Physics C: E&M · Unit 4: Magnetic Fields · 14 min read · Updated 2026-05-11
Ampère's Law — AP Physics C: Electricity and Magnetism
AP Physics C: Electricity and Magnetism · Unit 4: Magnetic Fields · 14 min read
1. The Ampère-Maxwell Law (Integral Form)★★☆☆☆⏱ 4 min
Ampère's Law relates the line integral of the magnetic field around a closed 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, avoiding messy Biot-Savart integrals.
On the left, $\oint$ is a closed line integral around the Amperian loop, $\vec{B}$ is the magnetic field vector, and $d\vec{l}$ is the infinitesimal tangential length vector. On the right, $\mu_0 = 4\pi \times 10^{-7} \text{ T·m/A}$ is the permeability of free space, $I_{\text{enclosed}}$ is net current through the loop, and $\epsilon_0 \frac{d\Phi_E}{dt}$ is the displacement current term 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 your right fingers along the integration direction; your thumb points to the positive current direction.
Exam tip: Always do a symmetry analysis before choosing an Amperian loop. If B cannot be pulled out of the integral, Ampère’s Law will not simplify to a solution for B.
2. Magnetic Fields of Solenoids★★★☆☆⏱ 3 min
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.
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 parallel to the axis, and two short sides perpendicular to the axis.
Exam tip: The $B = \mu_0 n I$ result only applies to infinite solenoids. AP questions often test that the field at the ends of a finite solenoid is half the center value.
3. Cylindrical Symmetry and Coaxial Cables★★★☆☆⏱ 4 min
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, simplifying the line integral to $B(2\pi r)$ for all regions.
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: $I_{\text{enclosed}} = I \frac{r^2}{R^2}$.
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.
4. Additional Exam-Style Worked Examples★★★★☆⏱ 3 min
Common Pitfalls
Why: The infinite solenoid result relies on infinite-length symmetry that cancels external fields and evens the internal field, which does not apply at the ends of finite solenoids.
Why: Students often focus on magnitude and ignore the sign convention for the line integral, leading to incorrect net enclosed current.
Why: Ampère's Law is always true for any closed loop, but simplifying the integral only works when $B$ is constant and parallel to $d\vec{l}$ everywhere on the loop.
Why: Students rush and forget that only current passing through the Amperian loop counts towards $I_{\text{enclosed}}$.
Why: Displacement current is omitted in most steady current examples, so students forget it is required when electric flux changes with time.