Ampère’s Law
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
Ampère’s law is the magnetic version of Gauss’s law: around any closed loop, ∮B · dl = μ0Ienc. When symmetry makes B constant along a well-chosen loop, it gives B in one line: wires, cylinders, solenoids and toroids.
1. Using the law
2. Choosing the loop
Wires and cylinders: use a circle centered on the axis. B is tangent and constant on it, so ∮B · dl = B(2πr).
Uniform current inside a wire: Ienc = I(r²/R²), so B = μ0Ir/(2πR²).
Solenoid: use a rectangle with one side inside. The outside field is nearly zero, so BL = μ0(nL)I.
Zero enclosed current means zero B only with symmetry: outside a coaxial cable carrying equal and opposite currents, B = 0; inside a hollow tube, B = 0.
3. Example, then try it
A 0.30 m solenoid has 900 turns. What current makes a 0.012 T field inside it?
n = 900 ÷ 0.30 = 3000 turns/m.
I = B/(μ0n) = 0.012 ÷ [(4π × 10⁻⁷)(3000)] ≈ 3.2 A.
A long wire of radius R carries current with density J = J0(1 − r/R): strongest at the center, zero at the surface. Find the total current and B inside the wire.
Enclosed current: Ienc = ∫0r J0(1 − r′/R) 2πr′ dr′ = 2πJ0(r²/2 − r³/3R).
Total (r = R): I = 2πJ0(R²/2 − R²/3) = πJ0R²/3.
Ampère: B(2πr) = μ0Ienc → B = μ0J0(r/2 − r²/3R).
Check: at r = R, B = μ0J0R/6 = μ0I/(2πR), matching the outside formula.
Answer: I = πJ0R²/3; B = μ0J0(r/2 − r²/3R) inside.
4. Common mistakes
A long solenoid’s length is doubled with the same number of turns and the same current. What happens to the field inside?
B = μ0nI, and n = N/L halves when L doubles.
Ready to practice?
5 questions on Ampère’s Law, each with a hint and an explanation for every wrong answer.