Ampère's Circuital Law from Biot–Savart

Derivation

Ampère's Circuital Law from Biot–Savart

D-052 Home PU-102 Threads fields · symmetry Depends on Biot–Savart Law from the Current Force Law, Continuity Equation and Charge Conservation, stokes-theorem

Statement

For steady currents, the Biot–Savart field B obeys ∇ × B = μ₀ J at every point, where J is the current density. Integrating over any surface S bounded by a closed loop C and applying Stokes' theorem gives the equivalent global form ∮_C B · dl = μ₀ I_enc, with I_enc the current threading S. This is Ampère's circuital law, obtained here as a theorem, not a postulate.

Why it matters

Ampère's law is the workhorse of magnetostatics: whenever a current distribution has enough symmetry, it lets you read off B from a single loop integral without ever evaluating the Biot–Savart integral. Showing it follows from Biot–Savart establishes that the two formulations carry identical physical content for steady currents.

The derivation also isolates exactly where charge conservation enters. The step ∇ · J = 0 is not cosmetic: it is the condition that makes ∇ · A = 0 and hence ∇ × B = μ₀ J consistent. When it fails — under time-varying charge — the naive Ampère law becomes contradictory, and repairing it is precisely what forces Maxwell's displacement-current term.

Assumptions

Magnetostatics: currents are steady, so all fields are time-independent. Drop it and Biot–Savart is no longer the correct field law; retarded contributions and induced electric fields appear, and ∇ × B = μ₀ J gains the μ₀ε₀ ∂E/∂t term. Charge conservation with no accumulation: ∇ · J = 0 everywhere. Drop it and ∇ · A ≠ 0, the reduction of ∇ × B fails, and the loop integral becomes surface-dependent (ill-defined). The current distribution is localized, so J → 0 faster than 1/r² at infinity. Drop it and the integration-by-parts surface term in ∇ · A need not vanish, so ∇ · A = 0 is not guaranteed and the vector-potential route stalls. The fields and sources are smooth enough that ∇²(1/|r−r′|) = −4π δ³(r−r′) may be used inside the integral. Drop the distributional identity and one cannot extract the local μ₀ J(r); only integrated (weak) statements survive.

Derivation

B(r) = (μ₀/4π) ∫ J(r′) × (r − r′)/|r − r′|³ d³r′

Biot–Savart law in volume-current form, the assumed prior result for steady currents. A

(r − r′)/|r − r′|³ = −∇(1/|r − r′|)

Vector identity; the gradient acts on the field point r. Substituting gives B = −(μ₀/4π) ∫ J(r′) × ∇(1/|r − r′|) d³r′. B

−J(r′) × ∇f = ∇ × ( f J(r′) ), f ≡ 1/|r − r′|

Because J(r′) is independent of r, ∇ × J(r′) = 0, so ∇ × (fJ) = (∇f) × J. The identity rewrites the integrand as a curl. B

B(r) = ∇ × A(r), A(r) = (μ₀/4π) ∫ J(r′)/|r − r′| d³r′

The curl acts only on r and the integration is over r′, so ∇× passes outside the integral. This defines the magnetostatic vector potential A. B

∇ × B = ∇ × (∇ × A) = ∇(∇ · A) − ∇²A

Standard "curl of a curl" identity for a vector field. Now the two pieces are evaluated separately. A

∇ · A = −(μ₀/4π) ∫ J(r′) · ∇′(1/|r − r′|) d³r′ = (μ₀/4π) ∫ (∇′ · J)/|r − r′| d³r′ = 0

Use ∇(1/|r−r′|) = −∇′(1/|r−r′|), integrate by parts in r′ (localization kills the surface term), then apply ∇′ · J = 0 from charge conservation. Hence the first bracket in Step 5 vanishes. C

∇²A = (μ₀/4π) ∫ J(r′) ∇²(1/|r − r′|) d³r′ = (μ₀/4π) ∫ J(r′)(−4π δ³(r − r′)) d³r′ = −μ₀ J(r)

The Laplacian acts on r; use the Green's-function identity ∇²(1/|r−r′|) = −4π δ³(r−r′) and collapse the delta. C

∇ × B = 0 − (−μ₀ J) = μ₀ J

Insert Steps 6 and 7 into Step 5. This is the differential form of Ampère's law. A

∫_S (∇ × B) · dA = μ₀ ∫_S J · dA ⟹ ∮_C B · dl = μ₀ I_enc

Integrate Step 8 over any surface S with boundary C and apply Stokes' theorem to the left side; the right side is the enclosed current by definition of J. A

Result

∇ × B = μ₀ J ⟺ ∮_C B · dl = μ₀ I_enc

Reading. The magnetostatic field circulates around current: its curl at a point is set entirely by the local current density there, and the net circulation around any loop counts only the current piercing that loop, weighted by μ₀. Regions with no current carry a curl-free (but not necessarily zero) field.

