Ampere's circuital law

Around every current-carrying wire, invisible circles of magnetic field silently form — you cannot see them, but a compass needle placed nearby will swing to trace them out. Hans Christian Oersted stumbled onto this in 1820 while lecturing to students, when a compass near his battery wire twitched every time he closed the circuit. Within weeks, Andre-Marie Ampere turned that qualitative surprise into a precise mathematical rule that lets you calculate the exact strength of the field, in one line, for any current arrangement with enough symmetry. That rule is Ampere's circuital law: the line integral of the magnetic field around any closed loop equals $\mu_0$ times the current enclosed by that loop. It sounds abstract, but it is really a shortcut — the magnetic twin of Gauss's law from electrostatics. Where Gauss's law lets you dodge a nasty Coulomb's-law integral by picking a symmetric surface, Ampere's law lets you dodge a nasty Biot-Savart integral by picking a symmetric loop. Choose that loop wisely — a circle around a straight wire, a rectangle straddling a solenoid — and a calculation that would otherwise need calculus collapses into simple algebra. This article treats the law exactly as CBSE Class 12 asks: state it, then apply it to two classic, high-yield situations. First, the field of an infinitely long straight wire, $B = \mu_0 I/(2\pi r)$ , worked out in full. Second, the field deep inside a long current-carrying solenoid, $B = \mu_0 n I$ , described and used qualitatively — its formal derivation is explicitly outside the syllabus, but the physical picture (strong uniform field inside, near-zero field outside) is exam-testable and genuinely useful, because solenoids are the working heart of electromagnets, relays, and MRI machines.

Key takeaways

Ampere's circuital law: $\oint \vec B\cdot d\vec l = \mu_0 I$ — the line integral of $\vec B$ around any closed loop equals $\mu_0$ times the current enclosed by that loop.
It is a symmetry shortcut, the magnetic analogue of Gauss's law in electrostatics — pick an Amperian loop where $\vec B$ is constant and either parallel or perpendicular to $d\vec l$ everywhere, and the integral becomes simple algebra.
Long straight wire: a circular Amperian loop of radius $r$ centred on the wire gives $B = \dfrac{\mu_0 I}{2\pi r}$ , field lines circling the wire, direction from the right-hand rule.
Long solenoid (qualitative only): field deep inside is uniform, axial, and given by $B = \mu_0 n I$ , where $n$ is turns per unit length; outside the solenoid the field is negligibly small. The derivation itself is out of syllabus — only the result and the picture are needed.
"Current enclosed" means the net current threading the loop's surface, counted with sign; a wire that misses the loop, or currents that cancel by symmetry, contribute nothing.
Ampere's law does not require the current to be at the centre or in any particular shape inside the loop — only the total enclosed current and the loop's own symmetry matter.
Key idea: Ampere's circuital law, $\oint \vec B\cdot d\vec l = \mu_0 I_{\text{enclosed}}$ , turns a symmetric magnetic field problem into simple algebra by choosing an Amperian loop on which B is constant, giving B = mu-zero I over 2 pi r for a straight wire and B = mu-zero n I inside a long solenoid.

The core insight

Ampere's circuital law states that the line integral of $\vec B$ around any closed Amperian loop equals $\mu_0$ times the net current enclosed, and choosing a loop that matches the current distribution's symmetry (a circle for a straight wire, a rectangle for a solenoid) turns this into simple algebra rather than a Biot-Savart integral.

Where it fits in the syllabus

In the CBSE Class 12 unit Moving Charges and Magnetism, Ampere's circuital law is presented as a law to be stated and applied, not derived from the Biot-Savart law — that derivation, along with the full toroid treatment, is explicitly out of scope this year. What you are expected to do is quote the law, then use it to obtain the field of an infinitely long straight wire in full, and the field inside a solenoid only qualitatively (state and use the result $B = \mu_0 n I$ , without reproducing the rectangular-loop derivation as an exam answer). This law sits right after Biot-Savart's law and its circular-loop application in the syllabus sequence, and immediately before the force on a moving charge — think of it as the "efficient" alternative to Biot-Savart for symmetric problems, not a replacement for it everywhere.

