12.4 Ampère's Law

Ampere's law relates the magnetic field along a closed loop to the current passing through it: $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{enc}$. You use it to find the magnetic field in symmetric setups like long straight wires, solenoids, and cylindrical conductors by choosing a smart Amperian loop.

Why This Matters for the AP Physics C: E&M Exam

Unit 12 carries a lot of weight on the exam, and Ampere's law is one of the cleanest tools for finding magnetic fields fast. On multiple-choice and free-response questions, you may need to set up the line integral, pick an Amperian loop that matches the symmetry of the current, calculate the enclosed current, and solve for $B$. The exam also expects you to derive symbolic expressions and apply the right-hand rule to get field direction, so understanding both the math and the physical meaning pays off.

This topic connects directly to the symmetry reasoning you built with Gauss's law for electric fields. When a setup has enough symmetry, Ampere's law turns a hard integral into simple algebra, which is exactly the kind of efficient problem solving the exam rewards.

Key Takeaways

• Ampere's law in integral form is $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{enc}$, where $I_{enc}$ is only the current passing through the loop.
• For a long straight wire, the field is $B_{wire} = \dfrac{\mu_0}{2\pi}\dfrac{I}{r}$, forming concentric circles set by the right-hand rule.
• For a long ideal solenoid, the inside field is $B_{sol} = \mu_0 n I$, uniform along the axis, with negligible field outside.
• The exam only expects quantitative Ampere's law work for symmetric cases: long straight wires, long solenoids, and conductive slabs or cylindrical conductors carrying current density.
• Use superposition to add the vector fields from multiple current sources.
• Ampere's law is Maxwell's fourth equation; with Maxwell's addition, a changing electric field also creates a magnetic field, but you are not expected to calculate with that term.

Magnetic Fields from Moving Charges

Ampere's Law and Magnetic Fields

Ampere's law ties electric currents to the magnetic fields they create. It is written as a line integral around a closed path:

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

• $\vec{B}$ is the magnetic field along the closed path, and $I_{enc}$ is the current passing through the area bounded by that path.
• $\mu_0$ is the permeability of free space, $4\pi \times 10^{-7}$ T·m/A.
• Only current that pierces the loop counts. Currents outside the loop do not add to $I_{enc}$, even though they still affect $\vec{B}$ at points along the loop.

For a long, straight current-carrying wire, Ampere's law gives:

$B_{\text{wire}} = \frac{\mu_0}{2\pi} \frac{I}{r}$

• $B_{\text{wire}}$ is the field strength a distance $r$ from the wire.
• The field forms concentric circles around the wire, with direction set by the right-hand rule (point your thumb along the current, and your fingers curl in the field direction).

For a long ideal solenoid, Ampere's law gives:

$B_{\text{sol}} = \mu_0 n I$

• $n$ is the number of turns per unit length.
• The field inside is uniform and parallel to the axis, and the field outside is treated as negligible. Unless told otherwise, assume solenoids are very long.

Choosing an Amperian Loop

An Amperian loop is an imaginary closed path you draw to apply Ampere's law. Picking it well is what makes the math easy.

• The loop is a closed path around a current-carrying conductor.
• Choose the loop shape to match the symmetry of the current so that $\vec{B}$ is either constant and parallel to $d\vec{\ell}$ or perpendicular to it.
• When $\vec{B}$ is constant in magnitude and parallel to the path, the integral simplifies to $B$ times the path length.
• Segments where $\vec{B}$ is perpendicular to $d\vec{\ell}$ contribute zero to the integral.

A reliable approach:

1. Identify the current distribution and its symmetry.
2. Choose an Amperian loop that uses that symmetry.
3. Determine which parts of the loop actually contribute to the integral.
4. Find the enclosed current $I_{enc}$.
5. Solve for the unknown magnetic field.

For a cylindrical conductor or conductive slab carrying a current density, remember that $I_{enc}$ is only the current inside your loop. Inside the conductor, that may be a fraction of the total current.

Superposition of Magnetic Fields

When several current-carrying conductors are present, the net magnetic field at a point is the vector sum of the individual fields.

• Each current source produces its own field.
• Add the fields as vectors: $\vec{B}_{total} = \vec{B}_1 + \vec{B}_2 + \vec{B}_3 + \dots$
• Direction matters as much as magnitude, so track field directions carefully with the right-hand rule.

Superposition applies to setups like multiple parallel wires, combinations of loops and straight segments, and current distributions in conductors.

