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

Taking as empirical input the force between two steady current elements (the Grassmann form of the current force law) together with the operational definition of the magnetic field through \(d\vec F = I\,d\vec l \times \vec B\), we derive that a steady current element \(I\,d\vec l'\) at \(\vec r'\) contributes at the field point \(\vec r\) a magnetic field \(d\vec B = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec l' \times \hat s}{s^2}\), with \(\vec s = \vec r - \vec r'\), \(s=|\vec s|\), and \(\hat s = \vec s/s\): an inverse-square field, transverse to both the current direction and the line of separation. Integrating over a complete circuit gives the Biot–Savart law.

Why it matters

The Biot–Savart law is the magnetostatic analogue of Coulomb's law: it converts an awkward action-at-a-distance force between currents into a local field that each current sources and to which each current responds. Once the field is in hand, superposition, flux, and circulation (Ampère's law) become available, and the whole apparatus of magnetostatics follows by integrating this one kernel.

Deriving it from the force law — rather than postulating it — exposes exactly which experimental fact does the work (the inverse-square, double-cross-product force between elements) and which features are conventions (the split of the two-element force into a "source" half and a "response" half). That bookkeeping is what makes the passage from forces to fields honest.

Assumptions

Steady currents (magnetostatics).The current is constant in time, so by charge conservation \(\nabla\cdot\vec J = 0\); dropping this makes the fields retarded and reintroduces the displacement current, so instantaneous Biot–Savart survives only as a quasi-static approximation.

Complete (closed) circuits.The measurable force is the double loop integral; the force per element is defined only up to a term that is a perfect differential around a closed loop. Dropping closure makes the single-element field ambiguous (Ampère's and Grassmann's element laws disagree), so "the field of one isolated element" has meaning only as the integrand of a closed-circuit integral.

Charge conservation supplies \(\nabla\cdot\vec J = 0\).Continuity is what guarantees the ambiguous element-force term integrates to zero around a closed steady circuit; without it the closure argument that fixes \(\vec B\) uniquely collapses.

Vacuum / linear non-magnetic medium.No bound currents or magnetization; dropping this replaces \(\mu_0\) by the medium response and forces the split into \(\vec B\) and \(\vec H\).

Non-relativistic sources (\(v \ll c\)).The current is a slow drift, not a fast point charge; dropping this sends the field of a moving charge to the Liénard–Wiechert form, which differs from naive Biot–Savart at order \((v/c)^2\).

Superposition holds.Field contributions add linearly; without it the integral over the circuit is meaningless and only the exact two-body force remains.

Derivation

\[ d^2\vec F_{1\leftarrow 2} = \frac{\mu_0}{4\pi}\,\frac{I_1 I_2}{s^2}\; d\vec l_1 \times \big( d\vec l_2 \times \hat s \big),\qquad \hat s = \frac{\vec r_1 - \vec r_2}{|\vec r_1 - \vec r_2|} \]

Empirical input: the Grassmann form of the force on current element 1 due to element 2, written as the integrand whose double loop integral reproduces the measured force between two steady circuits. B

\[ d\vec F_{1} = I_1\, d\vec l_1 \times \vec B(\vec r_1) \]

Operational definition of the magnetic field: the force on a current element is measured to be linear in the current, transverse to \(d\vec l_1\), and linear in an external agency we name \(\vec B\). This defines \(\vec B\). A

\[ \vec F_1 = \oint_1 I_1\, d\vec l_1 \times \vec B(\vec r_1),\qquad \vec F_{1\leftarrow 2} = \oint_1\oint_2 d^2\vec F_{1\leftarrow 2} \]

By superposition the total field at \(\vec r_1\) is the sum (integral) of contributions from every element of circuit 2, and the total force is the loop integral of the element force. Equating the two closed-circuit forces is legitimate because both express the same measured quantity. B

\[ \oint_1 I_1\, d\vec l_1 \times \vec B \;=\; \oint_1 d\vec l_1 \times \left[\frac{\mu_0}{4\pi} I_1 \oint_2 \frac{I_2}{s^2}\, d\vec l_2 \times \hat s \right] \]

