The Biot–Savart Law

A current makes a magnetic field, and the Biot–Savart law says how much each short piece of wire contributes: dB = (μ₀/4π) I dl × r̂/r², at right angles both to the wire and to the line to the point. Add the pieces and you have the field of any wire. Choose a shape, set the current, and compare the closed form with a sum of N straight elements.

The field of one current element	$d\vec B=\frac{\mu_0}{4\pi}\,\frac{I\,d\vec l\times\hat r}{r^{2}}$
The force on a current, and between two wires	$\vec F=I\vec L\times\vec B,\qquad\frac{F}{L}=\frac{\mu_0I_1I_2}{2\pi d}$

Worked example — The Biot–Savart Law

B at P	22.48 μT
The same, as 12 current elements	23.32 μT
B at the centre of the loop	31.42 μT
Dipole far field at this z	251.3 μT

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — The Biot–Savart Law

The Biot–Savart law gives the magnetic field of a current element I dl at a point a distance r away: dB = (μ₀/4π) I dl × r̂/r², with μ₀ = 4π × 10⁻⁷ T·m/A. The field of a whole wire is the sum of its elements, which is an integral along the wire.

What each symbol means

I dl: a short piece of wire, as a vector along the current
r̂: the unit vector from the element to the point
φ: the angle between dl and r̂; the element's field is proportional to sin φ

Where the formula comes from

One element. The field of I dl is at right angles to both dl and r̂ (right-hand rule) and has size (μ₀/4π) I dl sin φ/r². An element pointing straight at P contributes nothing.
$d\vec B=\frac{\mu_0}{4\pi}\,\frac{I\,d\vec l\times\hat r}{r^{2}},\qquad dB=\frac{\mu_0}{4\pi}\,\frac{I\,dl\,\sin\varphi}{r^{2}}$
A straight segment. Every element's field at P points the same way. Put x = d tan θ, so r = d/cos θ and sin φ = cos θ, and the integral becomes ∫cos θ dθ. An infinite wire runs from θ = −90° to 90°.
$B=\frac{\mu_0I}{4\pi d}\int_{\theta_1}^{\theta_2}\cos\theta\,d\theta=\frac{\mu_0I}{4\pi d}\left(\sin\theta_2-\sin\theta_1\right)\ \to\ \frac{\mu_0I}{2\pi d}$
A loop on its axis. Every element is the same distance √(R² + z²) away and at right angles to its line to P. Round the loop the components across the axis cancel, and each element leaves the fraction R/√(R² + z²) along it.
$B_z=\oint\frac{\mu_0}{4\pi}\,\frac{I\,dl}{R^{2}+z^{2}}\cdot\frac{R}{\sqrt{R^{2}+z^{2}}}=\frac{\mu_0IR^{2}}{2\left(R^{2}+z^{2}\right)^{3/2}}$
An arc at its centre. Each element is a distance R away and at right angles to the radius, so the integral is the arc's angle times μ₀I/(4πR).
$B=\int_{0}^{\theta_0}\frac{\mu_0}{4\pi}\,\frac{IR\,d\theta}{R^{2}}=\frac{\mu_0I\theta_0}{4\pi R}$
The force on a wire. A field pushes on a current element with dF = I dl × B, the force on its moving charges added up. Put one long wire in the other's field and the force per metre follows.
$d\vec F=I\,d\vec l\times\vec B,\qquad\frac{F}{L}=I_2\,\frac{\mu_0I_1}{2\pi d}$

Assumes: Steady currents in thin wires, in vacuum or air. Straight leads to an arc are taken to point at the centre, where their elements give nothing; a lead that passes P at a distance adds its own segment's field.

Problems solved in full

A straight segment, beside its middle 3 steps
A straight wire 20 cm long carries 5.0 A. Find B at a point 5.0 cm from its middle, and compare it with an infinite wire.

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1. Each end is 10 cm from the foot of the perpendicular:
  
  $\sin\theta_{2}=-\sin\theta_{1}=\frac{0{.}10}{\sqrt{0{.}10^{2}+0{.}05^{2}}}\approx0{.}8944$
2. The segment formula:
  
  $B=\frac{\mu_0I}{4\pi d}\left(\sin\theta_2-\sin\theta_1\right)=\frac{(10^{-7})(5{.}0)}{0{.}05}\,(1{.}789)\approx1{.}789\times10^{-5}\ \mathrm{T}$
3. An infinite wire at the same distance:
  
  $B_{\infty}=\frac{\mu_0I}{2\pi d}=\frac{(2\times10^{-7})(5{.}0)}{0{.}05}=2{.}0\times10^{-5}\ \mathrm{T}$
Answer

17.89 μT, against 20 μT for an infinite wire: the segment fills 126.9° of the 180° an infinite wire would.

