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Magnetic field of a dipole (Griffiths 5.33)

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Example: Problem 5.33
    Show that the magnetic field of a dipole can be written in the following coordinate free form:

    LectureNotesChapter5134.jpg

    LectureNotesChapter5135.jpg


    2. Relevant equations

    [URL]http://teacher.pas.rochester.edu/PHY217/LectureNotes/Chapter5/LectureNotesChapter5138.jpg[/URL]

    [URL]http://teacher.pas.rochester.edu/PHY217/LectureNotes/Chapter5/LectureNotesChapter5140.jpg[/URL]

    3. The attempt at a solution

    [URL]http://teacher.pas.rochester.edu/PHY217/LectureNotes/Chapter5/LectureNotesChapter5141.jpg[/URL]

    Okay, I'm stumped on the last step. I can see on the one before it that (m.r) is added and subtracted to keep it even, but how do the two negative terms (r-hat and theta-hat) equate to -m? If I use the trig terms, pull out the m, use a random angle and add them together, I don't get -1.
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Apr 26, 2010 #2
    Your axis is not a good choice. Your x-axis is labelled "r", yet you have a vector named "r" as well. It can get very confusing.

    Well when you describe a vector in 2D cartesian coords, you list it as:
    [tex]\vec{m} = m_x\hat{x}+m_y\hat{y}[/tex]

    Although it can be a bit more confusing, you can do the same in 2d polar coords for any general vector:
    [tex]\vec{m} = m_r\hat{r}+m_{\theta}\hat{\theta}[/tex]

    So you just need to understand how the dot products are related to the coefficients.

    Note: If you are afraid of this, then you can always be on the safe side and convert your polar unit vectors to cartesian unit vectors and get the same answer.
     
  4. Apr 26, 2010 #3
    Untitled-1.png

    Okay, so if I set it up like this, everything is the same except the coordinate system. So I get

    [tex] -(m \bullet \widehat{r}) \widehat{r} = -m cos ( \theta ) \widehat{r} [/tex]

    [tex] -(m \bullet \widehat{\theta}) \widehat{\theta} = m sin ( \theta ) \widehat{\theta}[/tex]

    ...I don't know why the \bullet or hats aren't showing up, but the left side of those equations are supposed to be dot products with the hat vectors.

    LectureNotesChapter5141.jpg

    I'm just not seeing how when you add those two things together, you get m.
     
    Last edited: Apr 26, 2010
  5. Apr 26, 2010 #4
    You could use the latex command "\cdot" for dot products.

    Also, when you take the dot product of a vector with a unit vector, then you are actually getting the component length of that vector in that direction. For example, the component of the m-vector in the x-direction is:

    [tex]\vec{m}\cdot\hat{x} = m_x[/tex]

    In your case you would have polar coordinates instead.

    EDIT: The proof is for cartesian (the same for polar as well)...

    [tex]\vec{m} = m_x \hat{x} + m_y \hat{y}[/tex]
    by definition...

    so taking the dot product with x-hat, we get:

    [tex]\vec{m}\cdot\hat{x} = m_x (\hat{x}\cdot\hat{x}) + m_y (\hat{y}\cdot\hat{x})[/tex]
    [tex]\vec{m}\cdot\hat{x} = m_x[/tex]
     
  6. Apr 28, 2010 #5
    Wait, I'm still not seeing this.

    Wouldn't that leave

    [tex] -(m \cdot \widehat{r}) \widehat{r} = -m_{r} \widehat{r}[/tex]

    and

    [tex] -(m \cdot \widehat{\theta}) \widehat{\theta} = -m_{\theta} \widehat{\theta}[/tex]
     
  7. Apr 28, 2010 #6
    Yes, and when you add those two values you get a vector in polar coordinates.
     
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