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
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.