$B\approx17{.}89\ \mu\mathrm{T},\qquad B_{\infty}=20\ \mu\mathrm{T}$
A wire that ends opposite the point 1 steps
A long straight wire carrying 5.0 A ends at the foot of the perpendicular from a point 5.0 cm away. Find B at the point.

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1. The wire runs from θ₁ = 0 to θ₂ = 90°:
  
  $B=\frac{\mu_0I}{4\pi d}\left(1-0\right)=\frac{(10^{-7})(5{.}0)}{0{.}05}=1{.}0\times10^{-5}\ \mathrm{T}$
Answer

10 μT, half of the infinite wire's 20 μT.

$B=\frac{\mu_0I}{4\pi d}=10\ \mu\mathrm{T}$
A loop, on its axis and at its centre 2 steps
A loop of radius 10 cm carries 5.0 A. Find B at its centre and 5.0 cm along its axis.

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1. At the centre, z = 0:
  
  $B_{0}=\frac{\mu_0I}{2R}=\frac{(4\pi\times10^{-7})(5{.}0)}{2\,(0{.}10)}\approx3{.}142\times10^{-5}\ \mathrm{T}$
2. At z = 5.0 cm:
  
  $B=\frac{\mu_0IR^{2}}{2\left(R^{2}+z^{2}\right)^{3/2}}=\frac{(4\pi\times10^{-7})(5{.}0)(0{.}01)}{2\,(0{.}0125)^{3/2}}\approx2{.}248\times10^{-5}\ \mathrm{T}$
Answer

31.42 μT at the centre, 22.48 μT at 5.0 cm.

$B_{0}\approx31{.}42\ \mu\mathrm{T},\qquad B(5\ \mathrm{cm})\approx22{.}48\ \mu\mathrm{T}$
A quarter circle 1 steps
A wire bent into a quarter circle of radius 10 cm carries 5.0 A; its straight leads point at the centre. Find B at the centre.

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1. The leads point at P, so sin φ = 0 for all their elements. On the arc, every element is R away with dl at right angles to r̂:
  
  $B=\frac{\mu_0I\theta_0}{4\pi R}=\frac{(10^{-7})(5{.}0)(\pi/2)}{0{.}10}\approx7{.}854\times10^{-6}\ \mathrm{T}$
Answer

7.854 μT, a quarter of the full loop's 31.42 μT.

$B=\frac{\mu_0I}{8R}\approx7{.}854\ \mu\mathrm{T}$
A hairpin 3 steps
A long wire carrying 5.0 A runs in, turns round a semicircle of radius 10 cm and runs back out parallel to itself. Find B at the centre of the semicircle.

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1. The semicircle is half a loop:
  
  $B_{1}=\frac{\mu_0I}{4R}=\frac{(4\pi\times10^{-7})(5{.}0)}{4\,(0{.}10)}\approx1{.}571\times10^{-5}\ \mathrm{T}$
2. Each straight wire ends level with the centre, a distance R from it, so each is half of an infinite wire, and both fields point the same way as the semicircle's:
  
  $B_{2}=2\cdot\frac{\mu_0I}{4\pi R}=2\cdot\frac{(10^{-7})(5{.}0)}{0{.}10}=1{.}0\times10^{-5}\ \mathrm{T}$
3. Add them:
  
  $B=\frac{\mu_0I}{4R}\Bigl(1+\frac{2}{\pi}\Bigr)\approx2{.}571\times10^{-5}\ \mathrm{T}$
Answer

25.71 μT, all three parts pointing the same way: out of the page when the current is seen to go round the semicircle anticlockwise.

$B\approx25{.}71\ \mu\mathrm{T}$
A square loop, at its centre 2 steps
A square loop of side 20 cm carries 5.0 A. Find B at its centre.

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1. Each side is a segment with d = 10 cm and ends at x = ±10 cm, so sin θ = ±1/√2:
  
  $B_{1}=\frac{\mu_0I}{4\pi d}\Bigl(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\Bigr)=\frac{(10^{-7})(5{.}0)}{0{.}10}\,\sqrt2\approx7{.}071\times10^{-6}\ \mathrm{T}$
2. The four sides give fields in the same direction:
  
  $B=4B_{1}=\frac{2\sqrt2\,\mu_0I}{\pi a}\approx2{.}828\times10^{-5}\ \mathrm{T}$
Answer

28.28 μT, a little less than the 31.42 μT of a circle of radius 10 cm, which the square surrounds.

