---
title: "Hysteresis loss and eddy current loss"
subject: "Electric Machine Design"
chapter: 1
key: "semester_5/electric_machine_design:hysteresis-eddy-loss"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/hysteresis-eddy-loss
site: BCT Engineering
---

# Hysteresis loss and eddy current loss

Electric Machine Design · Chapter 1: Materials used in electrical equipment

- Asked in: 2076 Chaitra, 4 marks
- Video: [AC's Hidden Enemies: Hysteresis & Eddy Current Loss Explained](https://www.youtube.com/watch?v=PO6buWiLwOU) — Electrology
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/hysteresis-eddy-loss
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-1

## Question

Explain hysteresis loss and eddy current loss.

## Short answer

Hysteresis loss arises from domain reorientation, proportional to f and B_m^x (Steinmetz exponent ~1.6), reduced by using narrow-loop silicon steel. Eddy current loss is from circulating currents induced by alternating flux, proportional to f^2, B_m^2 and square of lamination thickness, reduced by laminating the core with thin insulated sheets and adding silicon.

## Answer

Iron (core) loss in a magnetic core carrying alternating flux has two parts: **hysteresis loss** and **eddy current loss**.

#### Hysteresis loss
When the core is taken round a complete magnetisation cycle, the domains are repeatedly re-oriented. The energy spent on this, equal to the **area of the B-H loop** per cycle per unit volume, is lost as heat.

$$
P_h = k_h\, B_m^{x}\, f\, V \quad \text{watts}
$$

- $k_h$ = hysteresis constant of the material
- $B_m$ = maximum flux density
- $x$ = Steinmetz exponent (1.6 for older steels, about 1.6–2 for modern ones)
- $f$ = frequency, $V$ = volume of core

**Reduction:** use a soft material with a narrow loop (silicon steel, CRGO), work at a moderate $B_m$, and anneal the laminations.

#### Eddy current loss
The alternating flux induces emfs in the core itself. These drive circulating (eddy) currents in closed paths in the iron, causing $I^2R$ loss.

$$
P_e = k_e\, B_m^2\, f^2\, t^2\, V \quad \text{watts}
$$

For thin laminations of thickness $t$ and resistivity $\rho$, the loss per unit volume is

$$
p_e = \frac{\pi^2 B_m^2 f^2 t^2}{6\rho} \quad \text{W/m}^3
$$

**Reduction:**
- **Laminate** the core (0.23–0.5 mm sheets) insulated by varnish or oxide, so the eddy paths are broken up. The loss varies as $t^2$.
- **Raise the resistivity** by adding silicon.
- Use ferrite or powdered cores at high frequency.

```text
 solid core: large eddy loop    laminated: small loops
 +-------------+               +--+--+--+--+
 |  .-------.  |               |()|()|()|()|
 |  |  ->   |  |               |  |  |  |  |
 |  '-------'  |               |()|()|()|()|
 +-------------+               +--+--+--+--+
```

| Point | Hysteresis loss | Eddy current loss |
|---|---|---|
| Cause | Domain re-orientation | Induced circulating currents |
| Frequency dependence | $\propto f$ | $\propto f^2$ |
| Depends on thickness | No | Yes, $\propto t^2$ |
| Reduced by | Silicon steel, CRGO | Lamination, high resistivity |

## Check yourself

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

1. What is the hysteresis loss per cycle equal to?
   A. Area of the B-H loop per unit volume
   B. Peak B times H
   C. Zero for soft iron
   D. Resistance times current squared
   Answer: A. Energy spent re-orienting domains equals the loop area per cycle per unit volume. (Key point: Hysteresis loss origin)

2. How do hysteresis and eddy losses vary with frequency?
   A. Both ∝ f
   B. Hysteresis ∝ f, eddy ∝ f^2
   C. Hysteresis ∝ f^2, eddy ∝ f
   D. Both ∝ f^2
   Answer: B. P_h = k_h Bm^x f V and P_e = k_e Bm^2 f^2 t^2 V. (Key point: Frequency dependence)

3. If lamination thickness is halved, by what factor does eddy loss fall?
   A. 2
   B. 4
   C. 8
   D. unchanged
   Answer: B. Eddy loss ∝ t^2, so halving t gives one quarter. (Key point: Thickness)

4. Which reduces eddy current loss?
   A. Laminating and adding silicon to raise resistivity
   B. Solid core
   C. Lower resistivity
   D. Thicker sheets
   Answer: A. Lamination breaks eddy paths and silicon raises ρ. (Key point: Reducing eddy loss)

## Related in Electric Machine Design

- [No load current of single phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Atransformer-no-load-current) — Electric Machine Design, chapter 3

## Related in other subjects

- [Hysteresis and eddy current losses; loop area](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Ahysteresis-eddy-loss-area) — Electrical Machine, chapter 1
- [Hysteresis and eddy current losses, reduction](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Ahysteresis-eddy-losses-reduce) — Electrical Machines, chapter 1
- [Short note: Eddy current loss](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Ashort-note-eddy-current-loss) — Electrical Machine, chapter 1
- [Eddy current loss and its minimization](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aeddy-current-loss-minimize) — Electrical Machines I, chapter 1