Non-changing current with C-EMF

[XRZ]

@(*steve*) hope you could join!

 

[XRZ]

@Arouse1973 glade to hear it!
But thanks to you and @Gryd3 for making me realize I could avoid the "common" method, and finding our own... solution!

 

[Arouse1973]

Okay, consider it to be a copper square that is 20 cm wide, 0.4cm thick, 20 cm long. Its resistance is about 4.2x10^-5 Ω, inductance : 56 nH , magnetic field is a 0.5T uniform field from an electromagnet.
Voltage should be: 0.0000042Ω x 1000A = 0.0042VoltsDC. Imagine the magnetic field changing contniously(however you would like to imagine it going on/off or rotating next to conductor). However, the change is massive. delta(BA)/delta(t) = 0.5Tx0.4m^2/0.100s = -2VDC induced back-emf.

What is the factor of power jump here? From 0.0042VDC x 1000A = 4.2W to what...?

I hope this is a reasonable example, if you need to please adjust it! for frequency I dont know why it's needed since it's a DC circuit?

I need to know the rate of change of the magnetic flux, i.e the frequency. Only a changing magnetic field will induce a voltage.
Adam

 

[Arouse1973]

I make it 200 mV induced not 2. I chose the largest area, i.e. the largest face 20 cm * 20 cm. How did you get 0.4m^2?
Adam

 

[XRZ]

Ah, again with my simple math mistakes!
Okay, the area is 0.04 m^2, and the time should be 0.010s, not 0.1s.
The voltage will be -2V.

 

[XRZ]

@Arouse1973,The change in magnetic field, well it could be going from 0T to 0.5T in 0.010 seconds.

 

[Arouse1973]

Yes it could.

 

[XRZ]

Well then, that's the rate of change in the magnetic field , do you need anything else to aid the calculations?

 

[Arouse1973]

Nope I think we have gone as far as we can. I conclude that the origional design of a current source mentioned by Steve is the only option.
Adam

 

[XRZ]

That is one of the solution in my head @Arouse1973 , but what about the power required? How much of a massive difference is it? In my calculations due to the low resistance the power required in would be massive. However, @Gryd3 pointed out that the internal resistances of the supplies would be large, so I should no be concerned with it much.
@(*steve*) when using a constant current source, if the circuit has low resistance will it still require massive power to apply large voltages to the circuit?

I know that more power is need, but how much.
How can I figure this out?

 

[(*steve*)]

Its not a project at all...an experiment to prove the possibility of maintaining the same current on a conductor, without exhausting enormous power.

Have you googled "superconductor"?

I self-teach myself electronics and tropics in electrical engineering on my own free time(oh and Physics!), and really tend to think of abstract experiments!
Which teaches me a lot of things, and sometimes conceptualize more.

I'm still not convinced that an experiment with 100A couldn't be done with 100mA. Scale it down. Your life will be much easier.

 

[XRZ]

@(*steve*) superconductors are their own complications but its a good idea, but limited at larger magnetic fields, and not sure how it would react to "back-emf".

This is more of an experiment on paper, but certainly in a realistic one I won't risk myself and overcomplicate things with large current. But rather limit things to much smaller values.

But you haven't given me your opinion about the scale of power, and how "much" is needed to possibly maintain the same current with a large induced-back emf to a circuit that has low resistance?

 

[(*steve*)]

But you haven't given me your opinion about the scale of power, and how "much" is needed to possibly maintain the same current with a large induced-back emf to a circuit that has low resistance?

I would measure the resistance of the conductor, and assuming the current was constant, I would use V = IR to determine the voltage requires to maintain the current, and P = I²R to determine the power required.

"back-EMF" seems not to play any part at all.

 

[XRZ]

@(*steve*) given the example I proposed to @Arouse1973, there is something that will be missing.
Let me explain:
V = IR, so current is 1kA and resistance is 4.2x10^-5Ω that gives 0.0042VDC
Now the power needed: P = I2R = 1,000,000x0.000045Ω= 4.2W but that's not including the back-emf induced by the magnetic field that is 2V so we actually need 2V + 0.0042V = 2.0042VDC to maintain the current, and by ohms law we can predict the power to be: Vtot^2 / R = 2.0042/0.0000042 = 956385W that's a huge jump in power due to the low resistance to maintain the same current when back-emf is opposing the applied voltage.

Now the solution is certainly something that could avoid using a lot of power(just like above x277,000 times) solutions like a constant current source or a boost converter could solve the problem.
If the voltage was actually larger than the current, than a dc/dc convert could work and solves the problem very well.

 

[(*steve*)]

Why is there a back EMF? Is this in a motor? (You understand that back-EMF applies to motors?) Or do you mean counter EMF?

