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Magnetic Field of a Magnetic Dipole

  1. Mar 16, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm not sure how they got the components of the B-field in the r and θ direction. I know the B-field along the ∅ direction is zero though.

    10ono5g.png


    3. The attempt at a solution

    Here I have found the B-field at points P, Q and S that are very far along the z, x and y axis. But what about the field at point R, angle θ to the magnetic moment? How do the Bz,∞, By,∞ and Bx,∞ components contribute?

    121bswy.png
     
  2. jcsd
  3. Mar 16, 2013 #2

    TSny

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    The idea is to take the original dipole moment vector m and break it up into a component in the r direction, mr, and a component in the θ, mθ. Then think of mr as being produced by a little current loop oriented in a direction to produce a magnetic moment along r. You can then easily find the B field produced by this current loop out at the observation point. Similarly, imagine a little current loop producing the magnetic moment mθ. From the way this current loop is oriented, you can easily find B at the observation point.
     
    Last edited: Mar 16, 2013
  4. Mar 16, 2013 #3
    Sorry I don't know what you mean...m is simply defined as IA, which is a constant for a steady current, and that is all. I'm not sure how that is related to the B-field at any point at all...
     
  5. Mar 16, 2013 #4

    TSny

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    The B field is produced by the magnetic moment. You have already shown how to find the B field for two special cases: (1) observation point is along the axis of the magnetic moment (z-axis in your work) and (2) observation point is in a plane oriented perpendicular to the magnetic moment (on the x-axis, say).

    Now you have a problem where you want to find the field at a point located a distance r away and angle θ to the dipole moment vector.

    The problem is suggesting that you can reduce this problem to the two cases you have already worked out. Since the magnetic moment ##\vec{m}## is a vector, you can write ##\vec{m}## as a sum of components ##\vec{m} = m_r \hat{r} + m_{\theta} \hat{\theta}##. Note that ##m_r \hat{r}## can be thought of as a magnetic moment whose direction (axis) is in the direction to the field point where you want to find B. You know how to find B in this case. Likewise, ##m_{\theta} \hat{\theta}## is a magnetic moment vector whose axis is perpendicular to the direction that you want to find B. But that's the other case that you have worked out.
     
    Last edited: Mar 17, 2013
  6. Mar 16, 2013 #5
    ok i get that m can be dissolved into 2 components, r and θ.

    so along r, the component is m*cosθ.
    How do we find the B-field along r? is it B-field at infinite distance away ALONG M * cosθ as well?

    Then why isn't the B-field along θ = B-field at infinite distance away ALONG M * sinθ?
     
  7. Mar 16, 2013 #6

    TSny

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    Suppose you introduce a z-axis along the direction of r and with origin at the dipole moment. Then, ##m_r\hat{r}## points along your z-axis. So, you can use your result for ##B_z## to find the field produced by ##m_r\hat{r}## at the observation point.
     
  8. Mar 16, 2013 #7
    Ah, I think I finally understood it!

    First, we split the magnetic moment m into 2 components, one along r and one along θ.

    Along r
    Imagine a *smaller* (as cosθ ≤ 1) loop with same current (or it can be the same size but smaller current), but with its axis directed along r. Let's call this process augmentation.

    Now, to find the B-field along r, we simply go very far along it's axis and take the B-field there, which gives us the result, with a co-factor of cosθ due to the initial procedure.

    Along θ
    Imagine a *smaller* loop with same current, with it's plane along r. This makes it's axis directed perpendicular to r. Now, we don't want to go infinite distance along its axis, but rather along its plane and take the B-field there. Why? Because the B-field along its plane from very far away is perpendicular to the plane! Which is exactly the θ-component we are interested in.

    So at angle θ, very far away: Bθ,∞ = B*plane,∞ = Bplane,∞ sin θ

    where B* is the augmented loop.
     
    Last edited: Mar 16, 2013
  9. Mar 16, 2013 #8

    TSny

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    Yes, exactly. Good!
     
  10. Mar 16, 2013 #9
    Ah, it troubles me that sometimes I take so long to understand such easy things..
     
  11. Mar 16, 2013 #10

    TSny

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    No, don't be concerned. I probably didn't explain it very well. Sometimes it's hard to put things in words that would be better explained in diagrams.
     
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