---
title: "Circular coil field and magnetic dipole"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:circular-coil-axis-dipole"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/circular-coil-axis-dipole
site: BCT Engineering
---

# Circular coil field and magnetic dipole

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2074 Asoj, 5 marks; 2070 Chaitra, 5 marks; 2068 Chaitra, 5 marks (asked 3 times)
- Video: [Circular Current Loop as Magnetic Dipole | Class 12 Physics Moving Charges and Magnetism](https://www.youtube.com/watch?v=4MW1uVnq2Ug) — Scholarslearning Classes
- 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-dipole
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

Use the Biot-Savart law to find the magnetic field on the axis of a current carrying circular coil (radius R, at distance z from the center). Show that the circular current carrying coil behaves as a magnetic dipole for a large distance.

## Short answer

B = μ0 I R^2 / [2(R^2 + z^2)^(3/2)] along axis. For z >> R, B ≈ (μ0/4π) * 2m/z^3 with m = IπR^2, matching the dipole field form, so the coil acts as a magnetic dipole.

## Answer

**Biot–Savart law:** the field due to a current element I dl at displacement r from it is dB = (μ0/4π) I (dl × r̂)/r².

#### Field on the axis
Coil of radius R, current I; point P on the axis at distance z from the centre.

```text
            dl (element)
              o
             /|\
        s  /  | \   dB
         /    |R  \
   -----o-----O-----o-----
              |
              |  z
              P
```

1. Each element dl is perpendicular to s, where s = √(R² + z²). So dB = (μ0 I dl)/(4π s²).
2. dB is perpendicular to both dl and s. Components perpendicular to the axis cancel over the full ring. The axial component is dB_z = dB sinα, with sinα = R/s.
3. B = ∫ dB_z = (μ0 I/4π s²)(R/s) ∮dl = (μ0 I/4π)(R/s³)(2πR)

**B = μ0 I R² / [ 2 (R² + z²)^(3/2) ]**

#### Behaviour as a magnetic dipole
1. For z ≫ R: (R² + z²)^(3/2) ≈ z³, so

B ≈ μ0 I R² / (2 z³) = (μ0/4π) · 2 (I π R²)/z³

2. The magnetic dipole moment of the coil is m = I A = I πR². So

**B = (μ0/4π) · 2m / z³**

3. This has the same form as the axial field of an electric dipole, E = (1/4πε0)·2p/z³, and of a bar magnet. The field falls as 1/z³ and is along the axis, in the direction of m.

Hence a current-carrying loop acts as a magnetic dipole at large distance, with north pole on the side from which the current looks anticlockwise.

## Check yourself

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

1. Why do perpendicular components of dB cancel over the ring?
   A. Diametrically opposite elements give opposite components
   B. Each dB is zero
   C. The current alternates
   D. dl is parallel to the axis
   Answer: A. Only the axial components remain, dB sinα with sinα = R/s. (Key point: Biot-Savart)

2. The axial field of the coil is
   A. μ0 I R/[2(R^2+z^2)^(1/2)]
   B. μ0 I R^2/[2(R^2+z^2)^(3/2)]
   C. μ0 I/(2R)
   D. μ0 I R^2/[2(R^2+z^2)^2]
   Answer: B. B = (μ0 I/4π)(R/s^3)(2πR) with s^2 = R^2 + z^2. (Key point: Axial field)

3. For z ≫ R, B is
   A. (μ0/4π) 2m/z^3
   B. (μ0/4π) m/z^2
   C. μ0 I/(2z)
   D. (μ0/4π) m/z^3 perpendicular to axis
   Answer: A. With m = IπR^2, B ≈ (μ0/4π)·2m/z^3, like an electric dipole's axial field. (Key point: Far field)

4. The magnetic moment of the coil is
   A. I R
   B. I 2πR
   C. I π R^2
   D. μ0 I R^2
   Answer: C. m = I × area = IπR^2. (Key point: Dipole moment)

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