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Faraday's Law of EMI

  1. Aug 13, 2008 #1
    Using Faraday's law for a closed loop we say that the EMI is -d(phi)/dt .Now this emf is induced in the closed loop...So between which two points is the PD equal to -d(phi)/dt

  2. jcsd
  3. Aug 13, 2008 #2


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    Hi anantchowdhary! :smile:

    Imagine a circuit with a light bulb and nothing else.

    Now cut out a bit of wire between A and B and replace it with a 12V battery.

    Obviously, the PD between A and B is 12V, and it will light the bulb.

    Now start again, and cut 12 bits out of the wire, and replace each of them with a 1V battery (all the same way round!).

    Obviously, the PD between A and B is still 12V. But it's also 12V from one side to the other of any of the cuts.

    Similarly, if you make 12,000,000 cuts each with 0.000001V

    The closed loop in your question is like "infinitely" many very small batteries all the way round the loop … the PD is between any two "infinitely close" points.

    I expect you got that far yourself, and then wondered how do we know which way round the voltage is?

    Well, the practical answer is by using Lenz's law.

    The theoretical answer is that electric PD doesn't work like that:

    Electric PD isn't "conservative", like gravitational PD.

    The gravitational PD between two points doesn't depend on the path taken.

    The electric PD between two points does depend on the path taken.

    So knowing the actual electric PD isn't enough … you must specify the path also! :smile:
  4. Aug 13, 2008 #3
    umm....i appreciate your help...but i don't really understand what you are trying to say.
    if the EMF induced as calculated is between any two 'infinitely close' points ,then i dont get how this answers my question.
    And isnt electric PD conservative....if u see it using the integral(f.dr)...it is..

  5. Aug 13, 2008 #4
    The way I look a it is that the potential *around the closed loop* is 12V or whatever. In order to transport one coulomb of charge around the loop once along the path of the loop, you must do 12 joules of work as 1 volt = 1 joule/coulomb. Twice around the loop requires 24 coulombs. Thus the potential around the loop twice is 2 times 12 or 24 volts.

    The potential associated with time varying E fields is non-conservative. The potential depends on the path taken. However, with discrete charged particles, the E field is conservative and the potential is not path dependent. In the emi case, the E field has "curl", aka "rotation" or "circulation". This makes the potential path-dependent. In the discrete charged particle case, the E field has no curl.

    Have I helped or made matters worse? BR.

  6. Aug 13, 2008 #5
    i just dont understand the points betwwen which the emf is calculated...
  7. Aug 13, 2008 #6


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    It's calculated between any point and itself, once around the loop! :smile:

    Remember, in E = -d(phi)/dt, phi is the flux through the entire loop, so you wouldn't expect E to depend on any particular part of the loop, would you? :wink:

    Does http://members.lycos.nl/AmazingArt/E/artist7.html" help?
    Last edited by a moderator: Apr 23, 2017
  8. Aug 13, 2008 #7
  9. Aug 14, 2008 #8


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    Last edited by a moderator: Apr 23, 2017
  10. Aug 14, 2008 #9
    um....but while solving problems in case of say a fixed square semi-loop(duno what to call it) and a rod that is free to slide...we take the EMF between the two end points of the rod...now isnt this contradictory to your statement?:O
  11. Aug 14, 2008 #10


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    Hi anantchowdhary! :smile:

    I'm not certain what apparatus you're referring to, but I think the answer is that the rod has a very low resistance (R), so the voltage drop (potential drop) across it (IR) is very small, so measuring the EMF between the two end points of the rod almost gives the right figure, in practice. :smile:
  12. Aug 14, 2008 #11
    If we add the Potential drop across two 'infinitesmally' close points,then shoudlnt the PD across the two ends be HUGE?this is as the EMF across those close points is equal to d(phI)/dt

    sorry..i know this is a bit irritating
  13. Aug 14, 2008 #12


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    Hi, I think you're referring to something like qn 8 of this problem set:

    The d(phi)/dt refers to the rate at which magnetic flux lines are being cut by the moving rod. In that case you only have to work out the rate at which the magnetic flux lines are being cut when you're given the speed of the rod. I don't see why the EMF across 2 very close points should be very huge, if anything it should be really small since the area it covers when the infinitesimal segment ds of the rod slides across the field is also small.
  14. Aug 14, 2008 #13
    Not if for each of those infinitesimally close points, there is an infinitesimally small potential drop. One of Maxwell's equations for electromagnetism describes this phenomenon:
    If you notice, it's a path integral.
  15. Aug 15, 2008 #14
    thanks a lot everyone
  16. Aug 15, 2008 #15
    :confused: Well...i have a new doubt

    Please see this (the figure for the problem)

    Suppose that the rod ab moves with a constant speed v.Now there is a uniform magnetic field B..in the loop..

    So if we want to know the PD between a and b,as tiny_tim said
    we can think of a battery between two infinitesimally close points...the current in the circuit will be Blv/R(where R is the resistance of ab).

    Now with two different circuits we get different answers why?i mean...we assumed the emf to be induced between any two infinitesimally close points..
  17. Aug 15, 2008 #16
    This i assume is only for time varying fields...but isnt PD calculated at an instant...by moving a hypothetical test charge....otherwise we cant solve problems by applying ohm's law

    i suppose it is calculated over time also..am i correct?
    Last edited: Aug 15, 2008
  18. Aug 15, 2008 #17
    The problem with trying to compute the current is as follows. If the system is open circuited and the voltage is measured, Voc, then the circuit is terminated in a resistance R. The current is not necessarily Voc/R. When current is present its own magnetic field opposes the original magnetic field. The potential once around the loop changes as R decreases due to the bucking action of the additional magnetic field (law of Lenz) so that "d@/dt" gets reduced, hence V is reduced. If the R value is so high so that the induced current is small enough so that its magnetic field is tiny compared with the original mag field, the V is roughly Voc, and I is roughly Voc/R. However if R is reduced eventually the induced current's mag field is large enough to reduce the net flux so as to reduce the voltage. Does this help?

  19. Aug 15, 2008 #18
    im getting some of this...Could you please elaborate upon how electric field is non conservative here...and is potential difference taken as the integral including time or not....acc to me it is not over time...

  20. Aug 15, 2008 #19
    Lets say you had a circular loop of wire with a magnetic field running through the loop. The magnitude of the magnetic field is constantly changing over time and so an electric field is propagating outward due to the changing magnetic flux. This allows for a voltage to be established throughout the loop along with a current.
    If the electric field here was electrostatic, then the electric field would be dependent on position rather than path. But the electric field we're focusing on here is distributed across the wire and depends on the path the wire provides. Otherwise the closed loop integral provided would be 0 in value:
    http://upload.wikimedia.org/math/7/6...aed959500d.png [Broken]
    Also note that the electric field isn't restricted to the wire...its propagating outward.
    Last edited by a moderator: May 3, 2017
  21. Aug 15, 2008 #20
    how is possible to create such an electric field using charges????
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