AP Physics C: Electricity and Magnetism
Magnetic Forces and Fields
Ampère’s Law
Learning Objectives

By the end of this lesson, students should be able to:

  • State Ampère’s Law.

  • Explain the relationship between electric currents and magnetic fields.

  • Apply Ampère’s Law to symmetric current distributions.

  • Calculate magnetic fields of long straight wires, solenoids, and toroids.

  • Distinguish between Ampère’s Law and the Biot–Savart Law.

  • Solve AP Physics C problems involving Ampère’s Law.

Introduction
Why Ampère’s Law?

In the previous lesson, we studied the Biot–Savart Law:

$$
d\vec{B}=\frac{\mu_0}{4\pi}\frac{I,d\vec{\ell}\times\hat{r}}{r^2}
$$

Although the Biot–Savart Law is completely general, it often requires difficult integration.

For systems with high symmetry, a much simpler method exists:

$$
\text{Ampère’s Law}
$$

Ampère’s Law plays a role in magnetism similar to the role of Gauss’s Law in electrostatics.

Historical Background
André-Marie Ampère

Ampère’s Law is named after

André-Marie Ampère

who made fundamental discoveries concerning electric currents and magnetism.

His work helped establish the foundation of electromagnetism.

Statement of Ampère’s Law
Integral Form

Ampère’s Law states:

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

where:

  • (\vec{B}) = magnetic field

  • (d\vec{\ell}) = small path element

  • (I_{enc}) = current enclosed by the path

  • (\mu_0) = permeability of free space

Meaning of the Integral

The symbol:

$$
\oint
$$

indicates a closed-path integral.

The path is called an:

$$
\text{Amperian Loop}
$$

The loop may be any closed curve.

Permeability of Free Space
Constant Value

The permeability of free space is:

$$
\mu_0=4\pi\times10^{-7};T\cdot m/A
$$

This constant appears throughout magnetic field calculations.

Physical Meaning
Connection Between Current and Magnetism

Ampère’s Law shows that electric currents generate magnetic fields.

The larger the enclosed current:

$$
I_{enc}
$$

the stronger the magnetic circulation around the loop.

Comparison with Gauss’s Law

Gauss’s Law:

$$
\oint \vec{E}\cdot d\vec{A}=\frac{Q_{enc}}{\varepsilon_0}
$$

Ampère’s Law:

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

Comparison:

Electrostatics Magnetism
Charge creates electric fields Current creates magnetic fields
Gauss’s Law Ampère’s Law
Surface integral Line integral
When Ampère’s Law is Useful
Requirement of Symmetry

Ampère’s Law is most useful when the magnetic field has strong symmetry.

Examples:

  • Infinite straight wire

  • Long solenoid

  • Toroid

Without symmetry, the Biot–Savart Law is usually preferred.

Application 1: Infinite Straight Wire
Choosing an Amperian Loop

Consider a long straight wire carrying current (I).

Because of cylindrical symmetry:

  • Magnetic field lines form circles.

  • Magnetic field magnitude is constant on a circle centered on the wire.

Choose a circular Amperian loop of radius (r).

Applying Ampère’s Law

Because (\vec{B}) is tangent to the loop:

$$
\oint \vec{B}\cdot d\vec{\ell}=B\oint d\ell
$$

The circumference is:

$$
2\pi r
$$

Thus:

$$
B(2\pi r)=\mu_0 I
$$

Solving:

$$
B=\frac{\mu_0 I}{2\pi r}
$$

Result

The magnetic field decreases according to:

$$
B\propto\frac{1}{r}
$$

This matches the result derived from the Biot–Savart Law.

Application 2: Long Solenoid
Solenoid Geometry

A solenoid is a long coil of wire.

Inside the solenoid:

  • Magnetic field is nearly uniform.

  • Magnetic field lines are parallel.

Outside:

  • Magnetic field is approximately zero.

Applying Ampère’s Law

Choose a rectangular Amperian loop.

Only the section inside the solenoid contributes significantly.

Ampère’s Law becomes:

$$
BL=\mu_0(nL)I
$$

where:

  • (n) = turns per unit length

Solving:

$$
B=\mu_0 n I
$$

Result

The magnetic field inside an ideal solenoid is:

$$
B=\mu_0 n I
$$

This is one of the most important applications of Ampère’s Law.

Application 3: Toroid
What is a Toroid?

A toroid is a solenoid bent into a circular ring.

Examples include:

  • Transformers

  • Inductors

  • Magnetic storage devices

Applying Ampère’s Law

Choose a circular Amperian loop of radius (r).

