or $\quad B=\frac{2 \pi m v}{e}.........(i)$
As $\frac{m v^{2}}{R}=e v B$
${\text{or }}v = \frac{{eBR}}{m} = \frac{{e2\pi mvR}}{{me}}\,\,\,\,\,(U\sin g\,(i))$
$ = 2\pi vR{\text{ }}.........{\text{(ii)}}$
${\text{Kinetic energy, }}K = \frac{1}{2}m{v^2} = \frac{1}{2}m{(2\pi vR)^2}{\text{ }}$
$=2m\pi ^2v^2R^2$
If $\frac{x_0}{x_1}=3$, the value of $\frac{R_1}{R_2}$ is.

(Assume that the current is flowing in the clockwise direction.)