The physical picture

Every current produces a magnetic field that loops around it, and Ampere's circuital law is a precise statement of how much "looping" a given enclosed current produces, no matter how far away or what shape of loop you measure it on. The formal statement is a closed-loop integral:

\oint \vec B \cdot d\vec l = \mu_0 I

where: the small circle on the integral sign means the path is a closed loop (an Amperian loop, chosen by you, not a physical wire); $d\vec l$ is an infinitesimal displacement vector along that loop; $\vec B\cdot d\vec l$ is the component of the field along the loop's direction of travel, summed all the way around; $I$ is the net current enclosed by the loop (equivalently, passing through any surface bounded by the loop); and $\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ is the permeability of free space, the same constant that appears in the Biot-Savart law. The physical reading is that the left side only cares about the field pattern along a loop you get to choose, and the right side only cares about how much current is trapped inside it — nothing about the shape of the current distribution matters directly, only the total current linking the loop. The power of the law shows up only when you choose the Amperian loop to match the symmetry of the source. If, on your chosen loop, $\vec B$ has the same magnitude everywhere and is always parallel to $d\vec l$ , the dot product becomes an ordinary product and $B$ can be pulled outside the integral, leaving $B\oint dl = B\,L$ , where $L$ is the loop's total length. If instead $\vec B$ is everywhere perpendicular to $d\vec l$ on some stretch of the loop, that stretch contributes exactly zero to the integral. A badly chosen loop — one with no symmetry relative to the source — still satisfies the law, but the integral cannot be simplified, and you gain nothing over direct Biot-Savart integration.

Figure

A circular Amperian loop of radius r drawn around a straight current carrying wire with the field vector B shown tangent to the loop at several points and an arrow labelled current I along the wire, illustrating that B is constant in magnitude and parallel to dl everywhere on the loop

Application 1 — the infinitely long straight wire. Choose a circular Amperian loop of radius $r$ , centred on the wire, lying in a plane perpendicular to it. By the wire's cylindrical symmetry, $\vec B$ has the same magnitude at every point on this circle and is everywhere tangent to it, so $\vec B\cdot d\vec l = B\,dl$ and the integral collapses to $B(2\pi r)$ , giving the straight-wire result you will derive in full below.

Application 2 — the long solenoid. A solenoid is a long, tightly wound helical coil of wire. For a rectangular Amperian loop with one side buried deep inside the solenoid (parallel to the axis) and the opposite side far outside, Ampere's law shows that the field deep inside is essentially uniform and directed along the axis, while the field just outside a long solenoid is negligibly small. This syllabus asks you to use the resulting formula, not reproduce the rectangular-loop bookkeeping as a derivation.

B = \mu_0 n I

where: $n$ is the number of turns per unit length of the solenoid (turns/m), and $I$ is the current through the winding. Physically, every turn of wire adds its own small contribution, and because the turns are packed tightly and the solenoid is long compared to its radius, these contributions add up to a field that is the same strength anywhere well inside the coil, and drops off sharply near the open ends and outside the coil altogether.

Derivation AField of a long straight current-carrying wire from Ampere's circuital law

The goal here is to obtain the exact magnetic field magnitude at perpendicular distance $r$ from an infinitely long, straight wire carrying a steady current $I$ , using only Ampere's circuital law and the cylindrical symmetry of the situation — no Biot-Savart integration, no calculus beyond recognising a constant integrand. This is the single most tested application of the law at board level, and every step below is meant to be reproducible from memory in an exam.

Step 1. Identify the symmetry. An infinitely long straight wire looks the same from any point at a fixed perpendicular distance $r$ from it, in any rotational position around it — there is no way to distinguish one direction around the wire from another. This tells us the field magnitude $B$ can depend only on $r$ , not on the angular position around the wire, and by the right-hand rule the field direction at every such point must be tangent to a circle of radius $r$ centred on the wire, lying in a plane perpendicular to the wire.

Step 2. Choose the Amperian loop. Because the symmetry is circular around the wire, choose the Amperian loop to be exactly that circle: radius $r$ , centred on the wire, in a plane perpendicular to the wire. This loop is not a physical object — it is a mathematical path we are free to pick, and we pick it because $\vec B$ is guaranteed constant in magnitude and parallel to $d\vec l$ everywhere on it.

Step 3. Evaluate the integral on this loop. Since $\vec B$ is parallel to $d\vec l$ at every point of the circular loop and has the same magnitude $B$ throughout, $\vec B\cdot d\vec l = B\,dl$ , and $B$ can be pulled outside the sum: $\oint \vec B\cdot d\vec l = B\oint dl = B(2\pi r)$ , because the total length of a circle of radius $r$ is $2\pi r$ .

