Biot–Savart Law from the Current Force Law

Derivation

Biot–Savart Law from the Current Force Law

D-050 Home PU-102 Threads force · fields Depends on Continuity Equation and Charge Conservation, superposition-of-forces

Statement

A steady current element I dl′ located at source point r′ produces, at field point r, a magnetic field B(r) = (μ₀/4π) I dl′ × (r − r′) / |r − r′|³. This field is inverse-square in the source–field separation and transverse — perpendicular to both the current direction and the line of sight — and it is the field whose action on a second current element reproduces the empirical force law between two current-carrying circuits.

Why it matters

The Biot–Savart law is the magnetostatic analogue of Coulomb's law: it converts a specified steady current distribution directly into the field it sources, without solving a differential equation. Every downstream magnetostatic result — the field of a straight wire, a loop, a solenoid, and the identification of the magnetic dipole moment — is an integral of this one kernel.

Deriving it from the measured force between current elements, rather than postulating it, shows precisely which piece of the physics is an empirical inverse-square input and which pieces (transversality, the cross-product structure) are forced. It also exposes the well-known asymmetry of the element–element force, resolved only when one integrates over closed circuits where charge is conserved.

Assumptions

Steady currents (∂ρ/∂t = 0, ∇·J = 0).Drop it and the continuity equation forces ∇·J = −∂ρ/∂t ≠ 0; the field acquires the displacement-current and induction terms of full electrodynamics and the static kernel below is no longer complete. Linear superposition of forces.Without it the force from a distribution is not the sum over elements, and one cannot factor the two-element force into "field sourced here" times "response there." Empirical inverse-square, 1/|r−r′|², element strength ∝ I dl.This is the experimental input (Ampère's balance). Change the exponent and every geometric conclusion — flux conservation, the dipole far field — changes with it.

Non-relativistic sources; μ₀ defined constant.Drop the low-velocity restriction and the field of a moving charge picks up relativistic (retardation, velocity-dependent) corrections; the instantaneous Biot–Savart form is then only the leading term in v/c.

Vacuum, no magnetic media.In matter one must replace μ₀ by the response of bound currents (magnetization M); the free-space kernel then applies only to H or to microscopic fields.

Derivation

dF₂ = I₂ dl₂ × B(r₂)

Definition of the magnetic field as the agent of the force on a current element: we posit that the force on element 2 is linear in its own current and in an external field B sourced by everything else. This is the operational split we must justify. A

d²F_2←1 = (μ₀/4π) I₁I₂ [ dl₂ × ( dl₁ × ȓ ) ] / r²

Empirical element–element force law (Grassmann form), with ȓ the unit vector from element 1 to element 2 and r = |r₂ − r₁|. It encodes the two experimental facts: the force falls as 1/r² and each element contributes ∝ I dl. A

d²F_2←1 = I₂ dl₂ × [ (μ₀/4π) I₁ ( dl₁ × ȓ ) / r² ]

Compare with step 1: the empirical force already has the form I₂ dl₂ × [ … ]. We simply read off the bracket as the field dB that element 1 produces at element 2. Matching structure, not new physics. B

dB(r) = (μ₀/4π) I dl′ × ȓ / r², ȓ = (r − r′)/r

Drop the labels: any source element I dl′ at r′ contributes this to the field at the general field point r, with r = |r − r′|. The cross product with ȓ makes the field transverse to the line of sight by construction. B

dB(r) = (μ₀/4π) I dl′ × (r − r′) / |r − r′|³

Write ȓ/r² = (r−r′)/r³ to eliminate the unit vector in favour of the raw separation. Purely algebraic rearrangement; this is the standard integrable kernel. A

∮₁ ∮₂ dl₂ × ( dl₁ × ȓ )/r² = − ∮₁ ∮₂ ( dl₁·dl₂ ) ȓ/r²

Consistency check on closed circuits. Expand the triple product dl₂×(dl₁×ȓ) = dl₁(dl₂·ȓ) − ȓ(dl₁·dl₂). The first term is dl₁ ∮₂ (dl₂·ȓ)/r² = dl₁ ∮₂ (−d(1/r)) = 0 around a closed loop. So the asymmetric Grassmann force integrates to the symmetric Ampère form, and Newton's third law is restored for closed circuits — the regime where ∇·J = 0 holds. C

∇·B = 0, ∇×B = μ₀ J

Taking the divergence and curl of the integrated kernel B(r) = (μ₀/4π)∫ J(r′)×(r−r′)/|r−r′|³ d³r′ reproduces the magnetostatic Maxwell pair; the curl step uses ∇·J = 0 (steady currents, from continuity) to discard a boundary term. This confirms Biot–Savart is exactly the solution of the static field equations. C

Result

dB(r) = (μ₀/4π) I dl′ × (r − r′) / |r − r′|³

Reading. Each little length of current contributes a field that circulates around the wire: its magnitude falls as the inverse square of the distance and as the sine of the angle between the current and the line of sight, and its direction is set by the right-hand rule (dl′ crossed into the separation). Superpose (integrate) over the whole circuit to get the total field. The field is purely transverse — it has no component along ȓ.

