---
title: "Magnetic field on axis of circular coil"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:circular-coil-axis-field"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/circular-coil-axis-field
site: BCT Engineering
---

# Magnetic field on axis of circular coil

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2072 Chaitra, 5 marks; 2071 Chaitra, 5 marks (asked 2 times)
- Video: [Magnetic field due to a circular coil carrying current  || Class12 | Physics | Tamil](https://www.youtube.com/watch?v=QzWnyv4mO6I) — Mokka Physics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/circular-coil-axis-field
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

State Biot-Savart law. Derive an expression for the magnetic field (flux density) at a distance x on the axis of a circular coil of radius a carrying current I.

## Short answer

Biot-Savart law: dB = (μ0/4π) I (dl × r̂)/r^2. For a circular coil of radius a, the axial field at distance x is B = μ0 I a^2 / [2(a^2 + x^2)^(3/2)]. At the centre (x=0), B = μ0 I/(2a).

## Answer

**Biot–Savart law.** The magnetic field dB at a point due to a small current element I dl is directly proportional to I, dl and sinθ (θ = angle between dl and the line joining the element to the point), and inversely proportional to the square of the distance r:

dB = (μ0/4π) · I dl sinθ / r²

In vector form, dB = (μ0/4π) I (dl × r̂)/r².

#### Field on the axis of a circular coil
Coil of radius a, current I; point P on the axis at distance x from the centre O.

```text
          dl
           o
          /|\
      r  / |a \  dB
        /  |    \
   ----o---O-----o----
           |
           |  x
           P
```

1. Take an element dl on the ring. It is perpendicular to the line r joining it to P, with r = √(a² + x²). So θ = 90° and

dB = (μ0/4π) I dl / (a² + x²)

2. dB is perpendicular to r, making angle α with the axis, where sinα = a/r. Resolve it:
   - dB sinα along the axis
   - dB cosα perpendicular to the axis
3. By symmetry, the perpendicular components of diametrically opposite elements cancel. Only the axial components add:

B = ∮ dB sinα = (μ0 I/4π)(1/(a² + x²))(a/√(a² + x²)) ∮dl

4. ∮dl = 2πa, so

**B = μ0 I a² / [ 2 (a² + x²)^(3/2) ]**

For N turns multiply by N. At the centre (x = 0): B = μ0 I/(2a). For x ≫ a: B ≈ μ0 I a²/(2x³).

## Check yourself

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

1. dB is proportional to
   A. I dl sinθ / r^2
   B. I dl cosθ / r
   C. I dl / r^3
   D. I dl sinθ r
   Answer: A. dB = (μ0/4π) I dl sinθ/r^2. (Key point: Biot-Savart law)

2. For an element of the ring, the angle between dl and r is
   A. 0°
   B. 45°
   C. 180°
   D. 90°
   Answer: D. dl is tangent to the ring and r points to the axis point, so sinθ = 1. (Key point: Angle)

3. The field at distance x on the axis is
   A. μ0 I a^2/[2(a^2+x^2)^(3/2)]
   B. μ0 I a/[2(a^2+x^2)]
   C. μ0 I/(2a)
   D. μ0 I a^2/[2(a^2+x^2)]
   Answer: A. Integrating the axial component dB sinα with sinα = a/r over 2πa. (Key point: Axial result)

4. At the centre (x = 0), B equals
   A. μ0 I/(2a)
   B. μ0 I a/2
   C. μ0 I/(4πa)
   D. Zero
   Answer: A. Setting x = 0 gives μ0 I a^2/(2a^3) = μ0 I/(2a); for x ≫ a it falls as 1/x^3. (Key point: Centre and far)

## Related in Engineering Physics

- [Biot-Savart vs Ampere's law, coil axis](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Abiot-ampere-compare-coil-axis) — Engineering Physics, chapter 8
- [Circular coil field and magnetic dipole](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acircular-coil-axis-dipole) — Engineering Physics, chapter 8, asked 3×
- [Field on axis of a long coil](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Along-coil-axis-field-numerical) — Engineering Physics, chapter 8

## Related in other subjects

- [H of infinite filament by Biot-Savart](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Abiot-savart-infinite-filament-derive) — Electromagnetism, chapter 3
- [State Biot-Savart's law](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Abiot-savart-law-state-explain) — Electromagnetism, chapter 3, asked 2×
- [Two-coil magnetic circuit, current for 1.2 T](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atwo-coil-core-1-2t-current) — Electrical Machines, chapter 1
- [Flux density in circular iron core](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aflux-density-1m-core-50mm2) — Electrical Machines, chapter 1