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Magnetic Fields of Currents & Ampère's Law

Fields around wires and inside solenoids, forces between wires, and Ampère's law.

AP 2C: E&M
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The big idea

Moving charges don't just feel magnetic fields, they create them. A long straight wire is surrounded by circular field lines, and a solenoid (a coil) makes a strong, uniform field inside it. Ampère's law is the magnetic counterpart of Gauss's law: with the right symmetry, it turns field problems into a single line.

Key ideas

  • Long straight wire: B = μ₀I/(2πr), in circles around the wire. Right-hand rule: thumb along I, fingers curl along B.
  • Field lines form closed loops. There are no magnetic "charges".
  • Parallel currents attract, and opposite currents repel.
  • Solenoid: B = μ₀nI inside, uniform and along the axis, where n is the number of turns per metre. Outside, B ≈ 0.
  • Ampère (C: E&M): ∮B·dl = μ₀Ienc. Biot–Savart gives the field of any current element.

Equations

B = μ₀I/(2πr)long straight wire
B = μ₀nIinside a solenoid
F/L = μ₀I₁I₂/(2πd)force between parallel wires
∮B·dl = μ₀IencAmpère's law
μ₀ = 4π × 10⁻⁷ T·m/Apermeability of free space

Worked example

Find (a) the field 5.0 cm from a long wire carrying 10 A, and (b) the field inside a solenoid with 400 turns over 20 cm carrying 2.0 A.

  1. (a) B = μ₀I/(2πr) = (4π × 10⁻⁷)(10)/(2π × 0.050) = 4.0 × 10⁻⁵ T (40 µT), about the strength of Earth's field.
  2. (b) n = 400/0.20 = 2000 turns/m.
  3. B = μ₀nI = (4π × 10⁻⁷)(2000)(2.0) = 5.0 × 10⁻³ T (5.0 mT).
  4. A coil concentrates the field far more than a single wire does.
Answer: (a) 40 µT  (b) 5.0 mT

Common mistakes

  • Using the total number of turns N instead of turns per metre n.
  • Drawing the field lines from a wire pointing radially outward. They circle around it.
  • Getting attraction and repulsion between currents backwards (it's the reverse of charges).

Try it yourself

Two parallel wires 0.10 m apart each carry 10 A in the same direction. Find the force per metre and say whether it attracts or repels.

Show the answer

F/L = μ₀I₁I₂/(2πd) = (2 × 10⁻⁷)(100)/0.10 = 2.0 × 10⁻⁴ N/m, attractive.

AP Physics C level

Going deeper: AP Physics C

This part goes beyond the algebra course: calculus derivations, harder multi-step problems and exam-style practice, written at the level of the AP Physics C exam.

The Biot–Savart law

Each small piece of current I dℓ makes a field

dB = μ₀4πI dℓ × r̂r²

Integrating gives the standard results: at the centre of a circular loop, B = μ₀I/(2R). On the loop's axis, B = μ₀IR²/(2(x² + R²)3/2). Near a long straight wire, B = μ₀I/(2πr). For an arc subtending angle θ, use the fraction θ/2π of the loop result.

Ampère's law

For highly symmetric currents, use

∮B·dℓ = μ₀Ienc

The method mirrors Gauss's law: choose a loop on which B is constant and parallel to dℓ, so the integral becomes B × (length of loop). Results: inside a uniform wire, B = μ₀Ir/(2πR²). Solenoid: B = μ₀nI. Toroid: B = μ₀NI/(2πr). Outside a coaxial cable carrying equal and opposite currents, B = 0.

Force between parallel wires

Each wire sits in the other's field: F/L = μ₀I₁I₂/(2πd). Currents in the same direction attract, and currents in opposite directions repel.

More worked examples

Example 1. A circular loop of radius 5.0 cm carries 5.0 A. Find B at its centre and on its axis 5.0 cm from the centre.

  1. Centre: B = μ₀I/(2R) = (4π × 10⁻⁷)(5.0)/(0.10) = 63 µT.
  2. On axis at x = R: B = μ₀IR²/(2(2R²)3/2) = Bcentre/23/2.
  3. B = 62.8/2.83 = 22 µT.
Answer: 63 µT; 22 µT

Example 2. A long wire of radius 4.0 mm carries 20 A spread uniformly over its cross-section. Find B at 2.0 mm from the axis.

  1. Ampère loop of radius r inside: Ienc = I(r²/R²).
  2. B(2πr) = μ₀Ir²/R² → B = μ₀Ir/(2πR²).
  3. B = (2 × 10⁻⁷)(20)(0.002)/(0.004)² = 0.50 mT.
Answer: 0.50 mT

Example 3. A toroid has 500 turns carrying 2.0 A. Find B at radius 0.10 m inside its core.

  1. A circular Ampère loop inside the core encloses N turns: B(2πr) = μ₀NI.
  2. B = μ₀NI/(2πr) = (2 × 10⁻⁷)(500)(2.0)/0.10 = 2.0 mT.
  3. Outside the toroid the enclosed current is zero, so B = 0.
Answer: 2.0 mT

Exam-style practice

1. Two long parallel wires 5.0 cm apart each carry 10 A in the same direction. Find the force per metre.

Show the solution

F/L = (2 × 10⁻⁷)(100)/0.05 = 4.0 × 10⁻⁴ N/m, attractive.

2. A solenoid has 1000 turns per metre and carries 3.0 A. Find B inside.

Show the solution

B = μ₀nI = 3.8 mT.

3. Find B at the centre of a semicircular arc of radius R carrying current I.

Show the solution

Half of the full-loop value: μ₀I/(4R).

See it in action

Open the Magnetic Fields of Currents simulation and test every idea on this page yourself.

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