Units check. [μ₀] = T·m·A⁻¹. Then (T·m·A⁻¹)·(A·m)·(m)/(m³) = T·m·A⁻¹·A·m²/m³ = T. The result is a magnetic flux density in tesla, as required.

Limiting cases

Collinear: when dl′ ∥ (r−r′), the cross product vanishes — a straight wire produces no field on its own axis.
Perpendicular, θ = 90°: the contribution is maximal, |dB| = (μ₀/4π) I dl′/r².
Infinite straight wire: integrating gives B = μ₀I/(2πs), the 1/s falloff of a line source.
Far field of a closed loop: r ≫ loop size gives the magnetic dipole field B ∝ m/r³ with m = IA.
Single moving charge: I dl′ → qv recovers B = (μ₀/4π) qv×ȓ/r² (valid for v ≪ c).

Breaks when

Non-steady currents. If ∂ρ/∂t ≠ 0, ∇·J ≠ 0 and the naïve line integral no longer satisfies ∇×B = μ₀J; the missing displacement current μ₀ε₀∂E/∂t must be restored. Biot–Savart holds only in magnetostatics.
Relativistic sources / retardation. For charges with v not ≪ c, the instantaneous 1/r² form is wrong; the field is set by the retarded position and acquires acceleration (radiation) fields. Biot–Savart is the v/c → 0 limit of the Liénard–Wiechert field.
Open current segment in isolation. A single finite element is not a physical steady current (charge would pile up at its ends); its field is only meaningful as part of a closed-circuit integral. Applied to a lone open segment the third-law symmetry of step 6 fails.
Magnetic media. In magnetized matter bound currents contribute; the free-space μ₀ kernel must be replaced by the field of free plus magnetization currents.

Failure modes

Dropping the sine. Using |dB| = (μ₀/4π)I dl/r² for every element and forgetting the sin θ from the cross product — overcounts contributions from nearly-collinear elements.
Wrong distance in the denominator. Using the perpendicular distance s to the wire instead of the actual element–point separation r inside the integrand. Only the fully integrated straight-wire result contains s.
Right-hand-rule sign flip. Crossing (r−r′) into dl′ instead of dl′ into (r−r′), reversing B.
Applying it to a lone open segment and expecting a physically complete, third-law-obeying force. Only closed circuits are consistent.
Using it for fast-changing currents (AC at high frequency, antennas) where retardation and radiation dominate.
Forgetting the field point is fixed while dl′ ranges over the source — integrating over the wrong variable.

Discussion

The derivation makes precise what is empirical and what is structural. The single experimental input is the element–element force law of step 2: inverse-square in separation, linear in each I dl. Given that, the transverse cross-product form of B is not an extra postulate — it is forced by matching the empirical force to the defining relation dF = I dl × B. Biot–Savart is therefore the field whose existence lets a two-body force be rewritten as "source produces field, field acts on test element."

The asymmetry exposed in step 6 is physically important. The Grassmann element–element force does not obey Newton's third law element-by-element: d²F_2←1 ≠ −d²F_1←2 in general. Momentum bookkeeping is rescued only after integrating around closed circuits, where the offending term integrates to zero because ∮ d(1/r) = 0. Physically, the "missing" momentum in the open-segment case is carried by the electromagnetic field itself — a first hint that the field is a dynamical object, not just bookkeeping.

Structurally, Biot–Savart is the curl of a vector potential: B = ∇×A with A(r) = (μ₀/4π)∫ J(r′)/|r−r′| d³r′, the exact magnetostatic analogue of the Coulomb potential. This guarantees ∇·B = 0 identically (no magnetic monopoles) and reduces the two curl-and-divergence field equations to a Poisson equation ∇²A = −μ₀J in the Coulomb gauge. Seen this way the Biot–Savart kernel is nothing but the Green's function of the Laplacian, curled. The same 1/|r−r′| Green's function underlies both electrostatics and magnetostatics; the difference is only whether the source is a scalar charge density or a vector current density.

Common misconceptions. Biot–Savart is not "just Coulomb for magnetism" applied to a fictitious magnetic charge — there is no magnetic charge, and the field circulates rather than pointing radially. Nor is the current element a stand-alone physical entity: it is a formal integrand, meaningful only inside a closed-loop integral. And the law is not universally valid — it is the static (or slow) limit; for radiating systems the retarded, relativistic field replaces it.

