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I'm well aware that a solenoid wrapped around a solid ferromagnetic core produces a stronger magnetic field, but what about a hollow ferromagnetic core? Would the fields farther from the center near the hollow ferromagnetic core become more powerful, or would the middle of the field become more powerful? If something else happens do say so! I'm seeking to understand the fields produced by such a setup.

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The magnetic field would be concentrated in the areas occupied by the mass of the ferromagnetic core. The empty centre of the core would be largely devoid of magnetic lines of flux. This usually means that there will be a higher level of flux density in the material mass of the core compared to a solid core.

However, at the ends of the core (where air dominates), the flux lines would rapidly become very similar the those lines of a solid ferromagnetic core.

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  • \$\begingroup\$ can you point us to a derivation? Your answer above sounds very reasonable to me. \$\endgroup\$ – relayman357 Jul 10 at 21:33
  • \$\begingroup\$ @relayman357 which bit do you need deriving? There is this answer that discusses the flux in a core-tube where there is a significant distance between the core and the windings. It's all about reluctances in parallel. It represents very much a worse case example of the above question so it might be useful to you? \$\endgroup\$ – Andy aka Jul 10 at 21:36
  • \$\begingroup\$ Thanks Andy, that is helpful. I was just hopeful this was one of those cases where a nice derivation existed for the magnetic field at center of the hollow cylinder. \$\endgroup\$ – relayman357 Jul 10 at 23:19
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    \$\begingroup\$ The magnetic field will travel (in the main) via the path of lowest reluctance and the centre of the tube will have high reluctance and there'll be hardly any field in this region but, it depends how much permeability the ferrous material has. So, the flux path has a low reluctance path in parallel with a high reluctance path (air) and it's the same as a high resistance in parallel with a low resistance in that most of the current flows through the low resistance. \$\endgroup\$ – Andy aka Jul 10 at 23:59
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I'm adding this in an answer just so i can show the graphic - the results seem to agree with how Andy explains it in his answer and subsequent comments (Charles answer as well). I did this model in the free 2-d FEMM software. I started with the .fem file from this neat page and watched this video to learn how to use the simple program.

enter image description here

The clear rectangle is one side of the current carrying coil that wraps around the inner hollow cylinder. The inner hollow cylinder is M-19 steel. Inside the hollow cylinder (far left) is air. If you take the mirror image of this and place it on the left you can imagine a 2-d view of the full thing.

The magnetic field lines really pack into the low reluctance cylinder, much less in the air in the hollow. My first look at FEMM, seems like great tool. Also, all the blue is air.

EDIT: In the image below i created a mirror image and stuck them together. If you imagine a vertical field line in the center i think you would have a fair understanding of the field (at least this 2-d slice).

enter image description here

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Iron provides the easiest path for magnetic flux. The flux will be concentrated in that path the extent that is doesn't need to go too far in the air to get to that path. The magnetic field will try to move whatever iron is nearby to become a part of that path. Without the iron, the magnetic field would tend to go through the center and around the outside to the other end. Whatever you put in that path will cause a stronger force to close the part that you leave open. The strongest force would result from a layer of iron in the inside, connecting on one end to another layer on the outside of the coil. That would give the maximum force to pull something to close the open end.

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The magnetic field inside a long solenoid of homogenous material (all air, all ferro rod etc...) is homogenous: the field strength (density of field lines) is the same near/at the center of the core as it is near the edge of the core where the windings are.

If the core of the solenoid is not homogenous, and composed of a hollow cylinder, as in your case, then the H field is still homogenous all around, since it only depends on current that generates it (current density specifically, but that gets into the details of Maxwell's equations).

The H field depends on current only, and not material. The B field depends on H and the permeability plus any magnetization, and so where the permeability varies so does the B field.

The flux is the integral of the B field at a surface. By summing the B field across the inner of the solenoid we can calculate the total flux inside the solenoid. The flux is increased if measured through materials with higher permeability. The flux determines the inductance L of a coil inductor, even if it is an air inductor, or a straight wire.

That's why the "flux is concentrated" in a metal or ferro-metallic core. The B field, -and not H- is the bases for flux, and it is much stronger there, and relative to the surrounding air it appears "concentrated". Where the B field is stronger, the "B lines" are drawn closer together. Each line represents the direction of the B field, and their proximity represents the strength.

The total flux in a solenoid core that is part ferromagnetic and part air (as in the cross section of your cylinder) will be less than if it is all-ferromagnetic.

But the magnetic H field and the B field are not zero in the air inside the cylinder, and the H field is homogenous.

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