---
title: "Inductance of a toroid numerical"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:toroid-inductance-numerical"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/toroid-inductance-numerical
site: BCT Engineering
---

# Inductance of a toroid numerical

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2072 Kartik, 5 marks
- Video: [Calculation of Inductance of a Toroid, Toroidal Inductance](https://www.youtube.com/watch?v=Et8etQHHCgk) — Science Plus
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/toroid-inductance-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

A toroid has number of turns 1250, internal radius 52 mm, external radius 95 mm and thickness of the ring 13 mm, calculate the inductance.

## Short answer

Inductance L = (μ₀N²h/(2π)) ln(b/a). With N = 1250, inner radius 52 mm, outer 95 mm, thickness 13 mm, L ≈ 2.45 mH.

## Answer

For a toroid of rectangular cross-section (inner radius a, outer radius b, height h), the field at radius r is B = μ₀NI/(2πr). The flux through one turn is Φ = ∫ B h dr from a to b = (μ₀NIh/2π) ln(b/a). Hence

**L = NΦ/I = μ₀N²h ln(b/a)/(2π)**

**Given:** N = 1250, a = 52 mm, b = 95 mm, h = 13 mm (ring thickness), μ₀ = 4π×10⁻⁷ H/m.

```text
μ₀/(2π)    = 2×10⁻⁷ H/m
N²         = 1.5625×10⁶
ln(b/a)    = ln(95/52) = ln(1.827) = 0.6025
L = (2×10⁻⁷)(1.5625×10⁶)(0.013)(0.6025)
  = 4.0625×10⁻³ × 0.6025
  = 2.45×10⁻³ H
```

**Answer: L ≈ 2.45 mH**

## Check yourself

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

1. Inside a toroid of N turns carrying current I, the field at radius r is:
   A. B = μ₀NI/r
   B. B = μ₀NI/(2πr)
   C. B = μ₀NI/(2π r²)
   D. B = μ₀ I/(2πr) without N
   Answer: B. Ampere's law on a circle of radius r gives B(2πr) = μ₀NI. (Key point: Toroid field)

2. The inductance of a rectangular-section toroid is:
   A. L = μ₀N²h ln(b/a)/(2π)
   B. L = μ₀N h (b−a)/(2π)
   C. L = μ₀N² h (b/a)/(2π)
   D. L = μ₀N h ln(b/a)/(2π)
   Answer: A. Flux through one turn is (μ₀NIh/2π)ln(b/a); L = NΦ/I gives N² in the result. (Key point: Inductance formula)

3. In the numerical, which given value is used as h in the formula?
   A. 52 mm, the inner radius
   B. 95 mm, the outer radius
   C. 13 mm, the ring thickness
   D. 43 mm, the difference b − a
   Answer: C. The ring thickness is the height h of the cross-section; the radii give a and b. (Key point: Which dimension is h)

4. With N = 1250, a = 52 mm, b = 95 mm, h = 13 mm, the inductance is about:
   A. 24.5 mH
   B. 0.245 mH
   C. 2.45 H
   D. 2.45 mH
   Answer: D. L = (2×10⁻⁷)(1.5625×10⁶)(0.013)(0.6025) ≈ 2.45×10⁻³ H. (Key point: Result)

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