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Faraday's Law and Magnetic Flux

  1. Jul 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi, this is not a homework problem per say but I am just struggling with some concepts related to my homework. I am struggling with the concepts of flux linkage and flux cutting. Are these one and the same thing? The following statement in my book is where my confusion has come from:

    "The case of an isolated straight wire moving in a magnetic field can be treated as a special case of Faraday’s law (i.e. rate of change of flux linkage), rather than in terms of ‘cutting’ field lines, since the latter approach obscures the fact that there is really only one law underlying electromagnetic induction, irrespective of how it is brought about".

    Please could someone explain what they are trying to say here? It seems to imply that flux cutting is not really the true cause of induction?


    2. Relevant equations

    Emf (induced) = d (BAN) / dt


    3. The attempt at a solution
     
  2. jcsd
  3. Jul 10, 2014 #2

    marcusl

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    Here's what I think it means. Moving a wire loop through a non-uniform but static field (the wire "cuts through" field lines) is one way for the flux inside the loop to change, but there are others: the loop can be stationary and the field strength can change with time (by changing the current through a stationary electromagnet, for instance). Faraday's law relates the EMF to the time rate of change of the enclosed flux, as you have written, regardless of whether or not that flux change is caused by physical motion.
     
  4. Jul 10, 2014 #3
    Thanks for your answer. What happens with a single wire (not part of loop). In this situation there is no current induced as there is no complete circuit but there is an emf induced. My book seems to explain this in terms of the lorentz force on the free electrons as the wire is physically moved. How does this situation satisfy a changing magnetic flux as outline in Faraday's Law? There is no surface available for the flux to go through?
     
  5. Jul 11, 2014 #4

    marcusl

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    Ah, thanks for clarifying your question. The emf along a wire segment moving through a field is treated by the Lorentz force, as you have stated. On the face of it, the Lorentz force is independent of Faraday's Law--according to threads like this one
    https://www.physicsforums.com/showthread.php?t=322126
    the Lorentz force is not directly derivable from Maxwell's equations, which contain Faraday's Law, without the application of additonal independent constraints. Thus I'd say that the quotation from your book is somewhat misleading.
     
  6. Jul 11, 2014 #5

    Thanks for your answer, that has really helped. Could you please clarify - is my book trying to imply that a straight wire can be explained by flux linkage (which it can't as you have just shown it can only be explained by the Lorentz force)? Or am I misunderstanding the quote from my book?
     
  7. Jul 11, 2014 #6

    marcusl

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    Yes, that's how I read it. What book is this?
     
  8. Jul 11, 2014 #7
    Yeh me too. Its really confused me as it makes no sense to talk about flux linkage in the case of an isolated straight wire. I'm studying a level physics and it's the textbook for our course which is OCR advancing physics B (A2) which was apparently written by the institute of physics! I found a web link which contains the same information as the section in my textbook:

    http://tap.iop.org/fields/electromagnetism/414/file_46954.doc

    **It's the second paragraph from the bottom
     
  9. Jul 11, 2014 #8

    marcusl

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    Physics by committee--perhaps that explains it.
     
  10. Jul 11, 2014 #9

    rude man

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    An isolated bar moving in a B field does cut flux lines, and the emf so generated is given by the Blv law. l is length of bar, v is velocity of bar perpendicular to bar and B field.

    The complete equation of one of Maxwell's laws is

    ∇ x E =- ∂B/∂t + ∇ x (v x B). The Blv law derives from the second term on the RHS.

    The idea of cutting flux lines is crucial. Suppose you're sitting on the bar as it passes thru a uniform B field. You're not aware of the velocity since you have no outside reference. You do not see a ∂B/∂t either, just a constant, uniform field. The only datum you have is the emf generated in the bar. You conclude that you are cutting flux lines even though you don't see a v nor a change in B.
     
  11. Jul 11, 2014 #10

    Thanks for your answer. I thought that flux cutting and the rate of change of flux are the same? For example, if you move a magnet closer into a loop of wire then after 't' seconds there will be more flux lines inside the loop resulting in a rate of change of flux. You can also think of the flux lines (previously outside the wire) cutting across the wire as they enter. Therefore concluding these two ideas are the same just a different way of looking at it.

    But, if you imagine a loop in your example then the rate of change of flux through the loop (if the field is uniform) is zero. Flux would be cut but the number of flux lines wouldn't change with time so no emf. Therefore, how can an isolated wire give a different result using the same theory?
     
  12. Jul 11, 2014 #11

    marcusl

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    This equation does not give an EMF until a surface integral is taken, whereupon the LHS becomes a line integral that gives the Faraday's law in the usual form. The integral implies a closed loop, which is not exist for this problem, so this expression doesn't do the job.

    Looked at another way, you've generated an E field by moving the wire through a field, but you don't get the EMF until electrons in the wire respond by moving towards the end of the wire. That requires the Lorentz force to occur, which is what you are supposed to be deriving. Maxwell's equations don't give the Lorentz force, and you've shown that you need to apply it as a separate condition to generate the EMF in this wire.
     
