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Author Topic: Current produced to rate of change of magnetic field within an inductor  (Read 10130 times)

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I would like to see everyone's input on this.

A question was asked on OU.com
Quote: Do you not get more current if you move a magnet across a coil faster?

We are assuming a real coil that has a load across it.

So,if we assume a single cycle(one pass only across the coil),will more current be produced through that coil/load circuit,if the rate of change of the magnetic field is increased within that coil?.

Verpies(who i hope will join in here) said this-Quote: You do not - you only get more voltage (emf) if the coil is not ideal and not perfectly closed.

I then asked or stated the following for clarification
Quote: if the coil is not ideal,and not perfectly closed
,must mean that there is a load across that coil(not perfectly closed=partially closed) that has a resistance value to it,along with the resistance of the !non ideal! coil it self.


Verpies replied-Yes

So the question is clear--If we have a coil/inductor,that has a load placed across it,will more current be produced from that inductor, if the magnet is passed across that inductor faster?.

Remember,we are talking one single cycle-one pass across the inductor.

Brad


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I would like to see everyone's input on this.

A question was asked on OU.com
Quote: Do you not get more current if you move a magnet across a coil faster?

We are assuming a real coil that has a load across it.

So,if we assume a single cycle(one pass only across the coil),will more current be produced through that coil/load circuit,if the rate of change of the magnetic field is increased within that coil?.

Verpies(who i hope will join in here) said this-Quote: You do not - you only get more voltage (emf) if the coil is not ideal and not perfectly closed.

I then asked or stated the following for clarification
Quote: if the coil is not ideal,and not perfectly closed
,must mean that there is a load across that coil(not perfectly closed=partially closed) that has a resistance value to it,along with the resistance of the !non ideal! coil it self.


Verpies replied-Yes

So the question is clear--If we have a coil/inductor,that has a load placed across it,will more current be produced from that inductor, if the magnet is passed across that inductor faster?.

Remember,we are talking one single cycle-one pass across the inductor.

Brad

I feel that the EMF is raised at the output of the coil by the faster magnet pass.  The reaction to the higher EMF is increase current flow due to the Volt/Ohm ratio.  The increased EMF is the primary effect and the increased current flow is a secondary reaction.

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That's right.

Take a little hobby DC brushed motor, connect a little incandescent bulb (use one of those GOWs!) to the brushes and turn the armature. Does the bulb get brighter when you turn the armature faster?
QED.
   
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So the question is clear--If we have a coil/inductor,that has a load placed across it,will more current be produced from that inductor, if the magnet is passed across that inductor faster?.

Yes because it is the difference in potential  or voltage across the load which determines the current flow. Moving the magnet faster induces a higher voltage in the coil which produces more current across the load.

Whenever I have a question the first thing I do is go to the bench and prove it one way or another. At which point the context of the question comes into play. Imagine we move a magnet very slowly across a coil then very quickly and a light attached to the coil only comes on when we move the magnet quickly. Fair enough however we should remember only the induced voltage above a given threshold determined by the load resistance produces a current flow in the circuit. However if we added another switched low resistance inductor to the circuit such as a joule thief then even the slow movement of the magnet would light the bulb. In effect we have traded a small amount of induced low voltage current which could not light the bulb for a small amount of higher voltage which could light the bulb.

You see the question is not so clear when the context changes. Take a 10 turn inductor with an LED attached and move a magnet past it. Now attach a joule thief to the coil and we find even a very slow movement of the magnet still lights the LED. We should understand a great deal of the voltage which is induced is never actually applied to the load as a load current because of the load resistance. However we can boost the low voltage portion of the induced current which normally does nothing with respect to the load to increase the power output.

My best generator used boost conversion on the low voltage portion of the output waveform then transitioned to direct coupling once the induced voltage matched the load. In which case the modified output gives the appearance of a square wave versus a sine wave and the load current starts much sooner than expected ultimately generating more power.

Hands on boys... hands on is how we really learn new things.


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That's right.

