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Time dependent magnetic field

  1. Dec 7, 2014 #1
    1. The problem statement, all variables and given/known data
    I have a problem with the next situation. I have a magnetic dipole moving at constant speed on the z axis. On the plane z=0 I have a circular wire with a resistance R and radius a. I have to calculate the electromotive force on the wire as a function of the speed of the dipole, v, the position of the dipole ,z, and the dipolar moment m.

    2. Relevant equations
    ε=-dΦ/dt, where ε is the electromotive force and Φ is the magnetic flux.
    B=(μ0/(4π))*(3r(r*m)/r5-m/r3)
    Φ=∫Bds
    ds
    =( cosφ, sinφ, 0)
    r = (a cos φ, a sinφ, z)

    3. The attempt at a solution
    Once I calculate a huge expression for the magnetostatic field, B, and I integrate to find the flux, my problem comes; I have to make a time derivate, and the only time dependent factor here is z; but Φ(r(z(t))) do I have to derivate (∂Φ/∂z)(dz/dt)=(∂Φ/∂z)v or do I have to do something else?
     
  2. jcsd
  3. Dec 7, 2014 #2

    BiGyElLoWhAt

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    Why are you trying to calculate the magnetostatic field? Shouldn't the field be dynamic? Otherwise there would be no emf.

    I don't see a way around integrating spatially and then differentiating temporally, if that's what you're asking. Also, you're using cartesian coordinates? I think your life would be much easier if you worked in cylindrical. Just my opinion though.
     
  4. Dec 8, 2014 #3
    Well I Integrate in cartesian because I feel more comfortable doing operating with them; what is more, when I make the scalar product BdS I have an scalar multipied by sin2+cos2, so the dependance with the angle dissapears. My main concern here is not about the coordinates, but how to determine the electromagnetic foerce, -d/dt(∫BdS)
     
  5. Dec 8, 2014 #4

    BiGyElLoWhAt

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    One of the variables in B is a function of time.
     
  6. Dec 8, 2014 #5
    yes, B(r(z(t))), there is whem my doubt comes, do I simply derivate ∂B/∂z and them muptiply by dz/dt=v?
     
    Last edited: Dec 8, 2014
  7. Dec 8, 2014 #6

    BiGyElLoWhAt

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    Does r(t) equal z(t) for all points on the surface?
     
  8. Dec 8, 2014 #7
    No, r=Sqrt(z^2+a^2), where a is the radius of the wire
     
  9. Dec 8, 2014 #8

    BiGyElLoWhAt

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    Ok, so do you know how to chainrule derivatives? You either have to chainrule twice or substitute, frankly.
     
  10. Dec 9, 2014 #9
    That's it, I only wanted to know if I had to apply the chain rule twice.
    Thanks.
     
  11. Dec 9, 2014 #10

    BiGyElLoWhAt

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    No problem
     
  12. Dec 9, 2014 #11
    I realize you already solved it but I strongly suggest that you look up a youtube video on cylindrical/spherical coordinates. It makes E&M so much easier and if you ignore the derivation there are only two very easy formulas to remember.

    Just my two cents. It will be well worth a 30 min or-so investment of your time.
     
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