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 a steady (magnetostatic) current distribution with divergence-free current density \(\vec{J}\), the field produced by the Biot–Savart law satisfies \(\nabla\times\vec{B}=\mu_0\vec{J}\) at every point, and equivalently \(\oint_{C}\vec{B}\cdot d\vec{l}=\mu_0 I_{\text{enc}}\) for any surface \(S\) spanning the closed loop \(C=\partial S\), with \(I_{\text{enc}}=\int_S\vec{J}\cdot d\vec{a}\). Ampère's circuital law is thus obtained 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 \(\vec{B}\) 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 \(\nabla\cdot\vec{J}=0\) is not cosmetic: it is the condition that makes \(\nabla\cdot\vec{A}=0\) and hence \(\nabla\times\vec{B}=\mu_0\vec{J}\) consistent. When it fails — under time-varying charge — the naive 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 \(\nabla\times\vec{B}=\mu_0\vec{J}\) gains the \(\mu_0\varepsilon_0\,\partial\vec{E}/\partial t\) term.

Charge conservation with no accumulation: \(\nabla\cdot\vec{J}=0\) everywhere.Drop it and \(\nabla\cdot\vec{A}\neq 0\), the reduction of \(\nabla\times\vec{B}\) fails, and the loop integral becomes surface-dependent (ill-defined).

The current distribution is localized, so \(\vec{J}\to 0\) faster than \(1/r^2\) at infinity.Drop it and the integration-by-parts surface term in \(\nabla\cdot\vec{A}\) need not vanish, so \(\nabla\cdot\vec{A}=0\) is not guaranteed and the vector-potential route stalls.

The kernel is smooth enough that the distributional identity \(\nabla^2\!\left(\frac{1}{|\vec{r}-\vec{r}'|}\right)=-4\pi\,\delta^3(\vec{r}-\vec{r}')\) may be used inside the integral.Drop the identity and one cannot extract the local \(\mu_0\vec{J}(\vec{r})\); only integrated (weak) statements survive.

Derivation

\[ \vec{B}(\vec{r}) = \frac{\mu_0}{4\pi}\int \vec{J}(\vec{r}')\times\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^{3}}\,d^3r' \]

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

\[ \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^{3}} = -\,\nabla\!\left(\frac{1}{|\vec{r}-\vec{r}'|}\right) \]

Vector identity; the gradient \(\nabla\) acts on the field point \(\vec{r}\). Substituting gives \(\vec{B}=-\frac{\mu_0}{4\pi}\int \vec{J}(\vec{r}')\times\nabla\frac{1}{|\vec{r}-\vec{r}'|}\,d^3r'\). B

\[ -\,\vec{J}(\vec{r}')\times\nabla f \;=\; \nabla\times\big(f\,\vec{J}(\vec{r}')\big),\qquad f\equiv\frac{1}{|\vec{r}-\vec{r}'|} \]

Since \(\vec{J}(\vec{r}')\) is independent of \(\vec{r}\), \(\nabla\times\vec{J}(\vec{r}')=0\), so \(\nabla\times(f\vec{J})=(\nabla f)\times\vec{J}\). The identity rewrites the integrand as a curl. B

\[ \vec{B}(\vec{r}) = \nabla\times\vec{A}(\vec{r}),\qquad \vec{A}(\vec{r})=\frac{\mu_0}{4\pi}\int\frac{\vec{J}(\vec{r}')}{|\vec{r}-\vec{r}'|}\,d^3r' \]

The curl acts only on \(\vec{r}\) while the integration is over \(\vec{r}'\), so \(\nabla\times\) passes outside the integral. This defines the magnetostatic vector potential \(\vec{A}\). B

\[ \nabla\times\vec{B} = \nabla\times(\nabla\times\vec{A}) = \nabla(\nabla\cdot\vec{A}) - \nabla^2\vec{A} \]

Standard "curl of a curl" identity. The two pieces are now evaluated separately. A

\[ \nabla\cdot\vec{A} = -\frac{\mu_0}{4\pi}\int \vec{J}(\vec{r}')\cdot\nabla'\!\frac{1}{|\vec{r}-\vec{r}'|}\,d^3r' = \frac{\mu_0}{4\pi}\int\frac{\nabla'\!\cdot\vec{J}(\vec{r}')}{|\vec{r}-\vec{r}'|}\,d^3r' = 0 \]