Take a little hobby DC brushed motor, connect a little incandescent bulb (use one of those GOWs!) to the brushes and turn the armature. Does the bulb get brighter when you turn the armature faster?
QED.

No TK,that is different,as you are also increasing the number of cycles-more magnet passes across the coil per minute-so to speak.

We are talking about 1 single pass of a magnet across a coil/inductor.

Here is a thought experiment.

We have a coil with a FWBR attached to it's output leads,so as to rectify to DC.
We then have a cap across the FWBR output,so as to collect the energy produced by the coil.
Lets say that the cap is 100uF.

Test one
We now disconnect one end of the coil from the FWBR,so as we can do an open voltage test.
We move the magnet passed the coil as a set speed,and obtain an open voltage across the coil of say 5 volt's.
We now hook the coil back up to the FWBR,and move the magnet across that coil at the same speed as the open voltage test,and we end up with-say 3 volts across the 100uF cap. So our coil voltage has been limited to 3 volt's-plus the V drop across the FWBR.

Now we replace the 100 uF cap with a 200uF cap.
Now,we know that is we pass the magnet across the coil at the same speed as test one,we will not get 3 volt's across the cap.
BUT-if we pass the magnet across the coil faster,will we be able to get 3 volt's across the 200uF cap ?


Brad



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Surely you already know the answer to this, Brad.

Faraday's Law of Induction states that the induced EMF is proportional to the time rate of change of the flux. Faster change of flux, more EMF induced. More induced EMF, the higher the open circuit voltage, or equivalently the higher the current through a resistive load.

If you simply swing a magnet past a coil, you get one polarity of voltage induced as the magnet approaches the coil (flux increasing with time) and the opposite polarity as the magnet recedes after passing closest approach (flux decreasing with time.) You can easily see this on your oscilloscope, by connecting a coil to a high-impedance scope probe and swinging a magnet past it one time, using Single Shot trigger mode. Swing past slowly, you get less peak-to-peak voltage. Swing past more quickly, more p-p voltage and of course a shorter pulse duration.

And Ohm's law (and simple reasoning) tells you that if you have a higher voltage pulse into a fixed resistance, the current will be higher.

The capacitor question is similar, except that a capacitor charge curve isn't a straight line, because the energy on the capacitor depends not on voltage, but on the _square_ of the voltage.
   
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Verpies has pointed out that I should have used "phi" (the flux) rather than "B" (field strength) in my statement of Faraday's Law of Induction. This is a subtle distinction. Phi is equal to BA, where A is the area.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html



   

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Surely you already know the answer to this, Brad.

Faraday's Law of Induction states that the induced EMF is proportional to the time rate of change of the flux. Faster change of flux, more EMF induced. More induced EMF, the higher the open circuit voltage, or equivalently the higher the current through a resistive load.

If you simply swing a magnet past a coil, you get one polarity of voltage induced as the magnet approaches the coil (flux increasing with time) and the opposite polarity as the magnet recedes after passing closest approach (flux decreasing with time.) You can easily see this on your oscilloscope, by connecting a coil to a high-impedance scope probe and swinging a magnet past it one time, using Single Shot trigger mode. Swing past slowly, you get less peak-to-peak voltage. Swing past more quickly, more p-p voltage and of course a shorter pulse duration.

And Ohm's law (and simple reasoning) tells you that if you have a higher voltage pulse into a fixed resistance, the current will be higher.

The capacitor question is similar, except that a capacitor charge curve isn't a straight line, because the energy on the capacitor depends not on voltage, but on the _square_ of the voltage.

We are not talking EMF,we are talking current value.

The question again was-do we not get more current if the magnet is passed over the coil faster.

Now remember,the faster the magnet is passed over the coil,the higher the current value,but shorter the time value.

As i said,verpies answer was no-you do not get more current if the magnet is passed over the coil faster.

Also,reffering to my thought experiment-what is your answer?

