Hysteresis loop area and energy loss

The hysteresis loop is the closed B–H curve traced by a ferromagnetic material when the magnetising force H is taken through one complete cycle (+Hmax → 0 → −Hmax → 0 → +Hmax). Because B lags behind H, the curve for decreasing H does not retrace the rising curve, and the enclosed area represents energy lost as heat in every cycle.

The hysteresis loop

               B
               |      a (saturation)
           b   |  ____*
            *--|-'   /
           /   |    /
     c    /    |   /
  ---*---/-----+--/----*--- H
        /      |  /    f
       /   ____|_*
  d  *'        |  e
               |

• o–a: an unmagnetised sample is magnetised along the initial curve to saturation (point a).
• a–b: when H is reduced to zero, B does not fall to zero. The value ob is the residual flux density (retentivity).
• b–c: a reverse H is needed to make B zero. oc is the coercive force (coercivity).
• c–d: further reverse H gives reverse saturation at d.
• d–e–f–a: reducing H to zero gives the negative residual flux density oe, and a positive H (of) is needed to bring B to zero; the loop closes at a.

Reason: the magnetic domains line up with H but do not all turn back when H is removed. Turning them back needs extra energy, which appears as heat.

Proof: loop area ∝ energy loss per cycle per unit volume

Take a ring of mean length $l$, area $A$, wound with $N$ turns carrying current $i$.

$$H = \frac{Ni}{l} \Rightarrow i = \frac{Hl}{N}$$

If the flux density changes by $dB$ in time $dt$, the induced emf is

$$e = N\frac{d\phi}{dt} = NA\frac{dB}{dt}$$

The source must supply power $ei$ against this emf. Energy supplied in time $dt$:

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

Energy per unit volume (volume $= Al$): $dw = H\,dB$. Over one complete cycle:

$$w = \oint H\,dB \quad \text{J/m}^3 \text{ per cycle}$$

$\oint H\,dB$ is exactly the area enclosed by the B–H loop. Part of the energy given while B rises is returned while B falls; the difference, equal to the loop area, is lost as heat. Hence the hysteresis power loss is

$$P_h = (\text{area of loop}) \times \text{volume} \times f \quad \text{watts}$$

If the loop is drawn to scales of $x$ A/m per cm and $y$ T per cm, loop area (in J/m³) = measured area (cm²) × $xy$.

Practical result

Steinmetz found empirically that $P_h = \eta B_{max}^{1.6} f V$. Machines therefore use silicon steel, whose loop is narrow (small area), to keep hysteresis loss low.