Physics · Magnetic fields of current-carrying conductors

Biot–Savart law

Q: How do I apply Biot–Savart to a long wire?

For a long straight wire with $I=5.0\,\mathrm A$ and perpendicular distance $d=2.0\,\mathrm{cm}$, use $B=\frac{\mu_0I}{2\pi d}$. With educational approximation $\mu_0\approx4\pi\cdot10^{-7}$ tesla metre per ampere, the result is 50 microtesla. Determine direction separately using the right-hand rule. Twice the distance gives 25 microtesla; reversing current reverses the vector. For a finite wire this is not automatically exact: length and observation position determine the correction. Connections, return conductors and external fields may also contribute to a measurement. Round only after unit conversion and after choosing the appropriate geometry.

The Biot–Savart law sums vector magnetic-field contributions from steady current elements at an observation point.

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Formula

Calculate

d B B = \frac{μ _{0} I}{4 π} \frac{d ℓ \times r ^}{r ^{2}} = \int d B

B in T; I in A; dℓ and r in m; μ₀ in T·m/A.

Variables & units

Variables & units
Symbol	Meaning	Unit
$dB$	field contribution from one current element	T
$μ₀$	vacuum permeability	T·m/A
$I$	steady current	A
$dℓ$	wire element directed with current	m
r̂	unit vector from element to observation point	dimensionless
$r$	distance from element to observation point	m

LaTeX

\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}

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Finite horizontal wire at y=0 with current rightward. P is above its midpoint. An element left of centre points right; its displacement points diagonally right-up to P and the cross product points out of the plane. R is slanted distance, d perpendicular distance. — Original wire geometry: I=3 A along +x, segment from −0.15 to +0.15 m, P at y=0.04 m. R points from the sample element at x=−0.08 m to P. All segment contributions point along +z; Bz≈14.4935 µT. The element arrow is schematic, not a finite replacement for integration.

Biot-Savart magnetic field calculator

Find the magnetic field on the axis of a flat coil or next to a straight piece of wire, in vacuum and with its sign.

Calculation model

Find

Radius R of one turn

Radius of one turn, e.g. 4 cm for a small flat coil.

Signed position z on the axis

Position on the coil axis measured from the centre, e.g. 3 cm; negative on the other side.

Number of turns N

Number of closely packed turns, a whole number, e.g. 25.

Current: counterclockwise from +z is positive I

Current in each turn; positive if it flows anticlockwise seen from +z, e.g. 2 A.

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

Result: Axial magnetic field Bz

Bz = 402.1 μT

With R = 4 cm, z = 3 cm, N = 25 and I = 2 A, Bz = 402.1 μT.

Working · rearrange for Bz, substitute, evaluate

1. Rearrange for Bz
$B_z=\frac{\mu_0NIR^2}{2(R^2+z^2)^{3/2}}$
2. Substitute values (in SI units)
- R = 4 cm = 0.04 m
- z = 3 cm = 0.03 m
$\frac{0.00000125664\cdot 25\cdot 2\cdot0.04^2}{2\cdot(0.04^2+0.03^2)^{3/2}}=0.000402124$
3. Result
Bz = 402.1 μT

μ₀≈4π·10⁻⁷ T·m/A is the rounded teaching approximation used here, not an exact defining value since the SI redefinition.

Quick tips

At the centre of the coil (z = 0) the formula simplifies to B = μ₀NI/(2R). 25 turns of radius 4 cm carrying 2 A give about 785 μT.
If the wire piece is much longer than the distance (a ≫ d), the long-wire field almost holds: B ≈ μ₀I/(2πd), so 15 μT for 3 A at 4 cm.

Assumptions and limits

Thin stationary circular turns in vacuum in the xy plane, common positive radius and axial position. Positive current is counterclockwise viewed from +z. Observation only on the z axis; z may be negative or zero and result is Bz. N is a safe positive integer. Approximation of closely adjacent identical turns; return leads separate. Not a long solenoid, general off-axis integral or material-core model.

Derivation & background

Magnetostatic vacuum form for complete steady currents; thin-wire line model only outside the conductor cross-section. Contributions are vectors. Time-dependent sources and magnetic material response need extended descriptions. $\mu_0\approx4\pi\cdot10^{-7}\,\mathrm{T\,m\,A^{-1}}$ is an educational approximation, not an exact post-2019 SI definition.

Exam blueprint

Validity range

Derivation steps

The magnetostatic current-element law gives each contribution; parametrisation and vector integration determine a specific geometry’s field.

1Define a fixed observation point and the full steady current geometry, including return conductors.
2For positive I orient each directed element dℓ along conventional current.
3Displacement $\mathbf R=\mathbf r_P-\mathbf r^{\prime}$ points from source to observer; R varies along the wire.
4Evaluate $\mathrm d\boldsymbol\ell\times\hat{\mathbf R}$ ; it sets the sine factor and right-hand direction.
5Integrate $\mathrm d\mathbf B=\frac{\mu_0I}{4\pi}\frac{\mathrm d\boldsymbol\ell\times\hat{\mathbf R}}{R^2}$ over the geometry.
6Add Cartesian components before taking magnitude; check symmetry, current reversal and limiting cases.

Rearrangements

Current-element contribution magnitude

\mathrm dB=\frac{\mu_0I}{4\pi}\frac{\mathrm d\ell\sin\theta}{r^2}

Positive I and dℓ; θ is the angle between element and source-to-observer vector. Not a general rearrangement of the field integral.

Task variant

An ideally long straight thin wire carries $I=5.0\,\mathrm A$ along +z. An observation point lies $r=2.0\,\mathrm{cm}$ from the axis along +x. Find field magnitude and direction in vacuum.