Step 4. Apply Ampere's law and solve for $B$ . The current enclosed by this loop is simply the wire's current $I$ , since the wire pierces the loop exactly once. Setting the two sides equal, $B(2\pi r) = \mu_0 I$ , so

B = \dfrac{\mu_0 I}{2\pi r}

The field falls off as $1/r$ , circles the wire (never pointing toward or away from it), and its sense is fixed by the right-hand rule: point the thumb along the current, and the curled fingers give the direction of $\vec B$ . Verified: for $I = 15\ \text{A}$ , $r = 0.05\ \text{m}$ , this gives $B = (2\times10^{-7})(15/0.05) = 6\times10^{-5}\ \text{T}$ , matching Worked example — Board level below.

Worked examples

Board level - field near a long straight current-carrying wire

Find the magnetic field at a perpendicular distance of $5\ \text{cm}$ from a long straight wire carrying a current of $15\ \text{A}$ .

Step 1. Convert units to SI. $r = 5\ \text{cm} = 0.05\ \text{m}$ , $I = 15\ \text{A}$ .

Step 2. Write the straight-wire formula obtained from Ampere's circuital law.

B = \dfrac{\mu_0 I}{2\pi r}

Step 3. Substitute, using the handy grouped value $\mu_0/(2\pi) = 2\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ rather than carrying $4\pi\times10^{-7}$ and $2\pi$ separately.

B = 2\times10^{-7}\times\dfrac{15}{0.05} = 2\times10^{-7}\times300

Step 4. Compute the final value.

B = 6\times10^{-5}\ \text{T}

So the field at $5\ \text{cm}$ from the wire is $B = 6\times10^{-5}\ \text{T}$ (about $0.6$ gauss, comparable in size to Earth's own field), circling the wire in the sense given by the right-hand rule.

Competitive level - solenoid field, turns density, and a comparison distance

A solenoid has $2000$ turns wound uniformly over a length of $0.8\ \text{m}$ and carries a current of $3\ \text{A}$ . Find (a) the magnetic field deep inside the solenoid, well away from its ends, and (b) compare it with the field you would get at a distance of $2\ \text{mm}$ from a single long straight wire carrying the same $3\ \text{A}$ , to see why coiling a wire is such an effective way to build a strong field.

Step 1. Find the turns per unit length of the solenoid.

n = \dfrac{N}{L} = \dfrac{2000}{0.8} = 2500\ \text{turns/m}

Step 2. Apply the solenoid result from Ampere's circuital law.

B_{\text{sol}} = \mu_0 n I = (4\pi\times10^{-7})(2500)(3)

Step 3. Compute stepwise, keeping the multiplication clean.

4\pi\times10^{-7}\times2500 = 10\pi\times10^{-4}\ \text{T}\,\cdot\,\text{A}^{-1}, \qquad B_{\text{sol}} = 10\pi\times10^{-4}\times3 = 30\pi\times10^{-4}

Step 4. Evaluate numerically, using $\pi\approx3.1416$ .

B_{\text{sol}} \approx 94.25\times10^{-4}\ \text{T} = 9.425\times10^{-3}\ \text{T} \approx 9.4\ \text{mT}

Step 5. For comparison, find the field at $2\ \text{mm}$ from a single straight wire carrying the same current, using $\mu_0/(2\pi) = 2\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ and $r = 2\times10^{-3}\ \text{m}$ .

B_{\text{wire}} = 2\times10^{-7}\times\dfrac{3}{2\times10^{-3}} = 2\times10^{-7}\times1500 = 3\times10^{-4}\ \text{T}

So the solenoid's interior field, $B_{\text{sol}} \approx 9.4\times10^{-3}\ \text{T}$ , is roughly $31$ times stronger than the field just $2\ \text{mm}$ from a single straight wire carrying the identical current — the coiling of $2000$ turns concentrates and adds the contributions of every loop, which is exactly why solenoids (not single wires) are used to build practical electromagnets. Timing note: parts (a) and (b) together should take about 3-4 minutes in a competitive exam — the turns-per-length step and the grouped-constant trick ( $\mu_0/2\pi = 2\times10^{-7}$ ) are what save time; do not recompute $4\pi\times10^{-7}$ from scratch every time it appears.

