United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How does Ampere's law use symmetry to find the magnetic field of a current distribution?

Topic 12.4 Ampere's Law: apply Ampere's law with a chosen Amperian loop to find the field of wires, solenoids and toroids.

A calculus-based answer to AP Physics C E&M Topic 12.4, covering Ampere's law as a line integral, choosing an Amperian loop, and deriving the field of a long wire, a solenoid and a toroid.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Ampere's law states that the line integral of $\vec{B}$ around any closed loop equals $\mu_0$ times the current passing through the loop: $\oint \vec{B}\cdot d\vec{\ell} = \mu_0 I_{enc}$ . Like Gauss's law, it is always true but only solves for the field when symmetry makes $B$ constant along a chosen Amperian loop and either parallel or perpendicular to it. Three standard cases follow. A long straight wire: a circular loop gives $B = \dfrac{\mu_0 I}{2\pi r}$ outside, and $B = \dfrac{\mu_0 I r}{2\pi a^2}$ inside a uniform wire of radius $a$ . A solenoid ( $n$ turns per meter): a rectangular loop gives a uniform interior $B = \mu_0 n I$ and essentially zero field outside. A toroid ( $N$ total turns): $B = \dfrac{\mu_0 N I}{2\pi r}$ inside, zero outside.

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What this topic is asking
Ampere's law
Choosing an Amperian loop
The three standard cases
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What this topic is asking

The College Board (Topic 12.4) wants you to apply Ampere's law to find the magnetic field of current distributions with enough symmetry: a long straight wire, a solenoid, and a toroid. Ampere's law is the magnetic counterpart of Gauss's law, trading a hard Biot-Savart integral for a one-line calculation when the symmetry is right.

Ampere's law

The law is always valid, but it is useful for finding $B$ only when symmetry lets you pull $B$ out of the integral. Where it applies, it is far quicker than the Biot-Savart integral, just as Gauss's law shortcuts the Coulomb integral.

Choosing an Amperian loop

The skill is choosing a closed loop, the Amperian loop, matched to the symmetry so that on each part either:

$\vec{B}$ is constant in magnitude and parallel to the loop (so $\vec{B}\cdot d\vec{\ell} = B\,d\ell$ and the integral is $B\times\text{length}$ ), or
$\vec{B}$ is perpendicular to the loop (so $\vec{B}\cdot d\vec{\ell} = 0$ on that part).

This turns $\oint\vec{B}\cdot d\vec{\ell}$ into $B\times(\text{path length with field}) = \mu_0 I_{enc}$ , which you solve for $B$ .

The three standard cases

Geometry	Amperian loop	Field result
Long straight wire	Circle of radius $r$	$B = \dfrac{\mu_0 I}{2\pi r}$ (outside)
Solenoid ( $n$ turns/m)	Rectangle through the wall	$B = \mu_0 n I$ (uniform inside)
Toroid ( $N$ turns)	Circle of radius $r$ inside	$B = \dfrac{\mu_0 N I}{2\pi r}$

For the solenoid, the field outside is negligible and the interior field is uniform; this is the standard way to make a strong, controllable magnetic field.

Field of a solenoid from Ampere's law

A solenoid has $500$ turns over a length of $0.25$ m and carries $2.0$ A. Find the magnetic field inside. Use $\mu_0 = 4\pi\times10^{-7}$ T m/A.

step 1 Find the turns per meter

$n = \dfrac{N}{L} = \dfrac{500}{0.25} = 2000$ turns per meter.

step 2 Choose the Amperian loop

A rectangle with one long side inside the solenoid (parallel to $\vec{B}$ ) and one outside (where $B \approx 0$ ); the short sides are perpendicular to $\vec{B}$ and contribute nothing.

step 3 Apply Ampere's law

Only the inner long side contributes: $\oint\vec{B}\cdot d\vec{\ell} = B\ell = \mu_0 I_{enc} = \mu_0 (n\ell)I$ , so $B = \mu_0 n I$ .

step 4 Substitute

$B = (4\pi\times10^{-7})(2000)(2.0) = 5.0\times10^{-3}$ T, uniform throughout the interior.

Try this

Q1. State Ampere's law in symbols. [1 point]

Cue. $\oint\vec{B}\cdot d\vec{\ell} = \mu_0 I_{enc}$ .

Q2. A solenoid has $n = 1500$ turns/m and carries $3.0$ A. Find the interior field ( $\mu_0 = 4\pi\times10^{-7}$ ). [2 points]

Cue. $B = \mu_0 n I = (4\pi\times10^{-7})(1500)(3.0) = 5.7\times10^{-3}$ T.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)1 marksSection I (multiple choice). The magnetic field inside a long ideal solenoid with

n

turns per meter carrying current

I

is (A)

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

(B)

\mu_0 n I

(C)

\dfrac{\mu_0 I}{2R}

(D) zero. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the solenoid field. The answer is (B).

Ampere's law applied to a rectangular loop straddling the solenoid wall gives a uniform interior field $B = \mu_0 n I$ , independent of position inside. The trap is (A), the straight-wire field, or (C), the loop-center field; the solenoid result is $\mu_0 n I$ , with $n$ the turns per unit length.

AP 2024 (style)6 marksSection II (FRQ, derivation). A long straight wire of radius

a

carries a current

I

uniformly distributed over its cross-section. (a) State Ampere's law. (b) Use it to derive the field outside the wire (

r > a

). (c) Use it to derive the field inside the wire (

r < a

Show worked answer →

A 6-point FRQ applying Ampere's law inside and outside a wire.

(a) Statement (1 point): $\oint\vec{B}\cdot d\vec{\ell} = \mu_0 I_{enc}$ .
(b) Outside (2 points): a circular Amperian loop of radius $r > a$ encloses all of $I$ . By symmetry $B$ is constant and tangent: $B(2\pi r) = \mu_0 I$ , so $B = \dfrac{\mu_0 I}{2\pi r}$ .
(c) Inside (3 points): a loop of radius $r < a$ encloses the fraction $I_{enc} = I\dfrac{\pi r^2}{\pi a^2} = I\dfrac{r^2}{a^2}$ . So $B(2\pi r) = \mu_0 I\dfrac{r^2}{a^2}$ , giving $B = \dfrac{\mu_0 I r}{2\pi a^2}$ , which grows linearly with $r$ .

Markers reward the statement, the outside result, and the enclosed-fraction inside result.

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Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)