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Faraday's Law with square wire

  1. Aug 8, 2011 #1
    1. The problem statement, all variables and given/known data
    I have done a few problems with Faraday's law of induction ( closed line integral of E dot dl = -(d/dt)(magnetic flux)). Every time time I have done them, the , albeit simple, problems have used circular wires through which magnetic flux is passing through, and emf is induced into. I got to wondering why it would be more difficult to solve if the wire were a square. I know how to take line integrals and surface integrals (for the flux) of squares, but I know something would still not be right. I have 2 questions.

    1. Is the induced electric field rotational, so that for a square wire, E and dl are not parallel as they are for a circular one? This would be the reason the math would get more complicated and my book stuck to circular wires.

    2. Solving for E with a circular wire (or circular path of integration) can be interpreted as the electric field strength at that distance from the center of the circle. What would E represent for a square path of integration, seeing as not all the points are equidistant from the center.

    I am in calc-based second semester physics, and have had mathematics up to multivariable calc.

    Thank you
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 8, 2011 #2

    xts

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    That is a matter of symmetry. For all problems with circular wires you have translational symmetry along the wire and rotational symmetry for rotation by wire axis. So if you look at the problem in cylindrical coordinates, your solution depends only on distance from wire (r) but is independent of z and phi.

    For square wire you have no such symmetry and you must solve two-dimensional equations.
     
  4. Aug 11, 2011 #3
    Ok, so then instead of E being a function of R alone, it would be a function of XY (or some other two coordinates). Also, E would not be parallel to dl so the dot product would be more difficult?
     
  5. Aug 11, 2011 #4

    xts

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    More or less as you said ;)
    You may always choose such loop that dl along that loop will be parallel to E, but such loop won't be a nice circle. If your integration loop is circular - then E and dl are not parallel.

    PS. Your posts would be much better readable if you write your equations in [itex]\TeX[/itex] notation.
     
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