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Magnetic Dipole

  1. Apr 20, 2009 #1

    Matterwave

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    So I got a question on a E&M Midterm:

    There is a line current of I at the origin going in the Z direction. There is a magnetic dipole of:
    [tex]\vec{m}=m_r\hat{r}+m_\varphi\hat{\varphi}[/tex] located at [tex](r,\varphi)[/tex]. What is the force and torque on this magnetic dipole?

    So I used the equations:
    [tex]\vec{F}=-\nabla U=-\nabla(\vec{m}\cdot\vec{B})[/tex]
    [tex]\vec{\tau}=\vec{m}\times\vec{B}[/tex]

    I know from Ampere's law that the field is:
    [tex]\vec{B}=\frac{\mu_0 I}{2\pi r}\hat{\varphi}[/tex]

    So, I reason that the angle between the field and the dipole is:
    [tex]\theta=\frac{\pi}{2}-tan^{-1}(\frac{m_\varphi}{m_r})[/tex]
    Since the field is only in the [tex]\varphi[/tex] direction, I just drew a triangle.
    Therefore, I got that:
    [tex]\vec{F}=-\nabla(\vec{m}\cdot\vec{B})=-\nabla(mBcos(\theta))=-\nabla(\frac{\mu_0 I}{2\pi r}\sqrt{m_\varphi^2+m_r^2}sin(tan^{-1}(\frac{m_\varphi}{m_r}))=\frac{\mu_0 I}{2\pi r^2}\sqrt{m_\varphi^2+m_r^2}sin(tan^{-1}(\frac{m_\varphi}{m_r}))\hat{r}[/tex]
    [tex]\vec{\tau}=\vec{m}\times\vec{B}=mBsin(\theta)\hat{z}=\frac{\mu_0 I}{2\pi r}\sqrt{m_\varphi^2+m_r^2}cos(tan^{-1}(\frac{m_\varphi}{m_r}))\hat{z}[/tex]

    Unfortunately these are all wrong, as far as I can tell from looking at my Professor's solutions. What did I do wrong? =(
     
  2. jcsd
  3. Apr 21, 2009 #2

    Matterwave

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    I should say what my Professor's answers were:

    [tex]F_\varphi=-m_r\frac{B_\varphi}{r}[/tex]
    [tex]\tau_z=-rF_\varphi[/tex]
     
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