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Confusion with nature of induced electric field.

  1. Oct 9, 2009 #1
    The whole idea of an electric field induced due to a changing magnetic flux seems to run wildly counterintuitive to me. Consider this set up as shown in my physics textbook. A current carrying solenoid placed along the axis of a circular conducting loop. A changing current through the turns of the solenoid causes a changing magnetic flux linked with the coil , which induces a non-conservative electric field , whose direction is tangential to the coil everywhere, and whose magnitude is constant. So far so good. Now , symmetry arguments would have me believe that the potential energy of the electron as it moves around the coil remains unchanging(as it must be of the same value at any point on the loop, due to the symmetry of the loop about the solenoid).But the coil gets heated up due to the current. Where does the heat energy come from if the electrons dont lose any potential energy? More importantly , across what segment of the coil is the emf induced, i.e what portion of the coil effectively acts as the battery? In every other example of current in circuits , the electrons at some point along the circuit move from a lower potential region to a higher potential region due to the work done by the emf maintaining mechanism (like across a battery from the positive end to the negative end or across the moving conductor in the case of motional emf ) , but apparently that does not happen here. Why? I know i am missing something crucial here , but would somebody please elaborate.
  2. jcsd
  3. Oct 9, 2009 #2
    The heat is due to the coil resistance, I^2*R. The power source feeding the solenoid is providing this energy/power. When you say "potential energy of the electron", what is the point of reference? If I raise a ball 3 feet off the ground, the ball's PE is referenced to the earth. I don't understand what you mean by "potential energy of the electron".

    The emf appears around the entire loop, but the E field is distributed inside the conductor, in accordance with Ohm's law J = sigma*E. For a superconducting loop, E is zero inside the cross section of the wire. For a non-superconducting loop, like copper, there is a small E field inside the copper wire, again per J = sigma*E.

    The emf, or potential, is simply the work done transporting a charge around the loop once, divided by that charge. As far as which section of the loop is the "battery", I prefer to think in terms of a generator, where the entire loop is the said generator.

    Energy in the form of magnetic flux links the solenoid with the loop and is transferred. The entire system, solenoid and loop, form a generator, or for the non-moving case, a "transformer".

    Did this help?

  4. Oct 9, 2009 #3

    Yes , well it appears that i do not have a reference point. There is no charge build up in this case to provide the potential difference across the circuit. So what you are saying is that the 'potential energy of the electron ' is a meaningless expression here? But then if potential energy cannot be defined then potential cannot be defined, and that would mean that there is no 'potential difference' to speak of. Then what ultimately drives the current.
  5. Oct 9, 2009 #4
    The potential is defined to satisfy [tex]-\nabla \Phi = E[/tex], which is valid for a conservative field. This implies that [tex]\nabla \times E = 0[/tex], since the curl of a gradient is always zero. If the curl of the E field is not zero, as when you have a changing magnetic field, then you can't define the potential as a function of position (you can define something like it using the vector potential, but it does not end up being related to the energy).
  6. Oct 10, 2009 #5
    You don't necessarily need a potential difference to drive electrons through a wire.
    Electrons are always driven by an electric field.
    That field could be generated by a potential difference, i.e. a build up charge. But it could also be generated by a changing magnetic field. In this case there is a circular electric field which pushes at every single electron in the circuit.
    Those electrons will interact with the copper atoms. That means they will give some of their energy to a copper atom and slow down. Then they get accelerated again by the electric field until they "bump" into another atom and are slowed down again.
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