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Difficulty understanding EMF due to a changing magnetic flux

  1. Jun 25, 2014 #1
    Hello everybody,
    I'm studying classical electrodynamics and I'm having a difficulty grasping the concept of EMF.
    I'll try to explain what causes my confusion.

    Say I've got a conducting loop, an electrical wire, with some resistance R which contracts (or expands) with time.
    Let's put that loop in a constant magnetic field (call it B) so that the wire is perpendicular to the magnetic flux.
    Now I need a changing flux, so for simplicity, let's assume that the radius grows linearly.
    Obviously the flux changes with time, so according to Faraday, EMF occurs and current flows through that loop.

    My problem is this: I studied that in such cases, the flux is simply the magnetic field going through the area of the loop. Alright, but is the magnetic field only that constant B or is it B minus the magnetic field generated by the induced current (which is the result of the change in the flux itself!) ?


    This topic really confuses me :frown:
     
    Last edited: Jun 25, 2014
  2. jcsd
  3. Jun 25, 2014 #2

    UltrafastPED

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    Excellent question! And yes, it is not immediately clear how this should be handled, but since Maxwell's equations are linear in the fields, you would expect that there will be two simultaneous expressions that are coupled, and that there solution gives the complete answer.

    Most elementary treatments ignore this - at least during the initial presentation - due to the complexity. The problem is resolved by "self inductance". This set of lecture notes gets there on page 20 and following:
    http://www.phy-astr.gsu.edu/cymbalyuk/lecture24-25a.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
  4. Jun 25, 2014 #3

    vanhees71

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    The only caveat is that in the mentioned script they treat conductors at rest. It's very confusing that in almost all introductory textbooks they treat Faraday's Law in the wrong way. Feynman Lectures, vol. II, is an example for an exception. The point is that with time-dependent surfaces and its boundary, you cannot draw simply the time derivative in the flux integral out. It's very nicely described in Wikipedia, but the final formula is (in Heaviside-Lorentz units)
    [tex]\frac{\mathrm{d} \Phi}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{d} t} \int_{F} \mathrm{d}^2 \vec{F} \cdot \vec{B}=\int_{\partial F} \mathrm{d} \vec{r} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right ),[/tex]
    where [itex]\vec{v}(t,\vec{x})[/itex] is the velocity of the boundary curve of the surface. You find the derivation of this important result from the local form of the Maxwell Equations, which are always the save ground to start with in the Wikipedia:

    http://en.wikipedia.org/wiki/Faraday's_law_of_induction#Proof_of_Faraday.27s_law

    They use SI units, and thus the factor [itex]1/c[/itex] is missing under the line integral.
     
  5. Jun 25, 2014 #4

    WannabeNewton

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    The reason you have this confusion, and it exists for extremely good reason, is almost all introductory texts talk about how "a changing magnetic field induces a current" but by your correct reasoning this induced current would then induce its own magnetic field but if that magnetic field is also changing then why isn't that taken into account in the original flux? And to go further, the induced magnetic field will go on to induce a current of its own so why don't we take that into account in the magnitude of the original induced current? Hopefully you can see where I'm going with this ad infinitum.

    Consider the two Maxwell equations of interest: ##\vec{\nabla} \times \vec{B} = \partial_t \vec{E} + \vec{j}## and ##\vec{\nabla} \times \vec{E} = -\partial_t \vec{B}##; for simplicity I have set all fundamental constants to unity. Let's say, for your current loop setup, that we solve these two equations simultaneously and get ##\vec{B}(x,t)## and ##\vec{E}(x,t)##. This already takes into account the infinite "recursive induction" described above so yes when you calculate everything without approximation, all orders of induction of both magnetic and electric field (or current) will be taken into account.

    However in introductory problems what we do is make an approximation. We expand ##\vec{B}(x,t)## and ##\vec{E}(x,t)## (or, again, the current) in a power series with the expansion parameter loosely representing the amount of "recursive steps" we've taken in the induction process in the above sense. We then throw away all terms that are going to be extremely small. This allows us to treat the problem using only the zeroth order magnetic field (the constant one in the background in your example) and the first order induced current and ignore all other higher order contributions.
     
