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Consider a thin metal rod placed in a magnetic field whose direction is constant but whose magnitude is changing with time, with the length of the rod perpendicular to the direction of the magnetic field. The rod is stationary, so there is no motional emf. If the rod were part of a conducting loop, there would be an emf induced in the loop as the magnetic flux associated with the loop would change with time. But if I connected an ideal voltmeter (with infinite resistance) across the ends of the rod when it is not part of a conducting loop, would the voltmeter show any deflection?

If yes, what would be the magnitude of this emf?

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That is a good question because WP defines the rule of Electromagnetic induction as a production of a potential difference (voltage) across a conductor when it is exposed to a varying magnetic field. It does not say that the wire must be closed. Yet, elaboration proceeds with magnetic flux through the loop. And does not bother to explain which loop. What this loop has to do with the conductor in the def? Before I have discovered this question, I have asked how to measure this EMF physics.stackexchange.com/questions/62376/… –  Val Apr 27 '13 at 11:44

5 Answers 5

up vote 4 down vote accepted

This is more of an expansion of leongz's answer.

TL;DR: The situation is incomplete. There may be an emf, and there may be a deflection. Existence of emf and deflection are independent. We cannot calculate the value of the emf from the given data (i.e., from a given time-varying $\bf B$ field)

Your fundamental issue is that Maxwell's equations (of which Faraday's law is one) are not "cause and effect". You cannot "plug in" a value of magnetic field and get a corresponding value of $\bf E$ field induced by the $\bf B$ field. All Maxwell's equations tell you is "which kinds of $\bf E$ and $\bf B$ fields can coexist given so-and-so conditions".

Trying to solve the situation via Maxwell's equations

I remember solving a similar situation via Maxwell's equations and being surprised by the answer.

The "initial conditions" were $\mathbf {B}=\beta t\hat k$, $\rho=0$ (no charge), $\mathbf{J}=0$ (no current).

Solving{*} for $\mathbf{E}$, using the differential+microscopic form of Maxwell's equations(since the integral form can only get you the value of $\bf E$ at certain positions at many times), I got:

$$\mathbf{E}=\hat i (lx + \frac{\beta}{2}y+az+c_1)+\hat j(-\frac\beta{2}x+my+bz+c_2)+\hat k(ax+by+nz+c_3)$$

where $a,b,l,m,n,c_1,c_2,c_3$ are arbitrary constants subject to $l+m+n=0$

Note that this is a family of electric fields (Setting certain constants to zero, you get concentric ellipses IIRC). All this means is that any $\bf E$ field of this type can coexist with a $\bf B$ field.

Implication for your problem

This means that your initial conditions are insufficient/inconsistent. Along with such a magnetic field, any type of electric field satisfying the above equations can exist--and must exist.

So, in addition to knowing how your magnetic field is changing with time, you need to know:

  • Which one of these bajillion electric fields is present
  • Where is the rod in relation to this electric field?

These can usually be determined if you know the boundary conditions for the system. In a physical situation, these can be extracted from the setup.

Some more analysis

Let's choose a simple solution and analyse it. I'm taking the case where the coexisting electric field is just concentric circles.

enter image description here

In this diagram, the blue stuff is the $\bf B$ field, and the green stuff is the $\bf E$ field. Being lazy, I haven't added arrows to it (I also haven't spaced the circles properly. There should be more space between the inner ones and less space between the outer ones). The other things are just rods and wire loops.

To avoid confusion, when I refer to "emf", I mean "the energy gained/lost in moving a unit test charge along a given path". Mathematically, the path integral $\int_{path}\mathbf{E}\cdot \mathrm{d} \vec l$. I'll come to voltmeters and the like later.

Let's first look at the rods. The yellow rod $AB$ will have no emf across its ends, since the $\bf E$ field is perpendicular to its length at all points. On the other hand, the magenta rod $CD$ has an emf across its ends. This emf can be easily calculated via some tricks--avoiding integration--but let's not get into that right now.

You now can probably see why the second point "Where is the rod in relation to this electric field?" matters.

On the other hand, this second point is not necessary for a loop. Indeed, neither is the first point.

