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I am having troubles solving a question

The question is: A particle moves so that its position vector is given by:

r(t) = $Rcos(\omega t)$ i +$Rsin(\omega t)$ j, where R and $\omega$ are positive contants and i and j are unit vectors in the x and y directions.

Assume that the particle is a charge q with mass m subject to a magnetic force $q\dot{r}$ $\times$ B. Determine B assuming it is parallel to the unit vector k = i $\times$ j

I know the magnetic force is found from the Lorentz force F = q(E + v $\times$ B). In this case there is no Electric field. So F = q v $\times$ B, where v = $\dot{r}$. I'm not sure how I would find B. Do I just find a vector that is parallel to k = i $\times$ j ? Any help would be appreciated thanks

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  • $\begingroup$ I guess you need to compute $\ddot r$ and $\dot r$ and substitute into Lorentz formula? $\endgroup$ – mathshungry May 22 '17 at 10:48
  • $\begingroup$ @mathshungry I calculated ${r}''$ and $\dot{r}$ and substituted into the formula by setting F = m${r}''$ = q$\dot{r}$ $\times$ B. Not sure what I would do from there. (sorry, don't know how to format r double dot) $\endgroup$ – Sujjan Shaikh May 22 '17 at 11:04
  • $\begingroup$ Is there a problem in ${\bf r}(t)$? It appears to be $R\cos(\omega t)({\bf i}+{\bf j})$. $\endgroup$ – Jon May 22 '17 at 11:44
  • $\begingroup$ @Jon Ah thank you for that, it's meant to be Rsin($\omega$ t) j. I have edited the original question $\endgroup$ – Sujjan Shaikh May 22 '17 at 11:52
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This can be worked out in the following way. Take the first derivative of the position $$ \dot{\bf r}(t)=-\omega R\sin(\omega t){\bf i}+\omega R\cos(\omega t){\bf j} $$ and the second one $$ \ddot{\bf r}(t)=-\omega^2 R\cos(\omega t){\bf i}-\omega^2 R\cos(\omega t){\bf j} $$ and you realize that $\ddot{\bf r}(t)=-\omega^2{\bf r}$. Then you will use the Lorentz equation to get $$ m\ddot{\bf r}(t)=-m\omega^2{\bf r}(t)=q\dot{\bf r}(t)\times{\bf B}. $$ Choose $B{\bf k}$ and do the vector product. You will get $$ -m\omega^2 R\cos(\omega t)=q\omega R\cos(\omega t)B $$ $$ -m\omega^2 R\sin(\omega t)=q\omega R\sin(\omega t)B $$ and you see that both the equations are consistent provided $$ B=-\frac{m\omega}{q} $$ and, indeed, $$ \omega =\frac{qB}{m} $$ is the well-known Larmor frequency.

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  • $\begingroup$ Thanks for the reply, but shouldn't B =$\frac{ -m \omega} {q}$ ? $\endgroup$ – Sujjan Shaikh May 22 '17 at 13:07

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