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Induction, magnetism and conductivity‏

by MarkoniF
Tags: conductivity‏, induction, magnetism
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MarkoniF
#1
Dec14-12, 01:14 AM
P: 56
Relative motion between a permanent magnet and a conductor wire produces electric current in the wire. Would induced electric current be greater in a wire made of "magnetic" material like iron, "non-magnetic" material like copper, or it doesn't matter? In other words, is there relation between material magnetic properties and its inductivity? And similar but I think different question, is there relation between material magnetic properties and its conductivity?

How come magnetic field of a permanent magnet can interact with magnetic fields inside a wire that is overall magnetically neutral? Does the same thing happen with electric fields, so if instead of permanent magnet we had some electrically charged object, could we also induce electric current in a conductor wire by their relative motion?
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Andrew Mason
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Dec14-12, 08:57 PM
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Quote Quote by MarkoniF View Post
Relative motion between a permanent magnet and a conductor wire produces electric current in the wire. Would induced electric current be greater in a wire made of "magnetic" material like iron, "non-magnetic" material like copper, or it doesn't matter? In other words, is there relation between material magnetic properties and its inductivity? And similar but I think different question, is there relation between material magnetic properties and its conductivity?
I am not sure what you mean by "inductivity".

The induced emf in a conducting loop depends on the time rate of change of the magnetic field through the area enclosed by the conducting loop (Faraday's law). So induced emf has nothing to do with the type of conductor. The current that flows in the conducting loop in response to that induce emf does depend on the type of conductor. In order to maximize the induced current you would want to minimize the factors that impede current flow in the conductor. How would you do that?

AM
MarkoniF
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Dec15-12, 03:59 AM
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Quote Quote by Andrew Mason View Post
I am not sure what you mean by "inductivity".
I was referring to Faraday's law of induction. The question is whether the amount of induced current would be greater in a wire made of iron or copper, as measured in volts and amperes.


The induced emf in a conducting loop depends on the time rate of change of the magnetic field through the area enclosed by the conducting loop (Faraday's law). So induced emf has nothing to do with the type of conductor.
Does that mean induced EMF would be the same for copper and plastic wire? Electromotive force doesn't seem to be what I'm asking about. I'd like to know the difference we would measure in different types of conductors in units of volts and amperes.


The current that flows in the conducting loop in response to that induce emf does depend on the type of conductor. In order to maximize the induced current you would want to minimize the factors that impede current flow in the conductor. How would you do that?
That's the question I'm asking, but it doesn't seem to be simply proportional to material conductivity or resistance. Both conductivity and magnetic properties, such as diamagnetic or paramagnetic index, are related to unpaired electrons or "free electrons". So to rephrase the question, having two wires made of a material with the same conductivity would amount of induced voltage and/or current (volts/amperes) be greater in the material with greater paramagnetic index? The other, perhaps the same question, is whether two materials with different paramagnetic index can have the same conductivity?

Andrew Mason
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Dec15-12, 06:09 AM
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Induction, magnetism and conductivity‏

Quote Quote by MarkoniF View Post
Does that mean induced EMF would be the same for copper and plastic wire? Electromotive force doesn't seem to be what I'm asking about. I'd like to know the difference we would measure in different types of conductors in units of volts and amperes.
Welcome to PF by the way!

Yes. That means the induced emf around the wire loop is determined not by the type of wire but by the rate of change of the magnetic field and the area enclosed by the wire.
That's the question I'm asking, but it doesn't seem to be simply proportional to material conductivity or resistance.
The emf does not depend on the type of wire. The current produced by the wire does. If the wire is plastic, the conductivity is 0 so the induced emf produces no current.

Both conductivity and magnetic properties, such as diamagnetic or paramagnetic index, are related to unpaired electrons or "free electrons". So to rephrase the question, having two wires made of a material with the same conductivity would amount of induced voltage and/or current (volts/amperes) be greater in the material with greater paramagnetic index? The other, perhaps the same question, is whether two materials with different paramagnetic index can have the same conductivity?
I am not sure what you mean by diamagnetic or paramagnetic index.

If you are asking if the induced emf depends on the wire, the answer, as I have said, is no. The current that flows in the wire due to that emf does. The current can be affected by the magnetic properties of the conductor. In order to maximize current flow, the magnetic field inside the conductor must be constant (that is just from Farraday's law). There are quantum effects that actually eliminate the magnetic field inside the conductor in order to achieve superconductivity.