I am hoping verpies will join this discussion.
He also stated that if a magnet is withdrawn from the center hole of a super conductive loop,drawing out the magnet faster would not result in a higher current flow through that SC loop.


Brad


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Didn't I answer the questions in the last post? I thought I did....


As far as current amplitude goes, you could always just do the experiment....

Resistor connected across coil, scope probe across resistor, magnet passed by coil slowly then quickly. Probe is measuring voltage drop across resistor, ie current by Ohm's Law.
   
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....  But what does "more" current mean? Greater amplitude? Or more charge transferred? Certainly amplitude is greater the faster you go, and pulse duration is shorter. But over the total duration of the pulse, is there actually more _charge_ transferred, than with the slower but lower amplitude pulse? This I don't know. Looking at the areas of the two pulse curves above, it might be the case that the fast and slow pulses have approximately equal areas....  ??
   

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....  But what does "more" current mean? Greater amplitude? Or more charge transferred? Certainly amplitude is greater the faster you go, and pulse duration is shorter. But over the total duration of the pulse, is there actually more _charge_ transferred, than with the slower but lower amplitude pulse? This I don't know. Looking at the areas of the two pulse curves above, it might be the case that the fast and slow pulses have approximately equal areas....  ??

Ah-yes,now you are on track.

Like i stated in my last post-Now remember,the faster the magnet is passed over the coil,the higher the current value,but shorter the time value.

So that is the question-do we end up with a larger overall current flow value,if we pass the magnet over the coil faster.
You stated the above--But over the total duration of the pulse, is there actually more _charge_ transferred

So,this is directly related to my other question-->

We have a coil with a FWBR attached to it's output leads,so as to rectify to DC.
We then have a cap across the FWBR output,so as to collect the energy produced by the coil.
Lets say that the cap is 100uF.

Test one
We now disconnect one end of the coil from the FWBR,so as we can do an open voltage test.
We move the magnet passed the coil as a set speed,and obtain an open voltage across the coil of say 5 volt's.
We now hook the coil back up to the FWBR,and move the magnet across that coil at the same speed as the open voltage test,and we end up with-say 3 volts across the 100uF cap. So our coil voltage has been limited to 3 volt's-plus the V drop across the FWBR.

Now we replace the 100 uF cap with a 200uF cap.
Now,we know that is we pass the magnet across the coil at the same speed as test one,we will not get 3 volt's across the 200uF cap.
BUT-if we pass the magnet across the coil faster,will we be able to get 3 volt's across the 200uF cap ?



Brad


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Here is the link for OU.com,that started this conversation

http://overunity.com/17297/where-the-overunity-using-induction-coils-comes-from-eg-joule-thief/15/#.WVTKeNR95kg

Also the long discussion between verpies and MarkE-if anyone is interested.
Verpies reply to me after i questioned this subject was--  I already had this debate with the most difficult opponent (MarkE), and I eventually won it (including experimental confirmation).
Please read these links and reply to me if you still want to continue the debate about this subject in this thread.


These links are

http://overunity.com/14443/quantum-energy-generator-qeg-open-sourced-by-hopegirl/msg401342/#msg401342

and 

http://overunity.com/14443/quantum-energy-generator-qeg-open-sourced-by-hopegirl/msg402626/#msg402626

MarkE's final comment was--Yes the mistake that I made and acknowledged further in the thread is that I failed to account for the integration.  Faster speed yields a higher di/dt, but T is proportionally smaller.


I feel there is one thing both verpies and MarkE missed,that makes a big difference to the outcome of the question at hand.

I have finished my DUT,but i would like this discussion to go further before i post my result's,as i may be missing something here.
Perhaps some of those better versed in the art of !charge,voltage and current!,would care to put there understanding here.

But ATM,i am sticking with my answer to the question (Do you not get more current if you move a magnet across a coil faster?),and that is yes,you do.


Brad


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Believing something false does not make it true.
I think I understand what Brad is asking now.  Will the total energy stored in the cap be greater if the magnet is moved past the coil faster?  Is that correct Brad?  Current is a measure of an instantaneous value.  So a lower current for a longer time could also charge a cap to a value that is equal to a shorter time of higher current.

