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Dipole in non-uniform field

  1. Sep 14, 2016 #1
    1. The problem statement, all variables and given/known data
    I want to determine the net force and torqe on a moving electric dipole in non-uniform magnetic field.
    I suspect I should take some kind of a limit, but I'm not sure how to do so.
    Please help, I'd really like to understand this.
    2. Relevant equations
    ##\mathbf{F} = q \mathbf{v} \times \mathbf{B}, \quad \mathbf{\tau} = \mathbf{r} \times \mathbf{F}, \quad \mathbf{p} = q \mathbf{d}##

    3. The attempt at a solution
    I wrote the velocities as the average velocity ##\mathbf{v} = (\mathbf{v}_1+\mathbf{v}_2)/2## plus the difference (index 1 indicates the positive charge and 2 the negative charge):
    $$\mathbf{F} = q\mathbf{v}_1 \times \mathbf{B}(\mathbf{r} + \mathbf{p}/q) - q\mathbf{v}_2 \times \mathbf{B}(\mathbf{r})$$
    So what I got in the end was:
    $$\mathbf{F} = q \mathbf{v} \times (\mathbf{B}(\mathbf{r} + \mathbf{p}/q) - \mathbf{B}(\mathbf{r})) + \frac{q}{2} \dot{\mathbf{p}} \times (\mathbf{B}(\mathbf{r} + \mathbf{p}/q) + \mathbf{B}(\mathbf{r}))$$
    I'm stuck here and I'm not sure how to get a 'nicer' form that would only contain ##\mathbf{B}(\mathbf{r})##.

    Regarding the torque:
    $$\mathbf{\tau} = \frac{\mathbf{d}}{2} \times \mathbf{F}_1 - \frac{\mathbf{d}}{2} \times \mathbf{F}_2$$
    Rearraging a bit I got:
    $$\mathbf{\tau} = \frac{\mathbf{p}}{2} \times (\mathbf{v}_1 \times \mathbf{B}(\mathbf{r} + \mathbf{p}/q) + \mathbf{v}_2 \times \mathbf{B}(\mathbf{r}))$$
    From here on I have the same problem as before.

    If you can help me with a hint or two I'll be really grateful.
  2. jcsd
  3. Sep 14, 2016 #2


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    Hello Gertrude, :welcome:

    Did you work out the case for the homogeneous ##\mathbf B## field already ?
    An electric dipole generally has a small size ##\mathbf d## so I expect the derivative of ##\mathbf B## to appear for the non-homogeneous case (i.e. the limit ##\mathbf d \downarrow 0## seems useful to me.

    I would also work with the center of the dipole, not with one of the charges as central point.
  4. Sep 14, 2016 #3
    Hello, thanks for answering!

    As a matter of fact I did compute a net force and torque in uniform magnetic field before. I got:
    $$ \mathbf{F} = \dot{\mathbf{p}} \times \mathbf{B}, \quad \mathbf{\tau} = \mathbf{p} \times (\mathbf{v}_C \times \mathbf{B})$$
    with ##\mathbf{v}_C## being the velocity of the center of mass ##\mathbf{r}_C = (\mathbf{r}_1 + \mathbf{r}_2)/2##.

    Now for the non-uniform case: I can see that in the first summand of the force, I get a derivative (in three dimensions, thus a gradient in the direction of orientation):
    $$q \mathbf{v} \times (\mathbf{B}(\mathbf{r} + \mathbf{p}/q) - \mathbf{B}(\mathbf{r})) = q \mathbf{v} d \times \frac{(\mathbf{B}(\mathbf{r} + \mathbf{p}/q) - \mathbf{B}(\mathbf{r}))}{p/q} => p \mathbf{v} \times (\hat{\mathbf{d}} \cdot \nabla) \mathbf{B}(\mathbf{r})$$
    and in the second summand I get:
    $$\frac{q}{2} \dot{\mathbf{p}} \times (\mathbf{B}(\mathbf{r} + \mathbf{p}/q) + \mathbf{B}(\mathbf{r})) => q \dot{\mathbf{p}} \times \mathbf{B}(\mathbf{r})$$
    I believe ##\mathbf{v}## is the translation velocity of the center of mass. By the way, is it mathematically correct to take such limits as I did (especially the second term, where I had a sum of the fields)?
    So I see the force gets one extra term comparing to the uniform case, which seems logical.

    In computing the torque I did something similar:
    $$\frac{\mathbf{p}}{2} \times (\mathbf{v} \times (\mathbf{B}(\mathbf{r} + \mathbf{p}/q) + \mathbf{B}(\mathbf{r})) + \frac{\dot{\mathbf{p}}}{2q} \times (\mathbf{B}(\mathbf{r} + \mathbf{p}/q) - \mathbf{B}(\mathbf{r}))) =>$$
    $$=> \frac{\mathbf{p}}{2} \times (2\mathbf{v} \times \mathbf{B}(\mathbf{r}) + \frac{d}{2q} \dot{\mathbf{p}} \times (\hat{\mathbf{d}} \cdot \nabla) \mathbf{B}(\mathbf{r}))$$
    Since ##\dot{\mathbf{p}} = q(\mathbf{v}_1- \mathbf{v}_2)## only changes direction, it's always perpendicular on ##\hat{\mathbf{d}}## (and ##\mathbf{p}##) and thus the second term is zero.
    So the torque is the same as in uniform case? I didn't expect that.

    Am I on the right path of thinking?
    Last edited: Sep 14, 2016
  5. Sep 14, 2016 #4


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    I should think so (but haven't time to scrutinize all the steps).
    Turns out someone did a lot of work already; perhaps this article is a nice starting point ... complicated enough already without inhomogeneity (and they work in gaussian units, too. That's reasonable but an extra complication nevertheless o_O ). Your ## \mathbf{\tau} = \mathbf{p} \times (\mathbf{v}_C \times \mathbf{B}))## looks like their (19a) .
  6. Sep 14, 2016 #5
    I'll surely take a look at that, thank you for your time.
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