AM
MarkoniF
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Dec15-12, 09:07 AM
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Quote Quote by Andrew Mason View Post
I am not sure what you mean by diamagnetic or paramagnetic index.
It's susceptibility of a material to respond to external magnetic field, or more precisely said I think what I'm talking about is called "permeability". For example, copper has lower paramagnetic index than iron, which means iron would be more attracted to external magnetic field than copper.


If you are asking if the induced emf depends on the wire, the answer, as I have said, is no. The current that flows in the wire due to that emf does. The current can be affected by the magnetic properties of the conductor. In order to maximize current flow, the magnetic field inside the conductor must be constant (that is just from Farraday's law).
Let me put it this way. We have two equally long wires, one made of copper and the other made of iron, but the iron wire is thicker so that both of them have the same conductivity. We attach voltmeter and ammeter to each wire and then we move permanent magnet next to one and then next the other in exactly the same way. You say the reading of volts and amperes would be the same for both wires?


It seems to me there are more factors that would come into play than just conductivity, like the number and strength of magnetic fields per volume of substance that would be available (unpaired) to interact with this external magnetic field. But also not all of those magnetic fields would interact in the same way. Some would be due to electron spin and some would be due to electron orbits, some would repel while others would attract, so depending on how many of them there are, how "free" they are to move, and depending in what direction they prefer to move, on average and relative to that external magnetic field, it seems it would result in greater or less induced electric current. But then what I just described could be the same thing what defines conductivity as well, and if so it would all boil down to be the same after all. In any case the situation is very complex, especially considering the temperature and heating of the substance would be a factor too, so I'm afraid the answer I'm looking for is specific and experimental rather than general and theoretical.
Andrew Mason
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Dec15-12, 01:19 PM
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Quote Quote by MarkoniF View Post
Let me put it this way. We have two equally long wires, one made of copper and the other made of iron, but the iron wire is thicker so that both of them have the same conductivity. We attach voltmeter and ammeter to each wire and then we move permanent magnet next to one and then next the other in exactly the same way. You say the reading of volts and amperes would be the same for both wires?
In your example there would appear to be no current because there is no circuit. Let's use two same sized loops of each kind of wire. A magnet passes through the loops.


Are you suggesting that since the iron wire has a higher permeability, the magnitude of the magnetic field changes inside the iron wire will be somewhat greater than in the copper wire?

Just applying Faraday's law, one can see that this would tend to increase the induced emf in the iron wire, and, therefore the current. But the increase would be in proportion to the area of the wire loop itself (the diameter x length of the wire) compared to the area enclosed by the wire loop. If you are making the diameter of the loop much larger than the diameter of the wire, it should not be significant.

AM
cabraham
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Dec15-12, 07:30 PM
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Quote Quote by Andrew Mason View Post
I am not sure what you mean by "inductivity".

The induced emf in a conducting loop depends on the time rate of change of the magnetic field through the area enclosed by the conducting loop (Faraday's law). So induced emf has nothing to do with the type of conductor. The current that flows in the conducting loop in response to that induce emf does depend on the type of conductor. In order to maximize the induced current you would want to minimize the factors that impede current flow in the conductor. How would you do that?

AM
But if the loop resistance is very small its own magnetic field affects the loop emf. The flux in the loop is a combination of the external field plus the internal mag field inevitable when current exists. If the loop R value is small in comparison with the self inductance reactance of said loop, then the emf around the loop decreases. But if R >> XL, then R has negligible influence on the emf around the loop.

Claude
MarkoniF
#8
Dec15-12, 08:32 PM
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Quote Quote by Andrew Mason View Post
In your example there would appear to be no current because there is no circuit. Let's use two same sized loops of each kind of wire. A magnet passes through the loops.
The circuit would be closed by connecting voltmeter at the ends, but loop is fine too.


Are you suggesting that since the iron wire has a higher permeability, the magnitude of the magnetic field changes inside the iron wire will be somewhat greater than in the copper wire?
I think yes, if by that you mean what I mean.