To answer your question I believe the faster movement will actually charge the cap to a higher value because there is more energy being put into moving the magnet at the faster speed which will mean more energy available to charge the cap.


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I think I understand what Brad is asking now.  Will the total energy stored in the cap be greater if the magnet is moved past the coil faster?  Is that correct Brad?  Current is a measure of an instantaneous value.  So a lower current for a longer time could also charge a cap to a value that is equal to a shorter time of higher current.

To answer your question I believe the faster movement will actually charge the cap to a higher value because there is more energy being put into moving the magnet at the faster speed which will mean more energy available to charge the cap.

I will post the parameters of the test in my next post,as it will explain as to how the test should be carried out.

My last post(the bottom half in red),explains the DUT,and how i will carry out the test.

Brad


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The parameters of my test to be carried out

Test will be carried out in two parts
Part one
1-A magnet will be pasted across an inductor at a set constant speed
2-The distance between the magnet and inductor will remain unchanged in both parts of the test.
3-The inductor will retain the same peak value voltage across it for both parts of the test.
4-The capacitor will be charged in the first half of the cycle,and discharged(shorted)in the second half of the cycle,so as each cycle starts with a discharged cap.
5-The capacitor value for the first half of the test will be 220uF
6- 5 captured cycles will be recorded,and averaged out--the recording will be the voltage potential across the 220uF cap. From this we can calculate the average energy received from the coil for each single pass.

Part two
1-This part of the test is much the same as part one,but where we change the 220uF cap out for a 470uF cap.
2-The speed at which the magnet passes the inductor will be increased until such time as the peak voltage across that inductor is equal to that of what it was in part one of the test. This way we know that an increased voltage potential(EMF)across the inductor is not responsible for !a yet to be determined! outcome of the test.
3-Once again,5 captured cycles will be recorded ,and averaged out,so as we can calculate the energy received from the coil for each cycle.

From this test,where we have kept the EMF value across the inductor the same throughout each of the two test's,we can then arrive at 1 of two conclusions
 
Conclusions

1-If in each of the two test's,the value of the stored energy in the two cap's calculates out to be the same,then we can conclude that passing a magnet over a coil faster,dose not increase total current value delivered from that inductor.

2-If we end up with more stored energy in the 470uF cap,where we moved the magnet across the inductor faster,then we can conclude that you do get a higher total current value if you move a magnet across an inductor faster.

If anyone can see a flaw in these test parameters,please feel free to point them out.


I know many here may say--well if you follow this law,or that law,you will get your answer.

This is not me,as i prefer to find my answers on the bench.
Im just not into that dt/upsidedown*backtofront stuff.


Brad


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Just a note
verpies is traveling ATM in an area with Very poor Internet
There is a problem also with some links at OU.com which were linked to

Here is the issue and the fix

Quote
The absurd machine mentioned by me during the debate with MarkE about the current caused by induction, has errors in the blue links appearing therein.
here http://overunity.com/14443/quantum-energy-generator-qeg-open-sourced-by-hopegirl/msg401605/#msg401605

To fix them, just delete the www. prefix at the beginning of the link.  Please let the guys on the forum, know about this issue and the workaround.
end Quote
   

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Take a look at this scopeshot. I've used the scope's math to integrate the pulses over time. This gives a value for the area "under" the waveforms of the pulses -- amplitude times duration, expressed as area.

Note that the height of the integration trace peak above its slanted baseline is almost the same for both fast and slow pulses. (I counted pixels on an enlarged image.)

I think this means that the quantity of charge transferred is the same in both cases, or nearly so.  (Coulombs/second * seconds = Coulombs)

(The integration trace baseline is slanted because there is a slight DC offset on the CH2 trace baseline, and the integration is adding this up over the time slices on screen.)
   