Insert step 1 into step 3 and factor \(d\vec l_1\times\) out of the inner (circuit-2) integral, which is legal because \(d\vec l_1\) is constant over that integration. The bracket is a candidate field sourced entirely by circuit 2. B

\[ \oint_1 d\vec l_1 \times \Big(\vec B - \vec B_{\text{cand}}\Big) = 0 \;\;\Rightarrow\;\; \vec B(\vec r_1) = \frac{\mu_0}{4\pi}\oint_2 \frac{I_2\, d\vec l_2 \times \hat s}{s^2} \]

The identity must hold for every test circuit 1 and every orientation of its elements; a vector whose cross product with all directions of \(d\vec l_1\) integrates to zero around every closed path can differ from the stated field only by a term whose loop integral vanishes. Closure (assumption) removes exactly that ambiguity, fixing \(\vec B\) uniquely as the Biot–Savart integral. C

\[ d\vec B(\vec r) = \frac{\mu_0}{4\pi}\,\frac{I\, d\vec l' \times \hat s}{s^2} = \frac{\mu_0}{4\pi}\,\frac{I\, d\vec l' \times (\vec r - \vec r')}{|\vec r - \vec r'|^3} \]

Read off the differential contribution of a single source element from the integrand of step 5 (relabel circuit 2 as the source, \(\vec r_1 \to \vec r\)), and rewrite \(\hat s/s^2 = \vec s/s^3\) to eliminate the unit vector. Valid only inside a closed-circuit integral, per the closure assumption. A

\[ I\, d\vec l' \;\to\; \vec J(\vec r')\, d^3 r' \quad\Rightarrow\quad \vec B(\vec r) = \frac{\mu_0}{4\pi}\int \frac{\vec J(\vec r') \times \hat s}{s^2}\, d^3 r' \]

Promote the filamentary current to a volume current density using \(I\,d\vec l' = \vec J\, d^3 r'\); a distribution with \(\nabla\cdot\vec J = 0\) automatically consists of closed current paths, so the closure requirement is met by charge conservation. Taking \(\nabla\times\) of this integral returns \(\nabla\times\vec B = \mu_0\vec J\), confirming Biot–Savart solves the magnetostatic field equations. C

Result

\[ \boxed{\; d\vec B(\vec r) = \frac{\mu_0}{4\pi}\,\frac{I\, d\vec l' \times \hat s}{s^2}\,,\qquad \vec B(\vec r) = \frac{\mu_0}{4\pi}\oint \frac{I\, d\vec l' \times (\vec r-\vec r')}{|\vec r-\vec r'|^3}\; }\]

Reading. Each steady current element sources a field that (i) falls off as \(1/s^2\), the same inverse-square as Coulomb; (ii) is transverse — the cross product forces \(d\vec B\) perpendicular to both the current direction \(d\vec l'\) and the line of sight \(\hat s\), so the field circulates around the wire rather than pointing along it; and (iii) scales with \(\mu_0/4\pi\), the constant carrying the experimental magnitude of the current force law. Its size also carries a factor \(\sin\theta\), the angle between \(d\vec l'\) and \(\hat s\). The element form has physical meaning only summed around a closed circuit.

Units check. \([\mu_0] = \mathrm{N\,A^{-2}}\), \([I\,d\vec l'] = \mathrm{A\,m}\), \([s^2]=\mathrm{m^2}\). Then \(\dfrac{\mathrm{N\,A^{-2}}\cdot \mathrm{A\,m}}{\mathrm{m^2}} = \mathrm{N\,A^{-1}\,m^{-1}} = \mathrm{T}\), since \(1\,\mathrm{T} = 1\,\mathrm{N\,A^{-1}\,m^{-1}}\). The field is in tesla.

Limiting cases

Collinear element: when \(d\vec l' \parallel \hat s\), the cross product vanishes — a straight wire produces no field on its own axis.
Infinite straight wire: integrating over a full line gives \(B = \dfrac{\mu_0 I}{2\pi d}\) at perpendicular distance \(d\) — the result that also fixes \(\mu_0\) through the parallel-wire force.
On the axis of a circular loop: \(B_z = \dfrac{\mu_0 I R^2}{2(R^2+z^2)^{3/2}}\); at the centre \(z=0\), \(B = \dfrac{\mu_0 I}{2R}\).
Far field \(s \gg\) source size: a closed loop of area \(A\) looks like a magnetic dipole \(m = I A\), with \(\vec B \sim \mu_0 m/(4\pi s^3)\).
Single non-relativistic charge \(q\vec v = I\,d\vec l'\): \(\vec B = \dfrac{\mu_0}{4\pi}\dfrac{q\vec v \times \hat s}{s^2}\), the leading term of the moving-charge field.