Just applying Faraday's law, one can see that this would tend to increase the induced emf in the iron wire, and, therefore the current. But the increase would be in proportion to the area of the wire loop itself (the diameter x length of the wire) compared to the area enclosed by the wire loop. If you are making the diameter of the loop much larger than the diameter of the wire, it should not be significant.
I'm not sure what you just said. Let's make it a bit more clear and take both iron and copper wires to have the same length and the same thickness, but if permeability plays any role, could that then compensate for the lesser conductivity of the iron wire so that we get about the same amount of induced current in both wires?
Andrew Mason
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Dec15-12, 10:54 PM
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Quote Quote by MarkoniF View Post
The circuit would be closed by connecting voltmeter at the ends, but loop is fine too.
You cannot measure induced emf in a conductor with a voltmeter. You would be measuring the emf generated around the entire loop made by the conductor and the leads of the voltmeter, not the voltage across the ends of the conductor.

Faraday's law has some very subtle aspects that cause all sorts of confusion. Professor Lewin at M.I.T. has a very good lecture on this that can be found here.
. It can be found on Youtube here (it is easier to navigate in the Youtube version).
I'm not sure what you just said. Let's make it a bit more clear and take both iron and copper wires to have the same length and the same thickness, but if permeability plays any role, could that then compensate for the lesser conductivity of the iron wire so that we get about the same amount of induced current in both wires?
The magnetic fields inside the iron wire will have greater variation than in the copper. This will affect the emf in the wire and, hence, current. If the wire diameter/thickness is much smaller than the diameter of the loop, the effect will be rather small. To quantify the effect you would have to give all the dimensions and conductivities etc. and painstakingly apply Faraday's law.

AM
MarkoniF
#10
Dec16-12, 05:20 AM
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Quote Quote by Andrew Mason View Post
You cannot measure induced emf in a conductor with a voltmeter. You would be measuring the emf generated around the entire loop made by the conductor and the leads of the voltmeter, not the voltage across the ends of the conductor.
It wouldn't matter to measure the difference, but if you don't want leads of the voltmeter to play any part then you simply extend the ends of the wire further away from the loop(s), or make the leads parallel to the direction of magnet motion.


The magnetic fields inside the iron wire will have greater variation than in the copper.
What equation are you talking about?


This will affect the emf in the wire and, hence, current.
So having two equally thick wires with the same conductivity, would greater permeability of one of the wires lead to greater induced current in that wire?


To quantify the effect you would have to give all the dimensions and conductivities etc. and painstakingly apply Faraday's law.
Where do you see connection between Faraday's law and permeability?
Andrew Mason
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Dec16-12, 01:50 PM
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Quote Quote by MarkoniF View Post
It wouldn't matter to measure the difference, but if you don't want leads of the voltmeter to play any part then you simply extend the ends of the wire further away from the loop(s), or make the leads parallel to the direction of magnet motion.
It does matter! That point was very well demonstrated by Prof. Lewin. The voltage measured by the voltmeter depends very much on how the leads are configured. The measured voltage depends on how much flux the circuit - which includes the voltmeter leads - encloses! If you extend the leads farther away from the loops the induced voltage will actually increase because the flux enclosed by the circuit increases. If you make the leads parallel to the direction of magnetic motion so that the circuit encloses no flux, the measured voltage will be 0! No one said Faraday's law was simple or intuitive. It is very counter-intuitive.

What equation are you talking about?
I wasn't talking about an equation. I was talking about the way iron responds to an applied magnetic field. Since the iron atoms are strong magnetic dipoles, they will align with the applied field so that the magnetic field inside the iron conductor will be greater than the applied magnetic field. Consequently, any changes in the applied magnetic field will result in greater changes of the magnetic field inside the iron conductor.


So having two equally thick wires with the same conductivity, would greater permeability of one of the wires lead to greater induced current in that wire?
I seems to me that it would result in a slightly greater magnetic flux enclosed by the iron wire. This in itself would result in a greater induced voltage (slightly) but there are other effects of a magnetic field inside a conductor that tends to decrease conductivity, so I think you will have to do some experiments to determine whether the induced current is greater.

Where do you see connection between Faraday's law and permeability?
Faraday's law says that the induced voltage around a closed path is equal to the time rate of change of the flux enclosed by the path. The strength of the magnetic field depends on the permeability of the space. That is why you have an iron core in a solenoid or transformer - to create a strong magnetic field.

