centimetres is $0.020\,\mathrm m$ , not two metres. The existing result $5.0\cdot10^{-5}\,\mathrm T$ is fifty microtesla, not fifty millitesla. The vacuum permeability used is a rounded value of a constant with measurement uncertainty. As an additional check, retain the current-to-distance ratio: the same ratio gives the same field magnitude in the same model. Units alone cannot detect using surface distance instead of axis distance or overlooking a return conductor.

When do I need additional relations for a wire’s field?+

For multiple wires, first calculate every field vector at the same point, then add components. Between parallel equal-direction currents, contributions may cancel; adding positive magnitudes would be incorrect. To find force on a second wire, use the first wire’s field as external field in the conductor-force expression. Axis separation must match actual geometry. Inferring current from a field measurement additionally requires other contributions and measurement direction. Ampère’s law explains the simple relation but does not replace local field calculation for an asymmetric arrangement. Without a suitable symmetry path, direct Biot–Savart integration can provide the appropriate additional calculation.

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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, §12.2: Magnetic Field Due to a Thin Straight Wire

Related formulas

\begin{aligned}\mathrm d\mathbf B&=\frac{\mu_0I}{4\pi}\frac{\mathrm d\boldsymbol\ell\times\hat{\mathbf r}}{r^2}\\\mathbf B&=\int\mathrm d\mathbf B\end{aligned}

Biot–Savart Law

More formulas: Electricity and magnetism

\mathcal E=-N\frac{\mathrm d\Phi_B}{\mathrm dt}

Faraday’s law of induction

\oint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{\mathrm{innen}}}{\varepsilon_0}

Gauss's Law for Electrostatics

U = R \cdot I

Ohm's Law

P = U \cdot I

Electric Power

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Magnetic field of a long straight wire

The external field magnitude of a long straight current-carrying wire decreases inversely with shortest axis distance; the field vector circulates tangentially around the axis.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

B

Magnitude of magnetic flux density at the external measurement point

μ₀

Vacuum magnetic constant

N/A²

I

Current in the wire

r

Shortest perpendicular distance to the conductor axis, not its surface

Derivation & background

Exam blueprint

Validity range

Derivation steps

Cylindrical symmetry simplifies the magnetostatic Ampère integral; determine field direction separately.

1Specify a very long straight steady current and an external point in a vacuum/air model.
2Measure r as shortest distance to the axis, not to the surface.
3Cylindrical symmetry gives tangential field of constant magnitude on a concentric circle.
4Choose traversal consistently with current and use $\oint\vec B\cdot\mathrm d\vec\ell=\mu_0I_{\mathrm{enc}}$ .
5The integral becomes $B2\pi r$ ; outside, the full current is enclosed.
6Solving gives magnitude $B=\frac{\mu_0|I|}{2\pi r}$ .
7Check right-hand circulation, inverse-distance dependence and possible return-wire fields.

How do you calculate with Magnetic field of a long straight wire?

Here is how to work through a typical Magnetic field of a long straight wire (B = μ₀|I|/(2πr)) task step by step:

Task

A long straight wire carries 4.00 A. Find its external field magnitude 4.00 cm from the axis in the air model.

Solution path

Approach: $B=\frac{\mu_0|I|}{2\pi r}$ ; with $r=0.0400\,\mathrm m$ , $B\approx20.0\,\mathrm{\mu T}$ .

A compass beside a current-carrying wire

Prior knowledge: B is magnetic flux density in teslas. Current direction means conventional current.
Measure r from the axis. The exterior formula applies outside the wire cross-section, not down to its centre.
For current along +z and an observation point on +x, field points along +y. Reversing current reverses that direction.

[1][2]

How do you calculate field magnitude and direction outside a long wire?

