Electromagnet Strength

[Freethinker22]

Hi Guys and Gals! Ok so this is my first post on this site so bare with me here.

I'm trying to figure out how strong some of the electromagnets I've built are, and so far I can only get to the number of At, Gilberts, and Oersteds. What I want to figure out is the Gauss and or Webers. Now it's my understanding if you have an air core coil, 1 Gilbert or 1 Oersted ( H ) are equal to 1 Gauss ( B ). But if you use a ferromagnetic core, the intensity ( B ) is magnified based on the size and permeability of the core. Here is where I get stumped, how do you solve for B? I've been plugging around the inter tubes for a while now and have found 6 different ways to solve for B. Can anyone give me a simple explainable formula for calculating Gauss in my electromagnets? Here are the specs of one of my magnets if that helps.

2.5" x .5" soft iron core
719' of 22 Awg magnet wire
At = 2328
Applied Voltage = 12
Ω = 11.6
Amps = 1.04

Thank you for any help!

Matt

 

[saturn1bguy]

Ok so this is my first post on this site so bare with me here.

We're taking our shirts off, is that the deal?

 

[Freethinker22]

We're taking our shirts off, is that the deal?

Wire stripper: An electrician who takes their cloths off...

 

[MrAl]

Hi Guys and Gals! Ok so this is my first post on this site so bare with me here.

I'm trying to figure out how strong some of the electromagnets I've built are, and so far I can only get to the number of At, Gilberts, and Oersteds. What I want to figure out is the Gauss and or Webers. Now it's my understanding if you have an air core coil, 1 Gilbert or 1 Oersted ( H ) are equal to 1 Gauss ( B ). But if you use a ferromagnetic core, the intensity ( B ) is magnified based on the size and permeability of the core. Here is where I get stumped, how do you solve for B? I've been plugging around the inter tubes for a while now and have found 6 different ways to solve for B. Can anyone give me a simple explainable formula for calculating Gauss in my electromagnets? Here are the specs of one of my magnets if that helps.

2.5" x .5" soft iron core
719' of 22 Awg magnet wire
At = 2328
Applied Voltage = 12
Ω = 11.6
Amps = 1.04

Thank you for any help!

Matt

Hi,

B is calculated from H by the simple formula:
B=ua*H
where
ua is the absolute permeability.

We are usually interested in the relative (incremental) permeabilty ur, which is:
dB=ur*dH
where
dB is the change in B and
dH is the change in H.

The trick is knowing what the permeability of a given core is.
To find this out, you have to consult the BH curve for the given
core, but often that's a little hard to understand too because
there is hysteresis. You can get an idea what the core is doing
however by looking at the BH curve so if you have that data
you are one step ahead.
If you dont have that data and you want to know something
about the permeability, you have to do some measurements
which arent that straightforward.
Energizing the core with a DC level with an AC riding on top of
it helps, because then you can calculate the inductance for
each level of DC you apply, then calculate the BH for that point
using a formula for inductance.
This test requires swinging the DC supply current from say 0 to
some maximum and doing the calculation for say 10 points
in between.
Just to show how unusual these measurements can turn out,
near 0.00A DC the core might show very low ur, then as the
DC rises the ur hits some maximum, then starts to fall again,
until the coil starts to look like an air core again with very
little ur.

Here's a formula for V knowing other things:
V = 4.44 BNAf x 10^(-8) {rms, and B in gauss, A in square cm, f in Hertz, N is number of turns, sine wave}
and this can be solved for B, so knowing H and knowing B you can
solve for ur with:
B=ur*H
at the various points.

Good luck.

Here's a pic of a typical BH curve...
**broken link removed**

 

[MrAl]

Hello again,

I feel an example is in order...


Example:

Say we apply 1A dc and 0.1A ac at 100Hz to a coil of 20 turns on a core with 1 sq.
cm of area, and we measure 0.1V ac.

First, solving
V = 4.44 B*N*A*f x 10^(-8)
for B:
B=(V*10^8)/(4.44*f*A*N)

Since dV=0.1 and f=100Hz and A=1 and N=20 we get:

dB=(0.1*10^8)/(4.44*100*1*20)

and since dIac=0.1 and N=20 we get
dH=0.1*20

now we calculate ur:
ur=dB/dH
ur=1126/2
ur=563

It should be noted that this ur was obtained with a dc level of
1 amp, and if we increase this to 2 amps we may get a lower or
higher value for ur depending on the BH curve of the material.
This means we might want to do this for several levels of
current in steps of some increment such as 0.1 amps, which
would then give us a set of ur values:
ur(0.1)
ur(0.2)
ur(0.3)
etc.
and we would continue this until we reach a level where ur doesnt
change much and is close to zero, as this would be the far right
point on the curve which means the core is highly saturated.

This permeability is usually of interest because it greatly affects
the inductance of a given coil and core construction. As the
ur doubles, so does the inductance, and as the ur halves, so
does the inductance, so with very low ur the inductance of
a coil/core can drop significantly resulting in highly destructive
current levels, and this can happen even with inductors that
start out with a very high inductance with low current.

 

[Freethinker22]

Ok, so everything makes since except where you said "Energizing the core with a DC level with an AC riding on top of it helps, because then you can calculate the inductance for each level of DC you apply, then calculate the BH for that point using a formula for inductance."

I don't quite get how one would conduct that experiment , could you please explain how I could do that?

Thanks for the help!

 

[MrAl]

Hi again,

Yes, but it is not that easy because you need an amplifier that can put out
a dc current with a changing ac current, and both currents have to be of
the right level so you end up needing a high power amplifier. The amp
might have to put out 2 amps dc with 0.1 amps ac of some frequency.
One way to drive the amp would be to input an ac signal of some frequency
and amplitude and have an offset dc adjustment built into the amplifier.

An amplifier like this would be more or less like an audio power amplifier,
that also has an adjustment for dc offset, or else is dc coupled so that
it can pass dc as well, then you can input a dc signal with an ac signal
riding on that, which would then output to drive the coil.

You might also try a dc to dc down converter (buck), where you can
load the output to draw some average dc current (the dc component)
and either inject some ac or just look at the ripple to determine the
inductance from v=L*di/dt, but this requires a scope too.
I've done this before and it does work, mainly because a buck converter
does this anyway without doing anything special.

It might however be possible to inject some sine wave signal into
the converter to force it to put out a sine wave of some frequency
along with the average dc, and then measure the ac voltage.

Possibly introduce some small ac signal into the feedback by adding
it to the actual feedback, or better yet modulate the reference signal
(which is dc). This should produce an ac signal with the dc signal
through the coil. The ac frequency would have to be much lower than
the buck converter switching frequency of course, some 10 times lower
as a maximum.

The one catch with the dc to dc buck converter is that the switching
frequency has to be commensurate with the inductance so that there
is not too much ripple otherwise the inductor will be taken through
too much of its BH curve to get the kind of point readings we would want.