Homework Help: Electromagnet and core permeability

1. Jan 8, 2009

capthook

A magnet is attached (or over a small airgap) to the end of an electromagnet (EM) with a steel core, then the EM is pulsed with just enough juice to get the magnet to drop off. (just negating the attraction)

Which will require less energy input to get it to drop off?

1) a high permeability core
or
2) a low permeability core

(1-A) the magnet is very strongly attached to the high permeability core and almost all the domains of the core are aligned. This will require a large input of energy to negate the attraction. BUT, will the high permeability core more readily 'accept' the flux from the EM pulse meaning it will actually require LESS?

(2-A) the magnet is attached, but not quite as much as a much smaller % of the domains of the core are aligned, thus less energy input to the EM to get it to drop. BUT, will the core also be less 'accepting' to the EM pulse so it will require more input?

Or a small size core with a relatively large/strong magnet is going to fully saturate the core, so the high permeability core will require less.
But if the core is large with a relatively small/weak magnet this changes things?

Or what and why?

Tx

2. Jan 8, 2009

Staff: Mentor

You need to post some equations relating the energy input versus the flux generated by the electromagnet....

3. Jan 8, 2009

capthook

Hi Berkman - thanks for the reply.

Don't know why you moved this.... it's not a homework question..........

The equations of input vs. flux is not the issue, rather the relative performance of different core permeabilities. A higher permeability core in a motor application will perform better - more flux per watt input.
However, in this odd application, I'm unsure of how the permeability will affect the efficiency per my 1st post.

- -

flux density = Ampere-Turns/length of the core material X ur

Turns=number of turns of wire
length=measured in meters
u=permeabilty of free space/air/vacuum 4pi x 10^-7
r=relative permeability

A bolt that is .1 meters long (about 4 inches) with diameter of .01 meters(about .4 inches), the cross-sectional area of the bolt/nail would be .00007854.

For iron, r=50. r X u =.00006283. 1 Tesla divided by .00006283 is approximately 16000 Ampere-Turns/meter. 16000 X .1 meters(length of the core) is 1600 Ampere-Turns. If the coil resistance is 10 ohms and you wish to use 1 Amperes then you'll need 10 Volts. With 1 amperes you'll need 1600 turns of wire.(1 Amperes X 1600 Turns = 1600 Ampere-Turns).

The power consumption for the coil would be 10 Volts X 1 Ampere = 10 Watts.
At repulsion when the electromagnet is energized, force is calculated like this:

F=B^2A/2u

B=Total Tesla which in this case is 1(electromagnet) + 1(permanent magnet) =2
A=in this case .00007854
u=4pi X 10^-7

F=2^2(4) X .00007854/2 X 4pi X 10^-7=125 Newtons

(1) H in AT/m=NxI/length

(2) B in Tesla=H x (.000001256 x 800)

2000 Gauss=.2Tesla ur=.0010048
.2/.0010048=199 AT/m

H=N x I/5.08 199 x 5.08=1011 AT

4. Jan 8, 2009

Staff: Mentor

Sounds like a pretty complicated case. What shape is the permanent magnet? If it is a horseshoe magnet, with both poles exposed to the electromagnet face, will it ever fall off?

And if it is a slug magnet, will it fall off, or will it just flip as you get near to matching the interface flux? Remember that the electromagnet attracts the permanent magnet by virtue of the ferrous material that the permanent magnet is made of.... Just opposing the permanent flux may not be enough to make the permanent magnet "fall away".... You may have to define the geometry of the permanent magnet more closely in order to work toward a solution.

BTW, have you tried this experiment? Might be good to run a few trials to see what the mechanical behavior looks like, to help your calculations.