Use \(\nabla\frac{1}{|\vec{r}-\vec{r}'|}=-\nabla'\frac{1}{|\vec{r}-\vec{r}'|}\); integrate by parts in \(\vec{r}'\) (localization kills the surface term); then apply \(\nabla'\!\cdot\vec{J}=0\) from charge conservation. Hence the first bracket in Step 5 vanishes. C

\[ \nabla^2\vec{A} = \frac{\mu_0}{4\pi}\int \vec{J}(\vec{r}')\,\nabla^2\!\frac{1}{|\vec{r}-\vec{r}'|}\,d^3r' = \frac{\mu_0}{4\pi}\int \vec{J}(\vec{r}')\big(-4\pi\,\delta^3(\vec{r}-\vec{r}')\big)\,d^3r' = -\mu_0\,\vec{J}(\vec{r}) \]

The Laplacian acts on \(\vec{r}\); use the Green's-function identity \(\nabla^2\frac{1}{|\vec{r}-\vec{r}'|}=-4\pi\,\delta^3(\vec{r}-\vec{r}')\) and collapse the delta. C

\[ \nabla\times\vec{B} = \underbrace{\nabla(\nabla\cdot\vec{A})}_{=\,0} - \nabla^2\vec{A} = 0 - (-\mu_0\vec{J}) = \mu_0\,\vec{J} \]

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

\[ \int_S(\nabla\times\vec{B})\cdot d\vec{a} = \mu_0\int_S\vec{J}\cdot d\vec{a} \quad\Longrightarrow\quad \oint_C\vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}} \]

Integrate Step 8 over any surface \(S\) with boundary \(C\) and apply Stokes' theorem (prior result) to the left side; the right side is the enclosed current by definition of \(\vec{J}\). A

Result

\[ \nabla\times\vec{B} = \mu_0\,\vec{J} \qquad\Longleftrightarrow\qquad \oint_C\vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}} \]

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 \(\mu_0\). Regions with no current carry a curl-free (but not necessarily zero) field.

Units check. \([\nabla\times\vec{B}]=\mathrm{T\,m^{-1}}\). On the right, \([\mu_0]=\mathrm{T\,m\,A^{-1}}\) and \([\vec{J}]=\mathrm{A\,m^{-2}}\), so \([\mu_0\vec{J}]=\mathrm{T\,m\,A^{-1}}\cdot\mathrm{A\,m^{-2}}=\mathrm{T\,m^{-1}}\). Globally \([\oint\vec{B}\cdot d\vec{l}]=\mathrm{T\,m}\) and \([\mu_0 I_{\text{enc}}]=\mathrm{T\,m\,A^{-1}}\cdot\mathrm{A}=\mathrm{T\,m}\). Consistent.

Limiting cases

Current-free region \((\vec{J}=0)\): \(\nabla\times\vec{B}=0\), so \(\vec{B}\) is locally a gradient \(-\nabla\psi_m\) and the circulation around loops enclosing no current is zero.
Thin-wire limit: replacing \(\vec{J}\,d^3r'\) by \(I\,d\vec{l}'\) recovers the elementary \(\oint\vec{B}\cdot d\vec{l}=\mu_0 I\) for filamentary currents.
High symmetry (cylindrical, planar, toroidal): \(\vec{B}\) is constant along a chosen loop and the integral collapses to \(B\times(\text{loop length})=\mu_0 I_{\text{enc}}\), giving \(B\) directly.
Slowly varying limit: as \(\partial/\partial t\to 0\) the Ampère–Maxwell law reduces smoothly to this magnetostatic form.

Breaks when

Time-varying charge \((\nabla\cdot\vec{J}\neq 0)\). A charging capacitor is canonical: the current stops at the plate, so \(\oint\vec{B}\cdot d\vec{l}\) depends on whether the chosen surface passes between the plates or through the wire. The bare law becomes surface-ambiguous and must be replaced by the Ampère–Maxwell law with displacement current \(\mu_0\varepsilon_0\,\partial\vec{E}/\partial t\).
Non-localized or infinite current distributions. If \(\vec{J}\) does not fall off fast enough, the surface term in Step 6 survives, \(\nabla\cdot\vec{A}\neq 0\), 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 (arguments become \(t_r=t-|\vec{r}-\vec{r}'|/c\)) and the static curl relation no longer holds.
Magnetized matter with bound currents uncounted. Where \(\nabla\times\vec{M}\neq 0\), using \(\vec{B}\) with only free current is wrong; one works with \(\vec{H}\) and \(\nabla\times\vec{H}=\vec{J}_{\text{free}}\).