AM
MarkoniF
#12
Dec16-12, 07:39 PM
P: 56
Quote Quote by Andrew Mason View Post
It does matter! That point was very well demonstrated by Prof. Lewin. The voltage measured by the voltmeter depends very much on how the leads are configured.
It does not matter for the difference of induced current between copper and iron wire. If anything about leads changed the measurement, then that impact would be the same for both wires and so it would not matter.

Whether we measure voltage or amperes we would take the same measurement with both iron and copper wire, and the point in that lecture is about two DIFFERENT measurements. It's about two different measurements of the potential difference related to the direction of the current and two different resistors, which is indeed peculiar, so perhaps it's best to just measure amperes instead of voltage, although since we have no uneven distribution of resistance, like they did, we would not need to worry about anything like that.

The measured voltage depends on how much flux the circuit - which includes the voltmeter leads - encloses! If you extend the leads farther away from the loops the induced voltage will actually increase because the flux enclosed by the circuit increases. If you make the leads parallel to the direction of magnetic motion so that the circuit encloses no flux, the measured voltage will be 0!
No, voltmeter leads are not part of the loop, so at best they could contribute a little bit to the induced current, and if they are parallel to the direction of magnet motion they would be completely irrelevant. Also, nothing about voltmeter leads could ever make the voltage be zero if there is some induced current present in the loop, unless you disconnect them. Nothing like that was even addressed in that lecture.


I wasn't talking about an equation. I was talking about the way iron responds to an applied magnetic field. Since the iron atoms are strong magnetic dipoles, they will align with the applied field so that the magnetic field inside the iron conductor will be greater than the applied magnetic field. Consequently, any changes in the applied magnetic field will result in greater changes of the magnetic field inside the iron conductor.
Perhaps, however we are not interested in the changes of the magnetic fields in the conductor, we are interested only in the amount of induced current, and those two could be related, that's what I think too, but assumptions, either yours or mine, are not the answer I'm happy with.


I seems to me that it would result in a slightly greater magnetic flux enclosed by the iron wire. This in itself would result in a greater induced voltage (slightly) but there are other effects of a magnetic field inside a conductor that tends to decrease conductivity, so I think you will have to do some experiments to determine whether the induced current is greater.
That's what I think, and I could do experiment myself if I only had iron wire and more sensitive instruments.


Faraday's law says that the induced voltage around a closed path is equal to the time rate of change of the flux enclosed by the path. The strength of the magnetic field depends on the permeability of the space. That is why you have an iron core in a solenoid or transformer - to create a strong magnetic field.
That's not really what we are talking about, but I do agree it could be related. After all I have quite similar opinion, if not the same, which is why I ask the question in the first place.
Andrew Mason
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Dec16-12, 10:24 PM
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Quote Quote by MarkoniF View Post
It does not matter for the difference of induced current between copper and iron wire. If anything about leads changed the measurement, then that impact would be the same for both wires and so it would not matter.
But the point is that you are not measuring the voltage between the two points that the voltmeter leads are connected to - ie across the ends of your two wires. You are measuring the integral of [itex]E \cdot dl[/itex] around the loop that the voltmeter and leads form with the conductor.

This is a side issue to the issue you raise about the effect of the permeability of the conductor on the induced current, but it is important to appreciate it. Measuring voltages where there is a time dependent magnetic field present is very different than measuring voltages in a circuit containing a battery and resistance.

No, voltmeter leads are not part of the loop, so at best they could contribute a little bit to the induced current, and if they are parallel to the direction of magnet motion they would be completely irrelevant. Also, nothing about voltmeter leads could ever make the voltage be zero if there is some induced current present in the loop, unless you disconnect them. Nothing like that was even addressed in that lecture.

Perhaps, however we are not interested in the changes of the magnetic fields in the conductor, we are interested only in the amount of induced current, and those two could be related, that's what I think too, but assumptions, either yours or mine, are not the answer I'm happy with.
If the material did not have any effect on the magnetic field inside the wire then there would be no difference in the voltage that is induced in any same sized loop, whether it is made of copper, iron, aluminum or plastic. The currents will differ depending on the resistance/conductivity of the material, but that is all. As you study Faraday's law it will be easier to see why this is.

But iron does have an effect on the magnetic field, which is the basis of your original question. To determine how it would affect current is very complicated and involves quantum effects as well as Faraday's law.

Physics is not intended to make you happy.

