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The Nature of Constant EMF

  1. Aug 10, 2010 #1
    Electromotive force is usually defined as the work done in transporting a unit positive charge in a round trip in the circuit. That is,

    EMF= loop integral of E.dL

    Now let us examine the above statements in the light of a constant source of emf driving a steady current in the circuit.

    If the external magnetic field(say the earth's field) and the magnetic field produced by the current are constant(remembering that we are considering a steady current), we have net B=constant and hence,
    curl E= - (del_B)/(del_t) =0

    Curl E=0 indicates that the loop integral E.dl is again zero and hence the emf is zero!This appears to be a contradiction.
     
  2. jcsd
  3. Aug 10, 2010 #2

    Born2bwire

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    But there wasn't any work done. The magnetic field does no work. Since you are in a magnetostatic situation then the movement of a charge due to the magnetic field cannot generate/consume energy. You can of course work this out directly on your own for an easy problem like this. It is just cyclotron motion.
     
  4. Aug 10, 2010 #3
    Circuits are modelled assuming that there is no electromagnetic field outside the wires. Since the magnetic flux area enclosed by the loop is zero, curl E= - (del_B)/(del_t) =0 would hold only if E & B were continuous. This is not the case.
     
  5. Aug 13, 2010 #4
    It is true that a magnetic field does no work (especially if we think of the magnetostatic situation). But let us consider the source of emf which is supposed to do work in pushing the charges against the resistance in the circuit. This work is again of electrostatic nature if we consider the definition of emf--- closed loop integral of E.dL[=(d/dt)integral B.dA=0]

    The source of emf is performing work but mathematically,form the point of thedefinition of emf, it is coming out to be zero!
     
  6. Aug 13, 2010 #5

    Born2bwire

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    There is no resistance. The source here is the magnetic field. The magnetic field produces a Lorentz force on the moving charges in the wire (if there are any). We already know that the magnetic field alone can produce no work and thus the induced EMF is zero. The force is acting normal to the charges' trajectories, so it is not going to affect the way the charges flow through the wire. The charges' trajectories are constricted along the length and surface of the wire and so all this means is that they are going to "bunch" up along one edge of our thin wire.
     
    Last edited: Aug 13, 2010
  7. Aug 16, 2010 #6
    I didn't mean to say that.The source here is,say, a chemical cell of constant emf driving a steady current through a fixed resistance.Now the formula curl E=-del_B/del_t is universally applicable. We choose the current carrying wire as the boundary and consider a surface stretched over it. By applying Stokes law we have,

    loop integral E.dL=-d/dt integral B.dA

    For a steady current the induced magnetic field is zero. Of course the external field due to the earth is also constant.Therefore the right side is zero which indicates the the left hand side of the above relation is also zero.
    Now if we accept the definition of emf as the work done by a unit positive charge in completing a round trip ie,
    Emf= loop integral E.dL
    we have ,
    emf=0
    That is the key aspect of the problem
     
  8. Aug 16, 2010 #7

    Dale

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    I don't get it. Why is that a problem? The chemical cell does work on the resistor and heats it up. The current is steady so no energy is lost through induction, it just all goes to Ohmic heating.
     
  9. Aug 16, 2010 #8
    Now,
    P=Ei
    If E is zero[as I have tried to show in my previous thread] we have P= zero. How do we get the heating?
     
  10. Aug 16, 2010 #9

    Dale

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    By Ohmic heating. This is just normal resistance.

    See:
    http://en.wikipedia.org/wiki/Joule_heating

    If your source loop were a superconductor then there would be no resistance, no heating, and no need for any power input.
     
  11. Aug 16, 2010 #10
    Ohmic heating is given by I^2R. But there should be some source of energy from where the heat is coming. If P=0 the source cannot supply any energy[so the cell is not in a position to provide energy at least from the mathematical point of view,since P=0]

    So again there is a contradiction:
    The source is not in a position to supply energy but the current is running dissipating energy at the rate I^2 R
     
  12. Aug 16, 2010 #11

    Born2bwire

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    Again, the force from the magnetic field is always normal to the trajectory of the charges. Your charges are confined along the surface and path of the wire and thus the magnetic force is always going to be pointing normal to the surface of the wire. This means that the charges are not being moved through the resistive bulk by the magnetic fields since the magnetic force only serves to collect the charges on one side of the wire. So the energy that is being dissipated due to resistance is being supplied by the original current source since that is the source that is providing a force that moves the charges through the resistive bulk.

    If you are not assuming a power supply to maintain the current and are still assuming that they are moving through a resistive wire then the charges will eventually stop moving as the magnetic fields are not providing any energy.
     
  13. Aug 16, 2010 #12

    Dale

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    The battery. I don't know what you are not getting here. The battery supplies the energy to drive the current through the resistance to produce the heat.

    You are just talking about a loop of wire connected directly to a battery, correct? Or are you talking about a primary loop of wire directly connected to a battery where the primary loop of wire is inductively coupled to a secondary loop of wire?
     
  14. Aug 16, 2010 #13
    The best way to tackle this problem is to assume the existence of "non-potential" forces as suggested by Panofsky and Phillips in their well know text "ClassicaL Electricity and Magnetism" [Chapter 7,Steady Currents and their Interaction,Section 7.2,Electromotive Force]

    [Incidentally this problem is a very old one popular with the classical texts]

    We break up the field into a potential part and a non-potential part E+E'
    E is the potential part
    E' is the non-potential part
    Emf=loop integral (E+E')dL=
    = loop integral E.dL + loop integral E'.dL
    = 0 + loop integral E'.dl

    Thus,
    Emf=loop integral E'.dL
    But I am very much confused about the nature of this non-potential force.What could it be in the Electrostatic situation?
     
  15. Aug 16, 2010 #14

    Dale

    Staff: Mentor

    Can you clarify my question:

    You are just talking about a loop of wire connected directly to a battery, correct? Or are you talking about a primary loop of wire directly connected to a battery where the primary loop of wire is inductively coupled to a secondary loop of wire?
     
  16. Aug 16, 2010 #15

    Born2bwire

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    My understanding is that he has a loop of resistive wire connected to a battery that is driving a current. This loop of wire is then placed in a uniform static magnetic field and he is asking why the EMF is zero despite the resistivity of the wire. I feel that this is trivially shown by the fact that the magnetic force does not move the charges through the resistive material, that is done by the battery.

    The alternative could be just a loop of resistive wire in a uniform static magnetic field but that is trivially shown that there are no applied forces due to the fact that there are no moving charges.

    So I'm not sure where the confusion or misunderstanding lies.
     
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