# C-EMF and a coil?

Discussion in 'General Electronics Discussion' started by Dretron, Feb 26, 2013.

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1. ### Dretron

101
0
Jun 9, 2012
Hello everyone.

If a magnet passes through a coil it would produce C-EMF.
And lets say the coil is without any power inputed, so the only thing passing through that coil is C-EMF, can I immediately cancel C-EMF? With a resistor?
And if that "turned off" coil had C-EMF flowing through it, what would happen to that C-EMF? the coil is connected to a power source put switched off. So would that C-EMF turn into heat? What if we just used multiple resistors?

Would the resistors cancel the C-EMF and convert it to heat or what would happen in this situation?

2. ### Laplace

1,252
184
Apr 4, 2010
Not sure that C-EMF is the proper term to be used in this situation. While C-EMF is sometimes mentioned when discussing Lenz's Law, that topic is primarily concerned with the direction of induced currents and not EMF, counter or otherwise. C-EMF is a term primarily applied to the armature of DC motors as, "The voltage induced in the armature of a DC motor that opposes the applied voltage and limits armature current." What you are describing is a generator, a coil in a changing magnetic field. If there is a circuit path for this generated voltage to cause a current to flow through a resistor, then power will be dissipated in the resistor. But that does not "cancel" the voltage.

3. ### Dretron

101
0
Jun 9, 2012
So the voltage would dissipate in the resistor, would that happen instantly? Or would it take time?

I know it depends.
But lets compare a resistor to a capacitor.
What would charge quicker?

+ When the voltage dissipates that would lead to heat?

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
Resistors don't charge.

And you dissipate power, not voltage (you can have a voltage drop across a resistor -- and if so it is dissipating power according to (V^2)/R).

As soon as the magnet stops moving the current through the coil will start to decrease. The time it takes to do so is dependant on the inductance of the coil and the resistance (and capacitance) that is also in the circuit.

edit: yes, dissipated power is generally dissipated as heat.

5. ### Dretron

101
0
Jun 9, 2012
I think it will always be pretty fast(Close to instant).
If we had a generator and you stopped the magnet from moving instantly the power goes to 0 when the magnet stops.

I assume this might happen.
Not sure though.

6. ### Dretron

101
0
Jun 9, 2012
Another thing.

If a magnet passes through a coil.
And the coil is connected to nothing, it will only have induced voltage but no current.
Something has to draw the current.

But if there isn't a thing to draw that out.
Only voltage will be present.

Im I right here?

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
Generally this is true. Especially if we're talking about normally observed values.

No, not instantly. As you said above: "pretty fast (close to instant)"

Yes, that's right.

And because there's no current, the voltage will fall to zero immediately the magnet stops moving. If current was flowing, *then* when the magnet stops moving, the current will fall exponentially to zero (theoretically never quite reaching it)

8. ### Dretron

101
0
Jun 9, 2012
I meant close to instant matter of fact. SEMI-INSTANT.

Its good to know current = zero.
Voltage drops to zero semi-instantly

Thank you everyone.

9. ### Dretron

101
0
Jun 9, 2012
Steve,
Say we had a magnet "approaching" an electromagnet.
Distance between them is 10 cm.
As the magnet approches to the electromagnet the distance is dropping from 10 -> 0.
When the magnet reached to 0 cm, and C-EMF is already generated, but since the switch is off and there is no current, that means voltage will drop will drop to 0 very very fast, milliseconds after the magnet approached 0 cm distance. And the electromagnet is instantly turned on after the magnet is at 0 distance.

Will the electromagnet have any resistance at all? Since the magnet's motion is 0 and the switch was off before... C-EMF will not resist the input current + voltage.

I'm guessing the electromagnet will be powered 95%(including losses) and repel the magnet! With no C-EMF to worry about(So the full force of the electromagnet is applied on the magnet, without C-EMF that is possibile). As soon as the magnet gains force and moves the distance start to increase, the electromagnet will shut off again so C-EMF is not an issue here.

What do you think of this "process".

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
Remember that reading the voltage across a coil is placing a high resistance load onto it. Now things don't happen instantaneously (but still very fast)

Yeah the sign of the voltage will change at some point.

