Kinetic energy: $$K = 8.0\ \text{MeV} = 8.0\times10^{6}\times1.6\times10^{-19} = 1.28\times10^{-12}\ \text{J}$$ Speed of the alpha particle: $$v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(1.28\times10^{-12})}{6.4\times10^{-27}}} = \sqrt{4\times10^{14}} = 2\times10^{7}\ \text{m/s}$$ Radius of circular path: $$r = \frac{mv}{qB} = \frac{(6.4\times10^{-27})(2\times10^{7})}{(3.2\times10^{-19})(0.5)}$$ $$= \frac{1.28\times10^{-19}}{1.6\times10^{-19}} = 0.8\ \text{m}$$ (i) The particle describes a helical path when its velocity makes an angle (other than $0^\circ$ or $90^\circ$) with the magnetic field, so that there is a component of velocity along the field. (ii) The particle goes straight undeviated when its velocity is parallel (or anti-parallel) to the magnetic field, so that the magnetic force $qvB\sin\theta = 0$.