Induced Electric Field

1. Mar 17, 2010

kini.Amith

I am a high school student, so forgive me if my question appears stupid.
Consider a uniform magnetic field normal to the plane of the screen with varying magnitude. consider a point P on the screen in the magnetic field.the varying magnetic field will induce an electric field, with the field lines in the form of concentric circles centred around p. direction of field tangent to the circle .consider another point Q other than p. we can also think of concentric electric electric field lines around Q.
The circles around Q and those around P will intersect giving 2 directions of electric field at all points. if we think of more points other than P and Q, there will be electric field lines flying everywhere. So what will happen if i simply place a stationary electron in this varying mag field, without any circuit?
Thank you for reading.

2. Mar 17, 2010

marcusl

The electromotive force induced about each magnetic flux line drives a current that acts to shield the flux from the interior of the conductor. This is called an eddy current. By symmetry, as you point out, there is no net current in the region of uniform field. An intuitive picture of this is that the circular current around one flux line is canceled by those around neighboring flux lines. Here's the physical picture: ooo (if the sense is clockwise in each circle, then current to left and right of the center sums to zero. Similarly for currents above and below.) An electron at the center feels no force.

On the other hand, if the conducting sheet is large and the uniform field localized and smaller than the sheet, then at the edges where the field intensity falls away, the cancellation does not occur. Eddy currents flow round the outside of the region where the uniform field intersects the sheet. Similarly, if the field is uniform and the sheet finite and small, eddy currents will flow around the boundary of the sheet.

3. Mar 17, 2010

kini.Amith

But that is the case inside a conducting sheet.
according to Fundamental of Physics by Resnick, Halliday and Walker,
" A changing magnetic field produces an electric field. The striking feature of this statement is that the electric field is induced even if there is no conducting ring or sheet."
So does that mean that if i simply place a stationary electron in free space in the changing magnetic field, it will move along a circular path?

4. Mar 18, 2010

marcusl

Yes, that is correct. Mathematically, Faraday's law gives the electromotive force around a closed line path as

$$\varepsilon=\oint\vec{E}\cdot d\vec{l}=-\frac{d}{dt}\int_S \vec{B}\cdot d\vec{A}$$

where the line integral on the left is taken about the boundary to the surface S and where dA is the normal to the differential of surface area. The right-hand integral gives the magnetic flux enclosed by the boundary.

In the absence of a unique physical path such as a wire loop, you must consider the sum of all possible imaginary closed paths just as in the case of the conductive sheet that I already discussed. A test charge in the center of a time-varying but uniform field again feels no force.

If the field is contained within a finite volume, on the other hand, let's say a cylindrical region between two pole faces of a C shaped electromagnet, then a test charge located radially outside the cylindrical region will experience an electric force. We can draw a closed path through the test charge that fully surrounds the field, therefore generating an emf according to Faraday's law. All other loops through the charge surround field-free regions and have no E field or emf, so there is no cancellation. The charge sees a net E field and feels a force.

Last edited: Mar 18, 2010
5. Mar 18, 2010

kini.Amith

that's interesting. Thanks.

6. Mar 21, 2010

bjacoby

You have asked an excellent question extremely fundamental to the understanding of the critical details of magnetic induction. Resnick etc, is both right and wrong in their statement. A changing magnetic field does NOT produce an electric field. A changing current does that. The current ALSO produces a changing magnetic field related to the Electric field (both have the same source). But the correct part is that the changing current produces an E field in the space. IF the E field is located inside a conductor then a large current can easily be induced. If there is no conductor the E fields remains and can indeed influence free charges such as electrons.

The first thing you have to ask is what are the characteristics of the magnetic field, which is to ask in another sense what is the geometry of the source current? Just for drill let us take a long solenoid as our source current. Since our induced E field can ALSO be found from a time-changing Magnetic Vector Potential we note that the magnetic potential A inside the solenoid can be found to be equal to:

A=unIr/2L Where u is mu naught, n is the number of turns, L is the solenoid length, I is the current and r is the distance form the axis of the solenoid. The Vector A is in the azimuthal or theta direction.

