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Author Topic: Transformer Induction  (Read 26399 times)
Group: Experimentalist
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I disagree.  The image below shows the E field around a lossless core where the length of the arrows depict the field strength.  There is no sudden change to zero field going from "inside the core" position to outside the core.

Smudge

OK, can you give some field strength numbers for this lossless core verses a core with a minor loss of say 5%?

Pm
« Last Edit: 2025-12-19, 15:22:23 by partzman »
   

Group: Professor
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OK, can you give some field strength numbers for this lossless core verses a core with a minor loss of say 5%?

Pm
I will answer this when I get back from my Christmas break.  I will be away from my computer for the next 9 days.
Smudge
   
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...
Therefore IMO, a lossless core relative to it's flux magnitude, would result in zero E-Field outside the core.
...

Only if it had infinite permeability.
If it has finite permeability, then since the magnetic flux takes all possible paths, it also passes through the air, so E = -dΦ/dt is not zero.

The preferred path of the flux, with equal cross-section and length, is obviously the one with the highest permeability, in terms of the permeability ratio.
In the general case, the proportion is given by the ratio of reluctances R=µ.S/L, where S is the cross-section and L is the length of the path. In practice, the flux will never be zero outside the core.
« Last Edit: 2026-01-22, 10:56:52 by F6FLT »


---------------------------
"Open your mind, but not like a trash bin"
   

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Only if it had infinite permeability.
If it has finite permeability, then since the magnetic flux takes all possible paths, it also passes through the air, so E = -dΦ/dt is not zero.

The preferred path of the flux, with equal cross-section and length, is obviously the one with the highest permeability, in terms of the permeability ratio.
In the general case, the proportion is given by the ratio of reluctances R=µ.S/L, where S is the cross-section and L is the length of the path. In practice, the flux will never be zero outside the core.
Your u.S/L is not reluctance, it is the inverse, permeance.
   

Group: Professor
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OK, can you give some field strength numbers for this lossless core verses a core with a minor loss of say 5%?
Sorry about the late reply, I forget things these days.  I think we need to go back to your earlier post #169
Quote
It appears to me in my continued testing of charge separation in various cores, that the E-Field magnitude is dependent on the amount of flux retained in the core material.  IOW, any leakage flux outside the core creates a loss in the E-Field within the core window
It certainly will create a reduction from the value that occurs without flux leakage, but that is not a loss in the normal meaning of the word (power or energy loss).
Quote
and attributes to the apparent E-Field outside the core.
I took your “outside the core” to mean outside the core window.  You seem to consider voltage induction as occurring only on that part of the conductor within the core window (within the donut hole), and my reply post #171 shows that there is E field present everywhere external to the core, not just in the window.   I hope you now have moved away from that E field being “apparent” and from it being connected to flux leakage. 
Quote
  As a core saturates for example, more flux leaves the core and also creates a reduction in the measured E-Field within the core window.
Flux does not leave a ring core as it saturates.  Increased driving current gives a smaller increase of flux.  That is not flux leakage.  E field everywhere and your measured voltage are proportional to rate of change of flux, so they are reduced but this is not due to flux leakage. 
Quote
Therefore IMO, a lossless core relative to it's flux magnitude, would result in zero E-Field outside the core.
I hope I showed that to be nonsense.
Moving on to your latest request, what are the core dimensions and characteristics you want numbers for?  What is the driving waveform?  Do you want E field vector magnitudes at varying points.  Or are you just looking for a single number, the relative change in magnitude between a lossless core and a 5% lossy core that will apply everywhere external to the core?
Smudge
   

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so here im presenting 3 examples.

a toroid core. blue box is 4v ac input to 4 turn primary at the bottom of the core. at the top of the cutaway examples, 3 different secondaries. the yellow circle is a scope.

on the left the single turn is wound through the core.

middle just above the top of the core but same proximity to the core.

right, the secondary has close access to the top outer core and outer sides of the core.

what should we see in the 3 scopes?  not concerned with phase differences of the middle and right scopes compared to the left.


then if the yellow circle is a load, which secondary would produce the most output?


If you say the e field is inducing all of the wire of a normally wound secondary, in the window and outer sides, would your answer to what you believe the scopes would show, all be the same results? again,  not concerned with phase differences of the middle and right scopes compared to the left.

i know there looks like 5 turns for the primary..   lets say it is 4.

Mags
« Last Edit: 2026-01-24, 04:15:48 by Magluvin »
   

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I disagree.  The image below shows the E field around a lossless core where the length of the arrows depict the field strength.  There is no sudden change to zero field going from "inside the core" position to outside the core.

Smudge

if the core is lossless, why is the e field looking weaker around the outside of the core? or better to say, why is the e field looking stronger in the window of the core?

mags
   

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so here im presenting 3 examples.

a toroid core. blue box is 4v ac input to 4 turn primary at the bottom of the core. at the top of the cutaway examples, 3 different secondaries. the yellow circle is a scope.

on the left the single turn is wound through the core.

middle just above the top of the core but same proximity to the core.

right, the secondary has close access to the top outer core and outer sides of the core.

what should we see in the 3 scopes?  not concerned with phase differences of the middle and right scopes compared to the left.


then if the yellow circle is a load, which secondary would produce the most output?


