On the electromagnet (5 Volts DC, built upon a transformer core) setup pictured below , only the core part is magnetic (when powered, it attracts a piece of steel) but not the 2 "wings" outside that are supposed to channel the magnetic flux. Their magnetic force is about 1/10th of the central core

Why is that? Is there a geometry to improve channeling of magnetic flux on the outside of the coil ? enter image description here

  • \$\begingroup\$ BTW, the "why?" questions may not yield the desired answers if your real question is "how to make the magnet stronger?" \$\endgroup\$ Sep 7, 2016 at 13:02
  • \$\begingroup\$ Exact duplicate of a question on Engineering.SE \$\endgroup\$
    – Dave Tweed
    Sep 7, 2016 at 15:27

2 Answers 2


Learning to read a magnetic circuit is a skill that is not much taught nowadays.

You're right (in your comment to Dmitry's answer) that the same flux is present on the central pole and on the outer (split) pole.

However, notice that the total area on the central pole is one square (I'll guess one square inch) - flush with the bobbin.

Now measure the total area of the other pole - the whole back surface (about 3 square inches), the outer ends (about 2 square inches each), both sides (about 3 square inches each) and the two pole piece surfaces themselves (summing to 1 square inch). Total is somewhere around ... 14 square inches, so a very rough approximation to the flux density iwould be 1/14 of what you expect.

If you can read the currents and voltages in a circuit with low resistance wire, thinner high resistance wire and actual resistances, you can start to understand this circuit by imagining a thick iron bar as a low resistance, or having high conductivity - air or vacuum as a high resistance (i.e. low conductivity).

The actual term for resistance in magnetic circuits is "reluctance", and that for conductivity is "permeability". Air has a "relative permeability" of 1, iron in the thousands. So an iron bar conducts magnetic flux thousands of times better than an air path of the same length (until high flux densities - then it will saturate).

So the flux density is not equally distributed around the huge outer pole, it's proportional to the area of a section of that pole, and inversely proportional to the air gap length. So it'll be slightly stronger at the inner edge of those outer pole pieces where the air gap is only 1/2 inch, and a bit weaker on the bottom surface where the air gap (from the inner pole) is about 2-3 inches.

Calculating the exact flux densities can be done with calculus for simple shapes, but simulations and finite element analysis are more often used now.

Now, I hope you kept the "I" laminations? Use them as an iron bar spanning the top of the "E". As you bring them closer, you'll find the air gaps between E and I reduce - and as you reduce the gap, the flux will concentrate in those gaps - and as you reduce the air gaps, you reduce the "resistance" i.e. the reluctance, and so the "current" i.e. flux will increase dramatically, and so will the attractive force between the electromagnet and the bar. WARNING keep your fingers out of the way when you do this!

The magnetic flux can't increase infinitely high, eventually the iron will "saturate" at about 1.2 Tesla.

Now you can see how Dmitry's horseshoe magnet works, and how to improve it - bend the poles closer together to reduce the air gap. Also, look at toy electric motors - how the pole pieces are shaped to match the iron rotor (with the coil wound on it) to concentrate the flux in the small gap between the magnet's poles and the rotor.

EDIT: found quite a good introduction here...

Pay attention to the figures, read the words later... Note the following:

  • Figure 1.6 shows the relative flux density inside and outside a coil - even without an iron core to concentrate the flux inside, you can see how relatively dense it is.
  • Figure 1.7 shows how to make a horseshoe magnet with a very small air gap. (Note you can only fit thin objects in the gap where the field is strong) Also note they have drawn one line of "leakage flux" - all the exposed iron will radiate some leakage flux, but note how long the air paths are compared to the gap length.
  • Figure 1.10 shows how this evolves into an effective motor.

Having covered some of the "why", if you're really asking "what do I do about it?" add some context about what you want to achieve to the question. It should now be clear that magnetic circuits are designed for a specific purpose, and we don't know anything about what your purpose is.

  • \$\begingroup\$ I didnt get your 14 inch area calculation, the flux is supposed to mostly get out from the 2 split pole horizontal surface and go to the inner pole. Therefore the flux density should be the same in the inner pole and the outper split pole (their area is the same). \$\endgroup\$ Sep 7, 2016 at 12:34
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    \$\begingroup\$ No it's not "supposed to" come out of the split poles. You want it to. But it doesn't, and that's why your observed results don't match your expectation. Unless ... you provide a low-reluctance path between the central and outer split poles - such as the "I" pieces spanning all three pole pieces. \$\endgroup\$ Sep 7, 2016 at 12:41
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    \$\begingroup\$ You apparently haven't tried the horseshoe - the outside of a solenoid is fairly unimportant, the flux there is at a low concentration compared to the flux inside it. \$\endgroup\$ Sep 7, 2016 at 12:45
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    \$\begingroup\$ @ManudeHanoi Horseshoe magnets can have arbitrarily small gap between the poles. E.g. see HDD heads \$\endgroup\$ Sep 7, 2016 at 12:50
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    \$\begingroup\$ Let's look at that : mean path length to the "wings" is maybe 5x the path length to the intended poles - yes, that's a lot of air. But the surface area is maybe 13x the area of the intended poles, so I expect well under half the flux to go where intended, or well under 1/4 to each outer pole. Which roughly matches the observation as far as we can tell. \$\endgroup\$ Sep 7, 2016 at 13:18

This is due to the geometry of your magnet: one pole is concentrated in the middle of the coil, while the other one is distributed between the outside wings. Unless you provide enough current to saturate the whole core, the flux will be distributed unevenly, with a spot in the middle of the central pole channeling most of it.

If you need to have two poles with equal attractive force, you should use a U-shaped electromagnet (also called horseshoe electromagnet) like this one:

enter image description here

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    \$\begingroup\$ Surely, the total flux leaving the middle pole equals the total flux entering the two outer poles AND, given that the outer poles (together) have about the same cross-sectional area, the flux density will be the same. \$\endgroup\$
    – Andy aka
    Sep 7, 2016 at 9:44
  • \$\begingroup\$ @Andyaka Yes, the total flux on the inner pole and outer poles will be the same. The flux density, not necessarily. I think there is a spot in the middle of the inner pole which channels 90% of the flux, while the outer poles don't have such spots. \$\endgroup\$ Sep 7, 2016 at 9:54
  • \$\begingroup\$ my goal is to increase the strength of the magnet by decreasing the air gap, that means collecting the flux on the exterior of the solenoid \$\endgroup\$ Sep 7, 2016 at 12:38
  • \$\begingroup\$ @ManudeHanoi Actually, decreasing the air gap usually means moving the poles close together, and not collecting exterior flux which is about zero if the magnet is properly designed. Actually, horseshoe magnets work quite well with big gaps, because these gaps are closed by objects which are attracted by the magnet and stick to both poles. \$\endgroup\$ Sep 7, 2016 at 12:54
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    \$\begingroup\$ And in that respect, this transformer is the same, it should strongly attract a low reluctance object that bridges all three pole pieces. \$\endgroup\$ Sep 7, 2016 at 13:21

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