Field due to circular arc of wire

Note: the standard problem is an arc that is a quarter of a circle (angle θ = π/2) with the field found at the centre of curvature. For a general arc the result is B = μ₀Iθ/(4πR); with θ = π/2 this is μ₀I/(8R).

Derivation

The arc has radius R and carries current I. We want B at the centre C.

       dl
      __.__
    /       \  <- arc, angle θ
   |    C ----R-->

1. Take an element dl on the arc. Every dl is tangent to the arc and perpendicular to the radius vector from C, so the angle between dl and r is 90° and |dl × r̂| = dl.

2. Biot-Savart law: dB = (μ₀/4π) I dl/R².

3. All the dB point the same way (perpendicular to the plane), so just add magnitudes. Total arc length = Rθ:

   B = (μ₀ I/4πR²) ∫dl = (μ₀ I/4πR²)(Rθ) = μ₀ I θ/(4π R)

4. For a quarter circle, θ = π/2:

   B = μ₀ I (π/2)/(4π R) = μ₀ I/(8R)

Check: a full loop (θ = 2π) gives μ₀I/(2R), the known loop-centre field. A semicircle gives μ₀I/(4R). Straight sections that point toward C add nothing, because dl × r̂ = 0 for them.