Failure modes

Assuming \(\vec{B}\) is uniform on the Amperian loop when symmetry does not warrant it. The law always holds, but \(B\) factors out of the integral only when symmetry forces \(|\vec{B}|\) constant and \(\vec{B}\parallel d\vec{l}\) along the loop.
Counting current that does not pierce the surface. \(I_{\text{enc}}\) is the flux of \(\vec{J}\) through \(S\), not the total current nearby; a return conductor outside the loop contributes zero.
Sign / orientation errors. Loop orientation and surface normal must obey the right-hand rule of Stokes' theorem; flipping one flips the sign of \(I_{\text{enc}}\).
Dropping \(\nabla'\!\cdot\vec{J}=0\) as "obvious." Skipping Step 6's justification leaves students baffled by the capacitor paradox — the whole failure lives in that one condition.
Misreading Step 3 as a physical claim. \(\nabla\times\vec{J}(\vec{r}')=0\) holds only because \(\vec{J}\) is evaluated at \(\vec{r}'\) while \(\nabla\) differentiates \(\vec{r}\); it does not assert 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 \(\vec{B}\) as a nonlocal integral over all current; the differential form \(\nabla\times\vec{B}=\mu_0\vec{J}\) ties the field's curl at a point to the current at that same point. Together with \(\nabla\cdot\vec{B}=0\) (automatic from \(\vec{B}=\nabla\times\vec{A}\)) these are the complete field equations of magnetostatics, and Biot–Savart is their particular solution that vanishes at infinity.

Charge conservation is the hinge. The single condition \(\nabla\cdot\vec{J}=0\) makes the Coulomb-gauge potential transverse \((\nabla\cdot\vec{A}=0)\) 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 lets Maxwell read off the missing term: to keep the law consistent when \(\nabla\cdot\vec{J}=-\partial\rho/\partial t\neq 0\), one adds \(\varepsilon_0\,\partial\vec{E}/\partial t\) so that \(\nabla\cdot\!\left(\vec{J}+\varepsilon_0\,\partial\vec{E}/\partial t\right)=0\) identically.

Notice that \(\nabla\cdot\vec{A}=0\) emerges as a consequence here, not an imposed gauge: the Biot–Savart formula secretly selects the Coulomb gauge because the source is steady and localized. At the level of the retarded solution \(\vec{A}(\vec{r},t)=\frac{\mu_0}{4\pi}\int\frac{\vec{J}(\vec{r}',t_r)}{|\vec{r}-\vec{r}'|}\,d^3r'\), the magnetostatic potential of Step 4 is exactly the \(\partial_t\to 0\) limit, and the delta-function identity in Step 7 is nothing but the Green's function of \(\nabla^2\) inverting the curl-curl operator on transverse currents. This is why gauge freedom (adding \(\nabla\chi\) to \(\vec{A}\)) never touches \(\vec{B}\) or the final law.

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

Worked examples

Infinite straight wire, \(I=10\ \mathrm{A}\); find \(B\) at radial distance \(r=5.0\ \mathrm{cm}\).

Cylindrical symmetry: \(\vec{B}\) is azimuthal with constant magnitude on a circle of radius \(r\) coaxial with the wire.

\[ \oint\vec{B}\cdot d\vec{l} = B\,(2\pi r) = \mu_0 I_{\text{enc}} = \mu_0 I \;\Rightarrow\; B=\frac{\mu_0 I}{2\pi r} \]

Symbols first: \(|\vec{B}|\) constant along the circular loop and \(\vec{B}\parallel d\vec{l}\), so it factors out; the whole wire current is enclosed.

\[ B=\frac{(4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}})(10\,\mathrm{A})}{2\pi(0.050\,\mathrm{m})} = \frac{(2\times10^{-7})(10)}{0.050}\,\mathrm{T} \]

Insert numbers with units; \(\mu_0/2\pi=2\times10^{-7}\,\mathrm{T\,m\,A^{-1}}\).

\[ B = 4.0\times10^{-5}\ \mathrm{T} = 40\ \mu\mathrm{T}\ \text{(azimuthal)} \]

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, \(n=1000\ \mathrm{turns\,m^{-1}}\), \(I=2.0\ \mathrm{A}\); find the interior field \(B\).

Rectangular Amperian loop with one side of length \(L\) inside (parallel to the axis) and the opposite side outside where \(\vec{B}\approx 0\).

\[ \oint\vec{B}\cdot d\vec{l} = B\,L = \mu_0 I_{\text{enc}} = \mu_0 (nL) I \;\Rightarrow\; B=\mu_0 n I \]

Only the interior side contributes (exterior side \(\approx 0\); perpendicular sides give \(\vec{B}\cdot d\vec{l}=0\)). The loop encloses \(nL\) turns each carrying \(I\).

\[ B=(4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}})(1000\,\mathrm{m^{-1}})(2.0\,\mathrm{A}) \]

Insert numbers; \(nI\) has units \(\mathrm{m^{-1}\,A}=\mathrm{A\,m^{-1}}\).

\[ B = 8\pi\times10^{-4}\ \mathrm{T} \approx 2.5\times10^{-3}\ \mathrm{T} = 2.5\ \mathrm{mT}\ \text{(axial, uniform)} \]

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 cylindrical symmetry.

