---
title: "Three-coil core: air-gap flux"
subject: "Electrical Machine"
chapter: 1
key: "semester_4/electrical_machine:three-coil-core-air-gap-flux"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/three-coil-core-air-gap-flux
site: BCT Engineering
---

# Three-coil core: air-gap flux

Electrical Machine · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2071 Magh, 6 marks
- Video: [(E)EM(C) Q 1.8  || Core with Three Legs || Magnetic Circuits || Fringing Effect](https://www.youtube.com/watch?v=AKbpCc-3tg0) — Electrical Engineering Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/three-coil-core-air-gap-flux
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-1

## Question

A rectangular iron core is shown in figure 1. It has a mean length of magnetic path of [?] cm, cross-section of (2 cm × 2 cm), relative permeability of 1400 and an air-gap of 5 mm cut in the core. The three coils carried by the core have number of turns Na = 335, Nb = 600 and Nc = 600; and the respective currents are 1.6 A, 4 A and 3 A. The directions of the currents are as shown in the figure. Find the flux in the air-gap.
[Figure 1: a rectangular iron core with an air gap in the top limb; coil a (current Ia) on the upper left, coil b (current Ib) on the right limb, and coil c (current Ic) on the bottom limb; current directions as shown in the figure.]

## Short answer

Net mmf is the algebraic sum: 335×1.6 + 600×4 − 600×3 = 1136 AT. Iron reluctance is 1.414×10⁶ AT/Wb, air-gap reluctance 9.947×10⁶ AT/Wb, total 1.136×10⁷ AT/Wb. Flux = 1136 / 1.136×10⁷ = 1.0×10⁻⁴ Wb (0.1 mWb).

## Answer

The flux in the gap is the net mmf of the three coils divided by the total reluctance.

**Given and assumptions**
- The mean path length is unreadable in the question. The standard textbook version of this problem uses **l = 100 cm** (1 m), so that value is used here.
- Cross-section 2 cm × 2 cm = 4×10⁻⁴ m²; μr = 1400; gap lg = 5 mm.
- Current directions (from the figure): coils a and b magnetise the core in the same direction, and coil c opposes them.

**Step 1: net mmf**
```text
Fa = 335 × 1.6 = 536 AT   (+)
Fb = 600 × 4   = 2400 AT  (+)
Fc = 600 × 3   = 1800 AT  (−)
F  = 536 + 2400 − 1800 = 1136 AT
```

**Step 2: reluctances**
```text
Sᵢ = (1.0 − 0.005)/(4π×10⁻⁷ × 1400 × 4×10⁻⁴)
   = 1.414×10⁶ AT/Wb
Sg = 0.005/(4π×10⁻⁷ × 4×10⁻⁴)
   = 9.947×10⁶ AT/Wb
S  = 1.136×10⁷ AT/Wb
```

**Step 3: flux**
Φ = F/S = 1136/1.136×10⁷ = 1.0×10⁻⁴ Wb

**Answer: Φ ≈ 0.1 mWb** (B = Φ/A = 0.25 T). If instead all three coils aided each other, F = 4736 AT and Φ ≈ 0.42 mWb. Because the air gap dominates the reluctance, the result depends only slightly on the assumed length: with l = 60 cm, Φ ≈ 0.105 mWb.

## Check yourself

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

1. How are the three coil mmfs combined?
   A. Averaged
   B. Added with signs by direction (aiding positive, opposing negative)
   C. All added
   D. Only the largest used
   Answer: B. Coils whose currents oppose subtract: 536 + 2400 - 1800 = 1136 AT. (Key point: Net mmf)

2. How is the gap flux found?
   A. Mmf of coil a only
   B. Net mmf divided by area
   C. Net mmf divided by total reluctance
   D. Net mmf times reluctance
   Answer: C. Phi = F/S = 1136/1.0793x10^7. (Key point: Flux formula)

3. Why does the assumed path length matter only slightly?
   A. Length is not in the formula
   B. The air gap dominates the total reluctance
   C. Iron has high reluctance
   D. The area cancels it
   Answer: B. Sg = 9.9x10^6 versus Si = 8.5x10^5, so changing the iron length hardly changes S. (Key point: Gap dominance)

4. What is the flux in the gap?
   A. About 1.05 mWb
   B. About 0.105 mWb
   C. About 0.01 mWb
   D. About 0.44 mWb
   Answer: B. 1136/1.0793x10^7 = 1.05x10^-4 Wb = 0.105 mWb. (Key point: Result)

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