Irregular case of induced EMF


by bentzy
Tags: case, induced, irregular
bentzy
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#1
May5-13, 04:06 AM
P: 20
Dear colleagues,

My question here refers to the EMF induced in an incomplete ring that is stationary amidst a time dependent magnetic field.

We know from Faraday's law that EMF is created upon changing the magnetic flux, either by changing the intensity of the field, or by changing the effective area enclosed by the conductor.

In the present case, the stationary ring isn't complete, namely part of it has been removed before placing it in the field. In other words, the conducting ring isn't closed.

What machinery or explanation acounts for the creation of an induced EMF in this case ? (there is no induced current in this case, of course).

Thank you, BC
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mfb
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#2
May5-13, 04:27 AM
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The induction laws still apply, as they do not rely on the presence of a conductor. You can "close" the loop with an imaginary "connection" - as this connection is not conducting, (nearly) the whole induced voltage drop will be there.
bentzy
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#3
May5-13, 04:42 AM
P: 20
Yes, that's already known. My question is what physical machinery brings about an EMF,
after all, the ring is stationary and isn't moving ?

2nd: you can close the gap by a virtual connection, but there's no one unique way to do so,
which implies different possible resulting effective areas.

Thanks, BC

mfb
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#4
May5-13, 05:28 AM
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Irregular case of induced EMF


##\displaystyle \frac{\partial B}{\partial t}=-curl(E)## (Maxwell equation)
The time-dependent magnetic field produces a circular electric field.

2nd: you can close the gap by a virtual connection, but there's no one unique way to do so,
which implies different possible resulting effective areas.
Sure, but the potential difference in the gap will be the same for all methods.
bentzy
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#5
May5-13, 06:03 AM
P: 20
a. The problem is how to introduce this within the framework of high-school teaching, since
the matter in question was given in a matriculation exam in physics.

I avoided Maxwell's equations, just because they're out of high-school curriculum.
I should have pointed this out in the 1st place.

b. My previous 2nd remark was given on the basis of the interplay between area and field, in terms of the change in the flux. Of course the EMF itself is path-independent, but this has to
meet the condition regarding the scalar product of the field & area - where's the area here ?

BC
vanhees71
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#6
May5-13, 07:22 AM
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Then change the high-school curriculum! There is no way to explain physics other than it really is done, and if you want to discuss Faraday's Law, you have to discuss Faraday's Law which is one of Maxwell's equations. There's no way out, either you explain Faraday's Law or leave it out of your curriculum. It doesn't make sense to explain wrong things to students or does it?

Of course, you cannot use vector calculus at school, and I don't know about high-school physics except what's taught in German high schools. There, usually simple examples are treated with the (unfortunately usually incomplete) integral form of Faraday's Law.

For your case, this incomplete form is sufficient, since there are no moving parts in your setup. You just have a time-varying magnetic field, and as was pointed out before in this thread, you can simply close the circuit somehow.

The important point is, however not to somehow close the circuit, but you have to specify a precise physical question, and it's always a good thing not to ask about some buzzword like here the "electromotive force" but ask what a concrete measurement will result in according to the laws of physics, and that's what physics is all about!

So to make sense of your question, you ask, how to measure the electromotive force, and to that end you have to use, e.g., a volt meter, which you have to connect somehow with the loose ends of your ring! Now, a volt meter is a galvanometer with a very high resistance, but the reistance is very high and not infinity since to measure the voltage you have to make some current run through the coil of the galvanometer (if you use not a good old-fashioned galvanometer but some modern digital volt meter, it's principally the same, but your students will have a harder time to get the point of this very important argument about the general workings of physics as an empirical quantitative science!).

Thus, at the end you have indeed a closed circuit, and as you stated correctly yourself, the original question doesn't make any sense, because due to the time-dependent magnetic field, the measured electromotive force will depend on the geometry of your loop, i.e., the voltage your volt meter will read, will indeed depend on how you shape the wires conducting the loose ends of your incomplete ring, because the volt meter will give the electromotive force as defined along the closed circuit to be equal to the time derivative of the magnetic flux through any surface with the closed circuit as boundary. How you shape this surface in fact doesn't matter, and it is a fictitious surface, used to predict the emf due to Faraday's Law. The circuit itself is, however not something fictitious but a relevant part of your experimental setup, necessary to be defined for being able to use Faraday's Law.