Note in particular, this section:

The term Back electromotive force, or just Back-EMF, is most commonly used to refer to the voltage that occurs in electric motors where there is relative motion between the armature of the motor and the magnetic field from the motor's field magnets, or windings. From Faraday's law, the voltage is proportional to the magnetic field, length of wire in the armature, and the speed of the motor. This effect is not due to the motor's inductance and is a completely separate effect.

In a motor using a rotating armature in the presence of a magnetic flux, the conductors cut the magnetic field lines as they rotate. This produces a voltage in the coil; the motor is acting like a generator (Faraday's law of induction.) at the same time it is a motor. This voltage opposes the original applied voltage; therefore, it is called "back-electromotive force" (by Lenz's law). With a lower overall voltage across the armature, the current flowing into the motor is reduced.[4] One practical application is to use this phenomenon to indirectly measure motor speed and position since the Back-EMF is proportional to the armature rotational speed.
​

In this case back EMF is as a result of a changing magnetic field causing an induced voltage.

V = IR, so current is 1kA and resistance is 4.2x10^-5Ω that gives 0.0042VDC
Now the power needed: P = I2R = 1,000,000x0.000045Ω= 4.2W but that's not including the back-emf induced by the magnetic field that is 2V so we actually need 2V + 0.0042V = 2.0042VDC to maintain the current, and by ohms law we can predict the power to be: Vtot^2 / R = 2.0042/0.0000042 = 956385W that's a huge jump in power due to the low resistance to maintain the same current when back-emf is opposing the applied voltage.

If you have a constant current you also have a constant magnetic field resulting from it.

Do you have a varying current or a varying magnetic field, or motion through a constant magnetic field, or some combination of these? If so, where are these from, and how do you state them as part of your problem specification?

Or do you have a resistance of 4.2E-5Ω (let's just call it "R") which produces a voltage drop when a current of 1000A (let's just call that "I") which exceeds I*R? So are you trying to explain an experimental anomaly, or are you just asserting that V = I*R does not hold in your particular DC circuit without any actual evidence or theoretical prediction?

As far as I can tell, your 2V is a magic number that is obtained without any practical or theoretical justification. I can replace it with 1V, or 6V, or 1.4142136V without any need to justify it (it seems) in your statement. Why do we need this extra voltage? Mr Ohm says we don't.

 

[XRZ]

Why is there a back EMF? Is this in a motor? (You understand that back-EMF applies to motors?) Or do you mean counter EMF?

Well they all are the same thing, it just induced voltage the opposes you supply voltage.

Note in particular, this section:

The term Back electromotive force, or just Back-EMF, is most commonly used to refer to the voltage that occurs in electric motors where there is relative motion between the armature of the motor and the magnetic field from the motor's field magnets, or windings. From Faraday's law, the voltage is proportional to the magnetic field, length of wire in the armature, and the speed of the motor. This effect is not due to the motor's inductance and is a completely separate effect.

In a motor using a rotating armature in the presence of a magnetic flux, the conductors cut the magnetic field lines as they rotate. This produces a voltage in the coil; the motor is acting like a generator (Faraday's law of induction.) at the same time it is a motor. This voltage opposes the original applied voltage; therefore, it is called "back-electromotive force" (by Lenz's law). With a lower overall voltage across the armature, the current flowing into the motor is reduced.[4] One practical application is to use this phenomenon to indirectly measure motor speed and position since the Back-EMF is proportional to the armature rotational speed.
​

In this case back EMF is as a result of a changing magnetic field causing an induced voltage.

Well, consider it as counter-emf or -V. Yes c-emf is induced due to the conductor being next to a magnetic field. In my example specifically, the magnetic field is 0.5Tesla and it goes from 0-0.5T in 10 milliseconds, so that is the induced voltage would be equal to Faraday's law of induction.

If you have a constant current you also have a constant magnetic field resulting from it.

That's fine but it's possibly a small field and won't be something I'm focusing on the most.

Do you have a varying current or a varying magnetic field, or motion through a constant magnetic field, or some combination of these? If so, where are these from, and how do you state them as part of your problem specification?

Well, I have a fixed conductor next to a magnetic field that goes 0 to 0.5Tesla in 10 milliseconds, current should be constant(the goal when back-emf is induced).

Or do you have a resistance of 4.2E-5Ω (let's just call it "R") which produces a voltage drop when a current of 1000A (let's just call that "I") which exceeds I*R? So are you trying to explain an experimental anomaly, or are you just asserting that V = I*R does not hold in your particular DC circuit without any actual evidence or theoretical prediction?

Hm, this got me to scratch my head a bit... what I'm doing is trying to calculate the required power when back-emf is induced in my example in the previous post, but... the idea is really trying to maintain the same current, even if there is counter-emf without really using a lot of power.

