Ampère's Law — AP Physics C: E&M Study Guide

For: AP Physics C: E&M candidates sitting AP Physics C: E&M.

Covers: The Ampère-Maxwell law, integral form of Ampère’s Law, right-hand rule for enclosed current, symmetry analysis for Amperian loops, and magnetic fields from infinite wires, solenoids, toroids, and coaxial cables.

You should already know: Line integrals of vector fields. Right-hand rule for magnetic field direction. Magnetic field from infinitesimal current elements via the Biot-Savart law.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: E&M style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Ampère's Law?

Ampère's Law is a fundamental relation between electric current and the magnetic field it produces, analogous to Gauss’s Law for electric fields. Just as Gauss’s Law uses symmetry to simplify complex electric field calculations, Ampère’s Law lets us find magnetic fields for highly symmetric current distributions without messy Biot-Savart integrals. The AP Physics C E&M Course and Exam Description (CED) lists Ampère’s Law as making up ~15-20% of the Unit 4 (Magnetic Fields) exam weight, which corresponds to ~6-8% of the total E&M exam score. It appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with Gauss’s Law or Faraday’s Law in longer multi-part FRQs. The full exam-relevant form (including displacement current for time-varying systems) is the Ampère-Maxwell law, though the simpler original Ampère’s Law applies for all steady (time-independent) current scenarios. Ampère’s Law relates the line integral of the magnetic field around a closed loop (called an Amperian loop) to the net current enclosed by that loop. It is always true for any closed loop, but only simplifies to a solvable expression for symmetric current distributions.

2. The Integral Form of Ampère-Maxwell Law

The core equation for Ampère-Maxwell Law in integral form is: $$\oint \vec{B}\cdot d\vec{l}={\mu }_0\left(I_{\text{enclosed}}+{ϵ}_0\frac{d{\Phi }_E}{dt}\right)$$ On the left side, $\oint$ indicates a closed line integral around the Amperian loop, $\vec{B}$ is the magnetic field vector at each point on the loop, and $d\vec{l}$ is the infinitesimal length vector tangential to the loop at each point. On the right side, ${\mu }_0=4\pi \times 10^{-7}\text{ Tcdotpm/A}$ is the permeability of free space, $I_{\text{enclosed}}$ is the net current passing through the area bounded by the Amperian loop, ${ϵ}_0$ is the permittivity of free space, and $\frac{d{\Phi }_E}{dt}$ is the rate of change of electric flux through the bounded area. The second term ${ϵ}_0\frac{d{\Phi }_E}{dt}$ is called displacement current, required for time-varying electric fields. For steady currents, $\frac{d{\Phi }_E}{dt}=0$, so the equation reduces to the original Ampère’s Law: $\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_{\text{enclosed}}$.

The sign of $I_{\text{enclosed}}$ follows the right-hand rule: curl the fingers of your right hand along the direction of integration around the Amperian loop; your thumb points to the positive direction for current. Any current pointing opposite to your thumb is subtracted from the total.

Worked Example

Problem: Use Ampère’s Law to derive the magnetic field at a distance $r$ from an infinitely long straight wire carrying total current $I$ out of the page.

1. Symmetry analysis: The magnetic field circles the wire with constant magnitude at all points at the same radius $r$, so we choose a circular Amperian loop of radius $r$ centered on the wire, integrated counterclockwise (matches the direction of $\vec{B}$ from the right-hand rule).
2. Evaluate the line integral: $\vec{B}$ is parallel to $d\vec{l}$ everywhere on the loop, and $B$ is constant, so $\oint \vec{B}\cdot d\vec{l}=B\oint dl=B(2\pi r)$.
3. Calculate enclosed current: The entire current $I$ is enclosed, and it points out of the page, which matches the positive direction from our right-hand rule, so $I_{\text{enclosed}}=I$.
4. Apply Ampère’s Law: $B(2\pi r)={\mu }_{0}I$, so $B=\frac{{\mu }_{0}I}{2\pi r}$, matching the Biot-Savart result with far less work.

Exam tip: Always do a symmetry analysis before choosing an Amperian loop. If B cannot be pulled out of the integral (because its magnitude varies or direction is inconsistent relative to $d\vec{l}$), Ampère’s Law will not simplify to a solution for B.

3. Amperian Loops for Solenoids

A solenoid is a long coil of wire wrapped around a core, used to produce a uniform magnetic field, and is one of the most common geometries tested on the AP exam. For an ideal infinite solenoid, symmetry tells us the magnetic field is uniform and parallel to the solenoid axis inside the coil, and zero outside the coil (all non-axial components cancel out for infinite length). The standard Amperian loop for a solenoid is a rectangle with one side of length $L$ inside the solenoid parallel to the axis, one side of length $L$ outside the solenoid parallel to the axis, and two short sides perpendicular to the axis.

Worked Example

Problem: An infinite solenoid has $n$ turns per unit length, with each turn carrying current $I$. Find the magnetic field inside and outside the solenoid.