B=\frac{\mu_0|I|}{2\pi r}

Magnitude describes the external field of an ideally long straight wire; r is the shortest axis distance.[1]

Why a circle simplifies the integral

\begin{aligned}\oint_C\vec B\cdot\mathrm d\vec\ell&=\mu_0 I_{\mathrm{enc}}\\B(2\pi r)&=\mu_0|I|\\B&=\frac{\mu_0|I|}{2\pi r}\end{aligned}

Simplification requires cylindrical symmetry of current distribution, not merely a circular integration path.[2]

A finite segment gives less field

\begin{aligned}B_{\mathrm{seg}}&=\frac{\mu_0|I|}{4\pi r}\left(\frac{h_1}{\sqrt{r^2+h_1^2}}+\frac{h_2}{\sqrt{r^2+h_2^2}}\right)\\B_{\mathrm{mid}}&=\frac{\mu_0|I|}{2\pi r}\frac{h}{\sqrt{r^2+h^2}}\end{aligned}

The perpendicular foot lies between the ends; h₁ and h₂ are positive axial endpoint distances. At the midpoint segment length is 2h.[1]

Which distance belongs in the formula and how do you rearrange it?

\begin{aligned}|I|&=\frac{2\pi rB}{\mu_0}\\r&=\frac{\mu_0|I|}{2\pi B}\\{}[\mu_0]&=\mathrm{T\,m\,A^{-1}}\end{aligned}

Both rearrangements require known model conditions and nonzero denominators.[1]

Example: long-wire result and segment check

Examining a 24 cm segment

Given: I=7.50 A along +z; field point 3.00 cm along +x from the midpoint of a straight 24.0 cm segment. Vacuum/air model; other circuit contributions are treated separately.
Find: Find the ideally long result and finite-segment contribution.

Approach: first calculate the long-wire model, then multiply by the endpoint correction.
$\begin{aligned}B_\infty&=\frac{\mu_0I}{2\pi r}\\B_{\mathrm{mid}}&=B_\infty\frac{h}{\sqrt{h^2+r^2}}\end{aligned}$
Substitution: half the segment length is 0.120 m.
$\begin{aligned}B_\infty&=\frac{(1.25663706\cdot10^{-6})(7.50)}{2\pi(0.0300)}\,\mathrm T\approx50.0\,\mathrm{\mu T}\\\frac{h}{\sqrt{h^2+r^2}}&=\frac{0.120}{\sqrt{0.120^2+0.0300^2}}\approx0.970143\\B_{\mathrm{mid}}&\approx48.5071\,\mathrm{\mu T}\end{aligned}$

Result: The segment contributes about 48.5 µT along +y; the long-wire comparison is 50.0 µT.

Check: Relative to the long result, segment field is 2.99% smaller. The difference vanishes as h approaches infinity. A return conductor can still increase or decrease total field.

Example: unequal parallel currents

A zero-field point between two axes

Given: Long wires at x=−0.0400 m and +0.0400 m carry 8.00 A and 3.00 A, both along +z. Points lie on the joining line outside the wires; other fields are neglected.
Find: Find midpoint field and the zero-field point between the wires.

Approach: between equal-direction currents, field vectors oppose. Equate their magnitudes at the zero point.
$\begin{aligned}B_y(0)&=\frac{\mu_0}{2\pi a}(I_1-I_2)\\\frac{I_1}{d_1}&=\frac{I_2}{2a-d_1}\\d_1&=\frac{2aI_1}{I_1+I_2}\end{aligned}$
Substitution: d₁ is measured from the left axis.
$\begin{aligned}B_y(0)&\approx25.0\,\mathrm{\mu T}\\d_1&=\frac{(0.0800)(8.00)}{11.00}\,\mathrm m\approx0.0581818\,\mathrm m\\x_0&=d_1-0.0400\,\mathrm m\approx0.0181818\,\mathrm m\end{aligned}$

Result: Midpoint field is 25.0 µT along +y. The zero point lies at x≈+18.2 mm.

Check: The zero point is closer to the weaker current. Both individual fields there are approximately 27.5 µT and cancel vectorially. Adding positive magnitudes would miss this zero.

Inside the wire, only enclosed current counts

B(r)=\frac{\mu_0|I|r}{2\pi a^2}\quad(0\le r\le a)

This interior expression requires a long round wire of radius a with uniform current density.[2]

The same field magnitude inside and outside

Given: A round nonmagnetic wire of radius a=2.00 mm carries I=8.00 A uniformly over its cross-section along +z. Observe on +x at r=1.00 mm and r=4.00 mm.
Find: Find both fields and the surface value.

