Ampere's circuital law

Adding up the magnetic field of every tiny current element with the Biot-Savart law works, but for a long straight wire or a solenoid it is overkill — the geometry is far too symmetric to need that much bookkeeping. Ampere's circuital law is the shortcut: walk a closed loop around the current and the total field circulation around that loop is fixed by nothing but the current threading through it, ∮B⃗⋅dl⃗=μ0Ienclosed\oint \vec B\cdot d\vec l = \mu_0 I_{\text{enclosed}}. Choose the loop to match the symmetry of the wire, solenoid, or toroid, and the integral collapses to simple algebra. This article states the law, applies it to a straight wire and a long solenoid, and works one full numerical example so you can see the shortcut earn its keep.

The core insight

Ampere's law says the loop integral of B⃗\vec B around any closed Amperian loop equals μ0\mu_0 times the current it encloses, so a symmetric loop turns a hard Biot-Savart sum into simple algebra for a straight wire (B=μ0I/2πrB=\mu_0I/2\pi r) or a long solenoid (B=μ0nIB=\mu_0nI).

The core idea

Ampere's circuital law is a statement about circulation, not about any single point. Walk around a closed path — the Amperian loop — and add up B⃗⋅dl⃗\vec B\cdot d\vec l at every step. That sum, the line integral, depends only on the current passing through the surface bounded by the loop, never on the field's exact shape away from the loop:

∮B⃗⋅dl⃗=μ0Ienclosed\oint \vec B \cdot d\vec l = \mu_0 I_{\text{enclosed}}

where: B⃗\vec B is the magnetic field along the loop, dl⃗d\vec l is a tiny step along the loop's direction of travel, IenclosedI_{\text{enclosed}} is the net current threading through the loop (right-hand-rule sign convention), and μ0=4π×10−7 T ⋅ m/A\mu_0=4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A} is the permeability of free space. The law is always true, but it is only useful for a quick answer when you can pick a loop on which B⃗\vec B behaves simply — constant magnitude and parallel to dl⃗d\vec l everywhere (so the integral is just B×loop lengthB\times\text{loop length}), or perpendicular to dl⃗d\vec l (contributing nothing), or zero. Three standard geometries make this work.

Straight wire. A circular loop of radius rr centred on the wire, in a plane perpendicular to it, sees a field of constant magnitude circling tangentially all the way round (right-hand rule: thumb along the current, fingers curl the way B⃗\vec B points). Symmetry alone tells you this before you write a single equation.

Long solenoid. nn turns per unit length, current II: the field is uniform along the axis inside, and essentially zero outside. A rectangular Amperian loop with one side buried inside the coil and the opposite side far outside gives B=μ0nIB=\mu_0nI directly — the ISC syllabus asks you to know and apply this result, not derive it.

Toroid. A solenoid bent into a ring of NN total turns: at radius rr from the toroid's centre, B=μ0nIB=\mu_0nI with n=N/(2πr)n=N/(2\pi r) the turns per unit length along that circle. The field is zero outside the toroid and in the hollow space inside its core — again, stated and applied, not derived.

Figure
Circular Amperian loop of radius r around a straight current carrying wire with tangential field arrows
Amperian loop around a straight wire I (out of page) r B has the same size everywhere on the loop, tangent to it
Derivation AField of a long straight wire from Ampere's law

The goal is to reach B=μ0I/2πrB=\mu_0I/2\pi r for an infinitely long straight wire carrying current II, starting only from ∮B⃗⋅dl⃗=μ0Ienclosed\oint \vec B\cdot d\vec l=\mu_0I_{\text{enclosed}} and the symmetry of the geometry, so that the closed-loop integral collapses to a single algebraic step rather than a Biot-Savart sum over the whole wire.

Step 1. Choose the Amperian loop. Take a circle of radius rr, centred on the wire, lying in a plane perpendicular to the wire. By the cylindrical symmetry of a straight wire, the field at every point of this circle must have the same magnitude BB and must point tangentially along the circle — there is no reason for BB to be larger on one side than another, or to tilt out of this plane.

Step 2. Evaluate the integral. Because B⃗\vec B is tangent to the loop everywhere, B⃗⋅dl⃗=B dl\vec B\cdot d\vec l = B\,dl at every step, and BB can be pulled outside the sum since it is constant on the loop: ∮B⃗⋅dl⃗=B∮dl=B(2πr)\oint \vec B\cdot d\vec l = B\oint dl = B(2\pi r), since 2πr2\pi r is simply the loop's circumference.

Step 3. Apply Ampere's law. The current enclosed by this loop is just II, the current in the wire. So B(2πr)=μ0IB(2\pi r)=\mu_0 I.

Step 4. Solve for BB. Dividing both sides by 2πr2\pi r: B=μ0I2πrB=\dfrac{\mu_0 I}{2\pi r}. The field falls off as 1/r1/r, and its direction circles the wire according to the right-hand rule.

