Experiments with magnets and our surroundings


Designing Magnetic Circuits

Designing circuits with magnetic fields is very similar to designing electrical circuits.

With simple electrical circuits, we use an EMF (voltage source) and resistors, and can determine how much current will flow.
Similarly, for magnetic circuits, we use an MMF and reluctance of materials, and can determine how much flux will flow.
See Magnetic Fundamentals.

Here are links to help you learn about magnet circuit designs, and can provide you with details and examples.

http://www.magnetsales.com/
    http://www.magnetsales.com/Magnet_Design.htm
    http://www.magnetsales.com/Design/DesignG.htm
    http://www.magnetsales.com/Design/Glossary.htm

http://www.dextermag.com/
    http://www.dextermag.com/resource-center/document-library

http://www.usneodymiummagnets.com/
    http://www.usneodymiummagnets.com/calculator.html

Check out all the sections they have on design guidelines.  

This is a great tutorial, as all of the above sites are.
Raggio's Electromagnet Design Cookbook 

Magnetic Identities

Below are various identities for these magnetic terms, and their units. 
Use whichever one has the information you know.

 

 

A Typical Problem

Often, you need to create a circuit that will produce a certain amount of flux within an air gap.
This problem below can be used as a first pass.

 

A Solved Example:

Let
    A = 1 cm x 3 cm = 3*10-4 m2

    lc = 60 cm = 0.6 m

    lg = 1 mm = 1*10-3 m

    N = 500 turns

    B = 1.1 T

We will be using sheet steel for the core material, so we will need the B-H curve for this material.

 

From the chart, we can find that for this particular type of steel:

    Hc = 600 At/m when B = 1.1 T

    Hg = B/0 = 0.9 / 4*pi*10-7 = 716.2*103 At/m

    MMFc = Hc*lc = 600 * 0.6 = 360 At

    MMFg = Hg*lg = 716.2*103 * 1*10-3 = 716.2 At

    MMFt = 1076.2 At

    i = MMFt / N = 1076.2 / 500 = 2.15 A

So, with 2.15A flowing into the 500 turns of winding, with this particular type of steel, you would see something close to 1.1 T within the air gap.

If the circuit actually had two air gaps, simply add the two together and model it with one, larger air gap.  One thing that does also happen is that within the air gap, the flux tends to spread out a bit into the air, filling a slightly larger area.  So, the solution isn't exact, but is close, since we didn't take this fringing into account.

A larger air gap

What would happen if the air gap were made twice as large, but we still wanted 1.1 T within the air gap?
    lg = 2 mm = 2*10-3 m

    Hc is the same

    Hg is the same

    MMFc is the same

    MMFg would be twice as large = 1432.4 At

    MMFt = 1792.4 At

    i = 1792.4 / 500 = 3.58 A

The current would have to be 167% larger in order to create the same flux density of 1.1 T within the larger air gap!

A smaller air gap

What would happen if the air gap were made half as large, but we still wanted 1.1 T within the air gap?
    lg = 0.5 mm = 5*10-4 m

    Hc is the same

    Hg is the same

    MMFc is the same

    MMFg would be half as large = 358.1 At

    MMFt = 718.1 At

    i = 718.1 / 500 = 1.44 A

The current would only have to be 67% of the original current in order to create the same flux density of 1.1 T within the smaller air gap!

Lesson Learned

The air gap has the bigger effect on how much current (and MMF) is needed in order to create some kind of flux density within an air gap.
Every motor and generator has to have air gaps between the rotor and the stator in order to allow the rotor to rotate without rubbing against the stator.  However, the smaller the air gap, the lower the current for the same flux density.

 

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