Understanding Induced Current and EMF in Electromagnetic Induction

[sophiecentaur]

I dug out my old textbook (Panofski and Phillips) and found this passage, which, to me, suggests that the effect is there with or without a wire being present. That confirms my opinion that the emf is there without any current being needed.

 

[jsmith613]

...

going back to the EMF qustion could you quickly look at post #85 and tell me if the diagram there seems correct

 

[sophiecentaur]

going back to the EMF qustion could you quickly look at post #85 and tell me if the diagram there seems correct

Sorry- I though I'd already replied. Must have killed that window without actually posting. The diagram shows what happens for a coil (not tube) and that the peak EMF is greater when the magnet is falling faster. The area under both of those humps would be the same aamof. There is no current (scope is high impedance) so virtually no braking effect in this case.

Are you confusing the two situations or is this just 'for interest'?

 

[jsmith613]

Sorry- I though I'd already replied. Must have killed that window without actually posting. The diagram shows what happens for a coil (not tube) and that the peak EMF is greater when the magnet is falling faster. The area under both of those humps would be the same aamof. There is no current (scope is high impedance) so virtually no braking effect in this case.

Are you confusing the two situations or is this just 'for interest'?

most probably confusing the two!
i would have thought that it would be the same in both so simple said tube!

in the tube I would therefore presume the copper circlets are independent of each other. therefore the currents coult not cancel
here however it is one long coil so they are all interlinked and the two opposite currents DO cancel
right?

 

[sophiecentaur]

most probably confusing the two!
i would have thought that it would be the same in both so simple said tube!

in the tube I would therefore presume the copper circlets are independent of each other. therefore the currents coult not cancel
here however it is one long coil so they are all interlinked and the two opposite currents DO cancel
right?

I really don't understand what picture you have in your head. What currents "cancel". You are not using terms that I can understand. Do you mean that the mean current is zero?
When the magnet falls through a copper tube the currents are very high and vary, as I have already said, over time and distance.

 

[jsmith613]

I really don't understand what picture you have in your head. What currents "cancel". You are not using terms that I can understand. Do you mean that the mean current is zero?
When the magnet falls through a copper tube the currents are very high and vary, as I have already said, over time and distance.

hold on I will post a diagram...

see page 7...

 

[jsmith613]

see diagram

 

[cabraham]

I dug out my old textbook (Panofski and Phillips) and found this passage, which, to me, suggests that the effect is there with or without a wire being present. That confirms my opinion that the emf is there without any current being needed.

I disagree. The author presents his claim that E and B are there with or without a conductor. I agree with him so far. Then he writes the closed path line integral of E*dl, which is voltage, and relates it to B and area. But the integral E*dl, requires a specific path to have a value. It is a closed loop integral.

If the voltage is there in an imaginary closed loop in space, will charges in free space circulate in said loop? I don't think so. If a voltmeter, VM, were to have its probes placed in 2 points in empty space, would it read a non-zero voltage in the presence of a non-zero E field? What do you think?

The author correctly points out that E & B are there even in empty space. He then concludes that emf which is the integral of E*dl, must also be non-zero, which makes me wonder. In a physical conductor immersed in said fields, charges would move. In space they move, but not in a closed loop like they would in a conductor. Lorentz force is there, and the charges in free space do indeed move, but the path changes. The emf due to varying fields is path dependent. Since the path taken by free electrons in space differs from that taken in a conductor, the voltages are not equal.

To say that a voltage exists in free space can be supported by Maxwell et al. But I don't think it is the same value as the case w/ a conductor because electrons would move along a different path, and voltage value is path dependent. A CRT is an example. Two parallel plates have a charge, and an electron beam is projected between the plates. The electrons get attracted towards the positive plate and away from the negative plate.

In free space, at every point between the plates, it is safe to say there is indeed a potential. But the E field here is static. I've already stated that under static conditions, current can exist w/o voltage and vice-versa. This does not negate my earlier statement involving dynamic conditions.

Same problem, but the field between the plates is ac, sinusoidal for example. The sine curve plate voltage results in a sine curve E field. There is indeed a potential in between the plates, sine curve in nature. But the plates carry a current to maintain the sine E field. Free electrons in between the plates move back and forth, which is ac current.

Dynamic conditions, i.e. time-chaning, are different than static. No ac voltage exists w/o ac current. That is my point. In free space there can be a voltage w/o current, but only static, not dynamic.

