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Two wires with current - would this be correct?

  1. Feb 23, 2013 #1
    While chatting with a friend, I came up with something to think about, being, "if two parallel wires, with constant and equal current wrt position, are put next to each other, the charges in the wires can exert normal forces on the wires, and they're left at rest wrt one another to interact freely, what will happen from the point of view of the initial rest frame of the wires?"

    After a bit of RHR'ing, it turns out each wire feels a force away from the other. My friend and I having a dispute over if the current decreases.

    (EDIT: A bit more RHR tells me they attract, not repel.)

    I'm stating, well, two things. The first feels somewhat deprecated due to the fact that a particle with velocity u+v doesn't have kinetic energy equal to (the kinetic energy of a particle with velocity u + the kinetic energy of a particle with velocity v,) but hopefully the underlying point is still valid.

    Argument 1: As the wires accelerate away from each other, they gain kinetic energy. This obviously has to come from somewhere, and the logical place for it to come from is the motion of the charges along the wire. Thus current decreases.

    Argument 2: Using something like Euler's Method, after a short amount of time, the charges will have some of their velocity directed away from the other wire, and thus, assuming all components are real, less of their velocity will be directed along the wire. (Magnetic fields don't really do any work on particles.) Thus, as current is related to the component of velocity along the wire, current decreases.
    Last edited: Feb 23, 2013
  2. jcsd
  3. Feb 23, 2013 #2
    So, basically, my argument is that I don't see, physically, how the current should decrease just because of the given arguments. The fields, yes, decrease, but saying the current decreases is like saying "if you have two positive test charges" and you pull them away, the magnitudes of the charges decrease".

    Hopefully, this will help people to see what I'm having a problem with.
  4. Feb 25, 2013 #3
    With all due respect (which is quite a bit,) there isn't a potential field associated with magnetism, so you must be thinking of another field.
  5. Feb 26, 2013 #4
    Yes there is, it's called the magnetic field. The magnetomotive force is F = ni ampere turns.

    The magnetic field and its dynamics are directly related to current. Thus, if I assume the field decreases then there should be a corresponding decrease of current.


    Elementary lectures on Electric Discharges, Waves, and Impulses and other Transients by Charles Proteus Steinmetz

    Here is a nice table from the book:

  6. Feb 26, 2013 #5


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    If two wires attract each other and they actually move together, you have an electric motor. (Simple and only for a brief time, I know). Work is done in increasing the kinetic energy of the wires. This will involve a PD acting along the length of the wires, a 'back-emf' (Lenz's Law) which will decrease the current in the wires - as with all motors.
  7. Feb 26, 2013 #6
    Okay, now I'm confused. For a point charge, isn't magnetic force always perpendicular to velocity, thus perpendicular to a particle's path, and thus it doesn't do any work, so magnetic fields shouldn't have a scalar potential field associated with them (or, at least, one which varies in space?)

  8. Feb 27, 2013 #7
    Magnetic fields can and do perform work on magnetic dipoles because the Lorentz magnetic force acts in the correct direction for work, along the velocity of the moving dipole. Mag force (Lorentz) does NO work on charge carriers since the orientation is normal to the velocity.

    Menaus gave you the correct answer, there is a scalar potential associated with a mag field, called NI, or "magnetomotive force", mmf, measured in amp-turns. This mmf or potential is used when analyzing magnetic circuits such as transformers, motors, generators, solenoids, relays, etc. Any good energy conversion text should explain the development of the scalar magnetic potential aka "mmf".

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