As far as I can tell, your 2V is a magic number that is obtained without any practical or theoretical justification. I can replace it with 1V, or 6V, or 1.4142136V without any need to justify it (it seems) in your statement. Why do we need this extra voltage? Mr Ohm says we don't.

E = - d(BA)/dt = - (0.5x0.04)/0.010 = -2V opposing the supplied voltage. We don't need it, but based on Faraday's law of induction: If there is change in flux over a change in time, there is induced emf, and based on Lenz law, the "emf" opposes the circuit's voltage = -V.

 

[Merlin3189]

I'm not sure why I'm reading all this, as I don't understand what you are trying to do or why! Yes, it's just an experiment, but even an experiment needs a reason (even a thought experiment.) And if the reason is to find out how much power is needed to maintain a constant current (answer zero?), then I still wonder why and what is the reason for asking this question. When we get to the bottom of this (the reason) I suspect the answer will be easy.
Meanwhile, I'm prompted to write to add two thoughts.
1- You can't hold the current constant. However good your constant current circuit is, there must always be some change in current for it to detect, otherwise how does it know that it needs to change to counteract it? As people eventually pointed out, the simple solution to a constant current is a very large resistance, so that any changes in the load are tiny in comparison. But there will still always be a tiny change in current however insignificant (& we can't really tell what is significant until we get to the bottom of your reasons.)
Since you apparently need 100 Amp current, a very large resistance might be an Ohm or less, letting your supposed 2V emf change the current by an insignificant 2% or more.
2- The obvious way to hold the current constant is, as Arouse said, to balance the circuit. For example have a second identical slab of copper in the same field but wired into the circuit in the reverse direction. The emfs will exactly balance out and leave the current constant. If you can't get a whole slab in the same field, you could use a smaller slab (or coil) to detect the change in field and amplify the signal from that to control your current source (though there must still be a small error, as in 1.)
As for the power issue I'm not sure that there is any minimum power required to control current (other than the minute amount required on quantum mechanical grounds.) If you imagine controlling the flow of water in a pipe with a valve. With a very thin, light and strong valve, sliding in almost frictionless bearings, the power used to operate the valve should be very small and not obviously related to the quantity of water flowing.

 

[(*steve*)]

Well they all are the same thing, it just induced voltage the opposes you supply voltage.

Did you miss the part of the quote which said:

"This effect is not due to the motor's inductance and is a completely separate effect."​

So which of these two effects are you talking about?

Well, consider it as counter-emf or -V. Yes c-emf is induced due to the conductor being next to a magnetic field.

only if the magnetic field or the current is changing.

In my example specifically, the magnetic field is 0.5Tesla and it goes from 0-0.5T in 10 milliseconds, so that is the induced voltage would be equal to Faraday's law of induction.

What is causing this change in magnetic field? Remember that you're talking about a constant current and therefore a constant induced magnetic field.

Hm, this got me to scratch my head a bit... what I'm doing is trying to calculate the required power when back-emf is induced in my example in the previous post, but... the idea is really trying to maintain the same current, even if there is counter-emf without really using a lot of power.

Assuming you have some external magnetic field which is moving (say a rotating magnet) then this extra power you require is used to impart a force on the rotating magnet to alter its rotation.

E = - d(BA)/dt = - (0.5x0.04)/0.010 = -2V opposing the supplied voltage. We don't need it, but based on Faraday's law of induction: If there is change in flux over a change in time, there is induced emf, and based on Lenz law, the "emf" opposes the circuit's voltage = -V.

But what is causing this change in magnetic field? HINT - it's NOT the current through your wire, since the magnetic field from that is constant with a constant current.

Sure, as the current increases, the magnetic field changes, and this impedes the flow of current -- this is called inductance and is a very well known effect. Once the current is constant (inductance tries to prevent the change in current) there is no effect.

 

[XRZ]

@Merlin3189 Thanks for the post, reminded me of a few things...
If you check on post 74 where I stated that in order to maintain constant current while back-emf is present I require not 4.2W but almost 1MW of power check the numbers please, It might not make sense but give you the idea of what I'm talking about in terms of "power" needed to maintain constant current.

@(*steve*) Yes I have. The effect(counter-emf) is due to an electromagnet being next to a conductor switching on/off in 10 milliseconds.
The only thing that should change is the magnetic field, while the goal is to have the current constant while the magnetic field is changing.
The cuase of the change in magnetic field is intentionally, it's me turning a magnetic field off or on.

Please check post #74 to understand what I mean by "additional power".

 

[Arouse1973]

As far as an ideal current source is concerned it will do what ever is needed to maintain it's current. It does this by altering it's voltage, it can't do anything else as it has no direct control on the resistance. So to maintain this all it has to do it change it's voltage and does not need to absorb any power from anywhere. It's the reduction in the terminal voltage which maintains this current and this does not use any power. This is also an air cored transformer with a turns ratio of ??:1.

Adam