As far as joules per coulomb goes, the voltage across the car, bumperto bumper, is the joules per coulomb of charge transported from bumper to bumper. If a resistor were connected across the bumpers, large in value so that its own current generates a B field too small to cancel the external B field, then the voltage is equal to the joules of energy per coulombs transported through said resistor.

Is this clear? Do I need further clarification? BR.

Claude

 

[BruceW]

hmm. If we define emf as the closed-loop integral of E*dL then of course it is possible for an emf to exist even though there are no charges or currents.

 

[cabraham]

hmm. If we define emf as the closed-loop integral of E*dL then of course it is possible for an emf to exist even though there are no charges or currents.

But it becomes indeterminate. What is the value? What is the path of integration? When a conductor is present there is a definite path of integration meaning that we can compute a specific value. Without a conductor where is the path? How can we transport a charge along an arbitrary path without a conductor in place? Think about what you're saying.

Also, I've already explained that emf can exist w/o current, but only in the static case. For the dynamic case it can't happen. Feel free to offer an example if you believe currents and voltages can exist separately under dynamic conditions.

Claude

 

[sophiecentaur]

I disagree. The author presents his claim that E and B are there with or without a conductor. I agree with him so far. Then he writes the closed path line integral of E*dl, which is voltage, and relates it to B and area. But the integral E*dl, requires a specific path to have a value. It is a closed loop integral.

If the voltage is there in an imaginary closed loop in space, will charges in free space circulate in said loop? I don't think so. If a voltmeter, VM, were to have its probes placed in 2 points in empty space, would it read a non-zero voltage in the presence of a non-zero E field? What do you think?

The author correctly points out that E & B are there even in empty space. He then concludes that emf which is the integral of E*dl, must also be non-zero, which makes me wonder. In a physical conductor immersed in said fields, charges would move. In space they move, but not in a closed loop like they would in a conductor. Lorentz force is there, and the charges in free space do indeed move, but the path changes. The emf due to varying fields is path dependent. Since the path taken by free electrons in space differs from that taken in a conductor, the voltages are not equal.

To say that a voltage exists in free space can be supported by Maxwell et al. But I don't think it is the same value as the case w/ a conductor because electrons would move along a different path, and voltage value is path dependent. A CRT is an example. Two parallel plates have a charge, and an electron beam is projected between the plates. The electrons get attracted towards the positive plate and away from the negative plate.

In free space, at every point between the plates, it is safe to say there is indeed a potential. But the E field here is static. I've already stated that under static conditions, current can exist w/o voltage and vice-versa. This does not negate my earlier statement involving dynamic conditions.

Same problem, but the field between the plates is ac, sinusoidal for example. The sine curve plate voltage results in a sine curve E field. There is indeed a potential in between the plates, sine curve in nature. But the plates carry a current to maintain the sine E field. Free electrons in between the plates move back and forth, which is ac current.

Dynamic conditions, i.e. time-chaning, are different than static. No ac voltage exists w/o ac current. That is my point. In free space there can be a voltage w/o current, but only static, not dynamic.

As far as joules per coulomb goes, the voltage across the car, bumperto bumper, is the joules per coulomb of charge transported from bumper to bumper. If a resistor were connected across the bumpers, large in value so that its own current generates a B field too small to cancel the external B field, then the voltage is equal to the joules of energy per coulombs transported through said resistor.

Is this clear? Do I need further clarification? BR.

Claude

I think it would because there's a field there. A probe (infinite imdedance voltmeter) would register the appropriate voltage for the field strength and its length.

Because they have mass they would not follow any resulting curved field. I think introducing electrons is really a bad idea for this reason. It's only when in a metal (with almost zero speed,) that electrons will follow curved E field lines.
But DC is only the limit of decreasing frequency there can hardly be a step change in what happens when the current is not changing (in any case, there is no such thing as real DC because it was switched on at some time and will be switched off)

A resistor would also have an equal and opposite emf induced in it so no current would flow. I made this point before in a different context.

 

[cabraham]

I think it would because there's a field there. A probe (infinite imdedance voltmeter) would register the appropriate voltage for the field strength and its length.

