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Loop falling in a magnetic field

  1. Aug 31, 2014 #1

    ShayanJ

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    There is a conducting circular loop with resistance R falling in the magnetic field [itex] \vec{B}=B_\rho(\rho,z)\hat\rho+B_z(\rho,z)\hat z [/itex] and gravitational field [itex]\vec g=-g\hat z[/itex]. How does z and the current in the loop change in time?(assume the loop remains horizontal!)
    The flux through the loop is [itex] \int_0^a\int_0^{2\pi} B_z \rho d\varphi d\rho[/itex], Its time derivative is [itex] \int_0^a \int_0^{2\pi} \frac{\partial B_z}{\partial z} \frac{dz}{dt} \rho d\varphi d\rho [/itex] and so the induced current is [itex] I=\frac{1}{R} \int_0^a \int_0^{2\pi} \frac{\partial B_z}{\partial z} \frac{dz}{dt} \rho d\varphi d\rho [/itex]. Now we can write the z component of the magnetic force as [itex]-\frac{2\pi a B_\rho}{R} \int_0^a \int_0^{2\pi} \frac{\partial B_z}{\partial z} \frac{dz}{dt} \rho d\varphi d\rho [/itex]. So we have:
    [itex]

    \ddot z=-g-\frac{dz}{dt}\frac{4\pi^2 a B_\rho}{m R} \int_0^a \frac{\partial B_z}{\partial z} \rho d\rho

    [/itex]
    Which gives us z as a function of time and then I can be calculated easily.

    1- Is everything OK?
    2-Any hints or suggestions or further explanations?
    3-How does this change if the loop is superconducting?

    Thanks
     
  2. jcsd
  3. Sep 3, 2014 #2

    ShayanJ

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    Nothing?
     
  4. Sep 3, 2014 #3

    vanhees71

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    Looks good to me. I've not checked the details.
     
  5. Sep 3, 2014 #4

    ShayanJ

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    My main question is the third. What should I do if the loop is superconducting?
     
  6. Sep 10, 2014 #5

    ShayanJ

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    Is it right to say that the superconducting loop excludes the magnetic field and so there will be no induced current and so no magnetic force is applied to it?
     
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