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

# Three-coil core, flux in air gap

Electrical Machines · 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_machines/three-coil-core-airgap-flux
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

A rectangular iron core is shown in figure 1. It has a mean length of magnetic path of 100 cm, cross-section of (2 cm $\times$ 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 $N_a = 335$, $N_b = 600$ and $N_c = 600$; 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: rectangular iron core with an air gap in the top limb; coil $N_a$ with current $I_a$ on the left limb, coil $N_b$ with current $I_b$ on the right limb and coil $N_c$ with current $I_c$ on the bottom limb; the current directions are drawn in the figure and are not legible in the scan]

## Short answer

Net mmf is 1136 AT (assuming two coils aid, one opposes). Iron reluctance = 1.414×10⁶ AT/Wb, gap reluctance = 9.947×10⁶ AT/Wb, total = 1.136×10⁷ AT/Wb. Flux in gap = 1136 / 1.136×10⁷ = 1.0×10⁻⁴ Wb (0.1 mWb).

## Answer

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

**Assumption:** the current directions in the figure are not legible. I assume coils $a$ and $b$ magnetise in the same sense and coil $c$ opposes them. If the figure shows a different arrangement, only the net mmf changes and the method stays the same.

**Dimensions:** $l_g = 5$ mm, $l_i = 100 - 0.5 = 99.5$ cm, $A = 2\times2 = 4$ cm$^2$ $= 4\times10^{-4}$ m$^2$, $\mu_r = 1400$.

**Net mmf**

$$
\begin{aligned}
F_a &= 335\times1.6 = 536\ \text{AT} \\
F_b &= 600\times4 = 2400\ \text{AT} \\
F_c &= 600\times3 = 1800\ \text{AT} \\
F_{net} &= 536 + 2400 - 1800 = 1136\ \text{AT}
\end{aligned}
$$

**Reluctance**

$$
\begin{aligned}
S_i &= \frac{0.995}{4\pi\times10^{-7}\times1400\times4\times10^{-4}} = 1.414\times10^{6}\ \text{AT/Wb} \\
S_g &= \frac{0.005}{4\pi\times10^{-7}\times4\times10^{-4}} = 9.947\times10^{6}\ \text{AT/Wb} \\
S &= S_i + S_g = 1.1361\times10^{7}\ \text{AT/Wb}
\end{aligned}
$$

**Flux**

$$
\Phi = \frac{F_{net}}{S} = \frac{1136}{1.1361\times10^{7}} = 1.0\times10^{-4}\ \text{Wb}
$$

The flux density in the gap is $B = \Phi/A = 0.25$ T.

**Answer: $\Phi_{gap} \approx 0.1$ mWb ($1.0\times10^{-4}$ Wb)**

## Check yourself

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

1. With coils a and b magnetising in one sense and c opposing, what is the net mmf?
   A. 536 + 2400 + 1800 AT
   B. 536 + 2400 - 1800 AT
   C. 2400 - 536 - 1800 AT
   D. 1800 - 536 - 2400 AT
   Answer: B. Aiding mmfs add and the opposing one subtracts: 536 + 2400 - 1800 = 1136 AT. (Key point: Net mmf)

2. What iron path length is used with the 5 mm gap in a 100 cm mean path?
   A. 100 cm
   B. 105 cm
   C. 99.5 cm
   D. 95 cm
   Answer: C. The mean length includes the gap, so l_i = 100 - 0.5 = 99.5 cm. (Key point: Iron length)

3. How is the flux in the air gap found?
   A. Net mmf divided by total reluctance of iron and gap
   B. Sum of mmfs times reluctance
   C. Net mmf divided by gap reluctance only
   D. Net mmf divided by iron reluctance only
   Answer: A. Phi = F_net/(S_i + S_g) = 1136/1.1361e7 = 0.1 mWb. (Key point: Flux calculation)

4. Which is true of S_g and S_i here?
   A. S_i is much larger
   B. They are equal
   C. S_g (9.9e6) is much larger than S_i (1.4e6)
   D. S_g is zero
   Answer: C. The 5 mm of air has far higher reluctance than 99.5 cm of iron with mu_r 1400. (Key point: Gap dominates reluctance)

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## Related in other subjects

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