Breaks when

Currents vary in time. A changing current makes fields retarded; the correct expression (Jefimenko) carries \(\partial \vec J/\partial t\) terms and radiation. Instantaneous Biot–Savart then violates causality and omits the induced and radiated field — it is only the quasi-static limit.
Open circuits or non-conserved current (\(\nabla\cdot\vec J \neq 0\)). Charge piling up (a charging capacitor lead, a broken wire) breaks closure; the element law becomes ambiguous and the naive integral disagrees with experiment unless the displacement current is added.
Relativistic point charges. For \(v\) comparable to \(c\) the field of a single moving charge is the Liénard–Wiechert field, angularly compressed and retarded; naive Biot–Savart misses the \((v/c)^2\) and acceleration (radiation) contributions.
Magnetic media. Inside magnetized matter, bound currents contribute; one must use \(\vec H\), \(\vec M\), and the medium's permeability rather than \(\mu_0\) alone.

Failure modes

Dropping the cross product: writing \(dB \propto I\,dl/s^2\) as a scalar and losing both the \(\sin\theta\) factor and the transverse direction — the field then wrongly points along the wire and over-counts nearly collinear elements.
Reversing \(\hat s\): taking \(\hat s\) from field point to source instead of source to field point flips the field's sign and the handedness of circulation.
\(1/s\) instead of \(1/s^2\): confusing the potential-like falloff with the field falloff, or forgetting that \(\hat s/s^2 = \vec s/s^3\).
Wrong distance in the integrand: using the perpendicular distance \(d\) to the wire in place of the running element–point separation \(s\); only the fully integrated straight-wire answer contains \(d\).
Believing a lone element is measurable: quoting a force on a single isolated current element as physical, ignoring that only closed-circuit integrals are unambiguous.
Applying it to fast-switching or AC currents as if instantaneous, forgetting retardation and displacement current.
Self-field error: including an element's own \(d\vec B\) in the force on that same element (the integrand is singular there and does not act on its source).

Discussion

The derivation makes visible a fact often glossed over: the physically measured object is the force between closed circuits, and the "field of a current element" is a bookkeeping device. Ampère's original element force and Grassmann's (used here) differ by a term that is a total differential; it integrates to zero around any closed loop and so cannot be distinguished experimentally. The field concept is well-defined precisely because closure erases that ambiguity — which is why charge conservation, \(\nabla\cdot\vec J = 0\), is not decoration but the hinge of the argument.

The same closure resolves a puzzle about momentum. The Grassmann element force does not obey Newton's third law element-by-element: \(d^2\vec F_{1\leftarrow 2} \neq -\,d^2\vec F_{2\leftarrow 1}\) in general. Balance is restored only after integrating around closed circuits, where the offending term drops out because \(\oint d(1/s) = 0\). Physically, the momentum that appears "missing" in the open-segment case is carried by the electromagnetic field itself — an early hint that the field is a dynamical object, not merely a shorthand for forces.

Structurally, Biot–Savart mirrors Coulomb. Both are inverse-square; both are a source strength (charge, or \(I\,d\vec l'\)) times a geometric factor. The one difference is the cross product: electric fields point along \(\hat s\), magnetic fields point across it. That single vector rotation is the entire content of "magnetism as the relativistic partner of electricity" — boost a Coulomb field and the \(\vec v\times\) structure of Biot–Savart appears at order \(v/c\).

The deepest reading is that Biot–Savart is not fundamental but emergent. The truly local, causal statements are Maxwell's differential equations; Biot–Savart is their static Green's-function solution, \(\vec B = \nabla\times\vec A\) with \(\vec A(\vec r) = \frac{\mu_0}{4\pi}\int \vec J(\vec r')/s\;d^3r'\) — the exact magnetostatic analogue of the Coulomb potential, so \(\nabla\cdot\vec B = 0\) holds identically and \(\nabla^2\vec A = -\mu_0\vec J\) in Coulomb gauge. When currents vary, the vector potential picks up retardation, \(\vec A \to \frac{\mu_0}{4\pi}\int \vec J(\vec r', t_r)/s\;d^3r'\) with \(t_r = t - s/c\), and differentiating that yields Jefimenko's equations. Biot–Savart survives only as the \(c\to\infty\) shadow of a relativistic, retarded law — a reminder that the force-law starting point of this derivation was itself a low-velocity, steady-state distillation of electrodynamics.