Units check. ∇ × B has units T·m⁻¹. On the right, [μ₀] = T·m·A⁻¹ and [J] = A·m⁻², so μ₀J is T·m·A⁻¹ · A·m⁻² = T·m⁻¹. ✓ Globally, ∮B·dl is T·m and μ₀I_enc is T·m·A⁻¹ · A = T·m. ✓

Limiting cases

Current-free region (J = 0): ∇ × B = 0, so B is locally a gradient and Ampère's law gives zero circulation around loops enclosing no current.
Thin-wire limit: replacing J d³r′ by I dl′ recovers the elementary ∮B·dl = μ₀I for filamentary currents.
High symmetry (cylindrical, planar, toroidal): B is constant along a chosen loop and the integral collapses to B·(loop length) = μ₀I_enc, giving B directly.
Slowly varying limit: as ∂/∂t → 0 the Ampère–Maxwell law reduces smoothly to this magnetostatic form.

Breaks when

Time-varying charge (∇ · J ≠ 0). A charging capacitor is the canonical case: the current stops at the plate, so ∮B·dl depends on whether the chosen surface passes between the plates or through the wire. The bare Ampère law becomes surface-ambiguous and must be replaced by the Ampère–Maxwell law with displacement current μ₀ε₀ ∂E/∂t.
Non-localized or infinite current distributions. If J does not fall off fast enough, the surface term in Step 6 survives, ∇ · A ≠ 0 in the chosen gauge, and the clean local result cannot be extracted without extra boundary data.
Rapid time dependence / radiation zone. Retardation makes the instantaneous Biot–Savart law itself invalid; the field lags the source and the static curl relation no longer holds.

Failure modes

Assuming B is uniform on the Amperian loop when symmetry does not warrant it. Ampère's law always holds, but B factors out of the integral only when symmetry forces |B| constant and B∥dl along the loop.
Counting current that does not pierce the surface. I_enc is the flux of J through S, not the total current in the vicinity; a return conductor outside the loop contributes zero.
Sign / orientation errors. The loop orientation and the surface normal must obey the right-hand rule of Stokes' theorem; flipping one flips the sign of I_enc.
Dropping ∇′ · J = 0 as "obvious." Students often skip Step 6's justification, then are baffled by the capacitor paradox — the whole failure lives in that one condition.
Treating ∇ × J(r′) = 0 (Step 3) as a physical claim about the current. It is zero only because J is evaluated at r′ while ∇ differentiates r; it is not a statement that the current is irrotational.

Discussion

The derivation shows Ampère's law is Biot–Savart in disguise: no new physics is added between them. What the vector-potential route buys is locality. Biot–Savart expresses B as a nonlocal integral over all current; Ampère's differential form ∇ × B = μ₀ J ties the field's curl at a point to the current at that same point. Together with ∇ · B = 0 (from B = ∇×A) these are the complete field equations of magnetostatics.

Charge conservation is the hinge. The single condition ∇ · J = 0 makes the Coulomb-gauge potential transverse (∇ · A = 0) and simultaneously guarantees the integral form is surface-independent — both are the same statement that current has no sources or sinks in steady state. Recognizing this is what lets Maxwell later read off the missing term: to keep the law consistent when ∇ · J = −∂ρ/∂t ≠ 0, one adds ε₀ ∂E/∂t so that ∇ · (J + ε₀∂E/∂t) = 0 identically.

At the level of differential forms, the whole argument is that B = ∇×A is closed (dB = 0) and that inverting the Laplacian on a divergence-free source reproduces the source in the curl. The delta-function identity in Step 7 is the Green's function of ∇²; the entire content of Ampère's law is that the magnetostatic Green's function inverts the curl-curl operator on transverse currents. This is why gauge freedom (adding ∇χ to A) never touches B or the final law.

Common misconceptions. Ampère's law does not say B is zero wherever there is no current — only that its curl vanishes there; a solenoid's exterior field and a wire's external field are both curl-free yet nonzero. Nor does it say B points along J; it says B curls around J. And the loop integral does not require a symmetric geometry to be true — symmetry is needed only to make it useful for finding B.

Worked examples

Infinite straight wire — field from Ampère's law

Cylindrical symmetry: B is azimuthal with constant magnitude on a circle of radius r coaxial with the wire. Given I = 10 A, r = 5.0 cm.

∮ B · dl = B (2πr) = μ₀ I_enc = μ₀ I ⟹ B = μ₀ I / (2πr)

Symbols first: |B| constant along the circular loop and B∥dl, so it factors out; the whole wire current is enclosed.

