Understanding Induced Current and EMF in Electromagnetic Induction

[sophiecentaur]

so it would be more correct to say that no breaking effect occurs inside the coil
but as it enters / leaves we will get current and hence breaking effect?
(inside the tube we DO get breaking effect due to eddy currents though, right)

Only if there is a connection between the two ends of the coil. A scope won't make it happen. What path could the current take?

 

[cabraham]

Only if there is a connection between the two ends of the coil. A scope won't make it happen. What path could the current take?

Displacement current due to end to end capacitance of the coil. Once again, I hate to keep saying this, because it should be very well known. In the ac domain, we MUST DISCARD this archaic notion that in order for current to exist that a fully closed path is required. A single conductor open at both ends can and does carry a current.

I apologize for being repetitive, but this should be so well known that it needn't be necessary to remind everyone. Of course, this displacement current, depending on conditions like speed, frequency, etc. may be too small to be concerned about. In the case of the car, that is likely the case.

Also, you earlier mentioned that the scope connected in mid air can measure voltage using short dipoles. Please refer to any reference text and it will affirm that the routing of the leads will affect the reading displayed. When measuring voltage (or emf) due to time varying magnetic fields, the geometry of the leads determines the enclosed area and path of integration. This affects emf value.

Also, the integral E*dl is used to evaluate emf, but it is derived as follows. The work done transporting charge along a given path is the integral F*dl. Voltage is defined as work per charge. So emf is the integral of (F*dl)/q. But F/q is simply E, so we write integral E*dl. But there is a restriction.

If q = 0, then F/q is undefined. What is the work transporting no charge around a given path? Answer is zero. What is the charge? Answer is zero. What is the voltage/emf? Answer is work divided by charge which is 0/0! In math 0/0 is known as indeterminate. It is undefined. Although integral E*dl is defined, work/charge is 0/0. Integral E*dl is derived from integral (F*dl)/q, which is the most fundamental since work/charge is the very definition of voltage.

Anyway, I had to clarify that in space with time changing fields, I think emf has meaning only in the presence of charges,and path is defined by the conductor. To remove both and discuss a voltage value in empty space results in amiguous and indeterminate voltage values. Anyway, that's how voltage is defined. Like I said, under static conditions we can indeed define a voltage value in free space w/o charges and/or conductors if the fields are conservative. In this case they voltage is independent of path of integration.

The conditions and nature of the fields need to be considered. When quoting an equation, such as integral E*dl, we must know how it was derived, and the conditions where it applies. Math and science is typically applied this way. Comments welcome.

Claude

Claude

 

[BruceW]

the net emf (across the entire coil) is zero when the magnet is in the coil so the current in the coil is also zero...but the current at either end of the magnet (above and below it) is creating an instantaneous magnetic field opposing the motion of the magnet as it falls (due to the currents that are present - i.e: before they cancel). This is for when the oscilliscope is attached - it completes the circuit

When the magnet is in the middle of the coil, you are right that there is zero emf across the entire coil. But there is no current above or below the magnet (even if you 'short the ends together'). Remember that the current must be in the same direction over the entire coil, otherwise you'd get charge separation. (I am assuming here that the coil is too thin to allow significant eddy currents, which is the reason for using a coil in this experiment in the first place).

When the magnet is above the middle of the coil, the current will flow anticlockwise. This movement of charge will decrease the rate of change of flux below the magnet, but will increase the rate of change of flux above the magnet. But because the magnet is nearer the top half of the coil, the effect of the current flowing above the magnet is less than the effect of the current flowing below the magnet, so overall the rate of change of flux is decreased.

From this explanation, you can see that the braking effect will be zero when the magnet is in the middle of the coil, since there is as much coil above as there is below, so if there was a current, it would have no effect on the magnet's motion. So when the magnet goes through the middle of the coil, there is zero current. This is handy, since we know that once the magnet is below the middle of the coil, the current must be in the opposite direction to before. So it means that the current changes continuously from one direction to the other as the magnet passes through the coil.

When a magnet goes through a tube, the situation is different because eddy currents can exist, so the current can be different in different places in the tube. So the braking effect can exist even when the magnet is in the middle of the tube.

I hope this explanation has been of some help. To be honest, the most important thing is to get yourself ready for the kind of questions they will ask you (by reading past papers and practice questions). I guess you've done that and you're now looking to understand it more. And that is good, if they ask you an unexpected question, you will have an edge over other students.

 

[BruceW]

Also, the integral E*dl is used to evaluate emf, but it is derived as follows. The work done transporting charge along a given path is the integral F*dl. Voltage is defined as work per charge. So emf is the integral of (F*dl)/q. But F/q is simply E, so we write integral E*dl. But there is a restriction.

