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

# Field on axis of a long coil

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2069 Chaitra, 5 marks
- 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/long-coil-axis-field-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

A long circuit coil consisting of 50 turns with diameter 1.2m carries a current of 10Amp. (a) Find the magnetic field at a point along the axis 90cm from the center. (b) At what distance from the center, along the axis, the field is 1/8 greater as at the center.

## Short answer

Use B(x)=μ₀NIR²/[2(R²+x²)^(3/2)] with R=0.6 m, I=10 A, N=50. (a) At x=0.9 m, B≈8.9×10⁻⁵ T. (b) Setting B = B₀/8 gives (R²+x²)^(3/2)=8R³ → x=√3 R≈1.04 m.

## Answer

For a coil of N turns and radius R, the field on the axis at distance x from the centre is

**B(x) = μ₀ N I R² / [2 (R² + x²)^(3/2)]**, and at the centre B₀ = μ₀NI/(2R).

**Given:** N = 50, I = 10 A, diameter 1.2 m so R = 0.6 m.

#### (a) B at x = 0.90 m
```text
R² = 0.36 ;  x² = 0.81 ;  R² + x² = 1.17
(R² + x²)^(3/2) = 1.17 × √1.17 = 1.17 × 1.0817 = 1.2656
B = (4π×10⁻⁷ × 50 × 10 × 0.36) / (2 × 1.2656)
  = 2.262×10⁻⁴ / 2.531
  = 8.94×10⁻⁵ T
```
At the centre B₀ = (4π×10⁻⁷ × 500)/(2 × 0.6) = 5.24×10⁻⁴ T.

#### (b) Point where the field is 1/8 of the centre value
**Reading of the question:** on the axis the field is largest at the centre, so "1/8 greater as at the centre" is read as "1/8 of the centre value", B = B₀/8.

```text
B/B₀ = R³ / (R² + x²)^(3/2) = 1/8
(R² + x²)^(1/2) = 2R   →   R² + x² = 4R²
x = √3 R = 1.732 × 0.6 = 1.04 m
```

**Answer: (a) B ≈ 8.9×10⁻⁵ T (89 μT); (b) x ≈ 1.04 m from the centre.**

## Check yourself

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

1. A coil of diameter 1.2 m has the radius R used in the formula:
   A. 0.6 m
   B. 1.2 m
   C. 0.3 m
   D. 0.9 m
   Answer: A. R = d/2. (Key point: Radius)

2. The axial field of an N-turn coil is:
   A. B = μ₀NIR/[2(R² + x²)]
   B. B = μ₀NIR²/[2(R² + x²)^(3/2)]
   C. B = μ₀NI/[2(R² + x²)^(3/2)]
   D. B = μ₀NIR²/(2x³) at all x
   Answer: B. The general on-axis result. (Key point: Axis formula)

3. At x = 0.90 m the field is about:
   A. 5.2×10⁻⁴ T
   B. 8.9×10⁻⁷ T
   C. 8.9×10⁻⁵ T
   D. 0.89 T
   Answer: C. B = 2.262×10⁻⁴/2.531 ≈ 8.94×10⁻⁵ T. (Key point: Part (a))

4. B = B₀/8 requires (R² + x²)^(1/2) = 2R, giving x of about:
   A. 0.6 m
   B. 1.2 m
   C. 0.52 m
   D. 1.04 m (√3 R)
   Answer: D. R²+x² = 4R² gives x = √3 R = 1.73 × 0.6. (Key point: Part (b))

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