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Homework Help: Vector Potential of a Rotating Mangetic Dipole

  1. Dec 9, 2017 #1
    1. The problem statement, all variables and given/known data
    A rotating magnetic dipole is built by two oscillating magnetic dipole moments, one along the y-axis and one along the x-axis. Find the vector potential at a point: (0, 0, ##z_0##) along the z-axis. Then find the magnetic field at ##z_0## . As the magnetic field is a function of time, determine the induced Electric field from Faraday's laws. The magnetic dipole moment is given as a function of time as (##m_0## is a constant)
    ## \vec m ## = ##m_0 [cos(\omega t) \hat x + sin(\omega t) \hat y]##

    2. Relevant equations

    ##\vec A_{dip} (\vec r) = \frac {\mu_0} {4 \pi} \frac 1 {r^3} \vec m \times \vec r##

    ##\vec B = \vec \nabla \times \vec A##

    Induced electric field, Faraday's law:

    ##\vec E = \frac 1 {4 \pi}## ## \int ## ##[({\frac {\partial \vec B} {\partial t}} {\times \mathcal {\hat r}})## ##\frac 1 {\mathcal {r^2}}]##

    3. The attempt at a solution

    Since I am given the magnetic dipole moment, I attempted to find the vector potential by taking the cross product of ##\vec m## with ##\vec r##. I took ##\vec r## to be: ##{<x \hat x, y \hat y, z \hat z>}## which
    After taking the cross product I got:

    ##\vec m \times \vec r = m_0 z sin(\omega t) \hat x - m_0 z cos(\omega t) \hat y + (m_0 cos(\omega t) y - m_0 sin(\omega t) x) \hat z##

    Since I'm taking the potential at the point (0, 0, ##z_0##) I think I put that into the cross product which would mean my vector potential would be equal to:

    ##\vec A_{dip} (\vec r) = (\frac { \mu_0 m_0 z_0} {4 \pi r^3})(sin(\omega t) \hat x - cos(\omega t) \hat y)##

    I'm not entirely sure that this is right but when I try to take the cross product of ##\vec A_{dip} (\vec r)## to find the magnetic field, I came up with:

    ##\vec \nabla \times \vec A_{dip} =(\frac { \mu_0 m_0 z_0} {4 \pi r^3}) (2cos(\omega t) \hat x + 2sin(\omega t) \hat y)##

    which has no dependence on the z coordinate. I'm wondering if I maybe approached this problem wrong? Help would be much appreciated thank you!
     
  2. jcsd
  3. Dec 9, 2017 #2

    kuruman

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    I am not sure what you mean. Isn't there a z0 in your expression? It looks like you have a 1/z02 dependence because of the r3 in the denominator.
     
  4. Dec 9, 2017 #3
    Okay that seems to be what I was missing, I was not inputting ##z_0## back into my definition for ##\vec r##. By putting that into the equation I got:

    ##\vec A_{dip} (\vec r) = (\frac { \mu_0 m_0 } {4 \pi z_0^2})(sin(\omega t) \hat x - cos(\omega t) \hat y)##

    and

    ##\vec B = \vec \nabla \times \vec A##

    ##\vec B = (\frac { \mu_0 m_0 } {2 \pi z_0^3})(cos(\omega t) \hat x + sin(\omega t) \hat y)##

    I believe this makes more sense.

    To find the induced Electric field I have to take the time derivative of the magnetic field and cross it with ##\vec R## (the separation vector) I use a capitol R to distinguish it from the other r vectors used above.

    ##\vec E = \frac 1 {4 \pi}## ## \int ## ##[({\frac {\partial \vec B} {\partial t}} {\times \mathcal {\vec R}})## ##\frac 1 {\mathcal {r^2}}]##

    ##\vec R = \vec r - \vec r'##

    ##\vec R = < (x-x') \hat x, (y-y') \hat y, (z-z') \hat z > ##

    I'm fairly sure that (x, y, z) is equal to (0, 0, ##z_0##) the field point, which makes (x', y', z') the source point?

     
  5. Dec 10, 2017 #4

    Delta²

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    Can't we simply use the equation ##E=-\nabla V-\frac{\partial A}{\partial t}## for this problem V seems to be zero or spatially constant since there are no free charge densities.
     
  6. Dec 10, 2017 #5

    kuruman

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    That should work.
     
  7. Dec 10, 2017 #6
    Thanks for the help!
     
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