Because they have mass they would not follow any resulting curved field. I think introducing electrons is really a bad idea for this reason. It's only when in a metal (with almost zero speed,) that electrons will follow curved E field lines.
But DC is only the limit of decreasing frequency there can hardly be a step change in what happens when the current is not changing (in any case, there is no such thing as real DC because it was switched on at some time and will be switched off)

A resistor would also have an equal and opposite emf induced in it so no current would flow. I made this point before in a different context.

Well, let's clarify what I mean by "current must exist when a dynamic voltage exists". Fields are by their very nature distributed throughout space. An E field generated by a circuit, extends towards infinity, likewise for H. Likewise, the integral of E*dl can be defined for any path in space.

In order for a dynamic E/H to exist, there must be a current somewhere, but not everywhere. This is important. Take a 2 wire transmission line. An E field exists between the wires, plus some fringing, and the H field encircles each wire, extending towards infinity. An E field exists in the space between the 2 wires. But the current is confined to the interior of the 2 wires.

Thus we can state that even under dynamic conditions, that if we confine our region of examination to a space in between the 2 wires, but excluding both wires, that we have E & H field present in said region. We can integrate around a closed path that does not include either wire or portion thereof. Hence one can say that there is a field and an emf in this narrow spatial region, and at the same time without a current. But remember that fields are omnipresent, but current is confined to a locale.

Outside of the wires, the influence of the current is felt in the form of fields, although the current is contained wholly within the wires. So a clarification is in order. An ac voltage in a region of space, requires the existence of ac current, but not necessarily in the same region in space. Obviously current is not present in every point in the universe, whereas E & H fields are.

Do you see what I'm getting at? In a specific spatial volume we have a dynamic E & H, but there is no current present in said volume. But there is a current nearby in the 2 wires. This current is what makes E & H possible. The emf is the integral of E over a specific closed path. But E cannot be unless current is present somewhere, but not necessarily in the region in question. The current in the 2 wire t-line is what gives rise to the proagation of E & H. That is my point. Since current is confined to a narrow region in space, while E, H, and emf are defined everywhere, most closed regions in space have E, H, & emf, without a current in the enclosed region.

I guess I should state my point as follows. A dynamic emf somewhere in space cannot exist unless there is also a current, but not necessarily in the same spatial region. Hopefully we can all agree on that.

As far as the probe registering the right voltage, I say wait a minute. The voltage value is path dependent. The voltage value displayed on the infinite impedance VM would depend on how the leads are arranged. In free space with dynamic fields present, the emf from point a to point b, is path dependent. Re-arrange the voltmeter leads and the displayed reading changes.

So in order to define the emf, one must configure the leads along a specific path in question. But in doing so we have introduced conductors, i.e. the test probe leads, into the picture. Like I said, attempting to define a free space emf w/o conductors is arbitrary and ambiguous. Frankly, it is pretty arbitrary to say that a region in space has an emf w/o defining a conductor with a specific shape.

I believe I've supported my position with solid facts. Anything unclear can be discussed further.

Claude

 

[sophiecentaur]

I think we are going round in circles here and I wonder how you have arrived at some of your opinions. I am very rusty about a lot of this but, when I re-read textbooks, it usually makes sense.

Regarding my high impedance probe. It takes the form of the shortest 'electric dipole' you can get away with and the smallest meter, in situ. Lead lengths are not considered here, although, connecting to a larger meter with a twisted wire feed at right angles to the conductor shouldn't affect things.

But I'm afraid we have hijacked this thread, which was initially about a more mundane problem, if I remember right.

 

[jsmith613]

I think we are going round in circles here and I wonder how you have arrived at some of your opinions. I am very rusty about a lot of this but, when I re-read textbooks, it usually makes sense.

Regarding my high impedance probe. It takes the form of the shortest 'electric dipole' you can get away with and the smallest meter, in situ. Lead lengths are not considered here, although, connecting to a larger meter with a twisted wire feed at right angles to the conductor shouldn't affect things.

But I'm afraid we have hijacked this thread, which was initially about a more mundane problem, if I remember right.

yes
despite how interesting your discussion is..it goes right over my head
I hate to kill a party but I was wondering if we could return to my question post #97
it is the diagram explaining what i was talking about

although I may have cracked it...I think what I am getting at is actually the NET emf is zero but a current still flows...is this a more accurate description of what is going on?

 

[sophiecentaur]

see diagram

This is your set up (just as I imagined) but I want to know where these currents you refer to are flowing (as you see it). This must involve you putting some arrows with labels on that diagram. I still can't see how you are thinking with this.