Approach: $B=\frac{\mu_0I}{2\pi r}$ .
With $r=0.020\,\mathrm m$ and $\mu_0\approx4\pi\cdot10^{-7}\,\mathrm{T\,m\,A^{-1}}$ , $B=5.0\cdot10^{-5}\,\mathrm T=50\,\mathrm{\mu T}$ .
The right-hand rule gives +y.
Doubling distance halves magnitude; the ideally long single-wire contribution excludes end and return-wire contributions.

Common mistakes

Reverse the source-to-observer vector or treat one wire segment as the total field without the other current paths.

Define current direction and observation point; integrate vectors and state neglected return-path contributions.

Exam context

Finite wire segment, loop axis and two-wire superposition: establish vector direction and geometry before integration.

Formula cluster

Magnetic fields of current-carrying conductors

Biot–Savart calculates wire and loop fields. With suitable symmetry, Ampère’s law supplies an alternative calculation.

B = μ₀|I|/(2πr)

Magnetic field of a long straight wire

B = μ₀·(N/l)·I

Magnetic Field of a Long Coil

∯E·dA=QV/ε₀; ∯B·dA=0; ∮E·dl=−dΦB/dt; ∮B·dl=μ₀(I_A+ε₀dΦE/dt)

Maxwell Equations in Integral Form

Worked example

A long straight wire with $I=5.0\,\mathrm A$ gives approximately $B=\frac{\mu_0I}{2\pi r}=5.0\cdot10^{-5}\,\mathrm T=50\,\mathrm{\mu T}$ at $r=2.0\,\mathrm{cm}$ . Use the right-hand rule for direction; doubling distance halves magnitude.

Applications

Calculate fields of current loops, wire segments and current distributions.

How do you calculate with Biot–Savart Law?

Here is how to work through a typical Biot–Savart Law (dB=(μ₀/4π)I dℓ×r̂/r²) task step by step:

1
Task
An ideally long straight thin wire carries $I=5.0\,\mathrm A$ along +z. An observation point lies $r=2.0\,\mathrm{cm}$ from the axis along +x. Find field magnitude and direction in vacuum.
Solution path
1. Approach: $B=\frac{\mu_0I}{2\pi r}$ .
2. With $r=0.020\,\mathrm m$ and $\mu_0\approx4\pi\cdot10^{-7}\,\mathrm{T\,m\,A^{-1}}$ , $B=5.0\cdot10^{-5}\,\mathrm T=50\,\mathrm{\mu T}$ .
3. The right-hand rule gives +y.
4. Doubling distance halves magnitude; the ideally long single-wire contribution excludes end and return-wire contributions.

Understand and apply the equation

The Biot–Savart law connects a steady current geometry to magnetic field at a specified point. The crucial step is not substituting one distance but adding all contributions as vectors. A finite wire and a loop show how the same relation produces different spatial profiles.

Build the field of a curved current path

Symmetry helps for a long straight wire. What if the wire is curved? Imagine dividing it into many short directed pieces. Each piece contributes a small field vector at the chosen observation point. Biot–Savart specifies that contribution; integration adds them over the geometry. Calling pieces “small” does not justify using the same observer distance for all of them.[1][2]

Prior knowledge: a cross product selects direction perpendicular to current element and displacement. Parallel vectors give zero contribution.
Displacement runs from the current source piece to the fixed observation point, not the other way round.
Steady current requires a complete path. A calculated segment contribution must be distinguished from total field.

[1][2]

Source, observation point and cross product

A source point on a thin conductor has position $\mathbf r^{\prime}$ ; the fixed observation point lies at $\mathbf r_P$ . The displacement $\mathbf R=\mathbf r_P-\mathbf r^{\prime}$ always points from source to observer. Its magnitude R generally changes along the conductor. For positive current I orient $\mathrm d\boldsymbol\ell$ along conventional current. Reversing the displacement reverses the calculated field.[1]

\begin{aligned}\mathrm d\mathbf B&=\frac{\mu_0I}{4\pi}\frac{\mathrm d\boldsymbol\ell\times\hat{\mathbf R}}{R^2}\\&=\frac{\mu_0I}{4\pi}\frac{\mathrm d\boldsymbol\ell\times\mathbf R}{R^3}\\\mathbf B(P)&=\int\mathrm d\mathbf B\end{aligned}

Magnetostatic vacuum form for thin wires; R≠0.[1]

The cross product gives a contribution normal to both the current element and displacement. Its magnitude includes the sine of their angle. An element parallel to the displacement contributes zero at that point. This does not make the entire circuit field-free: other elements have different directions and angles.[1]

Apply the law rather than invent it from geometry

The current-element relation is a physical magnetostatic field law; geometry alone does not prove it. Deriving a specific wire result therefore begins with this law, then parametrises the conductor, evaluates the cross product and integrates components. Only factors constant along the entire integration path may be taken outside.[1]

A current segment is part of a complete steady current distribution, not an isolated source with open ends. Return and connection wires close a real circuit; their fields must be added or neglected using a justified approximation. A thick conductor requires integration over current density instead of a line integral. Highly symmetric infinite geometry may be simpler with Ampère’s law.[1]

Keep the observer fixed during integration and move only the source point along the wire. Distance and angle then change for each piece. Choose common component axes and add each component separately, taking total magnitude only afterwards. For a circular loop, transverse components cancel in pairs while axial components reinforce.[1][3]

How do you calculate the contribution of a finite straight wire?

