The Biot-Savart law

The electric field of a point charge points along r̂, straight out or straight in. A current's field does not, and the experiment that settled it is old: hold a compass near a straight wire and the needle sets itself across the wire, not towards it, reversing when you carry the compass to the other side. So the contribution of one element must be built from a vector that is perpendicular to the element and to the line joining it to the field point at the same time, and the cross product is the only product of two vectors that does that. Writing dB ∝ dl × r̂ then fixes everything: curl the right hand's fingers from dl round to r̂ and the thumb gives dB. Three consequences follow immediately. The field circles the element. An element sends nothing along its own line, forwards or backwards, because there dl × r̂ = 0. And moving the field point to the mirror-image position on the other side of the element's line reverses dB, which is why field lines close on themselves and no point in space acts as a magnetic source.

Strip the direction away and the magnitude is dB = (μ₀/4π) I dl sinθ/r², a product of three levers. It is linear in I and in dl, so doubling the current or taking twice as long a piece of wire doubles the contribution — that linearity is what lets the pieces be summed at all. It falls as 1/r², inherited from the same geometry that gives Coulomb's law its inverse square. And it carries sinθ, with θ measured at the element between dl and the line out to the field point. Put numbers on it: a 2.0 mm element carrying 10 A contributes 0.80 μT at 5.0 cm broadside, 0.40 μT at the same distance but 30° from the element's own direction, and 0.20 μT at 10 cm broadside. Distance is the sharper lever of the two — doubling r costs a factor of four, while swinging θ from 90° down to 30° costs only a factor of two.

A real field comes from summing elements, so the skill is turning a picture into a definite integral. Fix the field point first and keep it fixed. Choose one variable that runs along the wire — a length x, an angle φ, whatever the geometry names — and write everything in it: the position of the element, the distance r to the field point, the angle θ, and the direction of dl × r̂. Then find the limits the real conductor imposes, and integrate. For a straight wire on the x axis with the field point a perpendicular distance d away, r = √(x² + d²) and sinθ = d/√(x² + d²), so the integrand becomes d dx/(x² + d²)³⁄². The substitution x = d tanα turns that into cosα dα/d, because dx = d sec²α dα and (x² + d²)³⁄² = d³ sec³α — an integral of one line. Check the dimensions of the integrand before you evaluate: (A)(m)/(m³) times μ₀/4π must come out in tesla.

The integral is a vector integral, so before evaluating anything ask whether all the contributions point the same way. Sometimes they do. Every element of a straight wire, seen from a point off the wire, gives a dl × r̂ along the same normal, so their magnitudes add — that is why the straight-wire integral is so easy. Every element of a circular arc, seen from its centre, does the same. Often they do not, and then the honest move is to resolve dB into components and integrate each component separately, letting symmetry kill whichever sum cancels in pairs. On the axis of a circular loop, for instance, the perpendicular components from opposite elements cancel and only the axial component survives. Off that axis nothing cancels and the integral stops being elementary — it needs elliptic functions or a computer — so the working choices are numerical integration or, for the few geometries symmetric enough to allow it, Ampere's law in place of this integral altogether.

Nothing in a laboratory is a lone current element. A steady current carries the same amperage through every cross-section of its path, so a piece of current that simply stopped at both ends would be dumping charge there at a constant rate, and the fields would no longer be static. The element contribution is therefore a bookkeeping device, not a prediction: it is not separately measurable, and two integrands that disagree element by element can still agree on every closed loop, which is exactly what happens when a term integrates to zero around a circuit. Practically, this means quoting a field as due to part of a circuit is a statement about a contribution, never about what a probe would read at that point. What you may test is the integral round the whole loop — including the leads and the return path, which beginners often draw and then forget to include.

Four assumptions sit inside the law. Currents must be steady, or slow enough that the field arrives effectively instantly: a 10 cm circuit has a light-crossing time of 0.33 ns, so at 30 MHz, where one period is 33 ns, the quasi-static picture is still good to about a per cent, and at 3 GHz it is worthless and radiation must be treated. The wire must be thin compared with the distance to the field point, otherwise the line idealisation fails and the volume form with J dV is the honest one. The medium must be empty of magnetic material, since a magnetised sample adds bound currents of its own and a linear medium replaces μ₀ by μ = μᵣ μ₀. And the moving-charge version is a low-speed limit only: the exact field of a fast charge is not a simple inverse square.