← Back to course

Forces on Currents & Parallel Wires

You’ll be able to

A wire is a stream of moving charges

A current-carrying wire in a magnetic field feels a force because each drifting charge inside it does. Summing qv × B over every carrier in a straight segment of length L gives F = IL × B, where the vector L points along the current. The magnitude is F = BIL·sinθ: maximum when the wire is perpendicular to the field, zero when the wire runs parallel to it. For a bent wire, integrate dF = I dl × B along the shape.

Force on a current-carrying wire
→F = I→L × →B, F = BIL·sinθ
Right-hand rule: fingers along I, curl toward B, thumb gives F. A wire parallel to B feels nothing.

Wires exert forces on each other

Combine the two big ideas of this unit: wire 1 creates a field B₁ = μ₀I₁/(2πd) at wire 2’s location, and wire 2 feels F = I₂LB₁ in that field. The result is a force per unit length F/L = μ₀I₁I₂/(2πd) between parallel wires. Work the right-hand rules carefully and you find parallel currents attract, antiparallel currents repel — the opposite of the “likes repel” instinct from electrostatics. This mutual force once defined the ampere.

Force between parallel wires
F/L = μ₀I₁I₂/(2πd)
Currents in the same direction attract; opposite directions repel. Newton’s third law holds: each wire feels the same magnitude of force.
Worked example

Starting from the force on a single charge, derive F = BIL for a straight wire of length L carrying current I perpendicular to a uniform field B.

  1. 1.Model the wire: n carriers per volume, each with charge q, drifting at speed v_d through cross-sectional area A.
  2. 2.One carrier feels F₁ = qv_dB (with v ⊥ B). The segment of length L contains N = nAL carriers.
  3. 3.Total force: F = N·F₁ = (nAL)(qv_dB) = (nqAv_d)(LB).
  4. 4.Recognize the current: I = nqAv_d — charge per second crossing any section of the wire.
  5. 5.Therefore F = BIL. The microscopic details (n, q, v_d) all hide inside I, which is why the macroscopic formula is so simple.
Answer: F = BIL, perpendicular to both the wire and the field
Worked example

Two long parallel wires 0.10 m apart each carry 10 A in the same direction. Find the force per unit length between them and the force on a 2.0 m section.

  1. 1.F/L = μ₀I₁I₂/(2πd) = (4π × 10⁻⁷)(10)(10)/(2π × 0.10).
  2. 2.Simplify the constants: μ₀/(2π) = 2 × 10⁻⁷ T·m/A, so F/L = (2 × 10⁻⁷)(100)/(0.10).
  3. 3.F/L = (2 × 10⁻⁵)/(0.10) = 2.0 × 10⁻⁴ N/m.
  4. 4.For 2.0 m: F = (2.0 × 10⁻⁴)(2.0) = 4.0 × 10⁻⁴ N. Same-direction currents, so the wires pull toward each other.
Answer: F/L = 2.0 × 10⁻⁴ N/m, attractive; F = 4.0 × 10⁻⁴ N on a 2.0 m section
Tip

For wire-on-wire problems, always split the work into two right-hand rules: first find the field of wire 1 at wire 2 (thumb along I₁, fingers curl to get B₁), then find the force on wire 2 (fingers along I₂, curl toward B₁, thumb gives F). Trying to shortcut both steps at once is where directions go wrong.

Checkpoint

Two parallel wires carry currents in the same direction. The magnetic force between them is:

Checkpoint

If the current in *both* of two parallel wires is doubled, the force per length between them:

Answer the 2 checkpoints as you read.

Sign in to save your progress