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A particle of mass $m$ and charge $q$ has position vector $r(t)$ and moves in a constant, uniform magnetic field B so that its equation of motion is $\mathbf{\ddot r}=q\mathbf{\dot r\times B}$.

Let $\mathbf L = m\mathbf{r\times \dot r}$ be the particle’s angular momentum.

Show that $\mathbf{L\cdot B}+\frac12q|\mathbf{r\times B}|^2$ is a constant of the motion. Explain why the kinetic energy $T$ is also constant, and

show that it may be written in the form $T=\frac12m\mathbf{u\cdot((u\cdot v)v−}{r^2\mathbf{\ddot u}})$,

where $\mathbf{v=\dot r}, r=|\mathbf r| $ and $\mathbf u = \frac{\mathbf r}r$.

[Hint: Consider $\mathbf{u\cdot\dot u}$]

This may be slightly too physicsy to post on here but my issue is mathematical.

I have no issue with the first two parts, it is the "show that it may be written in the form $T=\frac12m\mathbf{u\cdot((u\cdot v)v−}{r^2\mathbf{\ddot u}})$" that I am having issues with.

I am looking at a solution and in the last part it uses that $\mathbf{\dot r\cdot r}=\dot rr$- where has this come from?

Any help is appreciated, thanks

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  • $\begingroup$ bad tripos student $\endgroup$
    – Harry Alli
    May 14 '19 at 8:12
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It is not entirely clear from your question what the final solution was but $$ \mathbf{\dot{r}}\cdot \mathbf{r} = \dot{\left(\matrix{x \\ y \\z}\right)}\cdot (\matrix{x & y & z}) = \dot{x}x + \dot{y}y + \dot{z}z = \frac{1}{2}\frac{d}{dt}(x^2+y^2+z^2) = \frac{1}{2}\frac{d}{dt}r^2 = r\dot{r} $$

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  • $\begingroup$ Fantastic- thank you very much $\endgroup$ Apr 10 '17 at 20:08
  • $\begingroup$ No problemo. This brings me back to my 1st and 2nd year Physics mods. $\endgroup$
    – Chinny84
    Apr 10 '17 at 20:09

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