For this reason one finds that the induced E field inside the solenoid E = - dI/dt (unr/2L) also in the azimuthal direction.

Ok. Now look at an electron placed in the exact center of the solenoid. At that point r = 0. And thus E = 0. And thus the electron will experience no force from the induction. Note that as the distance from the center, r, is increased so does the accelerating E field. The E field being in the theta direction is always tangential to a circular orbit about the center of the solenoid. The further out you are the greater the acceleration. F = qE = ma.

It is interesting to note that if one goes OUTSIDE the solenoid where the magnetic field is ZERO, the acceleration persists even though there is NO magnetic field present at the electron. Which is something of a hint that it is not the magnetic field creating the E field accelerating the electron!

For a really good time, I suggest doing a search on the subject "betatron"!

7. Mar 21, 2010

Phrak

I'd like to point out that bjacoby's adherence to the belief that magnetic fields have no independent existence from charge is his own theory, not consistent with accepted classical physics as born out by the generally accepted notion of propagating electromagnetic fields. The rest of it seem similarly constructed

This,

”It is interesting to note that if one goes OUTSIDE the solenoid where the magnetic field is ZERO, the acceleration persists even though there is NO magnetic field present at the electron. Which is something of a hint that it is not the magnetic field creating the E field accelerating the electron!”

is just plain wrong.

I suggest you ask more questions and pontificate less, bjocoby, or the powers-that-be will ban you from this forum. Personally, I find this category of misinformation extremely annoying.

8. Mar 21, 2010

Phrak

Every once in a great while here, kini.Amith, someone who is trying to learn such as yourself, asks a gem of a question. Thanks! Yours is one of them. A 'gem' is something I think I should know the answer to, but am shocked that I don't.

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How does a point charge behave in a uniform magnetic field changing with time?

I will use Maxwell's equation of induction in differential form.

We cannot use the integral form of inductance because it only tells us about the total path integral of the electric field over a loop, rather than at the coordinate of the particle.

In differential form,

$$\nabla \cdot E = \frac{\partial B}{\partial t} \ .$$

Make it a uniform magnetic field Bz along the z-axis for convenience.

$$\frac{\partial E_x}{\partial y} - \frac{\partial E_y}{\partial x} = \frac{\partial B}{\partial t} \ .$$

This tells us that the electric field is directed somewhere within the XY-plane, but not in what direction.

The uniformity of the magnetic field tells us that there is no preferred direction in the XY-plane.

Are the classical Maxwell equations complete or ambiguous in the specification of the electric and magnetic field vectors at spacetime coordinates?

Is the problem set-up erronious; are uniform time varying magnetic fields prelcuded on basis I'm not aware of?

9. Mar 21, 2010

Staff: Mentor

This is a pretty advanced question for a high school student. I can answer it, but I am not sure if it will be a comprehensible answer, so I apologize in advance if I cause more confusion than I clear up.

For most problems in electromagnetism, it is preferable to work in terms of the vector and scalar potentials, rather than in terms of the fields themselves: http://en.wikipedia.org/wiki/Mathem...lectromagnetic_field#Potential_field_approach

So for this problem, in units where c=1, we can specify the vector potential as
$$\mathbf{A}=\left( -y \, cos(\omega t),x \, cos(\omega t),0 \right)$$
which yields the desired B-field:
$$\mathbf{B}=\nabla \times \mathbf{A} = \left( 0,0,2 \, cos(\omega t) \right)$$

Everything else is free, so let us choose the simplest possible scalar potential
$$V=0$$
which yields the following E-field:
$$\mathbf{E} = \nabla V - \frac{\partial \mathbf{A}}{\partial t}=\left( -y \omega \, sin(\omega t),x \omega \, sin(\omega t),0 \right)$$