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Take a look at this scopeshot. I've used the scope's math to integrate the pulses over time. This gives a value for the area "under" the waveforms of the pulses -- amplitude times duration, expressed as area.

Note that the height of the integration trace peak above its slanted baseline is almost the same for both fast and slow pulses. (I counted pixels on an enlarged image.)

I think this means that the quantity of charge transferred is the same in both cases, or nearly so.  (Coulombs/second * seconds = Coulombs)

(The integration trace baseline is slanted because there is a slight DC offset on the CH2 trace baseline, and the integration is adding this up over the time slices on screen.)

TK

How did you get that trace?,as there is only some 250ms between wave form's.
Was this achieved by passing a magnet over a coil?.


Brad


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Ok,here is the video on the low speed test.

I will be building the test bed for the mid to high speed test tomorrow,as the capacitor shorting per cycle gets a little harder at high speeds.

https://www.youtube.com/watch?v=xANse1LZi74


Brad


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TK

How did you get that trace?,as there is only some 250ms between wave form's.
Was this achieved by passing a magnet over a coil?.


Brad

Yes, passing magnet over coil by hand, first slow then fast, just two movements forward then back. So the magnet approaches the coil then recedes, so you get a positive pulse rising from zero, a fast fall to negative as the magnet passes the coil's center axis ,then a return to zero. Coil is from a miniature relay, magnet is a strong NdBFe, load is a 10k resistor, scope probe across load.  Scope in single shot trigger mode. So the trace indicates Vdrop across load resistance, which by Ohm's law gives the value of the current through the load, right?

And using the scope's math to integrate the pulses over time, we get a math trace whose peak values indicate the amount of charge transferred, I think. The peaks form during the rising part of the pulse as charge flows one way as the magnet approaches, and then return to the baseline as the same amount of charge flows the other way as the magnet recedes from the coil. There's a little DC offset (error) in the CH2 trace that is constant, and this causes the math integration trace's baseline to have that upward slope. This doesn't interfere with interpretation, except to cause a bit of optical illusion.
   

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Below are a couple of scope shot's,along with test circuit.

The first scope shot shows results of slow pass of magnet over the coil.
The second scope shot shows a faster pass of the magnet over the coil.

Brad


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verpies asked me to put up some screenshots of:

"a capacitor getting charged through a diode from your FG with a sawtooth wave vs. a time-reversed sawtooth wave, that has the same peak and average amplitude ?"

So i toke a 470uF electrolylic capacitor, a schottky diode (Bat46) and a 100 Ohm resistor and hooked it up like TM shows above (FG for the coil)

The FG was set to a sawtooth wave with 2V all positive amplitude at 1Hz, as shown by the blue trace.
The resulting yellow trace was taken across the cap / 100 Ohm resistor.

First screenshot shows the sawtooth signal being from peak left,
second screenshot shows a reversed peak right signal.


Look at how the resulting yellow trace across the cap / 100 Ohm shows a difference in charge / discharge time

verpies will probably explain further when he is able to connect to the WWW.

Itsu
   

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Here some screenshots now with an AC (4Vpp) sawtooth signal so the diode can do its job.

Itsu
   

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Here some screenshots now with an AC (4Vpp) sawtooth signal so the diode can do its job.

Itsu

I am not sure what the two tests you just carried out,are suppose to show Itsu?,as i never had one channel of the scope across both the resistor/capacitor series part of the circuit.

Your scope shots show the voltage each side of the diode-blue trace one side,yellow trace the other  ???,where as mine show the current flowing into the cap,and the voltage across the cap.

Second.
My forward peak potential voltage available from the coil at the low speed pass,was 6 volt's,but where that voltage is clamped by the voltage across the capacitor + resistor+V drop of the diode.
So that's 2.16v + 1.36V +.2V=3.72V,which is still well under the potential voltage of the coils 6 volts.

Maybe you would answer the question everyone else seems to be avoiding ?
If a 220uF cap is charged to say 2.5 volts in one second,what would be needed to charge a 470uF cap to the same voltage in half a second?.


Brad


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