Does H field strength equal MMF divided by effective length?

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Hi all, I am trying to get my head around some experiment results but I am struggling a bit.
According to my notes H field strength is equal to MMF / effective length

However this doesn't take into account the effects of reluctance? Eg a toroid made from 2m of iron bar and a straight 1m length of iron bar will require the same mmf to generate the same H field regardless that in one case 50% of the path is through the low permitivitty of air?

can anyone shed some light on this for me please:confused:

thanks in advance guys and girls
 
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please -- someone's got to have a better understanding of this than me lol
 
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It's kind of similar to the way electric fields work. If you apply e.g. 1V to a parallel plate capacitor with a gap of 1mm you get 1000V/m independent of the permittivity of the dielectric.
Understanding the equations of electromagnetism becomes easier if you realize that the units and equations used in the SI system are defined such that the magnetic H field behaves mathematically equivalent to the electric E field. And B equivalent to D. That's just mathematically though and only in the SI system. The physical reality is basically the opposite of this, which can cause confusion. That has to do with the fact that matter interacts differently with magnetic fields than with electric fields. A dielectric material will weaken an E field. A ferromagnetic material will strengthen a B field.

Let's compare a few equations here.

Electric capacitance: [itex]C = q/V[/itex], measured in [itex]As/V[/itex]
But you could also write: [itex]C = \Phi_D/V[/itex]
Magnetic permeance: [itex]P = \Phi_B/NI[/itex], measured in [itex]Vs/A[/itex]

So the magnetic permeance is mathematically equivalent to the electric capacitance.

[itex]E = EMF/l[/itex]

[itex]H = MMF/l[/itex]

[itex]D = \epsilon E[/itex]

[itex]B = \mu H[/itex]

In fact when calculating a magnetic circuit you can usually look at the mathematically equivalent electric circuit ( by replacing B with D, H with E, C with P, V with A, etc. ), do all the calculations with that and in the end transform the results back.