This AP Physics C lesson on Ampère’s law notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on Ampère’s law and check yourself with the quiz.
Quick answer: The Biot–Savart law gives the field of a current element, dB = (μ₀/4π) I dl × r̂/r², and integrating it yields B = μ₀I/2πd for a long wire and μ₀I/2R at a loop’s centre. Ampère’s law, ∮B·dl = μ₀I_enc, gives symmetric fields quickly, including B = μ₀nI inside a solenoid.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:E&M Unit 12
Ampère’s law notes: the key ideas
What does the Biot–Savart law say?
Every small piece of current produces a small magnetic field around it, and the total field is the vector sum of all those pieces. The law gives the contribution of a single current element, and calculus adds them up for wires, loops and coils of any shape.
Biot–Savart law: dB = (μ₀/4π) I dl × r̂ / r², where dl is a length element along the current, r is the distance from the element to the field point and r̂ is a unit vector from the element to that point.
Infinite straight wire: B = μ₀I/(2πd) Circular loop, centre: B = μ₀I/(2R) Loop, on axis: B = μ₀IR²/[2(x² + R²)3/2]
B is the field (T), I is the current (A), d is the distance from the wire (m), R is the loop radius (m), x is the axial distance (m), and μ₀ = 4π × 10⁻⁷ T·m/A.
The cross product means dB is perpendicular to both dl and r. At a point on the line of the wire, dl is parallel to r, so sin θ = 0 and that element makes no field at all. For an infinite wire, the integral of sin θ gives B = μ₀I/2πd, and the right-hand rule gives the direction: thumb along the current, fingers curl along B.
Data: a loop of radius 5.0 cm carries 3.0 A. Find B at its centre.
Method: every element is a distance R from the centre and perpendicular to r, so B = μ₀I/(2R) = (4π × 10⁻⁷)(3.0)/(2 × 0.050).
Answer: B = 3.8 × 10⁻⁵ T (38 μT), directed along the loop’s axis.
How does Ampère’s law give magnetic fields?
When the current has a high degree of symmetry, you can skip the integration. Ampère’s law links the circulation of B around any closed path to the current that passes through the path.
Ampère’s law: the line integral of the magnetic field around a closed loop equals μ₀ times the net current through the loop: ∮ B · dl = μ₀Ienc. Currents are counted positive when they follow the right-hand rule for the direction of travel.
Choose a loop on which B has constant magnitude and is either parallel or perpendicular to dl. For a long cylindrical wire of radius R with uniform current density, the enclosed current inside the wire is I(r²/R²), so B = μ₀Ir/(2πR²) for r < R, and B = μ₀I/(2πr) for r > R.
Data: a wire of radius 2.0 mm carries 5.0 A uniformly. Find B at r = 1.0 mm and at r = 4.0 mm.
Method: inside, B = μ₀Ir/(2πR²) = (2 × 10⁻⁷)(5.0)(1.0 × 10⁻³)/(2.0 × 10⁻³)². Outside, B = μ₀I/(2πr).
Answer: B = 0.25 mT at 1.0 mm, and B = 0.25 mT at 4.0 mm, with a peak of 0.50 mT at the surface.
What is the field of a solenoid and the force between wires?
A long solenoid has n turns per metre. A rectangular loop with one side inside and one outside gives B ℓ = μ₀(nℓ)I, so the inside field is uniform and the outside field is almost zero. A toroid with N turns has B = μ₀NI/(2πr) inside its core.
| System | Magnetic field B |
|---|---|
| Long straight wire, distance d | μ₀I/(2πd) |
| Circular loop, centre | μ₀I/(2R) (times N for N turns) |
| Long solenoid, inside | μ₀nI |
| Toroid, at radius r inside the core | μ₀NI/(2πr) |
| Thick wire, inside (r < R) | μ₀Ir/(2πR²) |
| Two wires, force per metre | F/L = μ₀I₁I₂/(2πd) |
One long wire makes a field at the other; the other carries current in that field, so each feels a force F = ILB. Currents in the same direction attract and opposite currents repel.
Data: two parallel wires 0.10 m apart carry 20 A and 30 A in the same direction. Find the force per metre.
Method: F/L = μ₀I₁I₂/(2πd) = (2 × 10⁻⁷)(20)(30)/(0.10).
Answer: F/L = 1.2 × 10⁻³ N/m, attractive.
Ampère’s law gives B only when symmetry lets you pull B out of the integral. It is always true, but not always useful. Do not use B = μ₀nI near the ends of a short coil, where the field is not uniform.
Before any calculation, draw the current and the loop, then apply the right-hand rule. A correct direction catches sign errors long before the arithmetic does.
- dB = (μ₀/4π) I dl × r̂ / r²; μ₀ = 4π × 10⁻⁷ T·m/A.
- Long wire B = μ₀I/(2πd); loop centre B = μ₀NI/(2R); solenoid B = μ₀nI.
- Ampère’s law: ∮B·dl = μ₀I_enc, useful when symmetry fixes B.
- Thick wire: B = μ₀Ir/(2πR²) inside, μ₀I/(2πr) outside.
- Parallel wires: F/L = μ₀I₁I₂/(2πd); same direction attracts.
Biot–Savart, Ampère and magnetic sources: frequently asked questions
When should I use Biot–Savart rather than Ampère’s law?
Use Ampère’s law when symmetry makes B constant on a simple loop, as for a long wire or solenoid. Use Biot–Savart for less symmetric shapes, such as a loop off-axis or a finite wire segment.
Why is the field zero along the line of a straight wire segment?
At such a point dl is parallel to r, so the cross product dl × r̂ vanishes and every element contributes zero. The field is strongest perpendicular to the wire.
How do you get the direction of the magnetic field?
Point the thumb of your right hand along the current. Your curled fingers show how B circles the wire. For a loop or coil, curl the fingers along the current and the thumb gives B.
Why does B inside a thick wire grow with r?
The Amperian loop of radius r encloses only the fraction r²/R² of the current, so ∮B·dl = μ₀I(r²/R²). Dividing by 2πr gives B = μ₀Ir/(2πR²).
Why do parallel currents attract?
Wire 1 creates a field at wire 2 whose direction, combined with the current in wire 2, produces a force F = IL × B toward wire 1. The same argument gives an equal force back on wire 1.
