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Consider a small magnetic dipole of magnetic moment $\vec{\boldsymbol{\mu}} = (\mu_x, \; \mu_y, \; \mu_z)$ sitting at the origin. The magnetic field all around is \begin{equation}\tag{1} \mathbf{B} = \frac{\mu_0}{4 \pi} \Big( \frac{3 (\vec{\boldsymbol{\mu}} \cdot \vec{\mathbf{r}})}{r^5} \, \vec{\mathbf{r}} - \frac{\vec{\boldsymbol{\mu}}}{r^3} \Big). \end{equation} A particle of charge $q$ is moving from infinity with initial velocity $\vec{\mathbf{v}}_0 = (v_0, \; 0, \; 0)$ and impact parameter $b$ along the $y$ axis. The particle return to infinity with some deflection. What is the momentum variation, to lowest order?

The magnetic force is simply $\vec{\mathbf{F}} = q \, \vec{\mathbf{v}} \times \vec{\mathbf{B}}$. Integrating Newton's equation gives the momentum variation if we assume a slight deviation from a straight path: $\vec{\mathbf{r}}(t) \approx (v_0 \, t, \; b, \; 0)$ for the force. Calculations give this result: \begin{align} \Delta p_x &\approx 0, \tag{2} \\[12pt] \Delta p_y &\approx \frac{\mu_0 q}{2 \pi b^2} \, \mu_z, \tag{3} \\[12pt] \Delta p_z &\approx \frac{\mu_0 q}{2 \pi b^2} \, \mu_y. \tag{4} \end{align} Surprisingly, this momentum variation doesn't depend on the initial velocity $v_0$. These components aren't exactly a vectorial product, since $\vec{\mathbf{v}}_0 \times \vec{\boldsymbol{\mu}} = (0, -\, v_0 \mu_z, \; +\, v_0 \mu_y)$. I then have two questions:

  1. I need to confirm the results (2)-(4).
  2. More importantly, how can I write (2)-(4) in a simple closed vectorial form, using only the vectors $\vec{\boldsymbol{\mu}}$, $\vec{\mathbf{v}}_0$ (and probably the impact vector $\vec{\mathbf{b}} = (0, b, 0)$)?
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I confirmed your results (2)-(4). They can be written in vector form as

$$\Delta\vec{p}=\frac{\mu_0q}{2\pi b^2}\hat{v}_0\times\left[2(\vec{\mu}\cdot\hat{b})\hat{b}-\vec{\mu}\right].$$

It reduces to your result using $\hat{v}_0=\hat{x}$ and $\hat{b}=\hat{y}$.

I found this by doing the whole calculation in vector form rather than using components.

A handwaving argument for why the impulse is independent of the velocity is that, at twice the velocity, the Lorentz force is twice as large, but the time it acts over is half as large. It’s handwaving because of course in both cases the time is actually infinite. But if you put in a distance cutoff where the force becomes negligible, the time would be half as large.

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  • $\begingroup$ I'm surprised that this problem isn't shown in standard books on electrodynamics (I haven't found it yet). It's simple and instructive. $\endgroup$ – Cham Jun 25 at 23:45
  • $\begingroup$ Yes, it’s a nice problem that I had not seen before. $\endgroup$ – G. Smith Jun 25 at 23:46

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