Approach: inside, only current within the enclosed cross-section counts. Outside, the entire current is enclosed.
$\begin{aligned}B_{\mathrm{in}}(r)&=\frac{\mu_0I r}{2\pi a^2}\\B_{\mathrm{out}}(r)&=\frac{\mu_0I}{2\pi r}\\B(a)&=\frac{\mu_0I}{2\pi a}\end{aligned}$
Substitute radii in metres; directions follow the same circulation.
$\begin{aligned}B_{\mathrm{in}}(0.001)&=\frac{(1.25663706\cdot10^{-6})(8.00)(0.001)}{2\pi(0.002)^2}\,\mathrm T\approx0.400\,\mathrm{mT}\\B_{\mathrm{out}}(0.004)&\approx0.400\,\mathrm{mT}\\B(0.002)&\approx0.800\,\mathrm{mT}\end{aligned}$

Result: Both observation points have about 0.400 mT along +y. The model’s maximum is at the surface, about 0.800 mT.

Check: At half-radius inside, one quarter of current is enclosed, giving half the surface field. Outside, twice the radius halves field. Both expressions agree at r=a.

B grows inside the wire and decreases outside

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Five exercises: distance, direction and model choice

Calculate · Foundation

A long thin wire in air carries 12.0 A. Find field magnitude 5.00 cm from its axis; use μ₀ above.

Show hint

Convert the denominator distance to metres.

Learning goal: Calculate external field in SI.

Show solution and calculation

Model and approach: Observe outside the long wire, with r measured from its axis.
$B=\frac{\mu_0|I|}{2\pi r}$ Approach: use the external inverse-distance expression.
Result 48.0 µT.

Check: at 10.0 cm field would be 24.0 µT.

B=\frac{(1.25663706\cdot10^{-6})(12.0)}{2\pi(0.0500)}\,\mathrm T\approx48.0\,\mathrm{\mu T}

[1]

Rearrange · Application

The isolated field contribution of a long wire is 60.0 µT at axis distance 4.00 cm. What current magnitude is needed?

Show hint

Other fields have explicitly already been removed.

Learning goal: Infer current from an isolated field contribution.

Show solution and calculation

Model and approach: The supplied field is this wire’s isolated external contribution.
$|I|=\frac{2\pi rB}{\mu_0}$ Approach: solve for absolute current.
Result about 12.0 A.

Check: doubling distance at fixed target field requires twice the current. Current direction additionally requires field circulation.

|I|=\frac{2\pi(0.0400)(60.0\cdot10^{-6})}{1.25663706\cdot10^{-6}}\,\mathrm A\approx12.0\,\mathrm A

[1]

Interpret · Application

An external-field graph shows $B_0$ at radius $r_0$ and $\frac{B_0}{2}$ at $2r_0$ . What follows at $4r_0$ ? What changes under current reversal?

Show hint

The curve gives magnitude; circulation is additional information.

Learning goal: Separate radial decay from direction reversal.

Show solution and calculation

Approach: external inverse-distance dependence gives $B(4r_0)=\frac{B_0}{4}$ .
Reversal preserves the whole magnitude curve but reverses the vector everywhere.

Check: the curve alone does not reveal current sign.

\frac{B(4r_0)}{B(r_0)}=\frac14

[1]

Error analysis · Application

A long round wire has radius 3.00 mm. The measurement point is 1.00 mm outside its surface. A calculation uses r=1.00 mm. How does this alter external field?

Show hint

Draw radius and additional distance end to end.

Learning goal: Use axis distance rather than surface distance.

Show solution and calculation

Approach: correct axis distance is 4.00 mm.
The wrong value overestimates field by factor four.

Check: only the correct distance describes the physical measurement location.

\frac{B_{\mathrm{wrong}}}{B_{\mathrm{right}}}=\frac{4.00\,\mathrm{mm}}{1.00\,\mathrm{mm}}=4

[1]

Transfer · Challenge

A long round wire of radius 2.00 mm carries 10.0 A with uniform current density. Find B at 1.00 mm from its axis and compare the incorrectly applied external formula.

Show hint

The integration circle encloses only one quarter of cross-sectional area.

Learning goal: Calculate enclosed current inside a wire.

Show solution and calculation

Approach: enclosed current is 2.50 A.
Interior field is 0.500 mT; external form using total current would give 2.00 mT.

Check: correct interior field falls to zero on the axis rather than diverging.

\begin{aligned}I_{\mathrm{enc}}&=I\frac{r^2}{a^2}=2.50\,\mathrm A\\B&=\frac{\mu_0 I r}{2\pi a^2}\approx0.500\,\mathrm{mT}\end{aligned}

[2]

What to take away

Field circles the wire. Determine direction from current and observation point, not a positive magnitude alone.
Distance is measured from the axis. A finite segment and a complete circuit are different calculation objects.
Inside, the formula depends on enclosed current. For uniform current density, B rises linearly to the surface.