Furthermore, we can easily calculate the charge and current densities:
$$\rho = -\nabla^2 V - \frac{\partial}{\partial t} \left ( \nabla \cdot \mathbf A \right ) = 0$$
$$\mathbf J = -\left ( \nabla^2 \mathbf A - \frac{\partial^2 \mathbf A}{\partial t^2} \right ) + \nabla \left ( \nabla \cdot \mathbf A + \frac{\partial V}{\partial t} \right ) = \left( y \omega^2 \, cos(\omega t),-x \omega^2 \, cos(\omega t),0 \right)$$

And we can check back to verify that the obtained fields and densities satisfy Maxwell's equations:
$$\nabla \cdot \mathbf{E} = 0 = \rho$$
$$\nabla \cdot \mathbf{B} = 0$$
$$\nabla \times \mathbf{E} = \left( 0,0,2 \omega \, sin(\omega t) \right) = -\frac {\partial \mathbf{B}}{\partial t}$$
$$\nabla \times \mathbf{B} = 0 = \left( y \omega^2 \, cos(\omega t), -x \omega^2 \, cos(\omega t), 0\right) + \left( -y \omega^2 \, cos(\omega t), x \omega^2 \, cos(\omega t), 0\right) = \mathbf{J} + \frac{\partial \mathbf{E}}{\partial t}$$

So, the E field lines are not just "flying everywhere", but are quite well defined, and a charge at rest at any given location will be pushed in the direction of the local E field.

10. Mar 21, 2010

bjacoby

I suggest that instead of repeating errors you provide some justification for your dogmatic beliefs! Can you provide proof that magnetic and electric fields create each other in electromagnetic propagation? Please provide it. Do you agree that charges and their motions create both Magnetic and Electric fields? Do you not agree that these fields propagate away from that charge/current in free space at the speed of light? Do you not agree that therefore in Maxwell's equation that says the curl of E = the negative time rate of change of B both E and B are occurring simultaneously being equally retarded from the source? It is widely accepted that E and B are simultaneous in this equation in that time does not appear as a parameter! Do you not agree that magnetic and electric fields in electromagnetic propagation in free space are IN PHASE. And lastly do you not agree with the premise that actions which occur at the same time CANNOT "create" each other?

In addition you are saying that my assertion that there is no magnetic field outside a long solenoid is "just plain wrong", is that not true? So I take it your assertion is that the magnetic field that exists outside the solenoid creates the Induced E field there. Please explain this further. I'd really like to know more about this magnetic field OUTSIDE a long solenoid. Or how the magnetic field INSIDE the solenoid creates an E field "at a distance" outside it? In fact you have explained nothing here, simply going for "proof by assertion". Please explain exactly what you mean so we can all discuss it. It's a physics forum so it's relevant. These are all questions very fundamental to a study of electromagnetics.

Of course if you wish to turn this forum into a religious one with heresy suppressed through censorship, that is your opinion. I on the other hand believe that science is a matter of proof and data rather than opinion and "accepted" dogma. Let us hope that the "powers that be" have more respect for the scientific method than to ban discussions of widely accepted dogma. If I am making some kind of fundamental error here then I'd sure liked to hear about it in detail. The fact that I seem to annoy you is totally irrelevant to a sensible discussion of this issue.

It seems to me that questioning "accepted notions" is how science advances. I'm sure there is a lot of good done here by providing beginners an understanding of textbook principles, but to restrict this forum to being merely a rubber stamp for textbook dogma seems to me to restrict it's potential as a tool for the promotion of science and physics.

11. Mar 21, 2010

Phrak

Post one of your questions or assertions in a thread where it can be addressed.

Last edited: Mar 22, 2010
12. Mar 22, 2010

vanesch

Staff Emeritus
I think this is an inappropriate representation at this level of the discussion. Of course you can calculate the fields E and B by using retarded potentials, which you can see as "direct" effects of the sources (charges and currents). This is I think what you allude to. In this view, the E and B fields are not "dynamical variables" but just "directly" derived from the sources (charges and currents). Mind you that this is only a particular solution of the electromagnetic field equations, and that we took them to have boundary conditions of "zero at infinity".