That's not really what we are talking about, but I do agree it could be related. After all I have quite similar opinion, if not the same, which is why I ask the question in the first place.
?? It is exactly what we are talking about!! The magnitude of changes in a magnetic field, as well as the rate of change, determines the magnitude of the induced voltage in a closed path in that changing magnetic field. Anything that affects the magnitude of the magnetic field (such as the permeability of the region of space enclosed by and included in that path) will affect the induced voltage. That is just a natural consequence of Faraday's law.

AM
MarkoniF
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Dec17-12, 03:24 AM
P: 56
Quote Quote by Andrew Mason View Post
But the point is that you are not measuring the voltage between the two points that the voltmeter leads are connected to - ie across the ends of your two wires. You are measuring the integral of [itex]E \cdot dl[/itex] around the loop that the voltmeter and leads form with the conductor.
Leads don't form anything and have nothing to do with induction loop(s) where induction takes place.




This is a side issue to the issue you raise about the effect of the permeability of the conductor on the induced current, but it is important to appreciate it. Measuring voltages where there is a time dependent magnetic field present is very different than measuring voltages in a circuit containing a battery and resistance.
You are reading too much into that lecture making general conclusions from specific examples, thus applying them where they are not relevant at all. The difference pointed in that lecture is only due to having two different resistors and because voltage was measured at two points between them. We don't have any resistors and we do not measure anything directly on the loop. Our setup is like on that picture above, so given the same change in magnetic field we would measure the same amount of voltage regardless of how we connect the leads, only sign could be different.


If the material did not have any effect on the magnetic field inside the wire then there would be no difference in the voltage that is induced in any same sized loop, whether it is made of copper, iron, aluminum or plastic. The currents will differ depending on the resistance/conductivity of the material, but that is all. As you study Faraday's law it will be easier to see why this is.

But iron does have an effect on the magnetic field, which is the basis of your original question. To determine how it would affect current is very complicated and involves quantum effects as well as Faraday's law.
You are mixing inductor coil with induction loop. First one is to be substitute for moving magnet, you don't induce electric current in inductor coil but supply it and induce magnetic field. We are talking about the second one, induction loop, where the current is not supplied but induced with moving permanent magnet or inductor coil, and the fact that it then creates magnetic field as well is irrelevant for the question which is only about electric current induced as compared between induction loop made of copper and iron. This is not addressed by Faraday's law, and if it is addressed anywhere at all it would be classical electrodynamics, not quantum mechanics.


Physics is not intended to make you happy.
Answers, not physics. Physics always makes me happy, but answers can make me sad.


?? It is exactly what we are talking about!! The magnitude of changes in a magnetic field, as well as the rate of change, determines the magnitude of the induced voltage in a closed path in that changing magnetic field. Anything that affects the magnitude of the magnetic field (such as the permeability of the region of space enclosed by and included in that path) will affect the induced voltage. That is just a natural consequence of Faraday's law.
You are talking about inductor coil while the question is about induction loop. Faraday's law has nothing to do with magnetic permeability of induction loop, but only with external magnetic field or inductor coil which then induces current in induction loop. Forget inductor coil and iron cores, we don't need any of that as we have moving permanent magnet.
Andrew Mason
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Dec17-12, 07:38 AM
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Quote Quote by MarkoniF View Post
Leads don't form anything and have nothing to do with induction loop(s) where induction takes place.

My comments about a voltmeter were with respect to your original suggestion that you measure the induced voltage in a section of a conductor. You had posted:

"The circuit would be closed by connecting voltmeter at the ends, but loop is fine too."

I simply observed that measuring the induced voltage in a section of conductor like that will not give you induced voltage between the two ends of the conductor. It measures the emf around the whole loop comprised of the conductor and the voltmeter leads. That is why I suggested you use loops.

In the above example, the magnetic field enclosed by the voltmeter leads is not significant so a voltmeter will measure the induced emf across the ends of the coil, which is a function of the rate of change of the flux through the coil loops.

If you were replace the coil with a single straight conducting wire and pass a magnet in a direction perpendicular to the direction of the wire (replace the coil with a straight wire and have the magnet moving perpendicular to the page) as you were suggesting, a galvanometer connected as shown in your diagram would not measure the induced voltage in the conductor. That was my point.

