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Faraday's Law of Induction

  1. Nov 30, 2013 #1
    Hi, please could someone help explain how magnetic flux works in Faraday's law as I struggle with electromagnetism.

    From what I understand, if you have a loop of wire in a magnetic field then you get an induced current if the flux is both changing and perpendicular to the plane of the loop. What I don't understand is why they fix an imaginary surface to the loop. The area inside the loop is just empty space so how has that got anything to do with the induction process? The wire just needs to 'see' a changing magnetic flux so the area inside the loop seems to serve no role as far as I can see, what am I missing? Also, what area do you assign for the flux when the magnetic field is outside the loop? Thank you in advance for any help offered!
  2. jcsd
  3. Nov 30, 2013 #2


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    no, it's not empty space, it contains an electric field with a non-zero curl

    faraday's law (one of Maxwell's laws) has a microscopic form
    curlE = -∂B/∂t​
    and a macroscopic form
    C E.dl = - ∂/∂t ∫S B.dA

    the first equation applies in what you call the empty space

    the second equation applies (via stokes' theorem) to any surface S with the same boundary C (ie not just to the flat planar surface) :wink:
    but there's no flux :confused:
  4. Nov 30, 2013 #3


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    The contrary is the case! The local law is generally valid, i.e.,
    [tex]\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}[/tex]
    (in SI units).
    Using Stokes's Law in integrating this equation over an arbitrary surface then yields
    [tex]\int_{\partial F} \mathrm{d} \vec{x} \cdot \vec{E}=-\int_F \mathrm{d}^2 \vec{F} \cdot \partial_t \vec{B}.[/tex]
    This is, of course still generally valid.

    tiny-tim's formula, where the time derivative is taken out of the surface integrals, however, holds if and only if the surface is time-independent. A generally valid formula takes into account a possible time dependence of the surface. Let [itex]\vec{v}(t,\vec{x})[/itex] the velocity of the point [itex]\vec{x}[/itex] on the surface at time [itex]t[/itex]. Then the general formula reads
    [tex]\int_{\partial F} \mathrm{d} \vec{x} \cdot (\vec{E}+\vec{v} \times \vec{E})=-\frac{\mathrm{d}}{\mathrm{d}t} \int_{F} \mathrm{d}^2 \vec{F} \cdot \vec{B}.[/tex]
    There is a lot of confusion in the understanding of Faraday's Law, because this mathematical identities are often not clearly stated. For more details, see the very good article in Wikipedia:

  5. Nov 30, 2013 #4

    Thank you for your answer. When you do that experiment where you move a wire through a magnetic field and 'cut' the flux lines then you are only concerned with the wire cutting the magnetic field lines. However, it seems that when one is dealing with a loop of wire and a changing magnetic field inside the loop, it seems to be concerned with field lines inside the loop where there is no flux cutting of a wire taking place since there is no wire there. Would you not only be concerned with the field lines that are 'seen' by the wire as appose to those inside the space within the loop?
  6. Dec 1, 2013 #5


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    you mean a wire with two ends, ie not a loop?

    but then there's no current … you need a loop (a circuit) for current to flow

    in terms of current, you can check the potential difference between the ends of the wire by connecting them to a voltmeter outside the magnetic field

    that makes a loop whose intersection with the magnetic field is getting larger and larger …

    it is the fields lines cutting that loop which we count, not the field lines cut by the wire
    for current, it is always the field lines that are 'seen' by the loop :wink:
  7. Dec 1, 2013 #6

    Thank you tiny-tim that makes a lot more sense. In terms of flux cutting in a loop, how do you explain the experiment where you swipe a wire in between two bar magnets to generate a current? Most websites seems to imply that it is the number of field lines the wire cuts through but as you pointed out it should be the number of field lines cutting the loop shouldn't it?
  8. Dec 1, 2013 #7


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    Yes … "the number of field lines the wire cuts through" is a short-cut that only works (to find the current) if the magnetic field lines are unchanging. :smile:
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