1. Choose the rectangular Amperian loop described above, with integration direction matching the direction of $\vec{B}$ inside the solenoid.
2. Evaluate the line integral: The two sides perpendicular to the axis have $\vec{B}\cdot d\vec{l}=0$, because $\vec{B}$ is parallel to the axis. The side outside the solenoid has $B=0$, so it contributes nothing. Only the inside side contributes, giving $\oint \vec{B}\cdot d\vec{l}=BL$.
3. Calculate enclosed current: The number of turns enclosed by the loop is $nL$, each carrying current $I$ in the positive direction, so $I_{\text{enclosed}}=nLI$.
4. Apply Ampère’s Law: $BL={\mu }_{0}nLI$, cancel $L$ from both sides to get $B={\mu }_{0}nI$ inside the solenoid. Outside the solenoid, the net enclosed current is $nLI-nLI=0$ (current goes up each turn, returns the other way), so $B=0$ outside.

Exam tip: The $B={\mu }_{0}nI$ result only applies to infinite solenoids. AP questions often test whether you know the field at the ends of a finite solenoid is half the center value, not equal to ${\mu }_{0}nI$.

4. Cylindrical Current Distributions and Coaxial Cables

Cylindrically symmetric current distributions (solid wires, coaxial cables) are a staple of AP FRQs, because they require applying Ampère’s Law across multiple regions with different enclosed currents. For any cylindrically symmetric current aligned along an axis, we always use a circular Amperian loop centered on the axis, so $\vec{B}$ is tangential to the loop and constant in magnitude, simplifying the line integral to $B(2\pi r)$ regardless of the region. The only thing that changes between regions is the net enclosed current. For a uniform current distribution in a solid wire of radius $R$, the enclosed current at radius $r<R$ is proportional to the area enclosed: $I_{\text{enclosed}}=I\frac{\pi r^2}{\pi R^2}=I\frac{r^2}{R^2}$.

Worked Example

Problem: A coaxial cable has an inner solid wire of radius $R_1$ carrying total current $I$ out of the page, and an outer thin cylindrical shell of radius $R_2$ carrying total current $I$ into the page. Find $B$ as a function of $r$ for all $r$.

1. For all regions, the line integral is $\oint \vec{B}\cdot d\vec{l}=B(2\pi r)$ from symmetry.
2. Region 1: $r<R_1$ (inside the inner wire): Enclosed current is $I_{\text{enclosed}}=I\frac{r^2}{R_1^2}$. Ampère’s Law gives $B(2\pi r)={\mu }_{0}I\frac{r^2}{R_1^2}$, so $B=\frac{{\mu }_{0}Ir}{2\pi R_1^2}$, which increases linearly with $r$.
3. Region 2: $R_1<r<R_2$ (between inner wire and outer shell): All of the inner current is enclosed, so $I_{\text{enclosed}}=I$. This gives $B=\frac{{\mu }_{0}I}{2\pi r}$, which decreases as $1/r$.
4. Region 3: $r>R_2$ (outside the cable): Net enclosed current is $I-I=0$, so $B(2\pi r)=0\to B=0$ outside the cable.

Exam tip: Always check the direction of current when calculating net enclosed current. The outer current in a coaxial cable is almost always opposite the inner current, leading to zero field outside the cable.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using $B={\mu }_{0}nI$ for the magnetic field at the end of a finite solenoid. Why: The infinite solenoid result is easy to memorize, and students forget it relies on infinite-length symmetry that cancels external fields and evens the internal field. Correct move: If asked for a finite solenoid, state that the ${\mu }_{0}nI$ result is only a good approximation near the center, and the field at the ends is half the center value for a long finite solenoid.
• Wrong move: Adding all currents enclosed by the Amperian loop regardless of their direction. Why: Students focus on magnitude of B and ignore the sign convention for the line integral. Correct move: Always apply the right-hand rule to assign positive/negative signs to each enclosed current before summing.
• Wrong move: Pulling B out of the line integral for a non-symmetric Amperian loop. Why: Students assume Ampère’s Law is valid only for symmetric loops, so they incorrectly simplify the integral for any loop. Correct move: Always confirm B has constant magnitude and is parallel/antiparallel to $d\vec{l}$ everywhere on the loop before pulling it out of the integral.
• Wrong move: Using the full total current I for enclosed current when the Amperian loop is inside a current-carrying wire. Why: Students rush and forget that only current inside the loop counts. Correct move: For any $r$ inside a uniform current distribution, calculate $I_{\text{enclosed}}$ as $I$ times the ratio of the area inside the Amperian loop to the total cross-sectional area of the conductor.
• Wrong move: Omitting the displacement current term when analyzing a charging capacitor. Why: Displacement current is omitted in most steady current examples, so students forget it is required when electric flux changes with time. Correct move: If the problem involves a time-varying electric field, always add the displacement current term to the total enclosed current.