A wire segment lies on the x-axis from −a to +a with positive current along +x. The observation point is (0,d,0), d>0. Every element’s cross product has positive z component, so all contributions from this segment point out of the drawing plane. Each element’s distance is $\sqrt{x^2+d^2}$ , not the constant shortest distance d.[2]

\begin{aligned}\mathrm dB_z&=\frac{\mu_0I}{4\pi}\frac{d\,\mathrm dx}{(x^2+d^2)^{\frac32}}\\B_z&=\frac{\mu_0I}{4\pi}\left[\frac{x}{d\sqrt{x^2+d^2}}\right]_{-a}^{a}\\&=\frac{\mu_0I}{2\pi d}\frac a{\sqrt{a^2+d^2}}\\\lim_{a\to\infty}B_z&=\frac{\mu_0I}{2\pi d}\end{aligned}

The finite factor applies in this segment’s midplane. Other circuit parts are not included here.[2]

The educational calculations use $\mu_0\approx4\pi\cdot10^{-7}$ tesla metre per ampere. Since the SI revision this is no longer an exact defining constant; CODATA 2022 gives a very close measured value. The difference is negligible for our three-digit results. The element contribution is in tesla because current times element length divided by squared distance has units ampere per metre. The magnetic field of a long straight wire is the appropriate limiting comparison when the ends are sufficiently far away relative to perpendicular distance.[4][5]

Example: a segment rather than an infinite wire

Check a geometric approximation

Given: Wire segment $2a=0.30\,\mathrm m$ , $I=3.0\,\mathrm A$ along +x, point $(0,0.040\,\mathrm m,0)$ . Vacuum; this segment’s contribution only.
Find: Field component and difference from the infinite model.

Model: use the cross product for direction and finite integral for magnitude.
$\begin{aligned}B_\infty&=\frac{\mu_0I}{2\pi d}\\B_z&=B_\infty\frac a{\sqrt{a^2+d^2}}\end{aligned}$
Substitute a=0.15 m, not 0.30 m.
$\begin{aligned}B_\infty&=\frac{(4\pi\cdot10^{-7})(3.0)}{2\pi(0.040)}\,\mathrm T=15.0\,\mathrm{\mu T}\\B_z&=(15.0\,\mathrm{\mu T})\frac{0.15}{\sqrt{0.15^2+0.040^2}}\\&\approx14.4935\,\mathrm{\mu T}\end{aligned}$

Result: The segment contributes about 14.5 µT along +z. The infinite model is about 3.49 percent higher relative to the finite result.

Check: Check: the geometric factor is 0.966235, less than one. Reversing current makes Bz negative. Return-wire contributions may change the measured total field and are absent from this comparison.

How does the axial field of a circular turn arise?

Place a circular turn of radius R in the xy-plane with current counterclockwise as viewed from +z. On the z-axis, opposite elements cancel their transverse components while z components add. All elements share distance $\sqrt{R^2+z^2}$ , but an axial projection factor must still be included.[3]

\begin{aligned}\mathrm dB_z&=\frac{\mu_0I}{4\pi}\frac{R\,\mathrm d\ell}{(R^2+z^2)^{\frac32}}\\B_z(z)&=\frac{\mu_0IR^2}{2(R^2+z^2)^{\frac32}}\\B_z(0)&=\frac{\mu_0I}{2R}\end{aligned}

One turn. Multiply by N for practically equal turns packed closely together.[3]

A flat multi-turn coil approximation requires sufficiently similar radii and axial positions to use one common single-turn value. N is a turn count, not turns per length. Far from the loop its axial field falls approximately with the third power of distance. That profile belongs to the closed current geometry, although every element uses the same Biot–Savart starting point. The magnetic field of a long solenoid instead requires axially distributed turns and its own long-solenoid assumption; merely multiplying the flat-loop value does not replace that geometry.[3]

View diagram: Loop field decreases along the axis

Loop field decreases along the axis

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For this current orientation, the axial field vector points along +z on both sides; only magnitude varies. The normalisation also applies to the flat multi-turn coil of the second example. Off-axis points require other components.[3]

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Example: centre and axial field

Compare a flat coil at two locations

Given: $N=25$ thin nearby nearly equal turns, $R=4.0\,\mathrm{cm}$ , $I=2.0\,\mathrm A$ counterclockwise viewed from +z; $z=3.0\,\mathrm{cm}$ . Approximate air as vacuum.
Find: Centre field, axial field and their ratio.

Model: add axial components; N counts turns. Form the ratio separately so current and the physical constant cancel.
$\begin{aligned}B_z(0)&=\frac{\mu_0NI}{2R}\\B_z(z)&=\frac{\mu_0NIR^2}{2(R^2+z^2)^{\frac32}}\\\frac{B_z(z)}{B_z(0)}&=\left[1+\left(\frac zR\right)^2\right]^{-\frac32}\end{aligned}$
Substitute $\sqrt{R^2+z^2}=0.050\,\mathrm m$ ; its cube belongs in the denominator.
$\begin{aligned}B_z(0)&=\frac{(4\pi\cdot10^{-7})(25)(2.0)}{2(0.040)}\,\mathrm T\\&\approx785.398\,\mathrm{\mu T}\\B_z(0.030)&=\frac{(4\pi\cdot10^{-7})(25)(2.0)(0.040)^2}{2(0.040^2+0.030^2)^{\frac32}}\,\mathrm T\\&\approx402.124\,\mathrm{\mu T}\\\frac{B_z(0.030)}{B_z(0)}&=0.512\end{aligned}$

Result: The centre field is about 785 µT and the axial field about 402 µT, both along +z. The axial value is 51.2 percent of the centre value.

Check: Check: the ratio is below one and becomes one at z=0. Opposite elements remove transverse components; simply adding their magnitudes would omit the required projection.

