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I have a small project based around designing a small home-made DC actuator using a small DC electromagnet.

My final goal is to generate the strongest magnetic field possible with the minimum amount of current, but for prototyping purposes I used a 0.5mm diameter enamelled copper wire to generate lesser amounts of resistance while varrying the current with my 30V/5A DC power supply.

I know that the magnetic field B is dependent on the permeability of free-space (µ0), on the permeability of the core material (µr) and on the magnetomotive force divided by the length of the coil (NI/l=H).

Thus I started prototyping with what I had under the hand, namely a steel screw and laminated iron plates I cannibalised from an old nokia transformer, needless to say it was a lost cause to search for their B vs H hysteresis curve. In the end I made 3 increasingly powerful magnets, one with the steel screw core, one with a core made of several laminated iron plates and the last and most powerful one with E-shaped laminated iron plates.

But as I said, my goal isn't to generate the most powerful magnetic field possible but rather to find a decent compromise between B and I. Therefore I started looking for other materials online and found these ferrite rods (pdf warning).

These rods are 2.5cm long, have a medium permeability (initial permeability at µi=2300) and saturate at about 0.5T which was more or less exactly what I was looking for. Once I got them, I winded 100 turns of copper wire around them. And according to their B vs H curve and this small equation :

\$I = \frac{H.l}{N} = \frac{5*10^3}{4\pi}*2.5\times10^{-2}*\frac{1}{100} = 0.1A\$

The ferrite core should saturate at about 0.5T when a current of 0.1A is fed into the coil.

I tested this and half-surprised, a current of 100mA produced next to nothing and the magnetic field increased with the current at least up until 5A, which is my power supply upper current limit.

I mentioned that I was half-surpised because it was clearly indicated under each graph that all measurements had been done under a 10kHz AC signal, which finally brings us to my questions: first, do lower frequencies affect the permeability of the material or did I just misunderstand the datasheet and second, if the answer to the first question is "yes, lower frequencies affect the permeability of the material", does the datasheet somehow allow us to estimate the impact of these lower frequencies ?

Any help would be greatly appreciated. Thank you.

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  • \$\begingroup\$ The magnetic field will still increase with current even after saturation. \$\endgroup\$
    – Hearth
    Jul 26, 2021 at 14:45
  • \$\begingroup\$ Yes higher f in steel core causes more eddy current losses and higher impedance from 2pi*fL \$\endgroup\$ Jul 26, 2021 at 14:47
  • \$\begingroup\$ I’ve also been trying to figure out how to make 3Tesla ringing at 300 Hz with a Q of 2 at pulse rates up to 20 pps with 1kW from AC in a palm sized air cooled paddle. \$\endgroup\$ Jul 26, 2021 at 14:51
  • \$\begingroup\$ @Hearth yes, but it will increase at a rate proportional to the permeability of free space isn't it ? If this is the case this should be negligible considering the current involved but it definitely wasn't in my case. \$\endgroup\$
    – tampler
    Jul 26, 2021 at 14:53
  • \$\begingroup\$ @tampler Well, you didn't say anything about the magnitude of the increase, just that it continued to increase. \$\endgroup\$
    – Hearth
    Jul 26, 2021 at 14:53

2 Answers 2

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You are using a ferrite rod but you need to consider a closed path in your calculations. To calculate the magnetic flux or flux density you should first think about the closed magnetic circuit. One reluctance (magnetic resistance) of such a magnetic equivalent circuit would be the reluctance of the ferrite rod: $$R_\mathrm{m}=\frac{l}{\mu A}$$ To get a closed magnetic circuit, you still have to take the reluctance of the distance through the air into consideration:

enter image description here

Source

The magnetic path through the air is at least as long as the rod. The permeability \$\mu\$ is one. So the overall reluctance of the magnetic circuit is significantly larger than you think. There are certainly some formulas and even calculators online for a rod geometry that help you with these calculations. When you have an approximation of the overall reluctance you can calculate the magnetic flux density etc. based on: $$NI = \phi R_\mathrm{m} = B A R_\mathrm{m}$$

There are a lot of core shapes that are better suited for electromagnets than rods. The air reluctance should be as small as possible so that you can actually saturate the core material.

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  • \$\begingroup\$ Oooh got it, this also confirms my suspicions as to why the E-shaped electromagnet I made produced a lot more force than the rod-shaped one despite being made out of the same core material. Thanks a lot, I will delve deeper into the topic. \$\endgroup\$
    – tampler
    Jul 26, 2021 at 18:57
  • \$\begingroup\$ After some quick research, I understood that there are in grand total 3 reluctances in the system, the core reluctance Rc and the two air reluctances Rair. The total reluctance R is according to my sources R=Rc//Rair//Rair but considering that Rair>>Rc then R≃Rc. Which explains why the air reluctances are always disregarded in every examples I could find. Meaning that the overall reluctance of the magnetic circuit is not significantly larger than what I thought. Am I missing something ? \$\endgroup\$
    – tampler
    Jul 27, 2021 at 9:44
  • \$\begingroup\$ I'm not sure why the reluctance should be in parallel. You need a closed loop for the magnetic flux. About half of the path length is through the magnetic core of the rod, the other half is through the air as the picture should visualize. \$\endgroup\$ Jul 27, 2021 at 9:51
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If you want to generate the largest magnetic field with the lowest electrical power for a given core material and conductor material:

  1. Build a closed ring out of the magnet material e.g. Iron and cut a small gap into it, just big enough to fit whatever you want to expose to the field. This maximizes the permeance of the core. A stick bar core has a very bad permeance as explained very nicely by Lars Hankeln.

  2. Then wind your core with as much conductor as you can. The more volume your winding has, the more efficient it will create magnetic field.

When winding, consider to use a diameter of conductor that - in the end - will give you a coil resistance that you can drive well with your power source. Usually, that means a winding resistance in the 1 .. 100 Ohm ballpark.

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