Ampère's Law

AP Physics C: Electricity and Magnetism· AP Physics C: E&M CED — Magnetic Fields· 14 min read

1. The Ampère-Maxwell Law (Integral Form)★★☆☆☆⏱ 4 min

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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.

\oint \vec{B} \cdot d\vec{l} = \mu_0 \left( I_{\text{enclosed}} + \epsilon_0 \frac{d\Phi_E}{dt} \right)

On the left, $\oint$ is a closed line integral around the Amperian loop, $B$ is the magnetic field vector, and $d l$ is the infinitesimal tangential length vector. On the right, $μ_{0} = 4 π \times 1 0^{- 7} T \cdotp m/A$ is the permeability of free space, $I_{enclosed}$ is net current through the loop, and $ϵ_{0} \frac{d Φ _{E}}{d t}$ is the displacement current term for time-varying electric fields.

For steady currents, $\frac{d Φ _{E}}{d t} = 0$ , so the equation reduces to the original Ampère's Law: $\oint B \cdot d l = μ_{0} I_{enclosed}$ . The sign of $I_{enclosed}$ follows the right-hand rule: curl your right fingers along the integration direction; your thumb points to the positive current direction.

📐 Worked Example

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.

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1. Symmetry analysis: The magnetic field circles the wire with constant magnitude at all points at the same radius $r$ . We choose a circular Amperian loop of radius $r$ centered on the wire, integrated counterclockwise to match the direction of $B$ from the right-hand rule.
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1. Evaluate the line integral: $B$ is parallel to $d l$ everywhere on the loop, and $B$ is constant, so
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$\oint \vec{B} \cdot d\vec{l} = B \oint dl = B(2\pi r)$
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1. Calculate enclosed current: The entire current $I$ is enclosed, and it points out of the page matching the positive direction, so $I_{enclosed} = I$ .
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1. Apply Ampère’s Law and solve for $B$ :
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$B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$

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

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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.

📐 Worked Example

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

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1. Choose the standard rectangular Amperian loop, with integration direction matching the direction of $B$ inside the solenoid.
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1. Evaluate the line integral: The two perpendicular sides have $B \cdot d l = 0$ , the outside side has $B = 0$ , so only the inside side contributes:
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$\oint \vec{B} \cdot d\vec{l} = B L$
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1. Calculate enclosed current: The number of enclosed turns is $n L$ , each carrying current $I$ in the positive direction, so $I_{enclosed} = n L I$ .
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1. Apply Ampère’s Law and solve:
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$B L = \mu_0 n L I \implies B = \mu_0 n I \quad (\text{inside})$
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Outside the solenoid, net enclosed current cancels out to zero, so $B = 0$ (outside).

Exam tip:

The $B = μ_{0} n I$ 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

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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 $B (2 π r)$ 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: $I_{enclosed} = I \frac{r ^{2}}{R ^{2}}$ .

📐 Worked Example

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 $r$ .

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1. For all regions, symmetry gives the line integral result:
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$\oint \vec{B} \cdot d\vec{l} = B(2\pi r)$
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1. Region 1: $r < R_{1}$ (inside inner wire): Enclosed current is $I_{enclosed} = I \frac{r ^{2}}{R _{1}^{2}}$ . Apply Ampère's Law:
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$B(2\pi r) = \mu_0 I \frac{r^2}{R_1^2} \implies B = \frac{\mu_0 I r}{2\pi R_1^2}$
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$B$ increases linearly with $r$ in this region.
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1. Region 2: $R_{1} < r < R_{2}$ (between conductors): All inner current is enclosed, so $I_{enclosed} = I$ :
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$B = \frac{\mu_0 I}{2\pi r}$
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$B$ decreases as $1/ r$ in this region.
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1. Region 3: $r > R_{2}$ (outside cable): Net enclosed current is $I - I = 0$ , so $B = 0$ 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.

4. Additional Exam-Style Worked Examples★★★★☆⏱ 3 min

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📐 Worked Example

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 $r = 0.075$ m (center of the cross-section)?

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We use a circular Amperian loop of radius $r = 0.075$ m concentric with the toroid. Symmetry gives the line integral $\oint B \cdot d l = B (2 π r)$ , and enclosed current is $N I = 200 \times 2 = 400$ A.
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Rearrange Ampère's Law to solve for $B$ :
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$B = \frac{\mu_0 N I}{2\pi r} = \frac{(4\pi \times 10^{-7})(400)}{2\pi (0.075)} \approx 5.3 \times 10^{-4}\ \text{T}$

📐 Worked Example

A long solid conducting cylinder of radius $R$ carries a non-uniform current density given by $J (r) = k r$ , where $k$ is a positive constant. (a) Show total current $I = \frac{2}{3} k π R^{3}$ ; (b) Find $B (r)$ for $r < R$ ; (c) Find $B (r)$ for $r > R$ .

