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I am trying to calculate the magnetic field, Hg, and have been looking these formulas. I would like to know if I am on the right track, as something is not working out for me. From these with a uniform area, the area of the core, Ac is equal to the area of the air gap, Ag. The flux density in the core, Bc, is equal to the flux density of the gap, Bc?

And to work out Hg, it is Bg/Uo, Uo the permeability of free space? And similarly for Hc the magnetic field in the core is Bc/u, u the permeability of the core?

I am looking at the last formula for the magnetic motive force, Ni = Hclc + Hglg But in my workings I do not get it to match and I think I have gone wrong somewhere or interpreted something wrong.

Any help and explanation would be appreciated.

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  • \$\begingroup\$ Air, in your case it seems, has about 4k times the reluctance of your core material. So one thing that is immediately apparent, given the physical lengths you show, is that the air gap is almost entirely all that matters. Think of the air gap as a resistor and the core material as a wire. Do you worry about the resistance of the wire if the resistor itself has 100 times the resistance of the wire? Mostly, you don't. 4k*.01 m= 40 m. So your air gap "is like" 40 meters of your core. The air clearly dominates the problem. Do you need to worry over the remaining scraps? Or can you focus on air? \$\endgroup\$ – jonk Aug 16 '18 at 22:21
  • \$\begingroup\$ What I want to do is have a way to work out the magnetic field in the air gap reliably. So I can calculate the number of turns I need for my coil, for the amount of current I can put through. Or if needed to adjust the gap length as necessary. The last equation in the picture for Ni does not appear to work out for me, given the values already calculated. And I would like to know why and if I have made an error somewhere. \$\endgroup\$ – Dave Aug 17 '18 at 9:56
  • \$\begingroup\$ I think my error may be due to the permeability values, whether it is meant to be the permeability or relative permeability values used, I find this aspect a little confusing. \$\endgroup\$ – Dave Aug 17 '18 at 11:03
  • \$\begingroup\$ It's not precisely truth, but close enough: energy is stored only in vacuum space. Particles of matter cannot store it, they can only act as a short circuit across the vacuum space for flux lines. A substance with \$\mu_r=1000\$ merely means that \$99.9\$% of the measurable magnetic path length is bridged by atoms that can align themselves and create that short circuit. Which leaves \$0.1\$% as vacuum space that remains to be bridged, and where energy can be temporarily stored. The reality is more complicated, of course. \$\endgroup\$ – jonk Aug 17 '18 at 16:55
  • \$\begingroup\$ \$\mu_0\$ reflects the permeability of vacuum, where the energy can be temporarily stored and is an absolute number with units. \$\mu_r\$ is just a number that tells you what portion of the physical, measurable path length (measurable with a tape measure) is effectively vacuum and how much of it is short-circuit bridged by alignable atomic or electron "spin." It's "relative", in that sense. Again, it's more complicated as the reality is that there are no magnetic monopoles and Maxwell's equations are pre-Einstein work product and magnetic fields are only a relativistic side-effect of charges. \$\endgroup\$ – jonk Aug 17 '18 at 17:00
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Your formulas are correct. You copied those from a good source :-) As for the Hc, just note that the flux lines go round the entire magnetic circuit of core and gap, so that the flux density, which is B, will be equal in core and gap. Hence:

B = μ0 * Hg = μ0 * μr * Hc

Thus:

Hc = Hg / μr

As μr for a ferrite or iron core may be well over 100 or 1000, Hc is correspondingly much smaller than Hg. Most often you can neglect Hc * Lc compared to Hg * Lg, but this depends on your μr. In any case, the sum of all H * L parts equals your Ampere-Turns:

4 * 800 = Hc * 0.132 + Hg * 0.01

        = Hg * (0.132 / μr + 0.01)

You see that the term 0.132/μr may be overshadowed by 0.01 only if μr is in the thousands.

Now you will be able to compute your H in the air gap given the μr of your core.

As for the energy, that could not be stored in the core as noted by some commenters: the energy per unit of volume is about H * B / 2. So if there is no air gap, all energy is stored in the core. But because of μr being large, and B must be below saturation at e.g. 0.3 Tesla, H should not be very large, so you should not apply many turns and/or many amperes. On the other hand, if you apply lots of Ampere-Turns and get into saturation, the HB product and the stored amount of energy will rise as if you had an (expensive) air core.

With air gap, use the same math as above to arrive at the amounts of energy stored in core and gap.

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