Common misconceptions. The field does not "flow" radially outward from the wire like a Coulomb field, and there is no magnetic charge to make it do so; it wraps around the wire. The current element is not a stand-alone physical entity — it is a formal integrand, trustworthy only once the integral is taken over a closed loop (or a divergence-free current distribution).

Worked examples

Example 1 — Infinite straight wire. Find \(\vec B\) at perpendicular distance \(d\) from a long straight wire carrying steady current \(I\).

\[ d\vec B = \frac{\mu_0 I}{4\pi}\frac{d\vec l' \times \hat s}{s^2},\quad d\vec l' = dz\,\hat z,\quad s = \sqrt{z^2 + d^2},\quad |d\vec l'\times\hat s| = dz\,\sin\theta = dz\,\frac{d}{\sqrt{z^2+d^2}} \]

Place the wire on the \(z\)-axis, field point at distance \(d\) in the plane \(z=0\); all \(d\vec B\) point the same (azimuthal) way, so magnitudes add. A

\[ B = \frac{\mu_0 I}{4\pi}\int_{-\infty}^{\infty} \frac{d\,dz}{(z^2+d^2)^{3/2}} = \frac{\mu_0 I}{4\pi}\,d\left[\frac{z}{d^2\sqrt{z^2+d^2}}\right]_{-\infty}^{\infty} = \frac{\mu_0 I}{4\pi d}\,\big(1-(-1)\big) = \frac{\mu_0 I}{2\pi d} \]

Standard integral \(\int (z^2+d^2)^{-3/2}\,dz = z/[d^2\sqrt{z^2+d^2}]\); evaluate over the full line. B

\[ I = 10\ \mathrm{A},\ d = 0.05\ \mathrm{m}:\quad B = \frac{(4\pi\times10^{-7})(10)}{2\pi(0.05)} = \frac{(2\times10^{-7})(10)}{0.05} = 4\times10^{-5}\ \mathrm{T} \]

Insert numbers only after the symbolic result; \(\mu_0/2\pi = 2\times10^{-7}\,\mathrm{T\,m\,A^{-1}}\). A

\[ B = \frac{\mu_0 I}{2\pi d} = 4\times10^{-5}\ \mathrm{T} = 40\ \mu\mathrm{T} \]

Reading. The field circles the wire, falls as \(1/d\), and points azimuthally by the right-hand rule. Forty microtesla is comparable to the Earth's field.

Example 2 — Circular loop, on-axis field. Find \(B\) on the axis of a single circular loop of radius \(R\), current \(I\), at height \(z\), then evaluate at the centre.

\[ dB = \frac{\mu_0 I}{4\pi}\frac{dl'}{s^2},\qquad s = \sqrt{R^2 + z^2},\qquad d\vec l' \perp \hat s \;\Rightarrow\; |d\vec l'\times\hat s| = dl' \]

On the axis every element is perpendicular to its own \(\hat s\), so the cross product's sine is 1. A

\[ dB_z = dB\,\cos\alpha = \frac{\mu_0 I}{4\pi}\frac{dl'}{s^2}\,\frac{R}{s},\qquad \cos\alpha = \frac{R}{s} \]

Off-axis components cancel by symmetry around the loop; keep only the axial projection, where \(\alpha\) is the angle between \(d\vec B\) and the axis. B

\[ B_z = \frac{\mu_0 I R}{4\pi s^3}\oint dl' = \frac{\mu_0 I R}{4\pi s^3}(2\pi R) = \frac{\mu_0 I R^2}{2(R^2+z^2)^{3/2}} \]

Everything except \(dl'\) is constant on the loop; \(\oint dl' = 2\pi R\). B

\[ z=0:\quad B = \frac{\mu_0 I}{2R};\qquad I = 3\ \mathrm{A},\ R = 0.10\ \mathrm{m}:\quad B = \frac{(4\pi\times10^{-7})(3)}{2(0.10)} \]

Set \(z=0\), then substitute numbers. A

\[ B_{\text{centre}} = \frac{\mu_0 I}{2R} = 1.9\times10^{-5}\ \mathrm{T} \approx 19\ \mu\mathrm{T} \]

Reading. The centre field is strongest; moving off-centre along the axis it falls as \((R^2+z^2)^{-3/2}\), becoming the dipole \(1/z^3\) far away.