11. ### Dretron

101
0
Jun 9, 2012
So, the electromagnet will power up fully with no resistance from the C-EMF and repel the magnet!
Since we turned off the electromagnet before...
There should be no C-EMF to effect that force.

12. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
I'm not sure what you're getting at.

You have a magnet approaching a coil which is connected to nothing (so no current can flow)

And then you're saying it will "power up fully" <-- what does that even mean?

If you mean that you can now attach a voltage source to the coil and expect it to reach some certain current immediately, the answer is no. The coil still has inductance and will take some time to reach a steady current (theoretically infinite, but practically much sooner)

13. ### Dretron

101
0
Jun 9, 2012
I was saying that as the magnet approaches the off-electromagnet. The induced CEMF will be no problem because of the magnet being off... And when the electromagnet is turned on voltage will flow perfectly will and wont be effected because CEMF. That electromagnet will have full power.

Having the electromagnet off means CEMF will have no current. And it would not effect the input EMF at all.

Do you see where Im heading with this?

14. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
No current, but plenty of voltage.

And that voltage will be a second force acting to counter the flow of current.

Up a blind alley?

15. ### Dretron

101
0
Jun 9, 2012
I'll try all over again.

1) We have an electromagnet that is turned off and fixed on a table, and a magnet 10 cm away from the electromagnet.
2) We moved the magnet very quick closer to the electromagnet so the distance is approaching to 0cm.
3) As the magnet comes closer and closer to the electromagnet we know there will be induced C-EMF flowing.
4) Since the electromagnet is turned off the C-EMF's voltage will drop very very fast.
5) When the magnet is at distance 0cm. We turn the electromagnet on
6) EMF will flow without the C-EMF resistance.( Since we said in 4 the electromagnet is off and the voltage will drop rapidly)
7) The electromagnet repels the magnet
8) Instantly as the electromagnet is turned on and repels the magnet its turned back off to avoid C-EMF.
9) The magnet is repelled and C-EMF is strategically avoided.

My theory is that we can manipulate with the power source + the system to avoid C-EMF.
Ow yea, and all of that happened very fast!

Hope brings you clarity of my idea Steve.

+ All that amount of voltages means nothing, eventually it will drop (very quickly)
I believe its quicker than anything in the system too.

16. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,832
Jan 21, 2010
1 OK
2) OK
3) No we don't -- EMF doesn't flow.
4) Is the magnet still moving? Is it acting as a core to the inductor?
5) ok
6) No, the voltage will rise slowly. The metal will act as a core increasing the inductance. Force will be applied to the magnet.
7) Yes (or attract it)
8) And the energy you've put into the coil is either lost or imparted to the magnet
9) No.

Last edited: Mar 6, 2013
17. ### davennModerator

13,813
1,945
Sep 5, 2009
OK

OK

just induced EMF ... there is no counter EMF yet in this situation
reading through all your posts so far, you seem to be hung up on this counter EMF thing
its messing you up

No --- nonsense statement -- see previous comment

OK

NO, dont contradict yourself -- cuz have stated that you have turned the electro-magnet on see your #5
EMF doesnt flow, current flows. There is NO C-EMF yet as I stated earlier

yes ... maybe, depending of the orientation of each of them to each other ... they may attract !!

No again may repel or attract. and NO you cant avoid the C-EMF which is NOW produced by the collapsing magnetic field in the electromagnet coil when the power is shut off

NO -- your theory is flawed cuz of all the previous comments

Dave

Last edited: Mar 6, 2013

25,490
2,832
Jan 21, 2010
*SNAP*

19. ### Dretron

101
0
Jun 9, 2012
Hey floor.
Its me again, ow I wanted to let you know... This time, the fall is worse than before!

I need to re-think/study/solve this all over! My logic is flawed!

20. ### Dretron

101
0
Jun 9, 2012
You're right. There is only induced EMF since the electromagnet is off.
However, where is that flow of EMF going?

Wait, so EMF dosen't flow because?
Curent now flows because the electromagnet is on I know that... But what about EMF?

Since I said it would repel the magnet. That means they electromagnet is set to repel the magnet, and the magnet is fixed in a way that it can only be repelled besides the point of it being the same pole as the electromagnet.

It will repel.
Collapsing magnet field due to the induced EMF from the magnet approaching it?
If so, how can I calculate that?

Agreed. Need to rethink and reevaluate.