---
title: "Hysteresis and eddy current losses; loop area"
subject: "Electrical Machine"
chapter: 1
key: "semester_4/electrical_machine:hysteresis-eddy-loss-area"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/hysteresis-eddy-loss-area
site: BCT Engineering
---

# Hysteresis and eddy current losses; loop area

Electrical Machine · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2074 Bhadra, 5 marks
- Video: [Whats are Hysteresis Loss and Eddy Current Loss?](https://www.youtube.com/watch?v=ntrlOUCO6eM) — Regie Marinay (Regie)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/hysteresis-eddy-loss-area
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-1

## Question

Explain hysteresis and eddy current losses in electrical machines. Prove that hysteresis loss in a magnetic material is equal to the area of the hysteresis loop.

## Short answer

Hysteresis loss (P_h ∝ B_max^1.6 f V) comes from domain friction; eddy current loss (P_e ∝ B_max² f² t² V) from induced currents. Integrating e i dt gives energy/volume = H dB, so per cycle loss equals area of B-H loop.

## Answer

**Hysteresis loss.** When the flux in a magnetic material alternates, its magnetic domains are reversed again and again. They resist this because of internal friction, so the flux density *B* lags behind the field strength *H*. The energy spent in each cycle of reversal appears as heat in the core. This is the hysteresis loss.

Steinmetz's formula: **Pₕ = ηₕ · B_max^1.6 · f · V** (watts), where ηₕ is the Steinmetz coefficient of the material, *f* the frequency and *V* the volume of the core.

**Eddy current loss.** An alternating flux induces emf in the core itself. Since the core is conducting, this emf drives circulating currents (eddy currents) in planes perpendicular to the flux. They cause I²R loss and heat the core.

**Pₑ = kₑ · B_max² · f² · t² · V** (watts), where *t* is the thickness of the lamination and kₑ a constant depending on the resistivity of the material.

#### Proof: hysteresis loss = area of the B–H loop
Take a core of length *l*, area *a* and *N* turns, carrying current *i*, with field *H = Ni/l*.

```text
emf induced         e = N dφ/dt = N a dB/dt
energy in time dt   dW = e i dt
                       = N a (dB/dt) · (H l/N) · dt
                       = (a l) H dB
```
Here *a·l = V* is the volume of the core, so the energy per unit volume is **dW/V = H dB**.

For one complete cycle the total energy per unit volume is

**W = ∮ H dB** = area enclosed by the B–H loop (J/m³ per cycle)

At *f* cycles per second, **Pₕ = f · V · (loop area)** watts. A larger loop means a larger loss; this is why soft magnetic materials with a narrow loop are used for cores.

## Check yourself

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

1. What causes hysteresis loss in a core carrying alternating flux?
   A. Domains being repeatedly turned against internal friction, so B lags H
   B. Resistance of the winding
   C. Currents circulating in the core
   D. Leakage flux outside the core
   Answer: A. Repeated reversal of the magnetic domains against friction wastes energy as heat each cycle. (Key point: Cause of hysteresis loss)

2. Which expression gives eddy current loss?
   A. ke Bmax^2 f t V
   B. ke Bmax f^2 t V
   C. ke Bmax^2 f^2 t^2 V
   D. ke Bmax^1.6 f V
   Answer: C. Eddy loss is Pe = ke Bmax^2 f^2 t^2 V, depending on the square of thickness and frequency. (Key point: Eddy current loss formula)

3. In the proof, what is the energy per unit volume supplied in time dt?
   A. B^2/2
   B. B dH
   C. H dB
   D. H B
   Answer: C. dW = e i dt = (a l) H dB, so per unit volume it is H dB. (Key point: Energy per step)

4. Why does the hysteresis loss per cycle equal the B-H loop area?
   A. Because eddy currents add to it
   B. Because B is proportional to H
   C. Because the loop area equals the power in watts directly
   D. Integrating H dB over one full cycle gives the area enclosed by the loop
   Answer: D. W = closed integral of H dB is the area of the loop (J/m^3 per cycle); multiplied by f and V it gives watts. (Key point: Loss equals loop area)

## Related in Electrical Machine

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