The enclosed current is:

$$
NI
$$

where:

  • (N) = total number of turns

Ampère’s Law gives:

$$
B(2\pi r)=\mu_0 NI
$$

Thus:

$$
B=\frac{\mu_0 NI}{2\pi r}
$$

Important Feature

The magnetic field is concentrated inside the toroid.

Outside the toroid:

$$
B\approx0
$$

Current Enclosed by the Loop
Key Idea

Only currents enclosed by the Amperian loop contribute to:

$$
I_{enc}
$$

Currents outside the loop do not affect the right side of Ampère’s Law.

Example

If an Amperian loop encloses:

$$
I_1
$$

but not

$$
I_2
$$

then:

$$
I_{enc}=I_1
$$

Only enclosed currents matter.

Example 1
Magnetic Field Near a Straight Wire

A wire carries:

$$
I=15A
$$

Find the magnetic field at:

$$
r=0.050m
$$

from the wire.

Solution

Use:

$$
B=\frac{\mu_0 I}{2\pi r}
$$

Substitute:

$$
B=\frac{(4\pi\times10^{-7})(15)}{2\pi(0.050)}
$$

$$
B=6.0\times10^{-5}T
$$

Answer

$$
B=6.0\times10^{-5}T
$$

Example 2
Solenoid Field

A solenoid contains:

$$
n=1200
;turns/m
$$

and carries:

$$
I=3.0A
$$

Find the magnetic field inside.

Solution

Use:

$$
B=\mu_0 n I
$$

Substitute:

$$
B=(4\pi\times10^{-7})(1200)(3.0)
$$

$$
B=4.52\times10^{-3}T
$$

Answer

$$
B=4.52\times10^{-3}T
$$

Example 3
Toroid Field

A toroid contains:

$$
N=400
$$

turns and carries:

$$
I=2.0A
$$

Find the magnetic field at:

$$
r=0.10m
$$

inside the toroid.

Solution

Use:

$$
B=\frac{\mu_0 NI}{2\pi r}
$$

Substitute:

$$
B=\frac{(4\pi\times10^{-7})(400)(2.0)}{2\pi(0.10)}
$$

$$
B=1.6\times10^{-3}T
$$

Answer

$$
B=1.6\times10^{-3}T
$$

Ampère’s Law vs. Biot–Savart Law
Biot–Savart Law

$$
d\vec{B}=\frac{\mu_0}{4\pi}\frac{I,d\vec{\ell}\times\hat{r}}{r^2}
$$

Advantages:

  • Works for any geometry.

Disadvantages:

  • Often requires integration.

Ampère’s Law

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

Advantages:

  • Very simple.

Disadvantages:

  • Useful only when symmetry exists.

Common AP Exam Mistakes
Mistake 1

Using total current instead of enclosed current.

Ampère’s Law requires:

$$
I_{enc}
$$

not all nearby currents.

Mistake 2

Applying Ampère’s Law to asymmetric situations.

Without symmetry, Ampère’s Law may not simplify the problem.

Mistake 3

Choosing a poor Amperian loop.

Always choose a loop that matches the symmetry of the magnetic field.

AP Free-Response Strategy
Look for Symmetry

If the problem involves:

  • Long straight wires

  • Solenoids

  • Toroids

consider Ampère’s Law immediately.

Choose the Correct Loop

Your choice of Amperian loop often determines whether the problem becomes simple or difficult.

Match the loop to the symmetry of the field.

Memorize Core Results

Straight wire:

$$
B=\frac{\mu_0 I}{2\pi r}
$$

Solenoid:

$$
B=\mu_0 n I
$$

Toroid:

$$
B=\frac{\mu_0 NI}{2\pi r}
$$

These equations appear frequently on AP Physics C exams.

Summary
Key Takeaways

  • Ampère’s Law relates magnetic fields to enclosed current.

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

  • Ampère’s Law is the magnetic analogue of Gauss’s Law.

  • For a long straight wire:

$$
B=\frac{\mu_0 I}{2\pi r}
$$

  • For an ideal solenoid:

$$
B=\mu_0 n I
$$

  • For a toroid:

$$
B=\frac{\mu_0 NI}{2\pi r}
$$

  • Only enclosed current contributes to:

$$
I_{enc}
$$

  • Ampère’s Law is most useful when strong symmetry exists.

  • Ampère’s Law and the Biot–Savart Law together form the foundation for analyzing magnetic fields generated by electric currents.

PreviousNext