Does Core Material Influence the H-Field in a Toroidal Coil?

[Papikoss]

H-field depends on core material of toroid?

If I got this right, the H-field does not depend on the material it is applied on and is only determined by its sources (free currents).

So, suppose that we have a toroidal coil and some undefined core. As far as the H-filed is concerned, it would not make a difference if that core was air, a ferromagnetic material or some combination of those two. The H-field would remain the same.

However, if we apply Ampere's Law to a simple magnetic circuit, consisting of a ferromagnetic coil toroid having a small air gap, the result would be (taken from theory):

HfLf + HgLg = NI, where f, g refer to ferromagnetic material and air gap respectively.

Obviously here Hg is much greater than Hf

Now I'm confused! I know that the magnetic flux must remain constant throughout the circuit, but still I can't grasp how Hg and Hf could be different (the distribution of free currents did not change)!

Any help much appreciated!

 

[cortiver]

If I got this right, the H-field does not depend on the material it is applied on and is only determined by its sources (free currents).

No, this is not true except in special cases (despite what my not-quite-up-to-speed EM lecturer once tried to tell me!). If you have David Griffith's electromagnetism textbook, then he addresses this point in Section 6.3.2, "A Deceptive Parallel". The point he makes is that a vector field like H is specified by giving its curl and divergence at every point in space. The curl \nabla \times \mathbf{H} is indeed determined by the free current, by Ampere's Law \nabla \times \mathbf{H} = \mathbf{J}_f, but divergence \nabla \cdot \mathbf{H} is not. In the absence of magnetic materials, H has zero divergence, but otherwise it might not.

 

[Papikoss]

Thank you for widening my horizons! Let’s see if I got this right this time:

The magnetic field depends both on free and bound currents.

In the example above (toroid with a ferromagnetic core, having a small gap) we can assume uniform magnetization within the core region, so no bound current density can be present, as is deducted from: \nabla\times\mathbf{M} = \mathbf{J}_b

However, this is not the case with bound currents on the surface of the core, having a distribution similar to the distribution of the free currents and contributing to the final magnetic field.

Now if we assume a constant magnetic flux throughout the magnetic circuit (and a constant cross-sectional area) then the magnitude of magnetic flux density B is a constant too.

So within the ferromagnetic material it is :

\Large\mathbf{H}_f = \frac{\mathbf{B}}{\mu_0} - \mathbf{M}

And within the gap area, where no magnetization applies,

\Large\mathbf{H}_g = \frac{\mathbf{B}}{\mu_0}

Apparently, \mathbf{H}_g\gg\mathbf{H}_f.

Is that whole approach correct?

Thanks in advance!

PS How can I change the LaTeX font size of the whole document to Large?

 

[cortiver]

Yes, the H field inside the ferromagnetic material is much smaller because the effect of the B field is mostly canceled by the magnetization. In particular, if we forget about non-linearity, hysteresis, etc., and just assume the material is characterized by a large magnetic susceptibility \chi_m, then we get

<br /> H_f = \frac{1}{1 + \chi_m}\frac{B}{\mu_0} \ll \frac{B}{\mu_0} = H_g<br />

Another way of thinking about this, in light of your original question, is that, in addition to the free current, there is a contribution to H from the divergence of the magnetization M (since \nabla \cdot \mathbf{H} = -\nabla \cdot \mathbf{M}). The only place that M has nonzero divergence is at the edge of the gap, so this part of the H field is like the electric field from a parallel-plate capacitor. Thus the H field is much greater inside the gap due to this contribution.