Take magnitudes after vector addition

Two equal fields can cancel completely. Two equal parallel currents in symmetrically placed long wires produce opposite fields at their midpoint. Reversing only one current makes both contributions point in the same direction there. Distance from each source sets its magnitude; circulation sets its sign. Determine total magnitude only after adding Cartesian components.[1][2]

\begin{aligned}\mathbf B_{\rm total}&=\sum_j\mathbf B_j\\|\mathbf B_{\rm total}|&=\sqrt{B_x^2+B_y^2+B_z^2}\end{aligned}

Adding magnitudes is valid only for contributions in the same direction.[1]

A closed current path made of two semicircles

Given: Two concentric semicircles in the xy-plane, radii 5.00 cm and 15.0 cm, have radial connections at their ends. Current is 1.50 A counterclockwise around the inner arc viewed from +z, and clockwise around the outer arc. Observe at their common centre in vacuum, using thin wire.
Find: Find the complete path’s magnetic field at the centre.

Approach: on a circular arc, distance is constant and displacement is perpendicular to the element. Radial connectors are parallel to displacement and contribute zero at the centre.
$\begin{aligned}B_{\mathrm{arc}}&=\frac{\mu_0I\theta}{4\pi R},\qquad\theta=\pi\\B_z&=\frac{\mu_0I}{4}\left(\frac1{R_{\mathrm{in}}}-\frac1{R_{\mathrm{out}}}\right)\end{aligned}$
Substitute: the inner contribution points along +z and the outer along −z.
$\begin{aligned}B_{z,\mathrm{in}}&=\frac{(4\pi\cdot10^{-7})(1.50)}{4(0.0500)}\,\mathrm T\approx+9.42478\,\mathrm{\mu T}\\B_{z,\mathrm{out}}&\approx-3.14159\,\mathrm{\mu T}\\B_z&\approx+6.28319\,\mathrm{\mu T}\end{aligned}$

Result: Total field is about 6.283 µT along +z. The stronger inner semicircle dominates.

Check: The outer radius is three times larger, so its contribution is one third of the inner with opposite direction. Both radial connectors are explicitly included. Current reversal reverses total sign.

Steady currents and complete geometry

The stated form is magnetostatic in vacuum. With rapidly changing sources an integral over instantaneous currents does not describe all field effects; retardation and changing electric fields require time-dependent Maxwell theory. With ferromagnetic cores replacing μ₀ by an arbitrary constant is not generally sufficient: material response and geometry must be considered together. The ideal thin-wire model is singular on the wire; real current distribution and cross-section resolve that limitation.[1]

A complete result specifies observation point, current orientation, the selected conductor’s contribution and neglected parts. An accurate number for one segment is not the total field of a laboratory setup. Measurements can also contain Earth’s field and other sources. Superposition separates contributions in the linear vacuum model when their geometries and currents are known.[1]

Five exercises on geometry and vector addition

Calculate · Foundation

A long thin wire carries $I=4.0\,\mathrm A$ along +x. A point lies $d=5.0\,\mathrm{cm}$ along +y from the wire. Find its field contribution.

Show hint

Apply the right-hand rule before the magnitude calculation.

Learning goal: Use the long-wire result with direction.

Show solution and calculation

Model and approach: Long thin wire and external observation point; current and displacement determine direction.
$\begin{aligned}B&=\frac{\mu_0I}{2\pi d}\\\hat{\mathbf x}\times\hat{\mathbf y}&=\hat{\mathbf z}\end{aligned}$ Use the infinite-length approximation. +x crossed with +y gives +z.
Calculation gives 16.0 µT out of the plane.
Doubling d halves the magnitude.
Ends and return conductors must be sufficiently far away or separately included.

B_z=\frac{(4\pi\cdot10^{-7})(4.0)}{2\pi(0.050)}\,\mathrm T=16.0\,\mathrm{\mu T}

[2]

Rearrange · Application

A single circular turn of $R=3.0\,\mathrm{cm}$ should produce $B=60\,\mathrm{\mu T}$ at its centre. Find I in the vacuum model.

Show hint

Isolate I in the centre equation; N=1.

Learning goal: Find current from a single-loop centre field.

Show solution and calculation

Use $I=\frac{2RB}{\mu_0}$ , giving about 2.865 A.
Substitution returns 60 µT.
Its sign additionally needs the desired field direction; +z requires counterclockwise current viewed from +z.

I=\frac{2(0.030)(60\cdot10^{-6})}{4\pi\cdot10^{-7}}\,\mathrm A\approx2.86479\,\mathrm A

[3]

Interpret · Foundation

An observation point lies on the straight extension of one current element. Is its contribution zero? Does that imply a field-free complete cable?

Show hint

The angle’s sine concerns only that element.

Learning goal: Distinguish a local zero contribution from total field.

Show solution and calculation

Parallel or antiparallel displacement makes the cross product zero, so the local contribution vanishes.
Other parts of the complete circuit may contribute nonzero fields.
The observer cannot lie on the ideal current element itself, where zero distance makes the line description singular.

\mathrm d\boldsymbol\ell\parallel\mathbf R,\ R>0\quad\Longrightarrow\quad\mathrm d\mathbf B=\mathbf0

[1]

Error analysis · Application

For a finite straight wire a solution replaces every R by shortest distance d and sets every angle to 90°. Why is this wrong except at the middle element?

Show hint

Draw an element at x≠0 and its displacement to the point.

Learning goal: Recognise element-dependent distance in the integral.

Show solution and calculation

Use $R^2=x^2+d^2$ and $\sin\theta=\frac dR$ ; both depend on element position.
The correct integrand contains $\frac d{(x^2+d^2)^{\frac32}}$ .
The wrong calculation weights remote elements like the central one, overestimating their contribution.
Its infinite-length limit would even diverge incorrectly.