Problems

A long straight wire carries \(I=25\ \mathrm{A}\). Find \(B\) at a perpendicular distance of \(10\ \mathrm{cm}\).
Solution
\(B=\dfrac{\mu_0 I}{2\pi r}=\dfrac{(2\times10^{-7}\,\mathrm{T\,m\,A^{-1}})(25\,\mathrm{A})}{0.10\,\mathrm{m}}=\dfrac{5.0\times10^{-6}}{0.10}\,\mathrm{T}=5.0\times10^{-5}\,\mathrm{T}=50\ \mu\mathrm{T}\), azimuthal.
A solid cylindrical conductor of radius \(R=2.0\ \mathrm{mm}\) carries a uniform current density with total current \(I=8.0\ \mathrm{A}\). Find \(B\) at radius \(r=1.0\ \mathrm{mm}\) (inside).
Solution
Uniform \(\vec{J}\): \(I_{\text{enc}}=I\left(\dfrac{r}{R}\right)^2=8.0\left(\dfrac{1.0}{2.0}\right)^2=2.0\,\mathrm{A}\). Then \(B=\dfrac{\mu_0 I_{\text{enc}}}{2\pi r}=\dfrac{(2\times10^{-7})(2.0)}{0.0010}\,\mathrm{T}=4.0\times10^{-4}\,\mathrm{T}=0.40\,\mathrm{mT}\). Equivalently \(B=\dfrac{\mu_0 I r}{2\pi R^2}\), rising linearly inside.
A toroid has \(N=500\) turns and carries \(I=3.0\ \mathrm{A}\). Find \(B\) along the central circle of radius \(r=0.10\ \mathrm{m}\).
Solution
An Amperian circle of radius \(r\) encloses all \(N\) turns, so \(I_{\text{enc}}=NI\). \(B(2\pi r)=\mu_0 NI\Rightarrow B=\dfrac{\mu_0 NI}{2\pi r}=\dfrac{(2\times10^{-7})(500)(3.0)}{0.10}\,\mathrm{T}=\dfrac{3.0\times10^{-4}}{0.10}\,\mathrm{T}=3.0\times10^{-3}\,\mathrm{T}=3.0\,\mathrm{mT}\).
Two long parallel wires \(d=4.0\ \mathrm{cm}\) apart carry \(I_1=10\ \mathrm{A}\) and \(I_2=15\ \mathrm{A}\) 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 is \(B_1=\dfrac{\mu_0 I_1}{2\pi d}\). Force per length \(\dfrac{F}{L}=I_2 B_1=\dfrac{\mu_0 I_1 I_2}{2\pi d}=\dfrac{(2\times10^{-7})(10)(15)}{0.040}\,\mathrm{N\,m^{-1}}=\dfrac{3.0\times10^{-5}}{0.040}\,\mathrm{N\,m^{-1}}=7.5\times10^{-4}\,\mathrm{N\,m^{-1}}\). Parallel currents attract, so the force is attractive.
A parallel-plate capacitor (circular plates of radius \(R=6.0\ \mathrm{cm}\)) is charging so that the conduction current in the lead is \(I=2.0\ \mathrm{A}\). Using an Amperian loop of radius \(r=3.0\ \mathrm{cm}\) centred on the axis in the plane between the plates, find \(B\) on the loop. Which principle does the bare Ampère law violate, and what restores it?
Solution
No conduction current pierces the between-plates surface, so the bare law would give \(B=0\); yet a surface capping the same loop through the wire gives \(\oint\vec{B}\cdot d\vec{l}=\mu_0 I\neq 0\) — a contradiction, because \(\nabla\cdot\vec{J}\neq 0\) at the plate (charge accumulates). The Ampère–Maxwell law adds displacement current \(I_d=\varepsilon_0\,\dfrac{d\Phi_E}{dt}\), which between the plates equals \(I\) in total. With uniform \(\vec{E}\) over the plate area, the fraction threaded by the loop is \((r/R)^2\), so \(I_{d,\text{enc}}=I\,(r/R)^2\) and \[ B=\frac{\mu_0 I_{d,\text{enc}}}{2\pi r}=\frac{\mu_0 I r}{2\pi R^2}=\frac{(2\times10^{-7})(2.0)(0.030)}{(0.060)^2}\,\mathrm{T}=\frac{1.2\times10^{-8}}{3.6\times10^{-3}}\,\mathrm{T}\approx 3.3\times10^{-6}\,\mathrm{T}=3.3\ \mu\mathrm{T}. \] The restored principle is charge conservation: displacement current makes the total current solenoidal, so the loop integral is surface-independent again.