---
title: "Field due to circular arc of wire"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:curved-wire-arc-magnetic-field"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/curved-wire-arc-magnetic-field
site: BCT Engineering
---

# Field due to circular arc of wire

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2068 Shrawan, 5 marks
- Video: [Magnetic field due to arc of current carrying wire](https://www.youtube.com/watch?v=FX5eUM3hDPk) — Julie Alexander
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/curved-wire-arc-magnetic-field
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

Show that the magnetic field due to a curved wire segment carrying current I and circular arc of radius R is μ0I/(8R).

## Short answer

By Biot–Savart, the field at the centre of a circular arc of radius R subtending angle θ is B = μ₀Iθ/(4πR). For a quarter-circle arc (θ=π/2), this simplifies to B = μ₀I/(8R).

## Answer

**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.

```text
       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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. At the centre of curvature, the angle between dl and r for an arc element is:
   A. 0°, so the cross product is zero
   B. 90°, so |dl × r̂| = dl
   C. 45°
   D. 180°
   Answer: B. dl is tangent and r is radial. (Key point: Angle dl and r)

2. The field at the centre of an arc of angle θ and radius R is:
   A. B = μ₀Iθ/(4πR)
   B. B = μ₀Iθ/(2πR²)
   C. B = μ₀I/(4πR θ)
   D. B = μ₀Iθ R/(4π)
   Answer: A. B = (μ₀I/4πR²)(Rθ). (Key point: Arc formula)

3. For a quarter circle (θ = π/2) the field is:
   A. μ₀I/(4R)
   B. μ₀I/(2R)
   C. μ₀I/(8R)
   D. μ₀I/(16R)
   Answer: C. Substitute θ = π/2. (Key point: Quarter circle)

4. Straight sections pointing toward the centre contribute:
   A. the same as the arc
   B. half the arc field
   C. a field opposing the arc
   D. nothing, since dl × r̂ = 0
   Answer: D. dl is parallel to r̂ for them. (Key point: Straight sections)

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