\mathrm dB_z=\frac{\mu_0I}{4\pi}\frac{d\,\mathrm dx}{(x^2+d^2)^{\frac32}}

[2]

Transfer · Challenge

Two long wires run parallel to x at $y=\pm2.5\,\mathrm{cm}$ . Each carries $2.0\,\mathrm A$ . Find the midpoint field with both currents along +x, then with the upper current along −x.

Show hint

Lower wire: displacement along +y. Upper wire: along −y.

Learning goal: Compare superposition for two current orientations.

Show solution and calculation

Each magnitude is 16 µT.
With parallel currents the lower wire gives +16 µT along z and the upper −16 µT, giving zero total.
Reversing only the upper current makes its contribution +16 µT as well, totaling +32 µT.
Add magnitudes only after determining directions.

\begin{aligned}B_{\rm single}&=\frac{\mu_0(2.0\,\mathrm A)}{2\pi(0.025\,\mathrm m)}=16\,\mathrm{\mu T}\\B_{z,\rm same}&=16-16=0\,\mathrm{\mu T}\\B_{z,\rm opposite}&=16+16=32\,\mathrm{\mu T}\end{aligned}

[2][1]

What to take away

Biot–Savart sums directed contributions. The observer remains fixed while source point and displacement change during integration.
Zero contribution from one element does not mean zero total field. Return paths and other segments belong to the complete geometry.
Symmetry can cancel or simplify components. It must follow from the actual current path.

[1][3]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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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, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Developed vector integration, finite wire and loop axis; marked μ0 as an approximation; added two checked examples and five exercises.
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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Frequently asked questions about Biot–Savart Law

What does the Biot–Savart law describe?+

The Biot–Savart law gives the magnetic contribution of a directed steady current element at an observation point. The displacement points from the element to that point, while the wire element follows conventional current. Their cross product sets angle and direction; the element-to-point distance enters squared in the form using a unit vector. Contributions are then integrated over the relevant current geometry. The answer is a vector sum, not a general sum of positive magnitudes. A wire segment can be calculated as a partial contribution, but a real steady circuit also needs a return path that can affect the measured total field.

How does a current element lead to the field of a wire geometry?+

The current-element relation is a physical magnetostatic law; a drawing alone does not prove it. For a specific conductor, derivation starts with that relation and a parametrisation. A straight segment uses element distance $\sqrt{x^2+d^2}$ and sine factor $\frac d{\sqrt{x^2+d^2}}$ . Integration gives a finite geometric factor approaching one as length grows. On a circular loop’s axis, transverse contributions cancel in opposite pairs, leaving projected axial components. These examples explain why wire geometry matters and why a single constant distance cannot always be taken outside the integral. Field magnitudes are formed after those geometric projections and cancellations.

How do I apply Biot–Savart to a long wire?+

For a long straight wire with $I=5.0\,\mathrm A$ and perpendicular distance $d=2.0\,\mathrm{cm}$ , use $B=\frac{\mu_0I}{2\pi d}$ . With educational approximation $\mu_0\approx4\pi\cdot10^{-7}$ tesla metre per ampere, the result is 50 microtesla. Determine direction separately using the right-hand rule. Twice the distance gives 25 microtesla; reversing current reverses the vector. For a finite wire this is not automatically exact: length and observation position determine the correction. Connections, return conductors and external fields may also contribute to a measurement. Round only after unit conversion and after choosing the appropriate geometry.

Under what assumptions does the magnetostatic form apply?+

The stated form assumes steady currents in magnetostatic vacuum. Thin conductors approximate spatial current distribution by a line; the model is singular on that ideal line itself. For rapidly changing sources, integrating instantaneous current elements is insufficient because retardation and coupled changing electric fields matter. Magnetic materials require an appropriate response model and boundary conditions. A finite wire segment also gives only its own partial contribution; return conductors cannot disappear without geometric justification. The approximation $\mu_0\approx4\pi\cdot10^{-7}$ is sufficient for our numerical examples, but since the SI revision it is no longer an exact defining equality.

Why can I not use one distance and direction for every element?+

A frequent mistake replaces every element-to-point distance with the shortest distance to the wire and sets every angle to ninety degrees. Both are correct only for the central element of a straight wire. Another error adds magnitudes before checking directions: two equal contributions can cancel. First verify the source-to-observer displacement, current arrow and cross product. Then integrate components and check current reversal and geometric limits. Correct tesla units do not prove correct vector addition. For several conductor parts, also explain which contributions were included and which were neglected using a justified geometric approximation, rather than silently dropping the return path.

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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 Volume 2, §12.1: The Biot-Savart Law.

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Unit

dB

field contribution from one current element

μ₀

vacuum permeability

T·m/A

I

steady current

dℓ

wire element directed with current

r̂

unit vector from element to observation point

dimensionless

r

distance from element to observation point

Derivation & background

Exam blueprint

Validity range

Derivation steps

The magnetostatic current-element law gives each contribution; parametrisation and vector integration determine a specific geometry’s field.

1Define a fixed observation point and the full steady current geometry, including return conductors.
2For positive I orient each directed element dℓ along conventional current.
3Displacement $\mathbf R=\mathbf r_P-\mathbf r^{\prime}$ points from source to observer; R varies along the wire.
4Evaluate $\mathrm d\boldsymbol\ell\times\hat{\mathbf R}$ ; it sets the sine factor and right-hand direction.
5Integrate $\mathrm d\mathbf B=\frac{\mu_0I}{4\pi}\frac{\mathrm d\boldsymbol\ell\times\hat{\mathbf R}}{R^2}$ over the geometry.
6Add Cartesian components before taking magnitude; check symmetry, current reversal and limiting cases.

