---
title: "Fluxes in three-limb core with two air gaps"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:three-limb-core-two-airgaps-300turns"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/three-limb-core-two-airgaps-300turns
site: BCT Engineering
---

# Fluxes in three-limb core with two air gaps

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2078 Baisakh, 8 marks
- Video: [(English)EM ||  Question 1.6 || Magnetic Circuits || Core with Two Air Gaps](https://www.youtube.com/watch?v=P32UoFLVtXM) — 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-limb-core-two-airgaps-300turns
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

A ferromagnetic core with a relative permeability of 4000 is shown below. The dimensional area as shown and depth of core is 7 cm. The air gap is 0.05 cm on left and 0.07 cm on right side. The effective area of gap is 5% larger due to fringing effect. If current in coil is 10 A, calculate flux in left, right and middle part along with flux density in each air gap. [Figure: three-limb core; all limbs and yokes are 7 cm wide; the two windows are each 30 cm wide and 30 cm high; the centre limb carries a 300-turn coil with current i; the left outer limb has a 0.05 cm air gap and the right outer limb has a 0.07 cm air gap.]

## Short answer

Solve the magnetic circuit with 3000 AT MMF. Reluctances of left and right limbs (iron plus gap) are in parallel, then series with centre limb. Flux in middle = 36.1 mWb, left = 20.1 mWb, right = 16.0 mWb; gap flux densities 3.90 T (left) and 3.12 T (right).

## Answer

The centre limb (with the coil) carries the total flux, which divides between the left and right outer paths. Each outer path contains an air gap, so the two outer paths are in parallel and both are in series with the centre limb.

**Dimensions:** limbs 7 cm wide, depth 7 cm → $A = 49$ cm² $= 4.9\times10^{-3}$ m². Windows 30 × 30 cm, so the centre-line spacing is 30 + 7 = 37 cm.
- Centre limb: $l_c = 37$ cm.
- Each outer path: top yoke 37 + outer limb 37 + bottom yoke 37 = 111 cm (minus its gap).
- Gap area with 5% fringing: $A_g = 1.05 \times 49 = 51.45$ cm².
- MMF: $F = 300 \times 10 = 3000$ AT.

#### Reluctances ($\mu_0\mu_r A = 4\pi\times10^{-7}\times4000\times4.9\times10^{-3} = 2.463\times10^{-5}$)

$$
\begin{aligned}
S_c &= \frac{0.37}{2.463\times10^{-5}} = 15022\ \text{AT/Wb} \\
S_{L,iron} &= \frac{1.1095}{2.463\times10^{-5}} = 45047, \quad S_{L,gap} = \frac{0.0005}{4\pi\times10^{-7}\times5.145\times10^{-3}} = 77335 \\
S_L &= 45047 + 77335 = 122381\ \text{AT/Wb} \\
S_{R,iron} &= \frac{1.1093}{2.463\times10^{-5}} = 45038, \quad S_{R,gap} = \frac{0.0007}{4\pi\times10^{-7}\times5.145\times10^{-3}} = 108269 \\
S_R &= 45038 + 108269 = 153307\ \text{AT/Wb}
\end{aligned}
$$

#### Total reluctance and fluxes

$$
\begin{aligned}
S_L \parallel S_R &= \frac{122381\times153307}{122381+153307} = 68055\ \text{AT/Wb} \\
S_{tot} &= 15022 + 68055 = 83077\ \text{AT/Wb} \\
\phi_c &= \frac{3000}{83077} = 0.03611\ \text{Wb} \\
F_{LR} &= \phi_c \times 68055 = 2457.6\ \text{AT} \\
\phi_L &= \frac{2457.6}{122381} = 0.02008\ \text{Wb}, \quad \phi_R = \frac{2457.6}{153307} = 0.01603\ \text{Wb}
\end{aligned}
$$

#### Flux density in the gaps

$$
B_{gL} = \frac{0.02008}{5.145\times10^{-3}} = 3.90\ \text{T}, \quad B_{gR} = \frac{0.01603}{5.145\times10^{-3}} = 3.12\ \text{T}
$$

**Answer:** $\phi_{middle} = 36.1$ mWb, $\phi_{left} = 20.1$ mWb, $\phi_{right} = 16.0$ mWb; $B_{gap,left} = 3.90$ T, $B_{gap,right} = 3.12$ T. (These densities are far above saturation of real iron; the result holds only for the constant μr = 4000 given.)

## Check yourself

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

1. How are the paths arranged?
   A. All three in series
   B. Left and right outer paths in parallel, both in series with the centre limb
   C. Centre and left in parallel, right in series
   D. Left and right in series
   Answer: B. The centre limb carries the total flux, which divides between the two outer paths. (Key point: Circuit arrangement)

2. How is fringing handled for the gap area?
   A. Neglected
   B. Gap area = 1.05 x 49 = 51.45 cm^2
   C. Gap area = 0.95 x 49
   D. Doubled
   Answer: B. The effective gap area is 5% larger, so reluctance of the gaps is computed with 51.45 cm^2. (Key point: Fringing)

3. Which flux relation is right?
   A. Right limb carries more flux because its gap is larger
   B. Left limb carries more flux because its gap is smaller (lower reluctance)
   C. Both carry equal flux
   D. Flux in the centre equals the left flux
   Answer: B. Same mmf across both outer paths, smaller reluctance on the left gives phi_L = 20.1 mWb > phi_R = 16.0 mWb. (Key point: Flux division)

4. The centre limb flux is found by dividing 3000 AT by:
   A. S_L + S_R
   B. S_c + (S_L parallel S_R) = 83077 AT/Wb
   C. S_L x S_R
   D. S_c only
   Answer: B. Series of centre with the parallel pair gives phi_c = 3000/83077 = 36.1 mWb. (Key point: Total flux)

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