 

[BruceW]

But it becomes indeterminate. What is the value? What is the path of integration? When a conductor is present there is a definite path of integration meaning that we can compute a specific value. Without a conductor where is the path? How can we transport a charge along an arbitrary path without a conductor in place? Think about what you're saying.

If we define emf as the closed integral of E*dL, then you can choose whatever path of integration you want. That's the beauty of Maxwell's equations. An emf does not necessarily have to cause a movement of charge. Of course, you can define emf differently. I don't know what is the most commonly-used definition of emf.

 

[jsmith613]

This is your set up (just as I imagined) but I want to know where these currents you refer to are flowing (as you see it). This must involve you putting some arrows with labels on that diagram. I still can't see how you are thinking with this.

here is a labelled diagram
or a better image:

 

[BruceW]

yes, if we assume the magnet goes down smoothly, then the induced currents will circulate horizontally (which doesn't happen in practise, but it might be a useful approximation). I see where you're coming from now. So what was your question about emf? (And I'll assume we're talking about emf over a horizontal path around the tube)

 

[jsmith613]

yes, if we assume the magnet goes down smoothly, then the induced currents will circulate horizontally (which doesn't happen in practise, but it might be a useful approximation). I see where you're coming from now. So what was your question about emf? (And I'll assume we're talking about emf over a horizontal path around the tube)

well will either of them be zero for the time the magnet is in the tube?

 

[BruceW]

Were you wondering (in this idealised motion), if the currents going anticlockwise below the magnet are equal in magnitude to the clockwise currents above the magnet? I see no reason why not. The magnet produces the same amount of magnetic field on either side, so the magnetic flux change on both sides is the same.

 

[jsmith613]

Were you wondering (in this idealised motion), if the currents going anticlockwise below the magnet are equal in magnitude to the clockwise currents above the magnet? I see no reason why not. The magnet produces the same amount of magnetic field on either side, so the magnetic flux change on both sides is the same.

so you would agree then that the net EMF and current is zero?

 

[BruceW]

well will either of them be zero for the time the magnet is in the tube?

either of what be zero?

 

[BruceW]

so you would agree then that the net EMF and current is zero?

you mean that there is the same amount of current going clockwise as there is anti-clockwise? Again, I see no reason why not.

 

[jsmith613]

either of what be zero?

the current or emf

 

[BruceW]

the current or emf

They won't be zero. But as I said in my last post, I would guess that there is as much current going anti-clockwise as there is current going clockwise. The magnet is dipole, so I see no reason to break symmetry of clockwise and anti-clockwise.

 

[jsmith613]

They won't be zero. But as I said in my last post, I would guess that there is as much current going anti-clockwise as there is current going clockwise. The magnet is dipole, so I see no reason to break symmetry of clockwise and anti-clockwise.

but why won't they be zero
1 + (-1) = 0
??

 

[truesearch]

Phew! I wonder what Faraday would make of that lot!
Your falling magnet in the copper tube...the induced currents flow around the tube but the current at the bottom is induced to repel the S pole, the one at the top is induced to attract the N pole.
Both currents are induced to generate S poles... They are in the same direction.
This is the difference between the tube with a falling magnet and the (thin) coil with a magnet falling through.

 

[BruceW]

but why won't they be zero
1 + (-1) = 0
??

the charges above the magnet need to go clockwise and the charges underneath need to go anticlockwise (to slow the descent of the magnet). So there is not the same amount of charges going in either direction at all points in space.

 

[jsmith613]

Phew! I wonder what Faraday would make of that lot!
Your falling magnet in the copper tube...the induced currents flow around the tube but the current at the bottom is induced to repel the S pole, the one at the top is induced to attract the N pole.
Both currents are induced to generate S poles... They are in the same direction.
This is the difference between the tube with a falling magnet and the (thin) coil with a magnet falling through.

no but when produces a south pole up and the other a south pole down?

 

[jsmith613]

the charges above the magnet need to go clockwise and the charges underneath need to go anticlockwise (to slow the descent of the magnet). So there is not the same amount of charges going in either direction at all points in space.

so is this diagram wrong:
http://www.thestudentroom.co.uk/attachment.php?attachmentid=139647&d=1333538964

I don't want to show the image here as it is enourmous