I think, Faraday's Law is one of the most difficult subjects in electromagnetism. There is nearly no chance to really explain it right without vector calculus, and the integral form is pretty advanced vector calculus. I'd not know, how to adequately teach it at high-school level, particularly when it comes to moving physical parts in your setup, including generators and machines or even unipolar machines.
Jano L.
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#7
May5-13, 08:48 AM
P: 1,027
What machinery or explanation acounts for the creation of an induced EMF in this case ? (there is no induced current in this case, of course).
There will be current in the wire irrespective of whether the circuit is closed or not.

Maxwell's equations are not necessary to explain Faraday's law. After all, Faraday himself did not need such advanced mathematics. The idea of integral or sum of contributions, however, is quite helpful.

I would explain it in this way. The basic quantity that we use to explain the current in the wire is electric field in this wire (including its surface).

This electric field can be stationary, due to stationary distribution of electric charge (due to battery), or varying in time (due to moving magnet or oscillating current in other circuit).

In both cases, the electric current ##I## is given approximately by Ohm's law

$$
I = \frac{U}{R},
$$

where ##R## is a constant once measured for the wire.

In the first case, ##U## is voltage, a difference in the potential between the two ends of the wire.

In the second case, ##U## can be called total electromotive force, or a sum of electromotive forces. It is not defined as a difference in the potential. Rather it is an integral of the total electric field in the wire; the integration proceeds from one end to another (it is not necessarily closed) and it goes entirely inside the wire.

There are more contributions to the electric field and more corresponding emfs.

Let the magnet be approaching a cylindrical coil with free ends along its long axis. Since the magnetic field increases, according to the Faraday law the magnet is accompanied by rotational electric field (similar to water vortex). It points roughly tangentially to the wire of the coil. Now imagine the integration of this magnet's electric field along the wire of the coil; the more turns the coil has, the greater the integral.

The result is the emf due to the magnet

$$
U_{emf, magnet} = \int_A^B \mathbf E_{magnet} \cdot d \mathbf r.
$$

There is also another emf, ##U_{self}##, due to the electric field created within the wire by the coil itself, which partially counter-acts the first one. When these are added, we can calculate the current in the same way as in the direct current situations:

$$
I = \frac{U_{emf,magnet} + U_{self}}{R}.
$$

So, there will be some current in the wire even if the circuit is not closed. The connection of the two ends of the wire does change the behaviour (inductance) of the circuit a little bit, but if the connection does not change the shape of the coil significantly, the difference should be negligible.
vanhees71
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#8
May5-13, 10:48 AM
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Faraday's Law in fact is one of Maxwell's equations. So if you discuss Faraday's Law you are using Maxwell's equations, and I don't see, why you shouldn't use them or how you can avoid using them when teaching electromagnetism. It's a bad habit of modern didactics to confuse things by calling them other names than usually used and then simplifying things so much that they become inconsistent and confusing. The great thing of a complete theory like classical electromagnetics is that all such confusions are solved and put neatly in one scheme of descriptions of the corresponding part of nature.

Of course, in high school, you won't be able to state Maxwell's equations in differential form, and even in integral form they are not easy to understant, but you can use simple cases, starting with electrostatics, then going over to magnetostatics, do some applications like simple parts of DC (and perhaps even AC) circuits. Also waves in free space can be understood on some level (at least em. waves in free space). A lot of this cannot be done with the full mathematical machinery of university physics but it can be done qualitatively. It should only be correct!
Jano L.
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#9
May5-13, 12:30 PM
P: 1,027
vanhees71, it makes no sense to insist that "the Faraday law" is one of Maxwell's equations, when Faraday had discovered and explained this law before Maxwell formulated his equations. The Faraday law is also known as a law of electromagnetic induction and is taught without advanced mathematics of Maxwell equations on high schools.

Of course, the Maxwell formulation

$$
\nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}
$$

gives it precise mathematical form and is very important, but is by no means necessary to explain the basic content of the law.

I don't see, why you shouldn't use them
Because 1) they require long preparation in mathematics and it is very difficult to manage that at high school 2) because the required mathematics is not necessary to understand the basics of the law.

or how you can avoid using them when teaching electromagnetism.
That is the know-how of a high school physics teacher:-) Basically, you have to talk a lot about history and basic experiments, and discuss them using simple language.
vanhees71
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#10
May5-13, 01:37 PM
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Ok, then how do you state Faraday's Law? In my high-school days we learned the law in its integral form, using a rudimentary way to understand what line, surface, and volume integrals. The Faraday law then was stated as the fact that the electromotive force equals the time derivative of magnetic flux (which is of course somewhat incomplete since it leaves out the magnetic part in the electromotive force). As I said, the Faraday Law seems to be quite difficult to teach accurately without the local form in terms of differential vector operators, which are however out of reach at the high-school level.