Problems

(A) Single loop at its centre. A circular loop of radius \(R = 5.0\ \mathrm{cm}\) carries \(I = 2.0\ \mathrm{A}\). Find \(B\) at the centre.
Solution
\(B = \dfrac{\mu_0 I}{2R} = \dfrac{(4\pi\times10^{-7})(2.0)}{2(0.050)} = \dfrac{2.513\times10^{-6}}{0.10} = 2.5\times10^{-5}\ \mathrm{T} = 25\ \mu\mathrm{T}\).
(B) Force between parallel wires. Two long parallel wires \(d = 2.0\ \mathrm{cm}\) apart each carry \(I = 15\ \mathrm{A}\) in the same direction. Find the force per unit length and its sense.
Solution
Wire 1 makes \(B = \dfrac{\mu_0 I}{2\pi d}\) at wire 2, so the force per length is \(\dfrac{F}{L} = I B = \dfrac{\mu_0 I^2}{2\pi d} = \dfrac{(2\times10^{-7})(15)^2}{0.020} = \dfrac{(2\times10^{-7})(225)}{0.020} = 2.25\times10^{-3}\ \mathrm{N\,m^{-1}}\). Parallel currents attract, so the force is attractive.
(B) Square loop at its centre. A square loop of side \(a = 0.20\ \mathrm{m}\) carries \(I = 4.0\ \mathrm{A}\). Find \(B\) at the centre. (Use the finite-segment result \(B = \dfrac{\mu_0 I}{4\pi p}(\sin\theta_2-\sin\theta_1)\), with \(p\) the perpendicular distance.)
Solution
Each side sits at perpendicular distance \(p=a/2\) and subtends \(\theta = \pm45^\circ\), so per side \(B_1 = \dfrac{\mu_0 I}{4\pi(a/2)}(2\sin45^\circ) = \dfrac{\sqrt2\,\mu_0 I}{2\pi a}\). Four sides add: \(B = \dfrac{2\sqrt2\,\mu_0 I}{\pi a} = \dfrac{2\sqrt2\,(4\pi\times10^{-7})(4.0)}{\pi(0.20)}\). Numerically \(\dfrac{2.828\times(5.027\times10^{-6})}{0.628} = 2.3\times10^{-5}\ \mathrm{T} \approx 23\ \mu\mathrm{T}\), directed out of the loop plane by the right-hand rule.
(B) Semi-infinite wire. A straight wire runs from a point level with \(P\) out to infinity. \(P\) is a perpendicular distance \(d = 3.0\ \mathrm{cm}\) from the wire; \(I = 8.0\ \mathrm{A}\). Find \(B\) at \(P\).
Solution
The angle runs from \(0\) to \(90^\circ\): \(B = \dfrac{\mu_0 I}{4\pi d}(\sin90^\circ-\sin0^\circ) = \dfrac{\mu_0 I}{4\pi d} = \dfrac{(10^{-7})(8.0)}{0.030} = 2.7\times10^{-5}\ \mathrm{T} \approx 27\ \mu\mathrm{T}\). This is exactly half the infinite-wire value at the same distance, as expected.
(C) Rotating charged ring. A ring of radius \(R = 5.0\ \mathrm{cm}\) carries total charge \(Q = 1.0\ \mu\mathrm{C}\) spread uniformly and rotates at \(\omega = 100\ \mathrm{rad\,s^{-1}}\). Treating the rotating charge as a steady current, find \(B\) at the centre.
Solution
The rotation carries the whole charge past a point once per period \(T = 2\pi/\omega\), so the effective steady current is \(I = Q/T = Q\omega/2\pi\); it is steady precisely because charge is conserved on the ring. The centre field of a loop is \(B = \mu_0 I/2R = \mu_0 Q\omega/(4\pi R)\). Numerically \(B = \dfrac{(10^{-7})(1.0\times10^{-6})(100)}{0.050} = \dfrac{10^{-11}}{0.050} = 2.0\times10^{-10}\ \mathrm{T}\). Check: \(I = (1.0\times10^{-6})(100)/(2\pi) = 1.6\times10^{-5}\ \mathrm{A}\) gives the same \(B\).