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Query about induced voltage and open-circuit current

  1. Jun 14, 2015 #1
    When we move a wire through a magnetic field appears a magnetic force that distributes loads at one end of the wire ( following equation F = q⋅(vxB) ), until an opposite force is generated on the wire by the electric field (E) caused by charge distribution.

    200px-F2_GIA_induccion_barra_abierto.png
    This causes the EMF or induced voltage. When we close the circuit the current flows thanks to the electric field…


    but what happens when the circuit is open? in real generators the wires are coils in which a sinusoidal emf is induced, which means that charges flow sinusoidally from one end to another of the wire without load. If the intensity is dQ / dt through a section of the wire, why no intensity? has to do with the magnetic field does no work nor affect the kinetic energy of the electrons ? This shift in load distribution generates power losses?
     
  2. jcsd
  3. Jun 14, 2015 #2
    Why is there no load? Although one section of wire may not have a load, somewhere down the wire there is a load (at least in a closed circuit). That load causes a backup of charge carriers, like a traffic jam. The backup goes all the way back to the section in the magnetic field which provides the motive force for the charge carriers.

    BTW, the kinetic energy of the charge carriers is small. They typically move on the order of a few meters/second and have almost no mass. The energy is in the electric and magnetic fields (which "move" at the speed of light).
     
  4. Jun 14, 2015 #3

    davenn

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    because he is asking about a single bit of wire, with open ends
     
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