If you say the e field is inducing all of the wire of a normally wound secondary, in the window and outer sides, would your answer to what you believe the scopes would show, all be the same results? again,  not concerned with phase differences of the middle and right scopes compared to the left.

i know there looks like 5 turns for the primary..   lets say it is 4.

Mags
The left one is a 1 turn secondary with a 4 turn primary so with 4V ac input we get 1V ac output.  That assumes the core has high permeability so flux leakage (because the primary only occupies a small part of the core) is negligible.  With a poor low perm core the output will be less than 1V.
For the other two there will be zero volts except at high frequencies when capacitive coupling and RF coupling from source to scope take effect.
Smudge
   

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if the core is lossless, why is the e field looking weaker around the outside of the core? or better to say, why is the e field looking stronger in the window of the core?

mags

The E field is due to the flux in the core changing with time.  The flux lines are circular.  For any small length of flux (that is changing with time) the E field reduces with distance from it.  Within the circle the total contribution to the E field from the entire flux around the circle is greater than elsewhere.  Maximum E field is within the window close to the inner surface of the core.  Core loss plays no part in this geometric effect.
Smudge
   
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The E field is due to the flux in the core changing with time.  The flux lines are circular.  For any small length of flux (that is changing with time) the E field reduces with distance from it.  Within the circle the total contribution to the E field from the entire flux around the circle is greater than elsewhere.  Maximum E field is within the window close to the inner surface of the core.  Core loss plays no part in this geometric effect.
Smudge

I find this statement confusing!  Circular in the core?

OK, let's see if we can agree on some common truths with conventional theory!  We apply a DC voltage to a primary on a toroid core.  This results in a current ramp that starts from zero and increases to some peak value without reaching saturation of the core. During this current ramp, flux starts from zero and increases to a peak value in the core.  During this time, zero flux appears inside the hole of the core or outside the core and therefore zero E-Field can appear around the core.  We also have placed a single turn secondary completely around the core. We know that this completed turn will produce a voltage between it's open ends.  Why?  Because the surface bounded by the secondary loop cuts through the core where the flux B is varying with time.

If we agree thus far, then how does the potential across the secondary vary through it's perimeter? 

Regards,
Pm 
« Last Edit: 2026-01-24, 15:26:43 by partzman »
   

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I find this statement confusing!  Circular in the core?
You have a circular ring core and the B field forms circular lines within it.  Flux is that B field times the area so the flux is the total number of lines forming a circular ring of flux (the old measure of flux was lines per square cm).
Quote
OK, let's see if we can agree on some common truths with conventional theory!  We apply a DC voltage to a primary on a toroid core.  This results in a current ramp that starts from zero and increases to some peak value without reaching saturation of the core. During this current ramp, flux starts from zero and increases to a peak value in the core.  During this time, zero flux appears inside the hole of the core or outside the core and therefore zero E-Field can appear around the core.
There is zero flux external to the core material but there is the magnetic vector potential A field there.  And that A field changing with time (dA/dt) produces the E field.  I am surprised you say zero E field during this flux build-up because you know that V occurs while the flux is changing.
Quote
We also have placed a single turn secondary completely around the core. We know that this completed turn will produce a voltage between it's open ends.  Why?  Because the surface bounded by the secondary loop cuts through the core where the flux B is varying with time.
Which it does through this build-up phase, producing the E field that drives the electrons in the secondary wire to create the voltage.
Quote
If we agree thus far, then how does the potential across the secondary vary through it's perimeter? 
Ah, I think I can see where the problem lies.  You consider the surface bounded by the loop is where the magic happens and that is where the voltage happens.  That is incorrect.  The flux passing through that surface doesn't magically induce voltage at its periphery, it is more complex than that.  You have to integrate the A vector tangential component along the conductor and take rate-of-change of that sum to get the voltage.  That integral sum is equal to the flux passing through that surface, but that does not mean the electric field driving the electrons is equally distributed around its boundary.  V=dΦ/dt hides the E field necessary to create the voltage.  My image shows the A vectors, hence also the E vectors,  in their true colours.

Smudge
« Last Edit: 2026-01-24, 17:43:14 by Smudge »
   

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You have to integrate the A vector tangential component along the conductor and take rate-of-change of that sum to get the voltage.
That's right.
A bit of warning, though:  You cannot apply the Ohm's law to that voltage in order to determine the current flowing in the loop. i=ℰ/R does not work.
   

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That's right.
A bit of warning, though:  You cannot apply the Ohm's law to that voltage in order to determine the current flowing in the loop. i=ℰ/R does not work.
Yes, because the load current alters the magnetic field in the core and that alters the magnetic vector potential and that alters the E field that you used in the first place.  Many people find difficulty in getting out of this conundrum.  Your i=ℰ/R suggests the use of the E field (ℰ field?) there and I would have used the symbol V for voltage. 
   

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Yes, because the load current alters the magnetic field in the core and that alters the magnetic vector potential and that alters the E field that you used in the first place.  Many people find difficulty in getting out of this conundrum . 
Indeed

Your i=ℰ/R suggests the use of the E field (ℰ field?)
ℰ: induced EMF

Also, I'd like to add that A=0 implies that B=0 ....but B=0 does not imply that A=0 since superposition of multiple non-zero A fields can yield B=0.
   
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