The potential difference (p.d.) across the plates of the velocity selector is $V$. The plate separation is $d$ and the magnetic flux density is $B$. Show that the speed $u$ of ions that travel undeviated through the velocity selector is given by $u = \frac{V}{Bd}$.
Positive ions with kinetic energy $4.1 \times 10^{-17}\,\text{J}$ and mass $3.2 \times 10^{-27}\,\text{kg}$ travel undeviated through the velocity selector when $V$ is $980\,\text{V}$ and $d$ is $3.6 \times 10^{-2}\,\text{m}$. Determine $B$.
A proton passes undeviated through the velocity selector. An alpha particle enters at the same speed as the proton. State how the expression in (a) shows that the alpha particle also passes undeviated through the velocity selector.
Using Fig. 6.1 and the forces acting on a positive ion, determine the direction of the magnetic field. Explain your reasoning.
The positive ions in (b) enter the velocity selector with greater kinetic energy. On Fig. 6.1, sketch the path of these ions.