How do you calculate with Biot–Savart Law?

Here is how to work through a typical Biot–Savart Law (dB=(μ₀/4π)I dℓ×r̂/r²) task step by step:

Task

An ideally long straight thin wire carries $I=5.0\,\mathrm A$ along +z. An observation point lies $r=2.0\,\mathrm{cm}$ from the axis along +x. Find field magnitude and direction in vacuum.

Solution path

Approach: $B=\frac{\mu_0I}{2\pi r}$ .
With $r=0.020\,\mathrm m$ and $\mu_0\approx4\pi\cdot10^{-7}\,\mathrm{T\,m\,A^{-1}}$ , $B=5.0\cdot10^{-5}\,\mathrm T=50\,\mathrm{\mu T}$ .
The right-hand rule gives +y.
Doubling distance halves magnitude; the ideally long single-wire contribution excludes end and return-wire contributions.

Build the field of a curved current path

Prior knowledge: a cross product selects direction perpendicular to current element and displacement. Parallel vectors give zero contribution.
Displacement runs from the current source piece to the fixed observation point, not the other way round.
Steady current requires a complete path. A calculated segment contribution must be distinguished from total field.

[1][2]

Source, observation point and cross product

\begin{aligned}\mathrm d\mathbf B&=\frac{\mu_0I}{4\pi}\frac{\mathrm d\boldsymbol\ell\times\hat{\mathbf R}}{R^2}\\&=\frac{\mu_0I}{4\pi}\frac{\mathrm d\boldsymbol\ell\times\mathbf R}{R^3}\\\mathbf B(P)&=\int\mathrm d\mathbf B\end{aligned}

Magnetostatic vacuum form for thin wires; R≠0.[1]

Apply the law rather than invent it from geometry

How do you calculate the contribution of a finite straight wire?

\begin{aligned}\mathrm dB_z&=\frac{\mu_0I}{4\pi}\frac{d\,\mathrm dx}{(x^2+d^2)^{\frac32}}\\B_z&=\frac{\mu_0I}{4\pi}\left[\frac{x}{d\sqrt{x^2+d^2}}\right]_{-a}^{a}\\&=\frac{\mu_0I}{2\pi d}\frac a{\sqrt{a^2+d^2}}\\\lim_{a\to\infty}B_z&=\frac{\mu_0I}{2\pi d}\end{aligned}

The finite factor applies in this segment’s midplane. Other circuit parts are not included here.[2]

Example: a segment rather than an infinite wire

Check a geometric approximation

Given: Wire segment $2a=0.30\,\mathrm m$ , $I=3.0\,\mathrm A$ along +x, point $(0,0.040\,\mathrm m,0)$ . Vacuum; this segment’s contribution only.
Find: Field component and difference from the infinite model.

Model: use the cross product for direction and finite integral for magnitude.
$\begin{aligned}B_\infty&=\frac{\mu_0I}{2\pi d}\\B_z&=B_\infty\frac a{\sqrt{a^2+d^2}}\end{aligned}$
Substitute a=0.15 m, not 0.30 m.
$\begin{aligned}B_\infty&=\frac{(4\pi\cdot10^{-7})(3.0)}{2\pi(0.040)}\,\mathrm T=15.0\,\mathrm{\mu T}\\B_z&=(15.0\,\mathrm{\mu T})\frac{0.15}{\sqrt{0.15^2+0.040^2}}\\&\approx14.4935\,\mathrm{\mu T}\end{aligned}$

Result: The segment contributes about 14.5 µT along +z. The infinite model is about 3.49 percent higher relative to the finite result.

How does the axial field of a circular turn arise?

\begin{aligned}\mathrm dB_z&=\frac{\mu_0I}{4\pi}\frac{R\,\mathrm d\ell}{(R^2+z^2)^{\frac32}}\\B_z(z)&=\frac{\mu_0IR^2}{2(R^2+z^2)^{\frac32}}\\B_z(0)&=\frac{\mu_0I}{2R}\end{aligned}

One turn. Multiply by N for practically equal turns packed closely together.[3]

View diagram: Loop field decreases along the axis

Loop field decreases along the axis

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Example: centre and axial field

Compare a flat coil at two locations

Given: $N=25$ thin nearby nearly equal turns, $R=4.0\,\mathrm{cm}$ , $I=2.0\,\mathrm A$ counterclockwise viewed from +z; $z=3.0\,\mathrm{cm}$ . Approximate air as vacuum.
Find: Centre field, axial field and their ratio.

Model: add axial components; N counts turns. Form the ratio separately so current and the physical constant cancel.
$\begin{aligned}B_z(0)&=\frac{\mu_0NI}{2R}\\B_z(z)&=\frac{\mu_0NIR^2}{2(R^2+z^2)^{\frac32}}\\\frac{B_z(z)}{B_z(0)}&=\left[1+\left(\frac zR\right)^2\right]^{-\frac32}\end{aligned}$
Substitute $\sqrt{R^2+z^2}=0.050\,\mathrm m$ ; its cube belongs in the denominator.
$\begin{aligned}B_z(0)&=\frac{(4\pi\cdot10^{-7})(25)(2.0)}{2(0.040)}\,\mathrm T\\&\approx785.398\,\mathrm{\mu T}\\B_z(0.030)&=\frac{(4\pi\cdot10^{-7})(25)(2.0)(0.040)^2}{2(0.040^2+0.030^2)^{\frac32}}\,\mathrm T\\&\approx402.124\,\mathrm{\mu T}\\\frac{B_z(0.030)}{B_z(0)}&=0.512\end{aligned}$

Result: The centre field is about 785 µT and the axial field about 402 µT, both along +z. The axial value is 51.2 percent of the centre value.

