Biot–Savart law

Assume a magnetostatic current distribution: charge density and current do not vary with time, and current conservation gives ∇·J=0. For a thin filament carrying conventional current I in direction dℓ′, define R=r−r′ from SOURCE position r′ to OBSERVATION point r. The vacuum magnetic field is

The familiar decimal for μ₀ is approximate under the modern SI. A current element parallel to R contributes zero at that observation point. The cross product decides both direction and sign: do not replace it with an unsigned scalar 1/R² expression. This field-source relation and the Lorentz force law solve different physical questions.

A physical conductor has a finite cross-section. In the volume-current formulation the current element is J(r′)d³r′, not I dℓ′; in a thin sheet use surface-current density K(r′)dS′. The stationary vacuum field for localized sources is

The thin-wire form follows by integrating J over its narrow cross-section and assuming the observation distance exceeds that cross-section scale. Infinite uniform wires are idealizations and need a separate convergence/return-path discussion.

Put a wire along +z through x=y=0 and observe at r=(s,0,0), s>0. The source element is dℓ′=dz′ ẑ at r′=(0,0,z′), hence R=s x̂−z′ ẑ. The vector numerator simplifies before any integration:

For +I all source elements contribute +ŷ at the chosen observation point. On the opposite side x<0 the field reverses. The direction follows the right-hand geometry; reversing conventional current also reverses B.

Restrict that mathematical wire contribution to axial coordinates z′ from a to b, with a<b and s>0. The antiderivative is elementary:

The bracket is positive for b>a. Writing sinθ=z′/√(s²+z′²) with θ measured from the perpendicular to the wire gives the equivalent endpoint-angle formula. A finite isolated open segment cannot sustain a steady current indefinitely: a physical loop requires return conductors, whose fields must be added. This formula is the segment's contribution, not a complete standalone stationary circuit.

Let a→−∞ and b→+∞ in the exact segment formula. The two endpoint ratios tend to −1 and +1 respectively. Their difference is 2, so

This 1/s law follows here from DIRECT Biot–Savart integration, not from Ampère's law. It is singular at the axis of an ideal zero-radius filament. In an actual finite-radius wire the internal field depends on how current is distributed over its cross-section, and a long wire is only approximately infinite away from ends and return paths.

Take a circular filament of radius a in the xy plane, carrying I counterclockwise as seen from +z. Its center is the origin; observe at (0,0,z). Parametrize the source by r′=a(cosφ,sinφ,0) and dℓ′=a(−sinφ,cosφ,0)dφ. Every source point has distance √(a²+z²) to the observer. Upon integrating over φ, transverse components cancel pairwise and axial components add:

At the center Bz(0)=μ₀I/(2a). On the positive far axis z≫a, Bz≈μ₀Ia²/(2z³), with magnetic dipole moment magnitude m=Iπa². Current reversal changes the sign; the formula is ONLY for points on the symmetry axis, not an off-axis field.

For the filament integrand, μ₀I dℓ′/R² has units (N/A²)·A·m/m²=N/(A·m)=T. For a fixed +z wire and observer at +x, a +ŷ field has the expected direction; changing just I reverses B, but changing observation from +x to −x also reverses it. A zero-radius wire model is inappropriate at s=0.

The magnetostatic approximation is a source/time assumption. A rapidly varying current distribution cannot be treated as if every remote source element communicated its instantaneous change without retardation.

For a long straight wire I=2.00 A and s=0.0200 m,

For a circular loop a=0.0500 m and I=0.500 A, Bcenter≈6.28 μT. At z=a, the axial field is Bcenter/(2√2)≈2.22 μT, directed +z for counterclockwise current viewed from +z. These values assume thin wires and an observation away from their finite cross-section.

1. Declare time-independent current, geometry and conventional direction BEFORE writing dℓ′ or J.
2. Keep r fixed and parametrize r′: calculate R=r−r′, never r′−r by habit.
3. Compute the cross product as a vector; use symmetry to cancel components only when the entire source has that symmetry.
4. Integrate the correct source measure dℓ′, dS′ or d³r′ and declare singular points or endpoints.
5. Check dimensions, current reversal, reflection geometry and limiting cases.
6. Remember a finite steady current segment is part of a return circuit; add the other segments if computing the complete physical B.

1. For I=2.00 A along +z and an observer at (s,0,0), s=0.0200 m, compute magnitude and orientation of the infinite-wire B. Repeat after reversing I.
2. For a finite straight segment from z′=−0.100 m to +0.100 m at s=0.0200 m, I=2.00 A, compute the segment contribution and its ratio to the infinite-wire result.
3. For the same finite segment move the observer to x=−s. Which components change and why?
4. For a counterclockwise circular loop of a=0.0500 m and I=0.500 A, find B at the center and at z=a; state the direction and the loop's far-axis power law.
5. Why is it inconsistent to assume a permanently stationary current that terminates at an isolated open wire endpoint? Which equation expresses the needed condition?
6. Explain why an off-axis circular-loop field cannot be obtained by replacing z with an arbitrary distance in the axial formula.

Next, Ampère's law will turn the field circulation into a source relation and make explicit which symmetries allow the local B to be extracted from the circulation integral.