Going around the loop, both loops (cyan and red in diagram) will have an emf $-A\frac{\partial\mathbf{B}}{\partial t}$. It's an interesting exercise to try and verify this without resorting to Faraday's law--take an electric field $\mathbf{E}=kr\hat\tau$ and do $\int \mathbf{E}\cdot\mathrm{d}\vec l$ around different loops of the same area. You should get the same answer.

But, you cannot divide this emf by four and say that each constituent "rod" of the loop has that emf. For example, in the cyan loop $EFGH$, $EF$ has no emf, and the rest have different emfs. "dividing by four" only works (in this case) if the loop is centered at the origin.

Voltmeters

Voltmeters are an entirely different matter here. The issue with voltmeters is that, even for so-called "ideal" voltmeters, the p.d. measured depends upon the orientation of the voltmeter.

Reusing the same situation:

enter image description here

Here, the black wires are part of the current loop (and peripherals). The yellow/magenta wires are to be swapped in and out for using the voltmeters. Only one set of colored wires will be present at a given time.

Here, the "yellow" voltmeter will measure a pd three times that of the "magenta" one. This is due to the fact that it spans thrice the area and consequently has thrice the flux.

Remember, induced $\bf E$ fields are nonconservative, so voltmeters complicate things. They don't really tell you anything tangible, either.

If the voltmeter were an everyday, non-ideal galvanometer based voltmeter, there would be extra complications due to there being a second loop.

One more thing about rods

A rod can additionally cause the extra complication of being polarizable/magnetizable. Then, you have to consider the macroscopic Maxwell equations, with the $\bf D,P,M,H$ fields and bound/free charges/currents. But then you need to know about the material of the rod. Or, just find a hypothetical rod with $\mu_r=\varepsilon_r=1$ and use it.

Also, the charges in a rod will tend to redistribute, nullifying the electric field and thus the emf in the rod.

Conclusion

The given data is incomplete. There is a truckload of different $\mathbf E$ fields that you can use here, and you're not sure which one it is. Additionally, even if we knew which field it was, the orientation of the rod comes into the picture.

So, the rod will have a motional emf, but this emf may be zero. The exact value of this emf cannot be calculated if you only know $\bf B$.

An ideal voltmeter, again, may show deflection. Not necessarily, though.

*Solving simultaneous PDEs in four variables is not too fun, so I did make some assumptions regarding the symmetry of the situation to simplify stuff. So the given family of solutions is a subset of the actual solution set. That doesn't hamper this discussion though.

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Well I must admit that I was not very found of Classical electromagnetism (special theoretical subject in my studies), but as far as I know, you have to solve the $(\vec{A}, V)$ field, and this solution gives unique solution for $\vec{E}$ and $\vec{B}$ to a gauge value. But in this case the ambition of the answer far exceeds the ambition of the question. I pronounce myself as incompetent for future discussion and maybe I pull out of this problem completely (together with my answer) –  Pygmalion Apr 17 '12 at 8:47
    
@Pygmalion: I don't think V is definable here--not sure, I usually stay away from vector potential. Either way, Maxwell's laws are enough--and the given set of solutions satisfy the laws, so there is no unique solution. And no need to feel bad about it--I've had some major mea culpas here. Your answer is correct, it just answers something else. And remember, the question is incomplete. –  Manishearth Apr 17 '12 at 8:53
    
And not to mention you must consider material and "free" electrons within it. I think that general conclusions should be extracted without deep theory - I am certain that my colleague, which is into this stuff might find this discussion laughable, so I am not even inclined to present it to him. –  Pygmalion Apr 17 '12 at 8:54
    
@Pygmalion: True, though you can consider an imaginary rod and move an electron along it. Though its a valid point and I'll add it to the answer; thanks. –  Manishearth Apr 17 '12 at 8:57
    
I am sorry, I should use symbol $(\vec{A},\phi)$ instead of $(\vec{A},V)$. $\phi$ is indeed vector potential, but of another world than $V$. By the way $(\vec{A},\phi)$ is solution to Maxwell laws. –  Pygmalion Apr 17 '12 at 8:57

PLEASE NOTE: This answer is provided only as a student's reference. Following the principle of theoretical classical electromagnetism, solution might be much more complex and possibly exceed the ambitions of the question.