Check: Check: the ratio is below one and becomes one at z=0. Opposite elements remove transverse components; simply adding their magnitudes would omit the required projection.

Take magnitudes after vector addition

\begin{aligned}\mathbf B_{\rm total}&=\sum_j\mathbf B_j\\|\mathbf B_{\rm total}|&=\sqrt{B_x^2+B_y^2+B_z^2}\end{aligned}

Adding magnitudes is valid only for contributions in the same direction.[1]

A closed current path made of two semicircles

Given: Two concentric semicircles in the xy-plane, radii 5.00 cm and 15.0 cm, have radial connections at their ends. Current is 1.50 A counterclockwise around the inner arc viewed from +z, and clockwise around the outer arc. Observe at their common centre in vacuum, using thin wire.
Find: Find the complete path’s magnetic field at the centre.

Approach: on a circular arc, distance is constant and displacement is perpendicular to the element. Radial connectors are parallel to displacement and contribute zero at the centre.
$\begin{aligned}B_{\mathrm{arc}}&=\frac{\mu_0I\theta}{4\pi R},\qquad\theta=\pi\\B_z&=\frac{\mu_0I}{4}\left(\frac1{R_{\mathrm{in}}}-\frac1{R_{\mathrm{out}}}\right)\end{aligned}$
Substitute: the inner contribution points along +z and the outer along −z.
$\begin{aligned}B_{z,\mathrm{in}}&=\frac{(4\pi\cdot10^{-7})(1.50)}{4(0.0500)}\,\mathrm T\approx+9.42478\,\mathrm{\mu T}\\B_{z,\mathrm{out}}&\approx-3.14159\,\mathrm{\mu T}\\B_z&\approx+6.28319\,\mathrm{\mu T}\end{aligned}$

Result: Total field is about 6.283 µT along +z. The stronger inner semicircle dominates.

Steady currents and complete geometry

Five exercises on geometry and vector addition

Calculate · Foundation

A long thin wire carries $I=4.0\,\mathrm A$ along +x. A point lies $d=5.0\,\mathrm{cm}$ along +y from the wire. Find its field contribution.

Show hint

Apply the right-hand rule before the magnitude calculation.

Learning goal: Use the long-wire result with direction.

Show solution and calculation

Model and approach: Long thin wire and external observation point; current and displacement determine direction.
$\begin{aligned}B&=\frac{\mu_0I}{2\pi d}\\\hat{\mathbf x}\times\hat{\mathbf y}&=\hat{\mathbf z}\end{aligned}$ Use the infinite-length approximation. +x crossed with +y gives +z.
Calculation gives 16.0 µT out of the plane.
Doubling d halves the magnitude.
Ends and return conductors must be sufficiently far away or separately included.

B_z=\frac{(4\pi\cdot10^{-7})(4.0)}{2\pi(0.050)}\,\mathrm T=16.0\,\mathrm{\mu T}

[2]

Rearrange · Application

A single circular turn of $R=3.0\,\mathrm{cm}$ should produce $B=60\,\mathrm{\mu T}$ at its centre. Find I in the vacuum model.

Show hint

Isolate I in the centre equation; N=1.

Learning goal: Find current from a single-loop centre field.

Show solution and calculation

Use $I=\frac{2RB}{\mu_0}$ , giving about 2.865 A.
Substitution returns 60 µT.
Its sign additionally needs the desired field direction; +z requires counterclockwise current viewed from +z.

I=\frac{2(0.030)(60\cdot10^{-6})}{4\pi\cdot10^{-7}}\,\mathrm A\approx2.86479\,\mathrm A

[3]

Interpret · Foundation

An observation point lies on the straight extension of one current element. Is its contribution zero? Does that imply a field-free complete cable?

Show hint

The angle’s sine concerns only that element.

Learning goal: Distinguish a local zero contribution from total field.

Show solution and calculation

Parallel or antiparallel displacement makes the cross product zero, so the local contribution vanishes.
Other parts of the complete circuit may contribute nonzero fields.
The observer cannot lie on the ideal current element itself, where zero distance makes the line description singular.

\mathrm d\boldsymbol\ell\parallel\mathbf R,\ R>0\quad\Longrightarrow\quad\mathrm d\mathbf B=\mathbf0

[1]

Error analysis · Application

For a finite straight wire a solution replaces every R by shortest distance d and sets every angle to 90°. Why is this wrong except at the middle element?

Show hint

Draw an element at x≠0 and its displacement to the point.

Learning goal: Recognise element-dependent distance in the integral.

Show solution and calculation

Use $R^2=x^2+d^2$ and $\sin\theta=\frac dR$ ; both depend on element position.
The correct integrand contains $\frac d{(x^2+d^2)^{\frac32}}$ .
The wrong calculation weights remote elements like the central one, overestimating their contribution.
Its infinite-length limit would even diverge incorrectly.

\mathrm dB_z=\frac{\mu_0I}{4\pi}\frac{d\,\mathrm dx}{(x^2+d^2)^{\frac32}}

[2]

Transfer · Challenge

Two long wires run parallel to x at $y=\pm2.5\,\mathrm{cm}$ . Each carries $2.0\,\mathrm A$ . Find the midpoint field with both currents along +x, then with the upper current along −x.

Show hint

Lower wire: displacement along +y. Upper wire: along −y.

Learning goal: Compare superposition for two current orientations.

