---
title: "Minimum width of stator teeth"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:minimum-stator-teeth-width"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/minimum-stator-teeth-width
site: BCT Engineering
---

# Minimum width of stator teeth

Electric Machine Design · Chapter 4: Three phase induction motor design

- Asked in: 2071 Shrawan, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/minimum-stator-teeth-width
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4

## Question

Calculate the minimum width of stator teeth of induction motor.

## Short answer

The minimum stator tooth width limits tooth flux density to about 1.7 T. It is calculated as W_ts(min) = Φ / [1.7 × (S_s / P) × L_i], where Φ is flux per pole, S_s stator slots, P poles, and L_i the net iron length (stacked core) with stacking factor ~0.9.

## Answer

The **minimum width of a stator tooth** is the narrowest tooth that can carry the flux without its flux density exceeding about **1.7 Wb/m²**. Above this value the teeth saturate, and the magnetising current and iron loss rise sharply.

#### Derivation
- Flux per pole: $\Phi = B_{av}\tau L$.
- This flux passes through the teeth under one pole, and the number of teeth is $S_s/P$.
- Iron area of these teeth $= \dfrac{S_s}{P} \times W_{ts} \times L_i$, where $L_i = k_i(L - n_dw_d)$ is the net iron length ($k_i \approx 0.9$).
- Setting $B_t \le 1.7$ Wb/m²:
$$
\frac{\Phi}{\frac{S_s}{P}W_{ts}L_i} \le 1.7 \;\Rightarrow\; W_{ts(min)} = \frac{\Phi}{1.7 \times \frac{S_s}{P} \times L_i}
$$

#### Example
$\Phi = 4.51 \times 10^{-3}$ Wb, $S_s = 24$, $P = 4$, $L = 0.124$ m, $L_i = 0.9 \times 0.124 = 0.1116$ m:
$$
W_{ts(min)} = \frac{4.51\times10^{-3}}{1.7 \times 6 \times 0.1116} = 3.96\ \text{mm}
$$
The actual tooth width is the slot pitch minus the slot width. Here the slot pitch is $13.78$ mm and the slot width is $6.75$ mm, so the tooth is $13.78 - 6.75 = 7.0$ mm wide. This is larger than the minimum, so the design is safe.

## Check yourself

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

1. What fixes the minimum stator tooth width?
   A. Tooth flux density must not exceed about 1.7 Wb/m^2
   B. Slot pitch must be 12 mm
   C. Gap length must be 0.5 mm
   D. Bav must equal 0.5
   Answer: A. A narrower tooth would saturate and increase magnetising current and iron loss. (Key point: Basis of minimum width)

2. Which flux passes through the teeth under one pole?
   A. Phi = Bav D L shared among all teeth
   B. Phi = Bav tau L, shared among Ss/P teeth
   C. Twice the pole flux
   D. Zero, as flux goes through slots
   Answer: B. The flux per pole enters the stator via the teeth under that pole, numbering Ss/P. (Key point: Flux through teeth)

3. How is the net iron length found?
   A. Li = L + nd wd
   B. Li = L/ki
   C. Li = ki(L - nd wd), with ki about 0.9
   D. Li = L x ki x D
   Answer: C. Ducts and lamination insulation reduce the iron length. (Key point: Net iron length)

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