Nonlinear Time Harmonic Hysteresis Loss Model
\[ \mu_{eff} = \mu e^{-j \theta} = \mu \cos{\theta} - j \mu \sin{\theta}\]
\[L = \frac{\mu_{eff}a}{l}\]
\[Z = j \omega L = j \omega \frac{\mu a}{l} \cos{\theta} + \omega \frac{\mu a}{l} \sin{\theta}\]
Loss = \( \frac{1}{2}\omega \frac{\mu a}{l} \sin{\theta} i^2 \)
Defining \( H = i/l\),
Loss = \( \frac{1}{2}\omega \mu a l \sin{\theta} H^2 \)
Loss/Volume = \( \frac{1}{2}\omega \mu \sin{\theta} H^2 \)
If \(\left|B\right| = \mu \left|H\right|\),
Loss/Volume = \( \frac{1}{2} \frac{\omega}{\mu} B^2 \sin{\theta} \)
For small \(\theta\), \(\sin{\theta} \approx \theta\) so
Loss/Volume \( \approx \frac{1}{2} \omega \frac{\theta}{\mu} B^2 \)
If we desire for loss to be proportional to \(B^2\), even though \(\mu\) is amplitude-dependent for a nonlinear time harmonic problem, we need to define \( \theta/\mu \) define:
\[ \frac{\mu}{\theta} = \frac{\mu_{max}}{\theta_{hmax}} \]
where \(\theta_{hmax}\) is the hysteresis angle at the maximum permeability point of the material, and \(\mu_{max}\) is the maximum permeability achieved by the material. Solving for \(\theta\) as a function of permeability, \(\mu\), we get:
\[ \theta = \theta_{hmax} \left( \frac{\mu}{\mu_{max}}\right) \]
Inspecting the above, it is clear that \(\theta_{hmax}\) not only coincides with the maximum permeability of the material but also represents the biggest hysteresis angle experienced by the material.
For incremental permeability problems, we might still want the hysteresis loss to be proportional to \(B^2\) for minor loops. In these cases, the incremental permability is \(\mu_{inc}\), so we could then still say:
\[ \theta = \theta_{hmax} \left( \frac{\mu_{inc}}{\mu_{max}}\right) \]
where \(\mu_{max}\) is still the maximum permeability of the AC equivalent BH curve.