But if you consider the E and B fields as the "state variables" of the EM field, and you only allow for local field equations (and hence no "retarded" stuff), then you can't escape that the E-field IS dependent on the change in B-field.

If it is a straight, long, but finite solenoid, then of course there is a magnetic field outside of the solenoid! B-field lines are closed lines, so in order for them to be closed, they have to return on the outside.

For an infinite-length solenoid, that's not the case, and for a torus, either. But in these cases, you don't have your electron accelerating outside of the solenoid either.

So please stop nitpicking in this discussion as if it were a "dogmatic" point. You're on the verge of misinformation here.

13. Mar 22, 2010

kcdodd

DaleSpam, how can it end up with a current if we specify there is none? Perhaps a different choice of vector potential, limited to a first power of time. I would like to point out it still is not unique. If we make a simple example vector potential:

$$\vec{A} = xBt\hat{y}$$

then

$$\nabla\times \vec{A} = \partial_x A_y \hat{z} = Bt\hat{z}$$

and then

$$\vec{E} = -\partial_t \vec{A} = -xB\hat{y}$$

However, A is not unique. What if, for example, we had:

$$\vec{A} = -yBt/2\hat{x} + xBt/2\hat{y}$$

then

$$\nabla\times \vec{A} = (\partial_x A_y - \partial_y A_x)\hat{z} = Bt\hat{z}$$

But, now you see E would be completely different.

$$\vec{E} = yB/2\hat{x} - xB/2\hat{y}$$

Now, imagine a source. An infinite current sheet will produce a uniform magnetic field. But, notice it would not matter if we rotated the current sheet, magnetic field would still be uniform, unless you pass from one side of the sheet to the other where it flips direction. However, now you have a basis for which A to use (the one parallel to sheet). So, it seems you need the boundary conditions, even when B is uniform.

So, back to the origional question. It seems the electric field may not be uniform, depending on the boundary conditions. It is clearly zero at the center of a solenoid, but needs to form parallel to the boundary. So you get concentric circular e-field increasing in magnitude, until you reach the current source. Then outside it forms concentric rings of decreasing strength (this is how transformers work).

Last edited: Mar 22, 2010
14. Mar 22, 2010

Phrak

(As for infinite solenoids. 1) Classical charged particles are not deflected where the solenoid is constant current. (However, real charged particles undergo a phase change, measurable by self-interference, but where the average probability distribution is unchanged, that has no context withing the classical axioms of this folder .) 2) But the OP is asking about a B field changing over time, in which there are certainly exterior B fields present for an infinite solenoid.)

Last edited: Mar 22, 2010
15. Mar 22, 2010

Phrak

You're a darned genius Dale. I get it, that there is a constant that is lost upon differrentiating the vector potential so that E of B is not unique, but haven't you specified some specific boundry conditions to arrive at a particular solution?

In this case you've constrained the magnetic potential to zero at the origin. I haven't yet hashed it out: can we arbitrarily move the zero potential anywhere we wish by a gauge transformation of A?

16. Mar 22, 2010

Staff: Mentor

You are correct, the question itself is ill posed in that regard, it did not describe which of many possible current configurations (including no current) and time variations was to be considered.

This is an interesting example. You have no currents and no charge (for V=0), but a linearly increasing (in space) E-field leading to a linearly increasing (in time) B field. With sourceless fields it is in some sense a non-physical solution, but such solutions can be valuable in understanding anyway. Here we see that the usual intuition that the E field lines are curving around something is not necessarily the case. Here the E-field is $\mathbf{E}=(0,-B x, 0)$ so again the OP's concern about E-field lines "flying everywhere" does not arise.

Yes you would expect different fields since this change of A and V is not a gauge transformation, but the resulting fields do indeed satisfy Maxwell's equations and the E-field is still well-defined. This solution is closer to my original solution, just with a linear time variation rather than a sinusoidal one.

I agree fully.