The answer can be found using Lorenz force

$$\vec{F} = q \vec{E} + q \vec{v} \times \vec{B}.$$

Of course, there are always "free" electrons within metal rod. However, if there is no bar movement and $v = 0$, there is also no magnetic force on electrons and there will be no induced emf.

If rod, however, moves within constant magnetic field, so do all the electrons within it, magnetic force push them in one direction, concentrating electrons on one side of the rod. This creates electric field within the rod and consequently measureable emf.

(Back to your question: If you connected an ideal voltmeter (with infinite resistance) across the ends of the rod, then you do create a loop. But let's suppose for the sake of argument that we have some kind of loopless voltmeter based on some entirely new principle.)

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I think she wants it from the Faraday's law point of view--drawing an imaginary loop makes it appear to have a p.d. across the ends. Except that, in this case, different loops give different p.d's to the rod. –  Manishearth Apr 16 '12 at 6:29
    
I think confusion comes from the fact, that you can derive Faraday's law for loop of the variable size within constant magnetic field using moving rod (standard textbook procedure). You cannot however do that for the variable magnetic field and constant loop (therefore stationary rod). –  Pygmalion Apr 16 '12 at 6:31
    
It also comes from this: We can draw imaginary loops in, say, a solenoid and get pds/electric fields--which work. This does not work here. –  Manishearth Apr 16 '12 at 6:33
    
Indeed, but I think that Amrita is confused by non-symmetry in the derivation of two similar cases and my answer was directed there. However, I see that answering to green users is very ungrateful. You usualy don't get any points... –  Pygmalion Apr 16 '12 at 8:18
    
Green users? I'm sorry, I didn't quite get you. You haven't gotten any upvotes because (1) she hasn't logged in for a while, and/or (2) it doesn't answer the question. Actually I think the question is slightly vague; so your answer is valid. +1 –  Manishearth Apr 16 '12 at 11:22

1) Yes there is a deflection. 2) The voltmeter will measure the time rate of change of the magnetic flux enclosed by the conductor + voltmeter circuit (Faraday's law).

Elaboration: Faraday's law says that the line integral of the electric field around a loop (loop emf), is equal to the time rate of change of the magnetic flux enclosed by the loop \begin{equation} emf = -\oint \mathbf{E \cdot ds}=\frac{d\Phi}{dt} \end{equation}

For this particular loop, consisting of a conductor + a voltmeter:

  1. the contribution to the line integral from the part of the loop within the conductor is 0, by definition of a conductor. (The charges within the conductor distribute themselves so as to null its electric field.)
  2. therefore the remainder of the line integral (the voltmeter portion of the circuit) must be the full loop emf:

\begin{equation} -\oint \mathbf{E \cdot ds} = -\int_{cond} \mathbf{E \cdot ds} - \int_{v-mtr} \mathbf{E \cdot ds} = 0 -\int_{v-mtr} \mathbf{E \cdot ds}=\frac{d\Phi}{dt} \end{equation}

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This is wrong--it actually depends upon a whole bunch of stuff. See my answer above. –  Manishearth Apr 17 '12 at 8:31
1  
I disagree. 1) E=0 within a perfect conductor by definition, a fact that you neglect in your analysis. (leongz's answer describes how the conductor charges adjust to achieve this condition.) 2) Certainly you aren't denying Faraday's law (are you?). As for the "bajilion" of other consistent E fields that you posit, they just don't matter. Once you've specified the magnetic field, Faraday's law gives a unique answer for the loop emf. If you want a reference, check out Ramo et al "Fields and Waves in Communications Electronics" Chapter 4, especially the section on "Inductance element". –  Art Brown Apr 17 '12 at 21:14
    
oh whoops. This is correct. For some reason I read your $\bf E$ as $\mathcal{E}(emf)$. Tecnically, there is an E as well--it's just there at the ends. In my answer, I have avoided this by talking about imaginary rods. Faraday's law gives the correct answer for a loop, not a rod, and I've mentioned this. Note that to avoid voltmeter confusions, I have slightly redefined emf as work done initially, for clarity –  Manishearth Apr 18 '12 at 0:24
    
Edit your answer so I can undownvote –  Manishearth Apr 18 '12 at 0:24
    
Anyway, I'll edit it into my answer. I've avoided real rods and talked about imaginary paths, though there's one place where I haven't. –  Manishearth Apr 18 '12 at 0:43

According to the Maxwell's equation $$\nabla\times\mathbf{E}=-\frac{1}{c}\frac{\partial \mathbf{B}}{\partial t},$$ a time-dependent magnetic field $\mathbf{B}$ must co-exist with an electric field $\mathbf{E}$. So, even when the rod is stationary, the electric field will cause charges in the rod to move and gather near the ends of the rod, giving a potential difference across the rod.

Specifically, since the electric field inside a perfect conductor must be zero, the charges will move in order to exactly cancel out the electric field that comes with the magnetic field. This happens regardless of the presence of a voltmeter.

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OK, can you derive the value of that voltage? I think the problem is as follows: you can put tiny loop within the rod and electric field will be induced along this loop. However, such circular electric field would force electrons to circulate and not to concenctrate on the rod's ends. This circulating electrons would create magnetic field of their own and due to Lenz law you end up with a diamagnetic efect. –  Pygmalion Apr 17 '12 at 6:53
    
The very same principle was used for explaining diamagnetism of superconductors (Meissner efect), until it was proven that diamagnetism is not the consequence but part of supercondutivity. –  Pygmalion Apr 17 '12 at 6:59
    
@Pygmalion: You can't derive it, see my answer above –  Manishearth Apr 17 '12 at 8:32
    
@Manisheart so you actually believe that there would be some kind of non-zero $\int \vec{E} \text{d}\vec{s}$ between rod's ends? Well if there is, I think there should be explanation for it and theoretical calculation too. We still might not be able to measure it, because of practical limitations, but there are no theoretical limitations why something cannot be calculated. EDIT: ok I se your answer below –  Pygmalion Apr 17 '12 at 8:38
    
@Pygmalion: Yes, I believe that there is--but it depends. The system is incomplete, there isn't enough data to solve it. The fact is, different electric fields can coexist with a uniform time-varying magnetic field. –  Manishearth Apr 17 '12 at 8:39

I here give an other, perhaps more clear answer. Pygmalion is right in his answer http://physics.stackexchange.com/a/23825/16689, there should be no voltage drop when there is no loop.

The reason is pretty simple: you can always choose the gauge such that the magnetic field is zero when there is no loop. That the loop which enforces the magnetic field to be non-nul in the circuit. (More details: Having chosen a time dependent $A$, you should pay attention to your electric field, through the term $\partial A / \partial t$, but this one can be compensated by an exact gradient $\nabla \phi$, which indeed generate no voltage drop along a single connected circuit.)

As far as I can see, the incomplete situation found in http://physics.stackexchange.com/a/23895/16689 is precisely the gauge choice. It is well known that the Maxwell system (NB: Maxwell system consists in 5 equations: people usually forget the Newton law with Lorentz force, such that there is no charge displacement...) is incomplete until you fix a gauge. The physical (observable) variable (magnetic flux and electric voltage drop) does not depend on the gauge choice of course.

Of course adding a voltmeter creates a loop, and you would record a voltage drop thanks to the Faraday's law when the magnetic flux inside the loop is time dependent. This time dependency can be generated by a time dependent magnetic field, or by a time dependent loop (i.e. moving rod or wires).

One interesting effect that you should been able to see experimentally is about the geometry of the experiment. If you put your wires perpendicular to a time-dependent magnetic field (I discard the possibility to move the wire for commodity), you will record a voltage drop thank to Faraday. Perpendicular means in the plane of the experiment, as the figures in this answer http://physics.stackexchange.com/a/23895/16689 are showing. If you put your wire along the magnetic field (so, perpendicular to the plane of the figure) you will have no magnetic flux, and then no voltage drop.

The only way to have a non-vanishing voltage drop at the borns of a wire is, well, ... to apply it !

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