Worked examples

Field at the centre of a circular loop. Loop radius R = 5.0 cm = 0.050 m, current I = 3.0 A.

Set up: every element dl′ is perpendicular to ȓ (which points to the centre), so |dl′ × ȓ| = dl′ and every contribution points the same way (along the axis). A

B = (μ₀/4π)(I/R²) ∮ dl′ = (μ₀/4π)(I/R²)(2πR) = μ₀I/(2R)

Symbols first: with r = R constant and sin θ = 1, pull constants out and integrate the circumference. B

B = (4π×10⁻⁷)(3.0)/(2×0.050) = (1.2π×10⁻⁶)/0.10 = 3.77×10⁻⁵ T

Insert numbers: μ₀ = 4π×10⁻⁷ T·m/A. A

B ≈ 3.8×10⁻⁵ T (38 μT), along the axis by the right-hand rule

Reading. Comparable to Earth's field — a modest lab loop. Units check. (T·m/A)(A)/(m) = T. ✓

Field of a finite straight wire at perpendicular distance s = 2.0 cm from a segment carrying I = 10 A, with the point on the perpendicular bisector and the wire subtending half-angles θ₀ = 60° each side.

Set up: measure the element position by angle; dl′ = s dφ/cos²... use the standard integral for a straight segment. A

B = (μ₀I/4πs)(sin θ₁ + sin θ₂)

Symbols first: integrating the Biot–Savart kernel along a straight segment gives this closed form, with θ₁, θ₂ the angles from the perpendicular foot to each end. Here θ₁ = θ₂ = 60°. B

B = (4π×10⁻⁷ × 10)/(4π × 0.020) × (sin 60° + sin 60°) = (10⁻⁶/0.020)(2×0.866)

Insert numbers: μ₀/4π = 10⁻⁷; sin 60° = 0.866. A

B = (5.0×10⁻⁵)(1.732) = 8.66×10⁻⁵ T

Arithmetic. A

B ≈ 8.7×10⁻⁵ T, circulating around the wire

Reading. An infinite wire (θ → 90°) would give μ₀I/2πs = 1.0×10⁻⁴ T; the finite segment is smaller, as expected. Units check. (T·m/A)(A)/(m) = T. ✓

Problems

Show that on the axis of a circular loop of radius R, at distance z from the centre, B = μ₀IR²/[2(R²+z²)^3/2].
Solution
Each element gives |dB| = (μ₀/4π)I dl′/(R²+z²), since dl′ ⊥ ȓ. By symmetry only the axial component survives: multiply by cos α = R/√(R²+z²). Integrate ∮ dl′ = 2πR: B = (μ₀/4π)I(2πR)/(R²+z²) × R/√(R²+z²) = μ₀IR²/[2(R²+z²)^3/2]. At z=0 this reduces to μ₀I/2R. ✓
An infinitely long straight wire carries I = 20 A. Find B at s = 10 cm.
Solution
B = μ₀I/(2πs) = (4π×10⁻⁷)(20)/(2π×0.10) = (2×10⁻⁷×20)/0.10 = 4.0×10⁻⁵ T = 40 μT.
Two long parallel wires 5.0 cm apart each carry 15 A in the same direction. Find the force per unit length and its sign.
Solution
Wire 1 produces B = μ₀I/2πd = (2×10⁻⁷×15)/0.050 = 6.0×10⁻⁵ T at wire 2. Force per length F/L = IB = 15×6.0×10⁻⁵ = 9.0×10⁻⁴ N/m. Same-direction currents attract, so the force is attractive.
Using I dl′ → qv, find B from a proton (q=1.6×10⁻¹⁹ C) moving at v=1.0×10⁶ m/s, at 1.0 nm directly to its side (θ=90°).
Solution
B = (μ₀/4π)qv sin θ/r² = (10⁻⁷)(1.6×10⁻¹⁹)(1.0×10⁶)(1)/(1.0×10⁻⁹)² = (10⁻⁷)(1.6×10⁻¹³)/(10⁻¹⁸) = 1.6×10⁻² T = 16 mT. (Valid since v≪c.)
A square loop of side a = 8.0 cm carries I = 2.5 A. Find B at its centre.
Solution
Each side is a finite wire at perpendicular distance s = a/2 = 0.040 m subtending half-angles of 45° each side (sin 45° = 0.707). One side: B_side = (μ₀I/4πs)(2 sin 45°) = (10⁻⁷×2.5/0.040)(1.414) = (6.25×10⁻⁶)(1.414) = 8.84×10⁻⁶ T. Four sides add: B = 4×8.84×10⁻⁶ = 3.5×10⁻⁵ T ≈ 35 μT, directed out of the loop plane by the right-hand rule.