$B=\frac{2\sqrt2\,\mu_0I}{\pi a}\approx28{.}28\ \mu\mathrm{T}$
The force between two parallel wires 2 steps
Two long parallel wires 5.0 cm apart each carry 10 A in the same direction. Find the force on a metre of either wire, and say which way it points.

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1. Wire 1's field at wire 2:
  
  $B_1=\frac{\mu_0I_1}{2\pi d}=\frac{(2\times10^{-7})(10)}{0{.}05}=4{.}0\times10^{-5}\ \mathrm{T}$
2. The force on a length L of wire 2 is I₂L × B₁, at right angles to the wire and pointing towards wire 1:
  
  $\frac{F}{L}=I_2B_1=(10)(4{.}0\times10^{-5})=4{.}0\times10^{-4}\ \mathrm{N/m}$
Answer

0.40 mN per metre, attractive. Reverse one current and it repels with the same size.

$\frac{F}{L}=\frac{\mu_0I_1I_2}{2\pi d}=0{.}40\ \mathrm{mN/m}$
A Helmholtz pair 2 steps
Two coaxial loops of radius 10 cm, 10 cm apart, each carry 5.0 A the same way. Find B at the midpoint.

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Show solution
1. Each loop is s/2 = R/2 from the midpoint:
  
  $B=2\cdot\frac{\mu_0IR^{2}}{2\left(R^{2}+R^{2}/4\right)^{3/2}}$
2. Simplify:
  
  $B=\frac{\mu_0IR^{2}}{\left(\tfrac{5}{4}R^{2}\right)^{3/2}}=\frac{8}{5\sqrt5}\,\frac{\mu_0I}{R}=\frac{8}{5\sqrt5}\cdot\frac{(4\pi\times10^{-7})(5{.}0)}{0{.}10}\approx4{.}496\times10^{-5}\ \mathrm{T}$
Answer

44.96 μT, 1.431 times one loop's field at its own centre.

$B=\frac{8}{5\sqrt5}\,\frac{\mu_0I}{R}\approx44{.}96\ \mu\mathrm{T}$
A loop seen from far away 3 steps
Find the magnetic moment of the loop in problem 3 and estimate B 50 cm along its axis from the dipole formula. How good is the estimate?

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1. The moment:
  
  $m=I\pi R^{2}=(5{.}0)\,\pi\,(0{.}10)^{2}\approx0{.}1571\ \mathrm{A\,m^{2}}$
2. The dipole formula:
  
  $B\approx\frac{\mu_0m}{2\pi z^{3}}=\frac{(2\times10^{-7})(0{.}1571)}{(0{.}50)^{3}}\approx2{.}513\times10^{-7}\ \mathrm{T}$
3. The exact field:
  
  $B=\frac{\mu_0IR^{2}}{2\left(R^{2}+z^{2}\right)^{3/2}}=\frac{(4\pi\times10^{-7})(5{.}0)(0{.}01)}{2\,(0{.}26)^{3/2}}\approx2{.}370\times10^{-7}\ \mathrm{T}$
Answer

m = 0.1571 A·m²; the dipole formula gives 0.2513 μT against the exact 0.237 μT, 6% high at five radii.

$B_{m}\approx0{.}2513\ \mu\mathrm{T},\qquad B\approx0{.}2370\ \mu\mathrm{T}$
The force on a wire in a uniform field 1 steps
A straight wire 20 cm long carries 5.0 A at 30° to a uniform field of 0.30 T. Find the force on it.
Show solution
1. F = IL × B, of size ILB sin θ:
  
  $F=ILB\sin\theta=(5{.}0)(0{.}20)(0{.}30)\sin30^{\circ}=0{.}15\ \mathrm{N}$
Answer

0.15 N, at right angles to both the wire and the field.

$F=0{.}15\ \mathrm{N}$

Supplementary problems, answers only

A long straight wire carries 20 A. Find B 4.0 cm from it.

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100 μT.

$B=\frac{\mu_0I}{2\pi d}=100\ \mu\mathrm{T}$
A flat coil of 50 turns and radius 5.0 cm carries 2.0 A. Find B at its centre.

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1.257 mT: fifty times one turn's field.

$B=N\,\frac{\mu_0I}{2R}\approx1{.}257\ \mathrm{mT}$
Find B at the centre of a square loop of side a carrying I.

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2√2 μ₀I/(πa).

$B=\frac{2\sqrt2\,\mu_0I}{\pi a}$
Two long parallel wires 5.0 cm apart carry 10 A in opposite directions. Find B halfway between them.

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160 μT: the two fields point the same way there.

$B=2\cdot\frac{\mu_0I}{2\pi\,(d/2)}=160\ \mu\mathrm{T}$
A rectangular loop 10 cm by 5.0 cm carries 3.0 A in a uniform field of 0.20 T parallel to its plane. Find the torque on it.