If q = 0, then F/q is undefined. What is the work transporting no charge around a given path? Answer is zero. What is the charge? Answer is zero. What is the voltage/emf? Answer is work divided by charge which is 0/0! In math 0/0 is known as indeterminate. It is undefined. Although integral E*dl is defined, work/charge is 0/0. Integral E*dl is derived from integral (F*dl)/q, which is the most fundamental since work/charge is the very definition of voltage.

The integral E*dL is more fundamental than work per charge. It can be derived from Maxwell's equations, which do not require charges (the equations still work in free space).
\nabla \wedge \vec{E} = - \frac{\partial \vec{B}}{\partial t}
\int ( \nabla \wedge \vec{E} ) \cdot d \vec{A} = - \frac{\partial \int \vec{B} \cdot d \vec{A}}{\partial t}
\oint \vec{E} \cdot d \vec{L} = - \frac{\partial \Phi}{\partial t}
So if you define emf as E*dL, the definition does not require charges. But if you define emf as (F*dL)/q then the definition does require some kind of charge to exist. It just depends on which definition you prefer.

EDIT: woops, I forgot it should be partial differential with respect to time, keeping spatial coordinates constant. I've changed it to partial derivative symbols now.

 

[cabraham]

The integral E*dL is more fundamental than work per charge. It can be derived from Maxwell's equations, which do not require charges (the equations still work in free space).
\nabla \wedge \vec{E} = - \frac{d \vec{B}}{dt}
\int ( \nabla \wedge \vec{E} ) \cdot d \vec{A} = - \frac{d \int \vec{B} \cdot d \vec{A}}{dt}
\oint \vec{E} \cdot d \vec{L} = - \frac{d \Phi}{dt}
So if you define emf as E*dL, the definition does not require charges. But if you define emf as (F*dL)/q then the definition does require some kind of charge to exist. It just depends on which definition you prefer.

F/q is more basic than E. Force, F, can be observed between two charged bodies. But for this force to be felt, proagation time was observed. The field concept was created. By definition E field is a mathematical construct defined as F/q. Voltage was defined as work per unit charge even before fields were born. What is the voltage in an ac field if there is no charges nor conductor?

Per the definition of what voltage is, force per charge, it 0/0, which is indeterminate. But as soon as you introduce a conductor which posesses mobile charge carriers, all makes sense. You can then compute the work done per unit charge along a path which is defined by the size & shape of the conductor.

Since E was born out of the condition "F/q", we must apply the rules of math and stipulate the q can never equal 0. Otherwise we have no sense at all. That is my point. What is the work per charge along this path? No work is done moving no charge. Hence 0/0 is the only answer one can obtain. What is "0/0"? I don't know. I doubt anyone can know. I'm not trying to belabor this point, please understand that I just don't like things like "0/0". It doesn't sit right with me.

If you feel ok with defining a voltage in mid air which computes to "0/0", well who am I to argue? I just hope we can all agree that expressions like 1.3/24.6, -3.73/1.27, 0/1.52, etc., make sense, whereas "0/0" just doesn't do it for me. Am I being unreasonable here? Comments welcome.

Claude

 

[BruceW]

F/q is more basic than E. Force, F, can be observed between two charged bodies. But for this force to be felt, proagation time was observed. The field concept was created. By definition E field is a mathematical construct defined as F/q. Voltage was defined as work per unit charge even before fields were born. What is the voltage in an ac field if there is no charges nor conductor?

The concept of a field is still useful in free space, without charges or currents. For example, if we were interested to know about how a radiation-dominated universe would evolve. I think our own universe was radiation-dominant for a while, if I remember right.

 

[cabraham]

The concept of a field is still useful in free space, without charges or currents. For example, if we were interested to know about how a radiation-dominated universe would evolve. I think our own universe was radiation-dominant for a while, if I remember right.

Of course, a field is useful in free space. Fields were conceived to deal with the issue of propagation delay. Before the Coulomb force between 2 charges can take place, a disturbance, wave, field, whatever, had to propagate through space in the interim.

In free space there are definitely fields involved. But when we try to define a "work done per unit charge", i.e. "voltage", we run smack into that darn "0/0" indeterminate monster. But if we simply define a conductive path and introduce a very small quantity of charge, all is well.

When I think about it, what value does the voltage concept have if there are no conductors or charges? The whole concept of voltage is only useful when we observe charges moving under the influence of a field. Without charges and conductors, the fields are still there, but they have nothing to move. So the "work done divided by the charge" is literally 0/0!

But we seem to have resolved that E & B are always present even w/o conductors and/or charges. I don't think anybody would dispute that. I certainly don't, BR.