Common traps

Trying to apply Ampere's law to an arbitrary, asymmetric current distribution and expecting a simple answer — the law is always true, but it only becomes solvable by algebra when a loop can be chosen on which $B$ is constant (or zero); for irregular geometries you must fall back on the Biot-Savart law instead.
Forgetting that $I$ in $\oint\vec B\cdot d\vec l=\mu_0 I$ means the net current enclosed by the Amperian loop, counted with sign — a wire outside the loop contributes nothing at all, and two equal, oppositely directed currents both inside the loop cancel to give zero net enclosed current.
Treating the Amperian loop as if it must be a physical wire or boundary — it is a purely mathematical, freely chosen path, and its only job is to match the symmetry of the source.
Assuming the field outside a long solenoid is exactly zero everywhere — it is only negligibly small compared to the strong uniform field inside, not mathematically exactly zero for a real, finite-length solenoid, and it grows noticeably as you approach the open ends.
Confusing the straight-wire result $B=\mu_0I/(2\pi r)$ , which falls off with distance from the wire, with the solenoid's interior result $B=\mu_0nI$ , which is constant with position as long as you stay well inside and away from the ends — a frequent source of mixed-up formula recall under exam pressure.
Writing $\mu_0$ as $4\pi\times10^{-7}$ and then also separately writing another $2\pi$ or $\pi$ from the geometry without cancelling correctly — always simplify $\mu_0/(2\pi)$ to the memorised shortcut $2\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ before substituting numbers.

Common trapThe Amperian loop is a chosen mathematical path, not a real wire, and the law only simplifies to algebra when that loop matches the symmetry of the current

It is tempting to think Ampere's circuital law somehow requires a physical boundary or that any closed loop will make the calculation easy. Neither is true. The law $\oint \vec B\cdot d\vec l = \mu_0 I$ holds for every imaginable closed loop around any current distribution — but the left-hand integral can only be evaluated by hand when you deliberately pick a loop on which $B$ is constant and parallel (or perpendicular, or zero) to $d\vec l$ throughout. Pick a random, asymmetric loop around a symmetric wire, and the law is still true, but the integral becomes just as hard as starting from Biot-Savart. The entire skill this topic tests is loop selection, not the one-line algebra that follows it.

You should be able to…

State Ampere's circuital law $\oint\vec B\cdot d\vec l = \mu_0 I$ in words, and explain precisely what "current enclosed" means, including the sign convention for currents that partly cancel.
Explain why choosing a symmetric Amperian loop is the key skill this topic tests, and identify what shape of loop suits a given current distribution.
Derive, from Ampere's law, the straight-wire result $B = \mu_0 I/(2\pi r)$ , including the reasoning at each step, and use it in numerical problems.
State (without derivation) the solenoid result $B = \mu_0 n I$ , and describe the field pattern — uniform and axial inside, negligibly small outside — in words.
Use the right-hand rule to find the direction of $\vec B$ around a straight current-carrying wire.
Recognise Ampere's circuital law as the magnetic analogue of Gauss's law in electrostatics, and explain the analogy in one sentence.

Ask the tutor: Want to see, step by step, why choosing a rectangular Amperian loop straddling a solenoid's winding leads to $B=\mu_0nI$ inside and a negligible field outside — ask the AI tutor to walk through that loop argument.

Why can't I just use Ampere's law for every magnetic field problem instead of Biot-Savart's law?

Because Ampere's law only turns into a solvable equation when you can find an Amperian loop on which the field's magnitude is constant and its direction is either along or perpendicular to the loop everywhere — that only happens for highly symmetric sources like an infinite straight wire, an infinite sheet, or a long solenoid. For a finite wire, a single loop of wire, or any oddly shaped conductor, no such convenient loop exists, so you are stuck evaluating the messy integral from the Biot-Savart law instead, piece by piece.

If a current-carrying wire is completely outside my Amperian loop, does it really contribute nothing to the integral?

Correct — a wire that never passes through the surface bounded by your loop contributes exactly zero to the "current enclosed" term on the right-hand side, even though its magnetic field is technically still nonzero everywhere in space, including along your loop. What Ampere's law guarantees is that when you add up $\vec B\cdot d\vec l$ all the way around the closed loop, the contributions from any external current always cancel out to zero over the full loop, even though they are not zero pointwise.

Why is the solenoid derivation out of syllabus if the straight-wire one is not?

The straight-wire case has a single, clean cylindrical symmetry that makes the Amperian-loop argument short and fully rigorous at school level. The solenoid case needs a carefully constructed rectangular loop, an argument for why the field outside is negligible, and some care about what happens near the ends — CBSE has explicitly marked this fuller derivation as out of scope for 2025-26, keeping only the result $B = \mu_0 n I$ and the qualitative field picture as examinable content.

Is the field inside a solenoid really the same everywhere, no matter how far I am from the axis?

For an idealised long solenoid, yes — deep inside, away from the ends, the field magnitude and direction come out the same whether you are near the axis or near the winding, as long as you stay inside the coil. In a real, finite solenoid this holds well only in the central region; the field weakens noticeably as you approach either open end, which is exactly why the syllabus's formula $B=\mu_0nI$ is described as the field "well away from the ends."