6. Practice Questions (AP Physics C: E&M Style)

Question 1 (Multiple Choice)

A toroid with 200 total turns carries a current of 2 A. The inner radius of the toroid is 5 cm, and the outer radius is 10 cm. What is the magnitude of the magnetic field at the center of the toroid's cross-section (r = 7.5 cm = 0.075 m)? A) $1.1\times 10^{-3}\text{T}$ B) $5.3\times 10^{-4}\text{T}$ C) $1.1\times 10^{-4}\text{T}$ D) $0\text{T}$

Worked Solution: We use a circular Amperian loop of radius $r=0.075\text{m}$ concentric with the toroid. The line integral simplifies to $B(2\pi r)$, and the enclosed current is $NI=200\times 2=400\text{A}$. Rearranging Ampère's Law gives $B=\frac{{\mu }_{0}NI}{2\pi r}=\frac{(4\pi \times 10^{-7})(400)}{2\pi (0.075)}\approx 5.3\times 10^{-4}\text{T}$. Option A uses half the correct radius, option C omits the 200 total turns, and option D incorrectly cancels current like a coaxial cable. The correct answer is B.

Question 2 (Free Response)

A long solid conducting cylinder of radius $R$ carries a non-uniform current density given by $J(r)=kr$, where $k$ is a positive constant and $r$ is the distance from the cylinder axis. Current flows along the cylinder axis. (a) Show that the total current $I$ carried by the cylinder is $I=\frac{2}{3}k\pi R^3$. (b) Use Ampère's Law to find the magnetic field $B(r)$ for $r<R$. (c) Use Ampère's Law to find the magnetic field $B(r)$ for $r>R$, and describe how $B$ changes with $r$ for $0\le r\le 2R$.

Worked Solution: (a) Total current is the integral of $J$ over the cross-sectional area. For a thin ring of radius $r$, thickness $dr$, $dA=2\pi rdr$, so $dI=JdA=(kr)(2\pi rdr)=2\pi k{r}^{2}dr$. Integrate from $0$ to $R$: $I={\int }_0^{R}2\pi k{r}^{2}dr=2\pi k\frac{R^3}{3}=\frac{2}{3}k\pi R^3$, which matches the required result. (b) For $r<R$, enclosed current is $I_{\text{enclosed}}={\int }_0^{r}2\pi k{r}^{\prime 2}d{r}^{\prime }=\frac{2}{3}\pi k{r}^3$. Ampère's Law: $B(2\pi r)={\mu }_0{I}_{\text{enclosed}}={\mu }_0\left(\frac{2}{3}\pi k{r}^3\right)$. Cancel terms to get $B=\frac{{\mu }_{0}k{r}^2}{3}$ for $r<R$. (c) For $r>R$, $I_{\text{enclosed}}=I=\frac{2}{3}k\pi R^3$. Ampère's Law: $B(2\pi r)={\mu }_0\left(\frac{2}{3}k\pi R^3\right)$, so $B=\frac{{\mu }_{0}k{R}^3}{3r}$ for $r>R$. $B$ increases quadratically from $0$ at $r=0$ to a maximum of $\frac{{\mu }_{0}k{R}^2}{3}$ at $r=R$, then decreases inversely with $r$ for $r>R$.

Question 3 (Application / Real-World Style)

A power distribution coaxial cable carries 150 A of current from an offshore wind turbine to shore. The inner conductor has a radius of 2.0 cm, and the outer conductor is at a radius of 6.0 cm, with equal current flowing opposite directions on the inner and outer conductors. What is the magnetic field magnitude at a point 4.0 cm from the cable axis, between the two conductors? A navigation sensor placed near the cable has a detection threshold of $5\times 10^{-4}\text{T}$. Will the cable's magnetic field interfere with the sensor?

Worked Solution: The point of interest is at $r=0.04\text{m}$, between $R_1=0.02\text{m}$ and $R_2=0.06\text{m}$, so all of the inner current is enclosed. Using Ampère's Law for cylindrical symmetry: $B=\frac{{\mu }_{0}I}{2\pi r}=\frac{(4\pi \times 10^{-7}\text{Tcdotpm/A})(150\text{A})}{2\pi (0.04\text{m})}=7.5\times 10^{-4}\text{T}$. This field is larger than the sensor's $5\times 10^{-4}\text{T}$ detection threshold, so the cable's magnetic field will interfere with the navigation sensor.

7. Quick Reference Cheatsheet

8. What's Next

Ampère's Law is the foundational prerequisite for all subsequent topics involving magnetic fields and electromagnetic induction in AP Physics C: E&M. Next you will apply Ampère's Law to analyze displacement current in charging capacitors, which is critical for understanding Maxwell's equations, the unification of electricity and magnetism, and electromagnetic waves. You will also use Ampère's Law to derive the inductance of solenoids and coaxial cables, a core topic in Unit 5 (Electromagnetism). Without mastering symmetry analysis and Amperian loop selection from this chapter, you will not be able to solve inductance problems or handle Faraday's Law applications that require finding magnetic flux from symmetric current distributions. The big-picture role of Ampère's Law is as one of the four Maxwell equations that describe all electromagnetic phenomena.

• Displacement Current
• Faraday's Law of Induction
• Inductance
• Maxwell's Equations