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3.0 × 10⁻³ N·m, from τ = mB sin 90° with m = IA.

$\tau=IAB=3{.}0\times10^{-3}\ \mathrm{N\,m}$
A proton moves at 1.0 × 10⁵ m/s parallel to a long wire carrying 10 A, 2.0 cm from it, in the direction of the current. Find the force on it.

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1.602 × 10⁻¹⁸ N, towards the wire.

$F=qvB=qv\,\frac{\mu_0I}{2\pi d}\approx1{.}602\times10^{-18}\ \mathrm{N}$

Two currents running the same way attract

Press Two wires, same direction: 10 A in each, 5 cm apart, and each metre of wire is pulled towards the other with 0.4 mN. Now press Two wires, opposite directions and they push apart with the same force. Charges behave the other way round, with like charges repelling. The force on wire 2 is I₂ times its direction crossed with wire 1's field, μ₀I₁/(2πd), and for currents running the same way that cross product points back at wire 1. Until 20 May 2019 this force defined the ampere: 1 A was the current that gave 2 × 10⁻⁷ N per metre between two wires 1 m apart, which made μ₀ exactly 4π × 10⁻⁷ N/A². The ampere is now fixed by the charge of the electron, and μ₀ has become a measured number: CODATA 2022 gives 1.25663706127(20) × 10⁻⁶ N/A², the old value to within its uncertainty.

A loop of current is a magnet, and far away it looks like any other

Press A loop, seen from far away. 50 cm along the axis of a loop of radius 10 cm the field is 0.237 μT, and μ₀m/(2πz³), with the moment m = IπR², gives 0.2513 μT; further out the two converge. That is a dipole's field. An electric dipole p gives 2kp/z³ along its axis, and the loop gives (μ₀/4π)·2m/z³: the same law with μ₀/4π in place of k and m in place of p. Far away, all that survives of the loop is its moment, current times area; its shape and the details of the wire only correct the field closer in, which is why a small coil and a compass needle can stand in for each other.

Two loops one radius apart make the flattest field

Press A Helmholtz pair. With the loops 10 cm apart, equal to their radius, the field at the midpoint is (8/5√5)μ₀I/R = 44.96 μT, and moving 1 cm along the axis changes it by 0.011%. One loop alone changes by 1.5% over the same centimetre. One loop's field along its axis is curved downward near the loop and upward far from it, and changes over at z = R/2. Set s = R and the midpoint sits at that point of inflection for both loops, so the second derivative there is zero; the slope and the third derivative cancel between the two loops by symmetry, so the field departs from its midpoint value only as z⁴. Set s to 8 cm and the curve on the chart bulges in the middle; set it to 12 cm and it sags.

Learning path

Where fields come from

References (6)

The law dB = (μ₀/4π) I dl sin θ/r², and the field at the centre of an arc, μ₀Iθ/(4πR) (Example 12.2): S. J. Ling, J. Sanny and W. Moebs, University Physics Volume 2, OpenStax (2016), section 12.1, “The Biot-Savart Law”.
The infinite straight wire, μ₀I/(2πR), from the integral over θ (Eq. 12.8): S. J. Ling, J. Sanny and W. Moebs, University Physics Volume 2, OpenStax (2016), section 12.2, “Magnetic Field Due to a Thin Straight Wire”.
F/l = μ₀I₁I₂/(2πr), attractive for currents in the same direction and repulsive for opposite ones: S. J. Ling, J. Sanny and W. Moebs, University Physics Volume 2, OpenStax (2016), section 12.3, “Magnetic Force between Two Parallel Currents”.
The loop on its axis, its centre, the far field as a magnetic dipole (Eq. 12.19), and Helmholtz coils (Example 12.5): S. J. Ling, J. Sanny and W. Moebs, University Physics Volume 2, OpenStax (2016), section 12.4, “Magnetic Field of a Current Loop”.
Insight block 1: the pre-2019 definition (“…would produce between these conductors a force equal to 2 × 10⁻⁷ newton per meter of length”) and the new definition's date, 20 May 2019: National Institute of Standards and Technology, “SI Redefinition: Ampere”.
Insight block 1: μ₀ = 1.256 637 061 27(20) × 10⁻⁶ N A⁻², relative standard uncertainty 1.6 × 10⁻¹⁰, CODATA 2022: National Institute of Standards and Technology, CODATA Internationally Recommended 2022 Values of the Fundamental Physical Constants, “vacuum magnetic permeability”.

By the 7 Raven Room team · Reviewed against published sources. Corrections welcome.