---
title: "Choice of magnetic loading in induction machine"
subject: "Electric Machine Design"
chapter: 4
key: "semester_5/electric_machine_design:induction-magnetic-loading"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/induction-magnetic-loading
site: BCT Engineering
---

# Choice of magnetic loading in induction machine

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

- Asked in: 2072 Kartik, 8 marks
- Video: [Choice of Specific Magnetic Loading (Bav) #sdyelgatte #designofelectricalmachines](https://www.youtube.com/watch?v=xJ-B7L0Z5g0) — S D Yelgatte 📖 ⚡
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/induction-magnetic-loading
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-4

## Question

Discuss the factors to be considered for selection of magnetic loading in induction machine.

## Short answer

Specific magnetic loading (B_av, typically 0.3–0.6 T) is selected based on power factor (higher B_av raises magnetising current and lowers PF), iron loss and efficiency, overload capacity (higher B_av increases maximum torque), tooth saturation limit (~1.7 T), machine size and cost, and core material quality.

## Answer

**Specific magnetic loading** ($B_{av}$) is the average flux density over the air-gap surface:
$$
B_{av} = \frac{P\Phi}{\pi DL}
$$
For induction motors it usually lies between 0.3 and 0.6 Wb/m². The following factors decide its value.

#### 1. Power factor
The magnetising current depends mostly on the gap flux density. A higher $B_{av}$ needs more magnetising mmf, so the magnetising current rises and the power factor falls. Since an induction motor takes all its magnetising current from the supply, this is the main limit on $B_{av}$.

#### 2. Iron loss and efficiency
Iron loss rises roughly as $B^2$ (and with frequency). A high $B_{av}$ raises core loss, lowers efficiency and raises temperature rise. High-frequency machines use lower $B_{av}$.

#### 3. Overload capacity
A higher $B_{av}$ means fewer turns for the same voltage ($T \propto 1/\Phi$), so leakage reactance is lower and maximum torque is higher. A high value therefore improves overload capacity.

#### 4. Flux density in teeth
The teeth carry the gap flux. Tooth flux density must not exceed about 1.7 Wb/m² (to avoid saturation and high magnetising current). With narrow teeth, $B_{av}$ must be lower.

#### 5. Size and cost
From $Q = C_0D^2Ln_s$ with $C_0 \propto B_{av}$, a higher $B_{av}$ gives a smaller machine with less material and lower cost.

#### 6. Quality of core material
Better (low-loss, cold-rolled) steel allows a higher $B_{av}$ without excessive loss.

#### Summary
| Higher $B_{av}$ gives | Effect |
|---|---|
| Size and cost | Smaller, cheaper |
| Overload capacity | Higher |
| Magnetising current | Higher |
| Power factor | Lower |
| Iron loss, temperature | Higher |
| Efficiency | Lower |

So the designer chooses a value (about 0.4–0.5 Wb/m² for 50 Hz motors) that balances power factor and efficiency against size, cost and overload capacity.

## Check yourself

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

1. What is the main factor limiting a high Bav in an induction motor?
   A. Power factor, because magnetising current rises with flux density
   B. Overload capacity
   C. Cost of copper
   D. Starting torque
   Answer: A. The motor draws all magnetising current from the supply, so high Bav lowers power factor. (Key point: Main limit on Bav)

2. How does a higher Bav affect iron loss and efficiency?
   A. Iron loss falls and efficiency rises
   B. Iron loss rises (about B^2), efficiency falls and temperature rises
   C. Only copper loss changes
   D. No effect on temperature
   Answer: B. Core loss varies about as the square of flux density. (Key point: Iron loss)

3. Why does a higher Bav improve overload capacity?
   A. More turns give more leakage flux
   B. Magnetising current falls
   C. Fewer turns are needed for the same voltage, so leakage reactance is lower
   D. Slip falls
   Answer: C. T is proportional to 1/Phi, so fewer turns lower leakage reactance and raise maximum torque. (Key point: Overload capacity)

4. What limit applies to flux density in teeth?
   A. About 0.5 Wb/m^2
   B. About 3 Wb/m^2
   C. There is no limit
   D. About 1.7 Wb/m^2, above which teeth saturate
   Answer: D. Saturated teeth raise magnetising current and iron loss, so narrow teeth force a lower Bav. (Key point: Tooth density limit)

## Related in Electric Machine Design

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- [Effect of too wide air gap in induction machine](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Awide-air-gap-effects) — Electric Machine Design, chapter 4
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