---
title: "Air gap flux density in core with three windings"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:three-winding-core-airgap-flux-density"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/three-winding-core-airgap-flux-density
site: BCT Engineering
---

# Air gap flux density in core with three windings

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2080 Chaitra, 8 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_i/three-winding-core-airgap-flux-density
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

The magnetic core with three excitation windings is shown in figure below. The mean length lA, l1 and l2 is 50 cm, 20 cm and 10 cm respectively. The length of air gap g is 10 mm. The number of turns N1 and N is 1000 and 500. If iA, iB and i1 are 20, 15 and 10 amperes, what would be the magnetic flux density at the air gap. The core area Ac = 20 cm² and relative permeability of core is μr = 4000. [Figure: rectangular core with a centre limb; the centre limb carries the N1-turn winding (current i1) and contains the air gap g; the left and right outer limbs each carry an N-turn winding with currents iA and iB respectively; lA is the outer path length, l1 and l2 are the centre-limb lengths above and below.]

## Short answer

Treating the core as a magnetic circuit with node potential, and assuming all MMFs aid flux in the centre limb, the centre limb flux is 4.648 mWb. With area 20 cm², the air‑gap flux density is 2.32 T.

## Answer

Treat the core as a two-loop magnetic circuit. **Assumption** (polarities are not given): all three windings are connected so that their mmfs drive flux in the same sense through the centre limb, i.e. the outer coils push flux up the outer limbs and the centre coil pushes it down the centre limb. Each outer path has length $l_A = 50$ cm; the centre limb has iron $l_1 + l_2 = 30$ cm plus the 10 mm gap. Fringing is neglected.

```text
   +----------+----------+
   |  phiA ^  |  phic v  |  phiB ^
 [N iA]       [N1 i1]  [N iB]
   |          |   g      |
   +----------+----------+
```

#### MMFs

$$
F_A = 500\times20 = 10000\ \text{AT}, \quad F_B = 500\times15 = 7500\ \text{AT}, \quad F_1 = 1000\times10 = 10000\ \text{AT}
$$

#### Reluctances ($A_c = 20\times10^{-4}$ m², $\mu_r = 4000$)

$$
\begin{aligned}
S_A = S_B &= \frac{0.5}{4\pi\times10^{-7}\times4000\times20\times10^{-4}} = 4.974\times10^{4}\ \text{AT/Wb} \\
S_{core,c} &= \frac{0.3}{4\pi\times10^{-7}\times4000\times20\times10^{-4}} = 2.984\times10^{4}\ \text{AT/Wb} \\
S_g &= \frac{0.01}{4\pi\times10^{-7}\times20\times10^{-4}} = 3.979\times10^{6}\ \text{AT/Wb} \\
S_c &= S_{core,c} + S_g = 4.009\times10^{6}\ \text{AT/Wb}
\end{aligned}
$$

#### Node equation

Let $U$ be the magnetic potential of the top junction relative to the bottom. Then $\phi_A = (F_A - U)/S_A$, $\phi_B = (F_B - U)/S_B$, $\phi_c = (U + F_1)/S_c$, and $\phi_c = \phi_A + \phi_B$:

$$
\begin{aligned}
U &= \frac{F_A/S_A + F_B/S_B - F_1/S_c}{1/S_A + 1/S_B + 1/S_c} = 8634\ \text{AT} \\
\phi_c &= \frac{8634 + 10000}{4.009\times10^{6}} = 4.648\times10^{-3}\ \text{Wb}
\end{aligned}
$$

(The outer limbs carry $\phi_A = 27.46$ mWb up and $\phi_B = -22.81$ mWb, i.e. 22.81 mWb down; their sum is 4.65 mWb.)

#### Flux density in the air gap

$$
B_g = \frac{\phi_c}{A_c} = \frac{4.648\times10^{-3}}{20\times10^{-4}} = 2.32\ \text{T}
$$

**Answer: $B_g \approx 2.32$ T** (for the assumed winding polarities, linear core, no fringing). Because the gap reluctance is about 80 times that of the iron, $B_g \approx \mu_0(\bar F_{outer} + F_1)/g$, so the result is mainly set by the gap.

## Check yourself

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

1. How is the core treated in the solution?
   A. Single series loop
   B. Two-loop circuit: the two outer limbs feed the centre limb, which has the gap
   C. Three independent loops
   D. Only the gap matters
   Answer: B. The outer limbs A and B are in parallel with the centre branch, giving a node equation phi_c = phi_A + phi_B. (Key point: Circuit modelling)

2. Which are the mmfs of the three windings?
   A. F_A = 10000, F_B = 7500, F_1 = 10000 AT
   B. F_A = 20, F_B = 15, F_1 = 10 AT
   C. F_A = 500, F_B = 500, F_1 = 1000 AT
   D. F_A = 10000, F_B = 10000, F_1 = 7500 AT
   Answer: A. F = N x I: 500x20, 500x15 and 1000x10. (Key point: MMF values)

3. Why is the centre-limb reluctance about 4.0e6 AT/Wb?
   A. The iron reluctance is large
   B. The 10 mm air gap has a far larger reluctance than the iron (mu_r = 1 vs 4000)
   C. Because of N1
   D. Because of the fringing
   Answer: B. S_g = 3.98e6 against only 2.98e4 for the centre iron, so the gap dominates. (Key point: Gap dominance)

4. With phi_c = 4.648 mWb and Ac = 20 cm^2, the gap flux density is:
   A. 0.23 T
   B. 2.32 T
   C. 23.2 T
   D. 4.65 T
   Answer: B. B = phi/A = 4.648e-3 / 20e-4 = 2.32 T. (Key point: Flux density result)

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