Last edited: Mar 22, 2010
17. Mar 22, 2010

Staff: Mentor

Implicitly, yes. The boundary conditions were not specified in the problem, so I just picked whatever fell out easiest when I sat down to work it. As kcdodd has shown other choices lead to other equally valid solutions for different sources and boundary conditions.

Yes, but then you would have a non-zero scalar potential resulting in the same fields and currents in the end.

18. Mar 22, 2010

bjacoby

I'm not trying to escape the fact that Maxwell's equation is TRUE. The point I've tried to make is that while it's true (the VALUE of a changing magnetic field can be used to (usually) calculate the VALUE of the induced E field) the error is to assume that the equation implies that one side CAUSES the other. That is the important point. The fact that both sides of the equation happen simultaneously and transluminal action is prohibited proves what I contend.

I agree one can take a "state variable" approach and obtain useful answers, but I don't see where that has any bearing on causality. In fact by ignoring retardation it simply ignores the issue. We've already noted that E and B are related so that is not an issue but what "causes" what is an issue.

As for the field outside of a long solenoid, yes, you make a good point. I had indeed forgotten about the looping fields out there. Of course as you note, if the solenoid is "infinitely long" (which real solenoids can never be) those looping fields would tend to zero.

But the issue is easily addressed as you do by simply proposing a toroidal coil instead. And I assure you that the electron does indeed experience an accelerating E field outside both the theoretically "infinite" solenoid or the torus. Even if the field outside were relatively weak and not truly zero, how could one still explain the fact that the induced E field just inside the coils is nearly equal to the E field just outside the coils?

As for "nitpicking", I simply answered a question using my understanding of the problem. And then I was attacked for being "wrong". I am trying to understand just where in my understanding I have made false statements or false assumptions. So far nobody has provided any except your "state variable" approach which simply removes retardation from the calculations or your statement that an Induced E field does not exist outside a torus or long solenoid which is incorrect.

You have, however, raised a very important point which has to do with boundary conditions and the resolution of such conditions with propagation in source-free space. It's a matter that requires some thought.

On the other hand, I do not understand what being on the "verge of misinformation" means. I mean, if I'm right then theory should support it. If I'm wrong in some matter, then it certainly should be pointed out to me so I will not continue making statements that are not correct. I simply do not see where these arguments I've made are wrong.

19. Mar 22, 2010

bjacoby

Excellent suggestion! It is not my intention to highjack this thread with a different issue, even though the original question was very much related to what produces an E field when a changing magnetic field is present.

20. Mar 23, 2010

kcdodd

bjacoby, I think both your "position", and the "contrary" "position", are both right and wrong in some respects. Haha.

To Phrak's side, you are indeed mistaken here

This is a common misconception because there is no magnetic field, and yet you see an electric field. The fact is, Maxwell's equations are local quantities, but they actually define a global field! You cannot reliably construct the full E and B from a simple, local, view of maxwell equations, because they are simply differential equations. In a sense, you must still "integrate", or "solve", maxwell's equations over all of space (globaly), to get E and B.

As an example, look at a single point charge. The divergence of E only occurs at a single point, and is zero everywhere else. And yet you have E filling all of space.

The same view must is held here. Just because the curl of E is zero outside an infinite solenoid (or toroid), does not mean E is zero. If you use stokes theorem, and integrate around the solenoid, you can see that clearly. If you don't believe it, open up any transformer and see it work! The point is that maxwells equations do not give you a local solution to E, only local differential equation which must be solves somehow.

however, saying a changing B-field does not cause an E field, or rather vice versa, is a trickier subject. Consider the harmonic oscillator:

$$\ddot{x} = -\omega^2 x$$

Which side causes which in this case? This equation contains both simultaneous quantities. Your view on causality is, in some sense, subjective. But, we usually take the right side to cause the left side as a matter of our physical common sense (a spring "pulls on me"). And yet, if you where to instead specify an acceleration, I could tell you exactly what x is. The point is, If you say maxwells equations are non-causal, then you have to say all differential equations are non-causal.