[1][2]

Sources for this explanation

Quanta · Updated October 2, 2026

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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.

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Change history

October 2, 2026Highlighted field magnitude, direction and axis distance in question headings; added paragraph connections to Maxwell’s integral equations and Biot–Savart with primary-source attribution.
October 1, 2026Separated long external field, finite-segment correction and wire interior; wrote two original examples and five model/calculation exercises in DE/EN.
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 straight wire. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/magnetfeld-langer-gerader-leiter

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Magnetic field of a long straight wire

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Magnetic field of a straight wire calculator

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Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

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Common mistakes

Exam context

Magnetic fields from currents

Worked example

Applications

How do you calculate with Magnetic field of a long straight wire?

A compass beside a current-carrying wire

How do you calculate field magnitude and direction outside a long wire?

Why a circle simplifies the integral

A finite segment gives less field

Which distance belongs in the formula and how do you rearrange it?

Example: long-wire result and segment check

Examining a 24 cm segment

Example: unequal parallel currents

A zero-field point between two axes

Inside the wire, only enclosed current counts

The same field magnitude inside and outside

B grows inside the wire and decreases outside

Five exercises: distance, direction and model choice

A long thin wire in air carries 12.0 A. Find field magnitude 5.00 cm from its axis; use μ0 above.

The isolated field contribution of a long wire is 60.0 µT at axis distance 4.00 cm. What current magnitude is needed?

An external-field graph shows B0B_0B0​ at radius r0r_0r0​ and B02\frac{B_0}{2}2B0​​ at 2r02r_02r0​. What follows at 4r04r_04r0​? What changes under current reversal?

A long round wire has radius 3.00 mm. The measurement point is 1.00 mm outside its surface. A calculation uses r=1.00 mm. How does this alter external field?

A long round wire of radius 2.00 mm carries 10.0 A with uniform current density. Find B at 1.00 mm from its axis and compare the incorrectly applied external formula.

What to take away

Sources for this explanation

Frequently asked questions about Magnetic field of a long straight wire

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Magnetic field of a long straight wire for exams

Magnetic field of a long straight wire

Variables & units

Magnetic field of a straight wire calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Magnetic fields from currents

Worked example

Applications

How do you calculate with Magnetic field of a long straight wire?

A compass beside a current-carrying wire

How do you calculate field magnitude and direction outside a long wire?

Why a circle simplifies the integral

A finite segment gives less field

Which distance belongs in the formula and how do you rearrange it?

Example: long-wire result and segment check

Examining a 24 cm segment

Example: unequal parallel currents

A zero-field point between two axes

Inside the wire, only enclosed current counts

The same field magnitude inside and outside

B grows inside the wire and decreases outside

Five exercises: distance, direction and model choice

A long thin wire in air carries 12.0 A. Find field magnitude 5.00 cm from its axis; use μ0 above.

The isolated field contribution of a long wire is 60.0 µT at axis distance 4.00 cm. What current magnitude is needed?

An external-field graph shows B0B_0B0​ at radius r0r_0r0​ and B02\frac{B_0}{2}2B0​​ at 2r02r_02r0​. What follows at 4r04r_04r0​? What changes under current reversal?

A long round wire has radius 3.00 mm. The measurement point is 1.00 mm outside its surface. A calculation uses r=1.00 mm. How does this alter external field?

A long round wire of radius 2.00 mm carries 10.0 A with uniform current density. Find B at 1.00 mm from its axis and compare the incorrectly applied external formula.

What to take away

Sources for this explanation

Frequently asked questions about Magnetic field of a long straight wire

Subject references and further reading

Related formulas

A long thin wire in air carries 12.0 A. Find field magnitude 5.00 cm from its axis; use μ₀ above.

An external-field graph shows $B_0$ at radius $r_0$ and $\frac{B_0}{2}$ at $2r_0$ . What follows at $4r_0$ ? What changes under current reversal?

A long thin wire in air carries 12.0 A. Find field magnitude 5.00 cm from its axis; use μ₀ above.

An external-field graph shows $B_0$ at radius $r_0$ and $\frac{B_0}{2}$ at $2r_0$ . What follows at $4r_0$ ? What changes under current reversal?