The radius of the circular path or helical path followed by the test charge q_o moving in magnetic field B with some velocity v is

The circular motion of charged particle will be due to component of velocity v perpendicular to magnetic field, i.e., v sin$\mathrm{\theta }$
$\therefore \frac{\mathrm{m}(\mathrm{vsin}\mathrm{\theta }{)}^2}{\mathrm{r}}={\mathrm{q}}_0(\mathrm{vsin}\mathrm{\theta })\mathrm{B}$
or $\mathrm{r}=\frac{\mathrm{m}(\mathrm{vsin}\mathrm{\theta })}{{\mathrm{q}}_0\mathrm{B}}$