B = (4π×10⁻⁷ T·m·A⁻¹)(10 A) / (2π × 0.050 m)

Insert numbers with units: μ₀ = 4π×10⁻⁷ T·m·A⁻¹.

B = 4.0 × 10⁻⁵ T = 40 μT (azimuthal)

Reading. At 5 cm a 10 A wire produces about 40 μT, comparable to Earth's field; it circles the wire by the right-hand rule and falls as 1/r.

Long solenoid — interior field

Take a rectangular Amperian loop with one side of length L inside the solenoid (parallel to its axis) and the opposite side outside where B ≈ 0. Turns per length n = 1000 m⁻¹, current I = 2.0 A.

∮ B · dl = B L = μ₀ I_enc = μ₀ (n L) I ⟹ B = μ₀ n I

Only the interior side contributes (exterior side ≈ 0, perpendicular sides give B·dl = 0). The loop encloses nL turns each carrying I.

B = (4π×10⁻⁷ T·m·A⁻¹)(1000 m⁻¹)(2.0 A)

Insert numbers; nI has units m⁻¹·A = A·m⁻¹.

B ≈ 2.5 × 10⁻³ T = 2.5 mT (axial, uniform)

Reading. The interior field is uniform, set only by the surface current density nI and independent of the solenoid's radius — the hallmark of the Ampère-law result for planar/cylindrical symmetry.

Problems

A long straight wire carries I = 25 A. Find B at a perpendicular distance of 10 cm.
Solution
B = μ₀I/(2πr) = (4π×10⁻⁷)(25)/(2π×0.10) = (2×10⁻⁷×25)/0.10 = (5.0×10⁻⁶)/0.10 = 5.0×10⁻⁵ T = 50 μT, azimuthal.
A solid cylindrical conductor of radius R = 2.0 mm carries a uniform current density with total current I = 8.0 A. Find B at radius r = 1.0 mm (inside).
Solution
Enclosed current I_enc = I(r/R)² = 8.0×(1.0/2.0)² = 2.0 A. Then B = μ₀I_enc/(2πr) = (4π×10⁻⁷)(2.0)/(2π×0.0010) = (2×10⁻⁷×2.0)/0.0010 = (4.0×10⁻⁷)/0.0010 = 4.0×10⁻⁴ T = 0.40 mT. (Equivalently B = μ₀Ir/(2πR²), rising linearly inside.)
A toroid has N = 500 turns and carries I = 3.0 A. Find B along the central circle of radius r = 0.10 m.
Solution
An Amperian circle of radius r encloses all N turns, so I_enc = NI. B(2πr) = μ₀NI ⟹ B = μ₀NI/(2πr) = (4π×10⁻⁷)(500)(3.0)/(2π×0.10) = (2×10⁻⁷×1500)/0.10 = (3.0×10⁻⁴)/0.10 = 3.0×10⁻³ T = 3.0 mT.
Two long parallel wires 4.0 cm apart carry I₁ = 10 A and I₂ = 15 A in the same direction. Find the force per unit length between them and state whether it is attractive or repulsive.
Solution
Wire 1's field at wire 2: B₁ = μ₀I₁/(2πd). Force per length F/L = I₂B₁ = μ₀I₁I₂/(2πd) = (4π×10⁻⁷)(10)(15)/(2π×0.040) = (2×10⁻⁷×150)/0.040 = (3.0×10⁻⁵)/0.040 = 7.5×10⁻⁴ N·m⁻¹. Parallel currents attract, so the force is attractive.
A parallel-plate capacitor is charging so that the conduction current in the lead is I = 2.0 A. Using an Amperian loop of radius r = 3.0 cm centred on the axis, in the plane between the plates (no conduction current there), find B on the loop. Explain which physical principle the naive Ampère law violates and what restores it.
Solution
With no conduction current through the between-plates surface, bare Ampère's law would give B = 0, yet a surface capping the loop through the wire gives ∮B·dl = μ₀I ≠ 0 — a contradiction, because ∇·J ≠ 0 at the plate (charge accumulates). The Ampère–Maxwell law adds displacement current I_d = ε₀dΦ_E/dt, which between the plates equals I. Assuming a large plate (uniform E over the loop), the enclosed displacement current is I_d = I (r/R)² if r < plate radius R, or the full I if the loop spans the field region; taking the loop inside a uniform-field disk of radius R with the fraction (r/R)², B = μ₀I(r/R)²/(2πr). Numerically, if R ≥ r so the full flux fraction applies for a loop enclosing all the flux, B = μ₀I/(2πr) = (4π×10⁻⁷)(2.0)/(2π×0.030) = (2×10⁻⁷×2.0)/0.030 = (4.0×10⁻⁷)/0.030 ≈ 1.3×10⁻⁵ T = 13 μT. The restored principle is charge conservation: displacement current makes the total current solenoidal so the loop integral is surface-independent again.