Magnetic hysteresis and energy per cycle

Magnetic hysteresis is the lagging of the flux density $B$ behind the magnetising force $H$ in a ferromagnetic material. When $H$ is increased, reduced, reversed and brought back, the $B$–$H$ curve does not retrace itself but forms a closed loop.

      B
      |     ____ saturation
      |   /
   Br |--/--.
      | /    \
 -----+-+-----+------ H
     -Hc     Hc
      |  loop area = energy lost

Explanation: the material is made of small magnetic domains. When $H$ rises they rotate and align, and $B$ increases up to saturation. When $H$ falls to zero, some domains stay aligned, so a residual flux density $B_r$ (retentivity) remains. A reverse field $-H_c$ (coercive force) is needed to bring $B$ to zero. Overcoming this internal friction between domains costs energy, which appears as heat.

Energy per cycle equals loop area

Take a core of cross-section $A$, mean length $l$ and $N$ turns carrying current $i$.

$$\begin{aligned} e &= N\frac{d\phi}{dt} = NA\frac{dB}{dt} \\ H &= \frac{Ni}{l} \Rightarrow i = \frac{Hl}{N} \end{aligned}$$

Energy supplied by the source in time $dt$ is

$$\begin{aligned} dW &= e\,i\,dt = NA\frac{dB}{dt}\cdot\frac{Hl}{N}\,dt \\ &= (Al)\,H\,dB \end{aligned}$$

Here $Al = V$ is the volume of the core. Energy per unit volume is therefore

$$\frac{dW}{V} = H\,dB$$

For one full cycle,

$$w = \oint H\,dB = \text{area of the } B\text{–}H\text{ loop}$$

In the rising part of the cycle, part of this energy is stored in the field and is returned when the field falls. The rest is not returned and is lost as heat, which is the area enclosed by the loop. So the energy lost per cycle per unit volume equals the loop area (in J/m$^3$). For frequency $f$ and volume $V$, $P_h = f V \times (\text{loop area})$.