Show solution and calculation

Each magnitude is 16 µT.
With parallel currents the lower wire gives +16 µT along z and the upper −16 µT, giving zero total.
Reversing only the upper current makes its contribution +16 µT as well, totaling +32 µT.
Add magnitudes only after determining directions.

\begin{aligned}B_{\rm single}&=\frac{\mu_0(2.0\,\mathrm A)}{2\pi(0.025\,\mathrm m)}=16\,\mathrm{\mu T}\\B_{z,\rm same}&=16-16=0\,\mathrm{\mu T}\\B_{z,\rm opposite}&=16+16=32\,\mathrm{\mu T}\end{aligned}

[2][1]

What to take away

Biot–Savart sums directed contributions. The observer remains fixed while source point and displacement change during integration.
Zero contribution from one element does not mean zero total field. Return paths and other segments belong to the complete geometry.
Symmetry can cancel or simplify components. It must follow from the actual current path.

[1][3]

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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October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Developed vector integration, finite wire and loop axis; marked μ0 as an approximation; added two checked examples and five exercises.
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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Frequently asked questions about Biot–Savart Law

What does the Biot–Savart law describe?+

How does a current element lead to the field of a wire geometry?+

How do I apply Biot–Savart to a long wire?+

Under what assumptions does the magnetostatic form apply?+

Why can I not use one distance and direction for every element?+

Retain Biot–Savart Law for exams

Create a curated FSRS exam set for dB=(μ₀/4π)I dℓ×r̂/r²: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Biot–Savart law

Variables & units

Biot-Savart magnetic field calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Magnetic fields of current-carrying conductors

Worked example

Applications

How do you calculate with Biot–Savart Law?

Build the field of a curved current path

Source, observation point and cross product

Apply the law rather than invent it from geometry

How do you calculate the contribution of a finite straight wire?

Example: a segment rather than an infinite wire

Check a geometric approximation

How does the axial field of a circular turn arise?

Loop field decreases along the axis

Example: centre and axial field

Compare a flat coil at two locations

Take magnitudes after vector addition

A closed current path made of two semicircles

Steady currents and complete geometry

Five exercises on geometry and vector addition

A long thin wire carries I=4.0 AI=4.0\,\mathrm AI=4.0A along +x. A point lies d=5.0 cmd=5.0\,\mathrm{cm}d=5.0cm along +y from the wire. Find its field contribution.

A single circular turn of R=3.0 cmR=3.0\,\mathrm{cm}R=3.0cm should produce B=60 μTB=60\,\mathrm{\mu T}B=60μT at its centre. Find I in the vacuum model.

An observation point lies on the straight extension of one current element. Is its contribution zero? Does that imply a field-free complete cable?

For a finite straight wire a solution replaces every R by shortest distance d and sets every angle to 90°. Why is this wrong except at the middle element?

Two long wires run parallel to x at y=±2.5 cmy=\pm2.5\,\mathrm{cm}y=±2.5cm. Each carries 2.0 A2.0\,\mathrm A2.0A. Find the midpoint field with both currents along +x, then with the upper current along −x.

What to take away

Sources for this explanation

Frequently asked questions about Biot–Savart Law

Subject references and further reading

More formulas: Electricity and magnetism

Retain Biot–Savart Law for exams

Biot–Savart law

Variables & units

Biot-Savart magnetic field calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Magnetic fields of current-carrying conductors

Worked example

Applications

How do you calculate with Biot–Savart Law?

Build the field of a curved current path

Source, observation point and cross product

Apply the law rather than invent it from geometry

How do you calculate the contribution of a finite straight wire?

Example: a segment rather than an infinite wire

Check a geometric approximation

How does the axial field of a circular turn arise?

Loop field decreases along the axis

Example: centre and axial field

Compare a flat coil at two locations

Take magnitudes after vector addition

A closed current path made of two semicircles

Steady currents and complete geometry

Five exercises on geometry and vector addition

A long thin wire carries I=4.0 AI=4.0\,\mathrm AI=4.0A along +x. A point lies d=5.0 cmd=5.0\,\mathrm{cm}d=5.0cm along +y from the wire. Find its field contribution.

A single circular turn of R=3.0 cmR=3.0\,\mathrm{cm}R=3.0cm should produce B=60 μTB=60\,\mathrm{\mu T}B=60μT at its centre. Find I in the vacuum model.

An observation point lies on the straight extension of one current element. Is its contribution zero? Does that imply a field-free complete cable?

For a finite straight wire a solution replaces every R by shortest distance d and sets every angle to 90°. Why is this wrong except at the middle element?

Two long wires run parallel to x at y=±2.5 cmy=\pm2.5\,\mathrm{cm}y=±2.5cm. Each carries 2.0 A2.0\,\mathrm A2.0A. Find the midpoint field with both currents along +x, then with the upper current along −x.

What to take away

Sources for this explanation

Frequently asked questions about Biot–Savart Law

Subject references and further reading

A long thin wire carries $I=4.0\,\mathrm A$ along +x. A point lies $d=5.0\,\mathrm{cm}$ along +y from the wire. Find its field contribution.

A single circular turn of $R=3.0\,\mathrm{cm}$ should produce $B=60\,\mathrm{\mu T}$ at its centre. Find I in the vacuum model.

Two long wires run parallel to x at $y=\pm2.5\,\mathrm{cm}$ . Each carries $2.0\,\mathrm A$ . Find the midpoint field with both currents along +x, then with the upper current along −x.

A long thin wire carries $I=4.0\,\mathrm A$ along +x. A point lies $d=5.0\,\mathrm{cm}$ along +y from the wire. Find its field contribution.

A single circular turn of $R=3.0\,\mathrm{cm}$ should produce $B=60\,\mathrm{\mu T}$ at its centre. Find I in the vacuum model.

Two long wires run parallel to x at $y=\pm2.5\,\mathrm{cm}$ . Each carries $2.0\,\mathrm A$ . Find the midpoint field with both currents along +x, then with the upper current along −x.