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(a) Total current is the integral of $J$ over cross-sectional area. A thin ring of radius $r$ has area $d A = 2 π r d r$ , so $d I = J d A = 2 π k r^{2} d r$ . Integrate from 0 to $R$ :
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$I = \int_0^R 2\pi k r^2 dr = 2\pi k \frac{R^3}{3} = \frac{2}{3} k \pi R^3$
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(b) For $r < R$ , enclosed current is:
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$I_{\text{enclosed}} = \int_0^r 2\pi k r'^2 dr' = \frac{2}{3} \pi k r^3$
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Apply Ampère's Law: $B (2 π r) = μ_{0} I_{enclosed}$ , so:
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$B = \frac{\mu_0 k r^2}{3} \quad (r < R)$
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(c) For $r > R$ , all current is enclosed, so $I_{enclosed} = I = \frac{2}{3} k π R^{3}$ . Apply Ampère's Law:
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$B(2\pi r) = \mu_0 \left(\frac{2}{3} k \pi R^3\right) \implies B = \frac{\mu_0 k R^3}{3 r} \quad (r > R)$

📐 Worked Example

A coaxial power cable carries 150 A, inner radius 2.0 cm, outer conductor at 6.0 cm, with equal opposite currents. What is $B$ at 4.0 cm from the axis? Will it interfere with a sensor that has a $5 \times 1 0^{- 4}$ T detection threshold?

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The point of interest is $r = 0.04$ m, between the two conductors, so all inner current is enclosed.
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Use the Ampère's Law result for coaxial cables between conductors:
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$B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi \times 10^{-7})(150)}{2\pi (0.04)} = 7.5 \times 10^{-4}\ \text{T}$
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This is larger than the $5 \times 1 0^{- 4}$ T threshold, so the cable will interfere with the sensor.

5. Common Pitfalls

Wrong move:

Using $B = μ_{0} n I$ for the magnetic field at the end of a finite solenoid.

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.

Correct move:

For a long finite solenoid, state that $B = μ_{0} n I$ is only valid near the center, and the field at the ends is half the center value.

Wrong move:

Adding all enclosed currents regardless of their direction relative to the integration sign convention.

Why:

Students often focus on magnitude and ignore the sign convention for the line integral, leading to incorrect net enclosed current.

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:

Ampère's Law is always true for any closed loop, but simplifying the integral only works when $B$ is constant and parallel to $d l$ everywhere on the loop.

Correct move:

Always confirm that $B$ has constant magnitude and is parallel/antiparallel to $d l$ everywhere on the loop before pulling $B$ out of the integral.

Wrong move:

Using the full total current $I$ as $I_{enclosed}$ when the Amperian loop is inside a current-carrying wire.

Why:

Students rush and forget that only current passing through the Amperian loop counts towards $I_{enclosed}$ .

Correct move:

For any point inside a uniform current distribution, calculate $I_{enclosed}$ as total current 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 $ϵ_{0} \frac{d Φ _{E}}{d t}$ to the total enclosed current.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Ampère-Maxwell Integral Law	$\oint B \cdot d l = μ_{0} (I_{enc} + ϵ_{0} \frac{d Φ _{E}}{d t})$	General form for any closed loop, includes displacement current
Ampère's Law (Steady Current)	$\oint B \cdot d l = μ_{0} I_{enclosed}$	Use for time-independent currents, $d Φ_{E} / d t = 0$
Infinite straight wire	$B = \frac{μ _{0} I}{2 π r}$	$r$ = distance from wire axis, B circles the wire
Uniform solid infinite wire ( $r < R$ )	$B = \frac{μ _{0} I r}{2 π R ^{2}}$	$R$ = wire radius, uniform current distribution
Infinite solenoid (inside)	$B = μ_{0} n I$	$n$ = turns per unit length, uniform B along axis
Infinite solenoid (outside)	$B = 0$	Only applies to infinite length approximation
Toroid at radius $r$	$B = \frac{μ _{0} N I}{2 π r}$	$N$ = total number of turns
Coaxial cable (between conductors)	$B = \frac{μ _{0} I}{2 π r}$	Equal opposite currents on inner/outer
Coaxial cable (outside)	$B = 0$	Net enclosed current is zero

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · FRQ
Coaxial cable magnetic field calculation
2022 · MCQ
Solenoid field magnitude question

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 (Electromagnetic Induction). Without mastering symmetry analysis and Amperian loop selection from this sub-topic, you will not be able to solve inductance problems or handle Faraday's Law applications that require finding magnetic flux from symmetric current distributions. Ampère's Law is one of the four Maxwell equations that describe all electromagnetic phenomena.