Maxwell's Equations and Electromagnetism

Ampere's law is one of Maxwell's four equations describing electromagnetism. Maxwell extended it to include the effect of a changing electric field:

• Ampere's law: $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I$
• With Maxwell's addition: $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I + \mu_0 \varepsilon_0 \dfrac{d\Phi_E}{dt}$

The term $\mu_0 \varepsilon_0 \dfrac{d\Phi_E}{dt}$ shows that a changing electric field generates a magnetic field, much like a moving charge does. This completes the link between electricity and magnetism. For the exam, understand this idea conceptually, but you are not expected to calculate using the changing-electric-field term.

How to Use This on the AP Physics C: E&M Exam

Problem Solving

• Start by checking the symmetry. If the field is not symmetric enough for a clean loop, Ampere's law will not isolate $B$, and you may need the Biot-Savart law instead.
• Write the line integral, then use symmetry to pull $B$ out: for a circular loop concentric with a wire, $\oint \vec{B} \cdot d\vec{\ell} = B(2\pi r)$.
• Set that equal to $\mu_0 I_{enc}$ and solve. Keep units consistent and carry $\mu_0 = 4\pi \times 10^{-7}$ T·m/A.

Free Response

• Derive symbolic expressions before plugging in numbers when the question asks for a derivation.
• For a current density problem, write $I_{enc}$ as an integral of $J$ over the area inside the loop, then solve for $B$ both inside and outside the conductor.
• When you state a field direction, justify it with the right-hand rule, not just by naming the law.

Common Trap

• Do not confuse $I_{enc}$ with the total current. Only current passing through the loop counts.
• Watch the solenoid formula: $n$ is turns per unit length, not the total number of turns.

Common Misconceptions

• Ampere's law does not say $\vec{B}$ is zero where $I_{enc}$ is zero. It only says the line integral around the loop is zero. The field at points on the loop can still be nonzero due to outside currents.
• The field of a long straight wire falls off as $1/r$, not $1/r^2$. That $1/r^2$ behavior belongs to a single small current element in the Biot-Savart law, not the whole wire.
• Ampere's law is always true, but it is only useful for finding $B$ when there is enough symmetry. For asymmetric cases, you still need Biot-Savart.
• The field outside an ideal solenoid is treated as negligible, not because no field exists, but because the long-solenoid model approximates it as zero.
• In the solenoid formula, $B$ does not depend on the radius or on where you are inside, as long as you are well inside a long solenoid. The field is uniform.

Practice Problem 1: Magnetic Field of a Wire

Solution

Apply the formula for the magnetic field around a long, straight wire derived from Ampere's law:

$B_{\text{wire}} = \frac{\mu_0}{2\pi} \frac{I}{r}$

Given:

• Current $I = 5.0$ A
• Distance $r = 10$ cm = 0.10 m
• Permeability of free space $\mu_0 = 4\pi \times 10^{-7}$ T·m/A

Substituting these values:

$B_{\text{wire}} = \frac{4\pi \times 10^{-7}}{2\pi} \frac{5.0}{0.10}$

$B_{\text{wire}} = 2 \times 10^{-7} \times \frac{5.0}{0.10}$

$B_{\text{wire}} = 2 \times 10^{-7} \times 50$

$B_{\text{wire}} = 1.0 \times 10^{-5} \text{ T or } 10 \text{ μT}$

The magnetic field at 10 cm from the wire is 10 μT, directed tangentially around the wire according to the right-hand rule.

Practice Problem 2: Solenoid Magnetic Field

Solution

For a long solenoid, use the formula derived from Ampere's law:

$B_{\text{sol}} = \mu_0 n I$

Given:

• Number of turns per unit length $n = 200$ turns/m
• Current $I = 3.0$ A
• Permeability of free space $\mu_0 = 4\pi \times 10^{-7}$ T·m/A

Substituting these values:

$B_{\text{sol}} = 4\pi \times 10^{-7} \times 200 \times 3.0$

$B_{\text{sol}} = 4\pi \times 10^{-7} \times 600$

$B_{\text{sol}} = 7.54 \times 10^{-4} \text{ T or } 0.754 \text{ mT}$

The magnetic field inside the solenoid is 0.754 mT, directed along the axis of the solenoid.

Related AP Physics C: E&M Guides

• 12.2 Magnetism and Moving Charges
• 12.1 Magnetic Fields
• 12.3 Magnetic Fields of Current-Carrying Wires and the Biot-Savart Law