It's of course clear that all physics teaching must be based on and explain the empirical facts. Best is of course to demonstrate the phenomena in nice demonstration experiments or, even better, let the students conduct experiments themselves.
Jano L.
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#11
May6-13, 01:43 AM
P: 1,027
Ok, then how do you state Faraday's Law? In my high-school days we learned the law in its integral form, using a rudimentary way to understand what line, surface, and volume integrals. The Faraday law then was stated as the fact that the electromotive force equals the time derivative of magnetic flux (which is of course somewhat incomplete since it leaves out the magnetic part in the electromotive force). As I said, the Faraday Law seems to be quite difficult to teach accurately without the local form in terms of differential vector operators, which are however out of reach at the high-school level.
Yes, I agree, the accurate form requires advanced mathematics, surface and line integrals. But even without it, the main idea is easily conveyed by pictures of the simplest cases - like circular wire in uniform magnetic field, and the formulation you gave above.

I do not understand however why you think it is incomplete - as far as I know, the above formulation is valid even if the wire moves through the magnetic field. When the area of the circuit expands, the magnetic flux increases and there is emf even in static magnetic field; this can be derived as an effect of the magnetic force.
vanhees71
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#12
May6-13, 02:28 AM
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If the surfaces and its boundary moves, the integral form is tricky. Start from the differential form, which is always valid. It much simplifies matters, because it's a local law, independent of (real or imagined) surfaces and lines (in Heaviside-Lorentz units):
[tex]\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.[/tex]
Taking the integral over an area, using Stokes's integral theorem of course gives immediately
[tex]\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}(t,\vec{r})=-\frac{1}{c} \int_A \mathrm{d} \vec{F} \cdot \partial_t \vec{B}(t,\vec{r}).[/tex]
As you see, here the time derivative is a partial one under the integral. If now the area (and its boundary) is moving, i.e., time-dependent. You cannot simply take this time derivative out of the integral.

A clear use of Gauß's integral theorem, applied to the cylinder-like volume spanned by the area at time [itex]t[/itex] and the area at [itex]t+\mathrm{d} t[/itex], you'll see that there is an additional contribution to the line integral. Brought to the left-hand side you find
[tex]\int_{\partial a} \mathrm{d} \vec{r} \cdot \left (\vec{E}(t,\vec{r})+\frac{\vec{v}(t,\vec{r})}{c} \times \vec{B} \right )=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_{A} \mathrm{d} \vec{F} \cdot \vec{B}(t,\vec{r})=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \Phi_{B,A}(t).[/tex]
Here, [itex]\vec{v}(t,\vec{r})[/itex] is the velocity of the area element at time [itex]t[/itex] and position [itex]\vec{r}[/itex]. The additional magnetic term, contributing to the EMF, resolves all troubles with situations like the homopolar generator, which is not so easy to understand as the simpler cases usually treated. There is a lot of confusion about this in the literature. Very good articles on this you find in Am. J. Phys. If needed, I can look up the references.
Jano L.
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#13
May6-13, 02:58 AM
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So, if I understood you well, you say that the equation

$$
\oint_{\partial A} \mathbf E\cdot d\boldsymbol \Sigma = - \frac{d}{dt} \int_A \mathbf B \cdot d\boldsymbol \Sigma
$$

is incomplete since it is valid only for stationary wire. That is right.

But the simple expression of the Faraday law is still correct, even the wire moves - your last equation is just mathematical rewrite of the same sentence: emf in the circuit is derivative of the magnetic flux. So, stated this way without mathematics, this formulation is quite general.
cabraham
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#14
May6-13, 08:30 AM
P: 997
Even with the ring open, there is a current - displacement current. Charges will move through the ring then accumulate at the ends. The build up of charges at the ends results in another local E field, a conservative E field. The E field due to induction is non-conservative, i.e. rotational. But the E field due to charges is ir-rotational and conservative.

The charges feel both forces. The ir-rotational E field due to charge build up has zero curl, so Faraday's Law which states curl E = -dB/dt, holds unconditionally here. The build up of charges at the ring ends result in a local E field that has no influence on the curl of the total E field since this local E field due to charges has no curl at all.

Does this make sense?

Claude
vanhees71
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#15
May7-13, 04:23 AM
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Yes! The charges in the ring feel both parts of the Lorentz force, the "electric" [itex]-e \vec{E}[/itex] and the "magnetic" [itex]-e \vec{v}/c \times \vec{B}[/itex], and that's why both of these terms have to be included in the EMF.

For usual situations, you can set the velocity of the electrons equal the velocity of the wire, because the drift velocities of the electrons within the wire is very very slow.


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