You are mixing inductor coil with induction loop. First one is to be substitute for moving magnet, you don't induce electric current in inductor coil but supply it and induce magnetic field. We are talking about the second one, induction loop, where the current is not supplied but induced with moving permanent magnet or inductor coil, and the fact that it then creates magnetic field as well is irrelevant for the question which is only about electric current induced as compared between induction loop made of copper and iron. This is not addressed by Faraday's law, and if it is addressed anywhere at all it would be classical electrodynamics, not quantum mechanics.

Answers, not physics. Physics always makes me happy, but answers can make me sad.

You are talking about inductor coil while the question is about induction loop. Faraday's law has nothing to do with magnetic permeability of induction loop, but only with external magnetic field or inductor coil which then induces current in induction loop. Forget inductor coil and iron cores, we don't need any of that as we have moving permanent magnet.
Faraday's law is a fundamental part of classical electromagnetic theory. To progress any further you will have to study Faraday's law and induction. Your comments show that you are not inclined to do that. You seem to want answers that fit with how you already view things, so I am afraid I cannot help you.

AM
MarkoniF
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Dec17-12, 10:05 AM
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Quote Quote by Andrew Mason View Post
My comments about a voltmeter were with respect to your original suggestion that you measure the induced voltage in a section of a conductor. You had posted:

"The circuit would be closed by connecting voltmeter at the ends, but loop is fine too."

I simply observed that measuring the induced voltage in a section of conductor like that will not give you induced voltage between the two ends of the conductor. It measures the emf around the whole loop comprised of the conductor and the voltmeter leads. That is why I suggested you use loops.
I does not matter, just the same. Take that same image, it could be no loops up there, just a wire bent in 'U' shape, and leads of the whatever meter could be little horizontal wires at the bottom, far away from the magnet.


If you were replace the coil with a single straight conducting wire and pass a magnet in a direction perpendicular to the direction of the wire (replace the coil with a straight wire and have the magnet moving perpendicular to the page) as you were suggesting, a galvanometer connected as shown in your diagram would not measure the induced voltage in the conductor. That was my point.
I didn't say wire is straight, or that it has to be. But even if it was straight, there would again be induced current and galvanometer would measure. Try it.


Faraday's law is a fundamental part of classical electromagnetic theory. To progress any further you will have to study Faraday's law and induction. Your comments show that you are not inclined to do that. You seem to want answers that fit with how you already view things, so I am afraid I cannot help you.
As I said, you were mixing two different types of induction loops. I's exactly how I told you, check it out.
Andrew Mason
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Dec17-12, 10:53 AM
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Quote Quote by MarkoniF View Post
I does not matter, just the same. Take that same image, it could be no loops up there, just a wire bent in 'U' shape, and leads of the whatever meter could be little horizontal wires at the bottom, far away from the magnet.

I didn't say wire is straight, or that it has to be. But even if it was straight, there would again be induced current and galvanometer would measure. Try it.
There would be no induced current in a straight wire. You need a circuit. That is why I suggested a loop.

If you think it does not matter then apply Faraday's law and tell us what the induced voltage is in a straight conductor of length L as a function of dB/dt.

As I said, you were mixing two different types of induction loops. I's exactly how I told you, check it out.
You appear to be looking for someone to confirm your understanding of induction. In such circumstances, the only help anyone can give you is to suggest that you thoroughly study Faraday's law and then see if you still have the same questions.

AM
MarkoniF
#18
Dec17-12, 03:56 PM
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Quote Quote by Andrew Mason View Post
There would be no induced current in a straight wire. You need a circuit. That is why I suggested a loop.
You have no idea and yet you make statements as if you are certain about it, that's strange. There would be induced current in a straight wire. Try it out.


If you think it does not matter then apply Faraday's law and tell us what the induced voltage is in a straight conductor of length L as a function of dB/dt.
http://www.ndt-ed.org/EducationResou...inductance.htm
- Faraday's Law for an uncoiled conductor states that the amount of induced voltage is proportional to the rate of change of flux lines cutting the conductor. Faraday's Law for a straight wire is shown below.




http://www.sweethaven.com/sweethaven....asp?iNum=0201
- Faraday's Law for a Straight Wire: The amount of induced voltage is proportional to the rate of change of flux lines cutting the conductor.



You appear to be looking for someone to confirm your understanding of induction. In such circumstances, the only help anyone can give you is to suggest that you thoroughly study Faraday's law and then see if you still have the same questions.
Help yourself and learn some basics. My question stands, and as I explained it's not addressed by Faraday's law.


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