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Magnetic Field in Spherical Coordinate

  1. Jun 20, 2015 #1
    1. The problem statement, all variables and given/known data

    I am trying to use the equation ##B_{dip} (r) = \nabla \times A## to find the magnetic field due to a dipole at the origin pointing in the z direction (where A is the magnetic vector potential).

    The correct answer should be:

    ##B_{dip} (r) = \frac{\mu_0 m}{4 \pi r^3} \ (2 cos \theta \ \hat{r} + sin \theta \ \hat{\theta})##​

    But I'm unable to get this expression. I can't see how they got the part in the bracket. :confused:

    2. Relevant equations

    In spherical coordinates:

    ##A_{dip} (r) = \frac{\mu_0}{4 \pi} \frac{m sin \theta}{r^2} \hat{\phi}## .... (1)

    Curl:

    curl.jpg .... (2)


    3. The attempt at a solution

    So we only need to use the ##\hat{\phi}## term from the curl equation. Substituting (1) into (2), we obtain:

    ##\nabla \times A= \frac{1}{r}\left( \left( \frac{\partial}{\partial r} \left( \frac{r \mu_0 m \ sin \theta}{4 \pi r^2} \right) \right) - \left( \frac{\partial}{\partial \theta} \left( \frac{\mu_0 m \ sin \theta}{4 \pi r^2} \right) \right) \right)##​

    ##=\frac{1}{r} (- \frac{\mu_0 m \ sin \theta}{\pi r^2} - \frac{\mu_0 m}{4 \pi r^2} cos \theta)##​

    ##\therefore \ B_{dip} (r) = \frac{\mu_0 m}{4 \pi r^3} (-sin \theta - cos \theta)##​

    So how did they get from "-sinθ - cosθ" to "2 cosθ + sinθ"? Have I made a mistake somewhere? :confused:

    Any explanation would be greatly appreciated.
     
    Last edited: Jun 20, 2015
  2. jcsd
  3. Jun 21, 2015 #2

    blue_leaf77

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    This is the reason why you got wrong answer. Although the vector potential only has ##\phi## component, that doesn't mean its curl will also has only ##\phi## component. What you need to heed to in the expression of the curl of A is:
    ## A_r \rightarrow r## component of vector potential
    ## A_\theta \rightarrow \theta## component of vector potential
    ## A_\phi \rightarrow \phi## component of vector potential
    Now the vector potential does not have the ##r## and ##\theta## components, so any differential operation acting on these components should vanish. The differential acting on the ##\phi## component should, in general, not vanish depending on which variable it's differentiated against.
     
    Last edited: Jun 21, 2015
  4. Jun 21, 2015 #3
    What do you mean? The answer for ##B(r)## has a ##\hat{r}## and ##\hat{\theta}## component. How did they get that?

    If all the partial derivatives with respect to θ and r vanish, then I must only use the ##\partial/\partial \phi## terms. Namely:

    ##\frac{1}{sin \theta} \left( \frac{-\partial A_\phi}{\partial \phi} \right) + \frac{1}{r} \left( \frac{1}{sin \theta} \frac{\partial A_r}{\partial \phi} \right)##

    But since I do not have any ϕ variables this also vanishes...
     
  5. Jun 21, 2015 #4

    blue_leaf77

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    Ok if I rewrite the vector potential to ##A_{dip} (r) = \frac{\mu_0}{4 \pi} \frac{m sin \theta}{r^2} \hat{\phi} = A_{\phi}(r,\theta) \hat{\phi}##, will you know which terms in the curl should vanish, which one should not?
    If you calculate the fourth therm in the curl ##\frac{1}{r} \frac{(\partial rA_{\phi})} {\partial r}##, will it be zero?
     
    Last edited: Jun 21, 2015
  6. Jun 21, 2015 #5
    Thank you so much for the explanation, it makes perfect sense now. I got the right answer using the first and the fourth term. Thank you!
     
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