# Particle deflection in a magnetic dipole field

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 $$\vec{\mathbf{b}} = (0, b, 0)$$)?

$$\Delta\vec{p}=\frac{\mu_0q}{2\pi b^2}\hat{v}_0\times\left[2(\vec{\mu}\cdot\hat{b})\hat{b}-\vec{\mu}\right].$$
It reduces to your result using $$\hat{v}_0=\hat{x}$$ and $$\hat{b}=\hat{y}$$.