# Dipole in non-uniform field

1. Sep 14, 2016

### Gertrude

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.
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. Sep 14, 2016

### BvU

Hello Gertrude,

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.

3. Sep 14, 2016

### Gertrude

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
4. Sep 14, 2016

### BvU

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 ). Your $\mathbf{\tau} = \mathbf{p} \times (\mathbf{v}_C \times \mathbf{B}))$ looks like their (19a) .

5. Sep 14, 2016

### Gertrude

I'll surely take a look at that, thank you for your time.