  6. Jun 25, 2014 #5
    Thanks for the notes, UltrafastPED.
    vanhees71, I'll have to do some proper reading on that and try to understand it fully.

    EDIT:
    WannabeNewton, we posted almost at the same time so I haven't read your post before posting the text below...
    Thank you for the explanation and a way to approach this problem!


    For starters I'd like to be sure that my way of thought isn't wrong. I'll describe what I think about the setup in my original post:
    1. A "stretching" wire loop is placed perpendicular to a magnetic field B.
    2. The flux should be ∫∫BdS, where dS is an area element enclosed by the loop.
    3. The expression above isn't complete because the magnetic field isn't only B.
    4. The change in the area as a function of time results in a change in the magnetic flux.
    5. The change in magnetic flux induces EMF, due to which current flows through the wire.
    6. The current is flowing in a direction such that (according to Lenz's law) the magnetic field generated by that induced current is in the opposite direction relative to B. Let's call that generated magnetic field Bi. If B was "into the page", then Bi will be "out of the page". Obviously Bi can be calculated using Biot-Savart's law for instance. I won't type the full expression here.
    7. Going back to 2, the flux should be ∫∫(B-Bi)dS.
    8. That is to say, the flux is now dependent on its own derivative (via Bi), providing me with a differential equation.

    So, is the process above even correct? Am I thinking straight or is it a disaster?
     
    Last edited: Jun 25, 2014
  7. Jun 25, 2014 #6

    WannabeNewton

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    It's the correct thinking but it's not the fully story. In general, ##\vec{B}_1## will then go on to induce its own current so we have to augment the ##\vec{B}_0## induced current ##I_1## with the ##\vec{B}_1## induced current ##I_2##. But then ##I_2## will in general induce another magnetic field ##\vec{B}_2## which will induce a current ##I_3## which will induce a magnetic field ##\vec{B}_3## which will induce a current ##I_4##...ad infinitum. If you were to follow this to completion in your above process then you would have an infinite series for the true magnetic field ##\vec{B}## that is to appear in the flux integral. This ##\vec{B}## is equivalent to the magnetic field you would get by solving the coupled Maxwell equations simultaneously (the ones I wrote in post #4).

    As explained above, what we do instead is only take the first say ##n## terms in the series for ##\vec{B}## and the first ##n+1## terms in the series for ##I## and drop all higher order contributions to the true current and magnetic field because they are negligible compared to the lower order contributions. In typical introductory problems, such as the current loop in your example, we take ##n=0## i.e. we just take the zeroth order constant magnetic field in the background and the first order induced current.

    One way to calculate the higher order contributions is to first solve the lower order contributions and plug them back into Maxwell's equations for the higher order terms and solve again and keep repeating for as many terms as desired. In "Classical Electromagnetic Radiation"-Heald and Marion you will in fact find problems involving electromagnetic induction wherein one uses this prescription to calculate the second order contribution to the current (we call the above process a "perturbation theory" calculation).
     
  8. Jun 25, 2014 #7
    Yes, your first post indeed provided a good explanation and this post strengthens it. Thank you. My job will be to do some proper reading and exercises, in addition to examining the information given by UltrafastPED and vanhees71.
    It seems as though our Professor did not provided us with the full picture during the course, which resulted in my original question...
     
  9. Jun 25, 2014 #8
    ø = A*B, where A = area of loop, B = magnetic flux density, ø = flux linkage

    If A is varying with B constant, then ø is varying in relation to area A. Thus ø is changing and inductance takes place even though B is constant. Did I help?

    Claude
     
  10. Jun 25, 2014 #9
    Yeah but my original claim is that things aren't so simple. According to the guys who posted above, they indeed aren't. Meaning that B is really constant only if I ignore higher orders as WannabeNewton explained.
    Maybe I misunderstood you?
     
  11. Jun 25, 2014 #10
    Ok, yes I understand. I should have been more clear. B = Bext + Bint, where Bext is the external flux density in the loop, and Bint is the internal due to the loop's own current. Bext is constant, but Bint is directly related to I, the loop current. I agree with the above. I think we have consensus. Thanks, sorry if I confused matters.

    Claude
     
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