Worked example

A long straight wire carries a current of 10 A10\ \text{A}. Find the magnetic field at a point 0.02 m0.02\ \text{m} from the wire, and compare it with the field 0.5 m0.5\ \text{m} inside a solenoid of 10001000 turns and length 0.5 m0.5\ \text{m} carrying 2 A2\ \text{A}.

Step 1. Straight wire, apply B=μ0I/2πrB=\mu_0I/2\pi r with μ0=4π×10−7 T ⋅ m/A\mu_0=4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}, I=10 AI=10\ \text{A}, r=0.02 mr=0.02\ \text{m}.

B=4π×10−7×102π×0.02=4×10−7×102×0.02=4×10−60.04=1×10−4 TB = \frac{4\pi\times10^{-7}\times10}{2\pi\times0.02} = \frac{4\times10^{-7}\times10}{2\times0.02} = \frac{4\times10^{-6}}{0.04} = 1\times10^{-4}\ \text{T}

The π\pi cancels top and bottom, leaving a clean result: B=1×10−4 TB=1\times10^{-4}\ \text{T}, directed in circles around the wire.

Step 2. Solenoid, find turns per unit length first. n=N/l=1000/0.5=2000 turns/mn = N/l = 1000/0.5 = 2000\ \text{turns/m}.

Step 3. Apply B=μ0nIB=\mu_0nI. B=(4π×10−7)(2000)(2)B = (4\pi\times10^{-7})(2000)(2). Computing 4π×10−7×2000=8000π×10−7=2.513×10−34\pi\times10^{-7}\times2000 = 8000\pi\times10^{-7} = 2.513\times10^{-3}, then B=2.513×10−3×2=5.03×10−3 TB = 2.513\times10^{-3}\times2 = 5.03\times10^{-3}\ \text{T}.

Step 4. Compare. The wire's field, 1×10−4 T1\times10^{-4}\ \text{T} at 2 cm2\ \text{cm}, is about fifty times weaker than the solenoid's 5.03×10−3 T5.03\times10^{-3}\ \text{T} (roughly 5 mT5\ \text{mT}) — the solenoid's many close-packed turns reinforce each other along the axis, while a single wire's field is spread thinly around it. The wire gives 1×10−4 T1\times10^{-4}\ \text{T}; the solenoid gives about 5.03×10−3 T5.03\times10^{-3}\ \text{T}, uniform along its axis and essentially zero just outside the windings.

Common traps

  • Forgetting that IenclosedI_{\text{enclosed}} counts only current actually passing through the surface bounded by the loop — currents outside the loop contribute zero to the integral, even though they can still influence B⃗\vec B locally.
  • Believing Ampere's law only applies to a straight wire — it holds for any closed loop and any current distribution, but it is only a fast calculation tool when the geometry has enough symmetry to pull BB outside the integral.
  • Using B=μ0nIB=\mu_0nI for a short or non-ideal solenoid, where the field is not actually uniform inside or negligible outside — the formula strictly needs a long solenoid.
  • Mixing up nn (turns per unit length, units m−1\text{m}^{-1}) with NN (total turns, no units) — always check which one a given formula wants before substituting.

Ask the tutor: Want to see why a rectangular loop through a solenoid makes the outside-field terms vanish, or how Ampere's law connects back to Biot-Savart for a straight wire? Ask the AI tutor to walk through either derivation step by step.

Why do we need Ampere's law if Biot-Savart's law already gives the field from any current?

Biot-Savart's law is always correct, but it requires integrating the contribution of every tiny current element, which can be a genuinely hard integral even for simple shapes. Ampere's law is not a separate physical law from Biot-Savart — it is derivable from it — but for highly symmetric situations like a long straight wire, an ideal solenoid, or a toroid, it lets you skip the integration entirely and get the same answer from one line of algebra. Away from such symmetry, Ampere's law is still true but not easy to use, so Biot-Savart remains the general tool.

Does the shape of the Amperian loop matter, as long as it encloses the same current?

The value of ∮B⃗⋅dl⃗\oint \vec B\cdot d\vec l only depends on the enclosed current, so mathematically any loop enclosing the same current gives the same integral. But that does not mean any loop is useful. A loop that does not match the symmetry of the field will still satisfy the equation, yet you will not be able to pull BB out of the integral or know how it varies along the loop, so you cannot solve for BB itself. That is why we always choose a loop — circular for a wire, rectangular for a solenoid — where BB is known to be constant or zero along each segment.

If a wire passes through my Amperian loop at an angle, or the loop is not flat, does the law still work?

Yes — Ampere's law as stated is completely general: IenclosedI_{\text{enclosed}} is the net current passing through any surface bounded by the loop, and the loop itself can be any closed curve, flat or not. What changes is only how easy the calculation is. For the ISC syllabus you will only ever need the three standard symmetric cases — straight wire, long solenoid, toroid — where a sensibly chosen loop makes the algebra simple; you are not expected to handle tilted wires or non-planar loops.

Before this

Where this leads