Claude

 

[sophiecentaur]

@cabraham
You presumably don't subscribe, either, to the concept of a field then - which is a force on a mass / charge / current. The same "0/0" applies if the mass / charge / current isn't there? Is the field really not there when there's nothing to experience it? That's all a bit Zen for me, I'm afraid.

Did you do any calculation about who much this displacement current would be? I wonder if it is really relevant in the world of an A Level student really significant in comparison with the current that would flow if the two ends of a coil were joined together. I think you are allowing your preference for the way in which you choose to look at things to take over from a pragmatic approach that is used in most occasions where induction takes place.
[edit] Having just read your last post, I take it you don't subscribe, either, to the concept of Gravitational Potential?

 

[cabraham]

Just the opposite, I ascribe to fields in vacuum, and to a limited extent potentials in vacuum. Please reread. It's a question of how to compute/define. I can say that my opponents choose to view things in their preferred perspective. I'm just saying that voltage & current are equally important, neither one more or less than the other. In ac, neither is independednt nor causal or effectual. In dc, either one can exist alone. That is my point.

Claude

 

[sophiecentaur]

So, after all this, we are down to a matter of preference. We have also risked really confusing a poor A level student over whose head all this has sailed (I hope).
I think a separate thread would have been appropriate for this discussion of angels on pinheads.

 

[cabraham]

Also, FWIW, I do ascribe to gravitational potential. Remember that gravity is a conservative field. As such, it can be equated to the negative of the gradient of a scalar potential function. The potential at any point in space due to gravity is well-defined, computable as a unique single value, independent of path.

Likewise with an E field due entirely to charged particles the same scenario exists. Take a parallel plate capacitor with charge on the plates. The potential anywhere between the plates is well defined and path INdependent. THe field is IRrotational just like gravity. Also, E = -grad V, like gravity.

Now when dealing with rotational E fields, like that with induction, E is NOT the gradient of a potention scalar function. Across any 2 points in space, there is no single valued potentiol. Potential, or emf, have meaning only along a specific path of charge motion, i.e. charges moving through a conductor under the influence of fields per the Lorentz force equation. Here, E = -dA/dt, where A is magnetic vector potential.

Rotational fields, i.e. non-zero value of curl, do not behave like ir-rotational fields which have zero curl. Gravity is zero-curl ir-rotational as are E fields due to charges. Induction E fields are the opposite.

That is my point. Maybe I'm belaboring something not quite within the OP intentions. But several posters stated that an emf exists in free space. I felt compelled to point out that it is a very ambiguous ill-defined non-unique value of emf. In order to have any meaning to the emf concept, a conductor mast be placed in the region and the shape of said conductor determines the emf. Rearranging said conductor shape changes the emf. That is all I wish to emphasize. Sorry if I got carried away. BR.

Claude

 

[sophiecentaur]

Something we haven't actually stated explicitly in this thread is that the 'EMF' we are discussing is not really a single EMF but a set of EMFs in series, along the coil or around the tube. The line integral is what counts and all the EMFs around the tube go to make up the effective emf which is applied across the resistance of the metal area. So I agree with what you say about the path being important for the final answer. Kirchoff II doesn't translate directly to circuits where 'induction effects' can't be localised to a point in the circuit.

 

[sophiecentaur]

'A scientist' says that there may be much more water in the Earth's mantel than in all the Oceans.

 

[BruceW]

That is my point. Maybe I'm belaboring something not quite within the OP intentions. But several posters stated that an emf exists in free space. I felt compelled to point out that it is a very ambiguous ill-defined non-unique value of emf. In order to have any meaning to the emf concept, a conductor mast be placed in the region and the shape of said conductor determines the emf. Rearranging said conductor shape changes the emf. That is all I wish to emphasize. Sorry if I got carried away. BR.

That's alright. jsmith said he/she? would be away until Sunday anyway. You're right, the closed integral of E*dL does depend on the path. But this doesn't mean it is ill-defined, it just means we must specify the path. If we define the integral of E*dL as the emf, then the emf depends on the path specified (in other words, asking about the emf between two points is useless, unless we also state what path it is taken over). There is no problem with this. It is the more correct way of thinking. When we start talking about a wire, then the path is implied to be through the wire, but if we were speaking more accurately, we would also need to state explicitly what the path is.

I think in an earlier post you said something like 'what use is emf when there are no charges or currents?' And this is a good question. I can't think of any use in classical EM. And I doubt there is any use in EM without charges or currents in general relativity (although I'm not sure, because I don't know a lot about general relativity). This is probably why a lot of people define emf as existing only in a circuit - because there is not much use for the concept when there is no circuits. I would not impose this extra restriction, because I don't like to complicate a definition any more than is necessary, but it just depends on personal preference.

