An electron emitted by a heated cathode and accelerated through a potential difference of , enters…

Speed of the electron: $\frac12 mv^2 = eV$, so $v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2 \times 1.6\times10^{-19} \times 2000}{9.1\times10^{-31}}} \approx 2.65 \times 10^7\text{ m s}^{-1}$. Only the component of velocity perpendicular to the field, $v\sin\theta$, bends the path, so the radius is $r' = \frac{m v \sin\theta}{eB} = r\sin 30^\circ$, where $r = \frac{mv}{eB} \approx 1.0 \times 10^{-3}\text{ m}$ (the radius for case (a)). $r' = 1.0\times10^{-3} \times \frac12 \approx 5.0 \times 10^{-4}\text{ m}$. The component $v\cos 30^\circ$ along the field is unchanged, so the path is a helix (spiral) of radius $5 \times 10^{-4}$ m and pitch $2\pi r\cos 30^\circ \approx 5.5 \times 10^{-3}$ m.