  13. Jul 11, 2014 #12
    Please could someone kindly answer a final question. As I said in my previous thread rate of change of flux and flux cutting are supposedly synonymous. Can you explain the experiment I have attached in terms of a rate of change of flux? As far as I can tell when you pass the wire through the magnets you are only cutting flux lines and there seems to be no surface involved? Unless you treat the surface of the loop as your surface but then that doesn't seem to be experiencing any flux - at least you could make it that way. Unless I am picturing it wrong.....

    For example, you could orientate the loop so that it is parallel to the B field of the magnets and move the whole loop down together, thus keep the loop parallel to the field (hence no flux) but you would still be 'cutting' flux wouldn't you? hence you would still get an emf? So I don't get how this obeys Faraday's law which specifically states the rate of change of FLUX not flux cutting and to talk about flux without a surface makes no sense.
     

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    Last edited: Jul 11, 2014
  14. Jul 11, 2014 #13

    rude man

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    Oh, but it does:

    A surface integral is not taken if there is moving media.
    Look again at ∇ x E =- ∂B/∂t + ∇ x (v x B).

    In the case of an isolated bar moving in a uniform B field we use the right-hand term only. Obvioulsly, we can then say that
    E = v x B.

    So we integrate E along the bar from one end to the other and get
    E*dl = emf = (v x B)*l. Which is the Blv law.
     
  15. Jul 11, 2014 #14
    I have just been reading around various websites. Instead of using the Lorentz law for an isolated wire is it also satisfactory to alternatively treat an isolated wire as sweeping out an area and it is this area that is swept out that experiences a rate of change of flux as per this website:

    http://www.s-cool.co.uk/a-level/phy...x-flux-linkage-and-how-do-you-induce-an-elect
     
  16. Jul 11, 2014 #15

    rude man

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    Yes, but there are other circumstances ...

    See my post #8 attachment to the PF forum thread "Motion in a magnetic field and relativity". In that case the wire loop sees no dø/dt at all, yet an emf is generated by flux cutting.

    If you think of it, the area swept out per second is just dA/dt = lv and dø/dt, flux being area times B, is B dA/dt = Blv.

    BTW flux cutting is not synonymous with rate of change of flux. Think of a uniform but time-varying B field with a single-turn coil with its normal parallel to B, then there is no flux cutting at all yet emf = -dø/dt per Faraday.
     
  17. Jul 12, 2014 #16
    So if they are not synonymous then wouldn't that mean that flux cutting is not supported by Faraday's law? This would imply that induced emf's are not all explained by Faraday's law?
     
  18. Jul 12, 2014 #17

    rude man

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    Did you look at my attachment (from a previous post; PF won't allow multiple attachments of the same file)?

    The answer is yes. Faraday's law cannot explain the generated emf around the rectangular loop I describe in that attachment since dø/dt is zero inside the loop. For moving media the more general expression for ∇ x E I gave previously is required.

    In the conducting loop I describe in the attachment the emf generated around that loop is not derivable from dø/dt. For that loop, emf = Blv which is cutting flux. If you are persistent you can derive the Blv law by emf = ∫∫(∇ x E)*dA = ∫∫[∇x (v x B)]*dA which is a fun exercise for this case.
     
  19. Jul 12, 2014 #18
    I think I get the general point however some of the maths is beyond me! Could you kindly look at my post with the attached picture as this is bothering me. We were shown this as an example of Faraday's Law but I don't see how. Again, there is no surface involved. Even if you treat the surface of the loop as your surface you could still orientate it parallel to the field as you are cutting the wire through the flux hence no flux through the surface. How should the emf induced in this experiment be explained?
     
  20. Jul 12, 2014 #19

    rude man

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    There is a surface involved, the surface defined by your overall loop.

    Concentrate on the segment of your loop that runs across the magnets. If you start with that segment above the magnets and move the segment down as indicated on your image, and the normal to the loop is aligned with the B field, the induced emf can be explained either by flux cutting which is Blv, l being the length of the segment. Or, looking at the complete loop, you can see that initially there is no flux across the loop but as you descend into the B field there is d(phi)/dt inside the loop, which will also give you the emf. In either case all the emf is generated across the segment, although Faraday doesn't tell you that whereas Blv does.

    In the first case you are an observer seeing the segment move down, in the second you are sitting on the segment and see the rate of change of flux in the loop. In the first case there is no dB/dt. It's a bit complicated to be sure. But maybe you can see somehow that the two terms in the expression for del x E are needed, depending on where the observer sits.

    Now, if you reorient the loop so the normal is in the direction of v, then you have the situatiin very much like my attachment. As you say, there is no d(phi)/dt in that case. A small section of the area is continuously covered by B, the rest is not. And, B is not aligned with the normal as Faraday requires. So Farady falls flat, but the Blv law comes to the rescue as then you are cutting flux lines.

    What's interesting is if you now are sitting on the segment. You see no v and you see no d(phi)/dt but you would still measure the emf on your voltmeter. You would conclude that you're cutting flux lines which is the only explanation of the voltage.

    BTW when you speak of orientation of the loop, I suggest always specifying the direction of the normal. Saying a loop is parallel or perpendicular to a field is ambiguous. The direction of an area (or part of an area if it's not flat) is always the normal to that area.
     
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