 

[BruceW]

About this idea of a displacement current which causes a current to flow despite a very large resistance of the oscilloscope: This would mean we are saying the scope effectively acts as a capacitor, right? At first guess, I would have thought that the scope is designed so that it does not act as a capacitor. Also, there is the issue that the induced electric field might not change quickly enough to allow a significant displacement current. Both these issues are difficult to calculate.

I Looked it up on wikipedia, found: "General-purpose oscilloscopes usually present an input impedance of 1 megohm in parallel with a small but known capacitance such as 20 picofarads." So our equation for the current is:
I = N C \frac{\partial^2 \Phi}{\partial t^2}
(where N is the number of turns of the coil, and C is 20 picofarads). So our answer depends on the twice differential (with respect to time) of the magnetic flux. This is a bit more difficult to estimate. We can say we are using a magnetic dipole, which is falling under gravity, and the magnetic flux will be the integral of the magnetic field over the area inside the coil (which will vary at different points in the area, and depend on the distance of the area from the dipole). Difficult :(

Anyway, I should go to bed. I should have been asleep already!

 

[jsmith613]

When the magnet is in the middle of the coil, you are right that there is zero emf across the entire coil.

When the magnet is above the middle of the coil, the current will flow anticlockwise. This movement of charge will decrease the rate of change of flux below the magnet, but will increase the rate of change of flux above the magnet. But because the magnet is nearer the top half of the coil, the effect of the current flowing above the magnet is less than the effect of the current flowing below the magnet, so overall the rate of change of flux is decreased.

From this explanation, you can see that the braking effect will be zero when the magnet is in the middle of the coil, since there is as much coil above as there is below, so if there was a current, it would have no effect on the magnet's motion.

back a little early!

(from the oscilliscope we can see the emf is zero for the entire time the magnet is in the coil)

so let's say we connected an ammeter instead of an oscilliscope as a current would definately flow then. you seem to be saying we would only get no breaking effect when the magnet is EXACTLY in the centre of the long coil.
using your logic (when the magnet is above the centre there is a greater change in flux beneath the magnet) there should be an emf.
I think a better way to think of it is the amount of flux linking with the coil. When the entire magnet is inside the coil there is no change of flux linking with the coil so no emf is induced.

Having never performed the experiment with a coil I actually don't know the results but I would presumed, base on my way of thinking we would only get magnetic breaking at the ends of the magnet (i.e: as it enters and as it leaves).

I may be getting terribly confused in which case I apologise.

 

[BruceW]

I may be getting terribly confused in which case I apologise.

No, you're not confused, you're pretty much right. But the emf would be slightly non-zero when the magnet was in the coil. If you think about it, the graph of emf is smooth, this is because there is still a slight emf even when the magnet has entered the coil.

The other way to think about it, is that when the magnet is inside the coil, there is less flux linkage than if the magnet was right in the middle of the coil, so the flux linkage is still changing even when the magnet is inside.

But I think you are right, that the emf when the magnet is inside the coil is much less than when the magnet is entering/leaving the coil. (Especially if the coil was quite long, the emf would be very small when the magnet was inside).

So for an exam, I would just say there is an emf when the magnet is entering/leaving. (Since the emf when the magnet is inside will be an order of magnitude smaller). I hope I haven't made things unnecessarily confusing.

 

[jsmith613]

No, you're not confused, you're pretty much right. But the emf would be slightly non-zero when the magnet was in the coil. If you think about it, the graph of emf is smooth, this is because there is still a slight emf even when the magnet has entered the coil.

The emboldened statement does not make sense but the rest of it does

so there will be no magnetic breaking effect (for A-level purposes), right?
thanks :)

 

[sophiecentaur]

so there will be no magnetic breaking effect (for A-level purposes), right?
thanks :)

That's definitely the right A level answer for the coil.

 

[jsmith613]

That's definitely the right A level answer for the coil.

cheers
once again I must express my gratitude to EVERYONE who has helped me in this discussion
I do hope that the tangent discussion interested other people as well...you can now discuss to your hearts content :)

 

[sophiecentaur]

No problem.
The one big risk on this forum is over-answering questions. I'm sure it has given many A level students real headaches. "I only wanted to know . . . . . ."
Even when the questioner states his or her level of interest, that doesn't necessarily deter the enthusiastic contributor.
Good luck with the A level stuff.

 

[BruceW]

The emboldened statement does not make sense but the rest of it does

so there will be no magnetic breaking effect (for A-level purposes), right?
thanks :)

For A-level purpose, yes, there is only a braking effect when the magnet enters and leaves the coil. I think you've got a pretty good grasp of this stuff. Good luck anyway, even though I don't think you'll need it :)