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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 Bd=μ0Ienc\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(2πr)=μ0IencB(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=μ0nIB = \mu_0 n I, and for a toroid it is B=μ0NI2πrB = \frac{\mu_0 N I}{2\pi r} inside the core.

Key Facts

  • Ampere's law is Bd=μ0Ienc\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}, where IencI_{\text{enc}} is the net current passing through the loop.
  • For a long straight wire, a circular Amperian loop gives B(2πr)=μ0IB(2\pi r) = \mu_0 I, so B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}.
  • For multiple wires through the loop, use signed current: Ienc=IoutIinI_{\text{enc}} = I_{\text{out}} - I_{\text{in}} if out of the page is chosen positive.
  • For an ideal long solenoid, B=μ0nIB = \mu_0 n I, where n=NLn = \frac{N}{L} is the number of turns per meter.
  • For an ideal toroid inside the windings, B=μ0NI2πrB = \frac{\mu_0 N I}{2\pi r}, where rr is the distance from the center of the toroid.
  • If the Amperian loop encloses no net current, then Bd=0\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 B\vec{B} tangent to the path and constant in magnitude along useful parts of the loop.
  • The permeability of free space is μ0=4π×107 Tm/A\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 μ0\mu_0 times the net current enclosed.
Amperian loop
A closed path chosen to apply Bd=μ0Ienc\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 μ0=4π×107 Tm/A\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=μ0nIB = \mu_0 n I for an ideal solenoid.
Toroid
A doughnut-shaped coil whose magnetic field inside the core is approximately B=μ0NI2πrB = \frac{\mu_0 N I}{2\pi r}.

Common Mistakes to Avoid

  • Using B=μ0I2πrB = \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 IencI_{\text{enc}}.
  • Assuming Bd=0\oint \vec{B} \cdot d\vec{\ell} = 0 means B=0B = 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 BB may not be constant or parallel to dd\vec{\ell}, making the integral hard to simplify.
  • Using total turns NN instead of turns per length nn for a solenoid is wrong because the ideal solenoid formula is B=μ0nIB = \mu_0 n I, with n=NLn = \frac{N}{L}.

Practice Questions

  1. 1 A long straight wire carries I=8.0 AI = 8.0\ \text{A}. Find the magnetic field magnitude at r=0.040 mr = 0.040\ \text{m} using B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}.
  2. 2 An ideal solenoid has N=600N = 600 turns, length L=0.30 mL = 0.30\ \text{m}, and current I=2.5 AI = 2.5\ \text{A}. Find n=NLn = \frac{N}{L} and B=μ0nIB = \mu_0 n I.
  3. 3 A toroid has N=250N = 250 turns and current I=1.8 AI = 1.8\ \text{A}. Find the magnetic field at r=0.075 mr = 0.075\ \text{m} using B=μ0NI2πrB = \frac{\mu_0 N I}{2\pi r}.
  4. 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.

Understanding Ampere's Law Worked Examples

The direction of the magnetic field is as important as its size. Around a straight current carrying wire, the field forms circles centered on the wire. Use the right hand grip rule to find the direction.

Point your right thumb with the conventional current. Your curled fingers show the magnetic field direction. This rule helps assign signs to currents passing through an imagined surface bounded by the loop.

Choose a surface direction first. Your curled fingers around the loop set the positive surface direction through the right hand rule. Currents through that surface in one direction count positive, while currents in the opposite direction count negative.

A loop does not have to be a real wire or a physical boundary. It is a mathematical path selected to match the shape of the field. For a straight wire, circles are useful because every point on one circle is equally far from the wire.

The field has the same magnitude all around that circle and points along the path. For a solenoid, a long thin rectangle is useful. One long side lies inside, where the field is nearly uniform.

The other long side lies outside, where the ideal field is very small. The short sides contribute nothing because the field crosses them rather than following them.

Be careful about what enclosed current means. It means current that passes through a surface stretched across the chosen loop, not current that merely lies nearby. A wire can be close to the path yet contribute zero if it does not pass through that surface.

In problems with several wires, draw dots for currents out of the page and crosses for currents into the page. Add them with signs before doing any calculation. If the total is zero, the circulation of the field around the loop is zero.

That does not prove that the field is zero at each point. Fields from separate wires may be present but cancel only when their contributions are added around the full path.

Real devices show why these ideas matter. Power cables produce magnetic fields that become weaker as distance increases. A coil in a relay, doorbell, speaker, or electric motor produces a stronger field because many turns carry current together.

A toroid is used in transformers and inductors because its circular core keeps much of the magnetic field within the ring. Its field is not equally strong at every radius. Points nearer the center have a stronger field than points farther out.

In worked examples, first sketch the current directions and field shape. Then state why a chosen loop works before inserting numbers. Check units at the end.

A magnetic field result should be in teslas. Finally, remember that these simple results rely on ideal symmetry. Short coils, bent wires, gaps, and nearby magnetic materials can make the actual field less uniform.