1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted