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Faraday's Law of Induction; Why is it the *enclosed* magnetic flux?

  1. May 22, 2013 #1
    Hello Reader,

    Basically in Faraday's law, a change of flux over time induces an EMF. I was wondering why is it the enclosed magnetic flux and NOT the magnetic flux through the conductor that we consider.

    More specifically, I understand that the flux through a rectangular surface would be B.A. However when looking at a rectangular coil under the under the influence of a uniform magnetic field, despite the fact that the rectangular coil is 'empty' in the middle, we still consider the entire area of the rectangular when calculating the flux. Intuitively, I would only consider the area the flux lines are touching, which is this case is the area of the wire. Below is just a visual aid of the image in my head of the rectangular coil.

    | 'Emptiness' |_____
    | 'Emptiness' ______ Rest of circuit
    | 'Emptiness' |

    I just cannot understand how the field lines within the area of the rectangular would interact with the wire. I do understand though that the field lines touching the coil would influence the wire. Also, since the field lines in accordance to faraday's law inside the rectangular coil cause an EMF, what about if there was field lines outside the coil area, shouldn't that influence the wire as well?

    Perhaps this might help (or maybe not), the equation F = Idl x B behaves as I would intuitively expect; we only consider the field lines that intersect/go through the conductor/wire. Although, if the wire was very thick, then shouldn't we also consider its area so F = IdA x B or along those lines? I was attempting to understand why F = Idl x B uses only the intersected field lines while Faraday's law doesn't.

    Any assistance would be greatly appreciated =)
  2. jcsd
  3. May 22, 2013 #2
    Connect this with Stokes' theorem from your math class. The line integral of a vector around a boundary can be found by integrating its curl over the whole enclosed surface area. Your question becomes "how can the curl of E at the center of the the loop have any influence on the line integral of E just around the boundary?". In other words its not a physics thing but a math thing.
  4. May 22, 2013 #3
    Thanks for the very speedy reply =).

    I see. Unfortunately, we have yet to cover Stoke's Theorem and curls until next year (which is the 2nd year in undergraduate electrical engineering).

    Your reply links in with explanation on this forum (which hopefully will make more sense by next year): https://www.physicsforums.com/showthread.php?t=663673&highlight=faraday+enclosed+flux

    I got a bit doubtful that my question was not related to Stoke's Theorem and curls; thanks for clearing that up =)
  5. May 22, 2013 #4

    jim hardy

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    Science Advisor
    Gold Member

    For me, it was realizing that the field is a continuous, fluid thing not discrete lines. Else induction would not be smooth(continuous?) but discrete bumps.
    After that I abandoned "Flux Cutting" and became a "Flux Linker Thinker", though the two are equivalent.

    Check out episode 414 here:


    old jim
  6. May 26, 2013 #5
    My personal view - that the changing flux - is cutting through the conductor ( Eg changing flux - OR the conductor moving through the M field) - is inducing the EMF. Agreeing with JH that thinking of flux as "lines" could be misleading - but that had not really been a hang up to me.

    As for the enclosed area - consider a loop of wire out and back - with effectively NO space between them - as the EMF is created in one wire - it is also created in the other wire - since this is a Loop - to the length of wire these two induced EMF cancel out.

    If the loop is large ( infinitely large) as in a straight piece of wire, but looped far enough away that the flux does not touch the "return wire" -- all of the flux cutting across this wire induce the EMF - none is cancelled by the return wire.

    So the Area of the loop- and change of total flux inside the "loop" creates the NET emf on total loop.
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