---
title: "Three-limb core, ampere-turns for flux"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:three-limb-core-amp-turn"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/three-limb-core-amp-turn
site: BCT Engineering
---

# Three-limb core, ampere-turns for flux

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2078 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
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/three-limb-core-amp-turn
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

For the magnetic circuit shown below, calculate the Amp-turn (NI) required to establish a flux of 0.75 Wb in the central limb. Given that $\mu_r = 4000$ for iron core.
[Figure: double-window (three-limb) iron core, overall width 44 cm and height 20 cm, with 4 cm thick top and bottom yokes; bottom dimensions from left to right 4 cm, 15 cm, 6 cm, 15 cm, 4 cm (outer limbs 4 cm, central limb 6 cm, windows 15 cm); a coil of N turns carrying current I is on the central limb; three hatched regions at the bottom of the limbs [?]]

## Short answer

Assuming 5 cm core depth, central limb reluctance = 1.061×10⁴ AT/Wb, each side path reluctance = 5.57×10⁴ AT/Wb. Flux divides equally. NI = Φ S_c + (Φ/2) S_side = 0.75×1.061e4 + 0.375×5.57e4 = 2.88×10⁴ AT for 0.75 Wb; if flux is 0.75 mWb, NI ≈ 28.8 AT.

## Answer

The coil is on the central limb, so the central flux divides equally between the two outer limbs, which are identical. This gives a parallel-series circuit, solved with reluctances.

**Assumptions (the figure is incomplete):**
- The core depth (thickness) is not shown, so I take $d = 5$ cm throughout. The result is inversely proportional to $d$.
- Mean paths are taken through the centre of each member. Central limb and each outer limb: $20 - 4 = 16$ cm. Top or bottom yoke between centre limb and outer limb: $15 + 3 + 2 = 20$ cm.
- The flux is taken as given, 0.75 Wb. The answer is proportional to flux, so for 0.75 mWb divide by 1000.

```text
   +---+---------------+------+---------------+---+
   |   top yoke 4 cm                              |
   |  outer      window    centre    window outer |
   |  4 cm       15 cm     6 cm      15 cm   4 cm |
   |   bottom yoke 4 cm                           |
   +----------------------------------------------+
```

**Areas**

$$
A_c = 0.06\times0.05 = 3\times10^{-3}\ \text{m}^2,\qquad A_o = A_y = 0.04\times0.05 = 2\times10^{-3}\ \text{m}^2
$$

**Reluctances** with $\mu = 4\pi\times10^{-7}\times4000 = 5.027\times10^{-3}$ H/m:

$$
\begin{aligned}
S_c &= \frac{0.16}{5.027\times10^{-3}\times3\times10^{-3}} = 1.061\times10^{4}\ \text{AT/Wb} \\
S_o &= \frac{0.16}{5.027\times10^{-3}\times2\times10^{-3}} = 1.592\times10^{4}\ \text{AT/Wb} \\
S_y &= \frac{0.20}{5.027\times10^{-3}\times2\times10^{-3}} = 1.989\times10^{4}\ \text{AT/Wb}
\end{aligned}
$$

One side path (outer limb + top yoke + bottom yoke) has $S_{side} = 1.592\times10^4 + 2(1.989\times10^4) = 5.570\times10^{4}$. Flux in each side path is $\Phi_c/2 = 0.375$ Wb.

**Mmf**

$$
\begin{aligned}
NI &= \Phi_c S_c + \frac{\Phi_c}{2}S_{side} \\
&= 0.75\times1.061\times10^{4} + 0.375\times5.570\times10^{4} \\
&= 7957 + 20\,889 = 28\,847\ \text{AT}
\end{aligned}
$$

**Answer: $NI \approx 2.88\times10^{4}$ AT for 0.75 Wb (about 28.8 AT if the flux is 0.75 mWb), with $d = 5$ cm assumed.**

## Check yourself

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

1. The coil is on the central limb. How does its flux divide?
   A. All flux goes to one outer limb
   B. Equally between the two identical outer limbs
   C. The outer limbs carry twice the central flux each
   D. It does not divide
   Answer: B. The outer paths are identical and in parallel, so each carries half the central flux. (Key point: Flux division)

2. How is the total mmf found?
   A. Phi_c S_c + (Phi_c/2) S_side
   B. Phi_c (S_c + S_side)
   C. Phi_c S_side only
   D. (Phi_c/2) S_c + Phi_c S_side
   Answer: A. The central limb carries full flux; the series side path (outer limb + yokes) carries half. (Key point: Circuit structure)

3. Why is the central limb reluctance S_c less than S_o, despite equal length?
   A. Its area is larger (6 cm vs 4 cm width)
   B. Its permeability is higher
   C. It has an air gap
   D. It is shorter
   Answer: A. S = l/(mu A) falls as area rises; A_c = 3e-3 against 2e-3 m^2. (Key point: Dependence on area)

4. The answer assumes core depth 5 cm. How does NI change with depth d?
   A. Proportional to d
   B. Proportional to d^2
   C. Inversely proportional to d
   D. Independent of d
   Answer: C. Reluctance is inversely proportional to area and so to depth, so NI for a given flux is proportional to 1/d. (Key point: Dependence on assumptions)

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