Ampere's Law Worked Examples Cheat Sheet

A printable reference covering Ampere's law, magnetic fields of wires, solenoids, toroids, symmetry, and enclosed current for grades 11-12.

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Ampere's law connects magnetic fields to the electric current enclosed by a closed path called an Amperian loop. This cheat sheet helps students choose the right loop, identify symmetry, and solve common worked examples for wires, solenoids, and toroids. It is especially useful because many mistakes come from using the formula before checking whether the magnetic field is constant along the path. The core equation is $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$ , where the left side adds the magnetic field around a closed loop. In highly symmetric cases, this becomes $B(2\pi r) = \mu_0 I_{\text{enc}}$ for a circular loop around a long straight wire. For an ideal solenoid, the field is approximately $B = \mu_0 n I$ , and for a toroid it is $B = \frac{\mu_0 N I}{2\pi r}$ inside the core.

Key Facts

Ampere's law is $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$ , where $I_{\text{enc}}$ is the net current passing through the loop.
For a long straight wire, a circular Amperian loop gives $B(2\pi r) = \mu_0 I$ , so $B = \frac{\mu_0 I}{2\pi r}$ .
For multiple wires through the loop, use signed current: $I_{\text{enc}} = I_{\text{out}} - I_{\text{in}}$ if out of the page is chosen positive.
For an ideal long solenoid, $B = \mu_0 n I$ , where $n = \frac{N}{L}$ is the number of turns per meter.
For an ideal toroid inside the windings, $B = \frac{\mu_0 N I}{2\pi r}$ , where $r$ is the distance from the center of the toroid.
If the Amperian loop encloses no net current, then $\oint \vec{B} \cdot d\vec{\ell} = 0$ , but the magnetic field does not have to be zero everywhere.
Ampere's law is easiest to use when symmetry makes $\vec{B}$ tangent to the path and constant in magnitude along useful parts of the loop.
The permeability of free space is $\mu_0 = 4\pi \times 10^{-7}\ \text{T}\cdot\text{m/A}$ .

Vocabulary

Ampere's law: A law stating that the circulation of the magnetic field around a closed path equals $\mu_0$ times the net current enclosed.
Amperian loop: A closed path chosen to apply $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$ .
Enclosed current: The net current passing through the surface bounded by an Amperian loop.
Magnetic permeability: A constant or material property that measures how strongly a medium supports magnetic fields, with free space value $\mu_0 = 4\pi \times 10^{-7}\ \text{T}\cdot\text{m/A}$ .
Solenoid: A long coil of wire that produces an approximately uniform magnetic field inside, given by $B = \mu_0 n I$ for an ideal solenoid.
Toroid: A doughnut-shaped coil whose magnetic field inside the core is approximately $B = \frac{\mu_0 N I}{2\pi r}$ .

Common Mistakes to Avoid

Using $B = \frac{\mu_0 I}{2\pi r}$ for every situation is wrong because that formula only applies to a long straight wire with circular symmetry.
Forgetting signs on enclosed current is wrong because currents in opposite directions subtract, so the correct value is the net $I_{\text{enc}}$ .
Assuming $\oint \vec{B} \cdot d\vec{\ell} = 0$ means $B = 0$ is wrong because the integral can cancel even when the field is nonzero at points on the loop.
Choosing an Amperian loop without symmetry is ineffective because $B$ may not be constant or parallel to $d\vec{\ell}$ , making the integral hard to simplify.
Using total turns $N$ instead of turns per length $n$ for a solenoid is wrong because the ideal solenoid formula is $B = \mu_0 n I$ , with $n = \frac{N}{L}$ .

Practice Questions

1 A long straight wire carries $I = 8.0\ \text{A}$ . Find the magnetic field magnitude at $r = 0.040\ \text{m}$ using $B = \frac{\mu_0 I}{2\pi r}$ .
2 An ideal solenoid has $N = 600$ turns, length $L = 0.30\ \text{m}$ , and current $I = 2.5\ \text{A}$ . Find $n = \frac{N}{L}$ and $B = \mu_0 n I$ .
3 A toroid has $N = 250$ turns and current $I = 1.8\ \text{A}$ . Find the magnetic field at $r = 0.075\ \text{m}$ using $B = \frac{\mu_0 N I}{2\pi r}$ .
4 Explain why a circular Amperian loop is a good choice for a long straight wire but not usually a good choice for an irregular current distribution.

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