Question

Which of the following particles will describe the smallest circle when projected with the same velocity perpendicular to a magnetic field?
(a) Electron
(b) Proton
(c) He⁺
(d) Li⁺

Answer 1

(a) Electron

When a particle moves in a magnetic field, the necessary centripetal force, for the particle to move in a circle, is provided by the magnetic force acting on the particle.
So, equating the Lorentz force with the cetripetal force for a charged particle of charge q describing a circle of radius r, we get:

$$\begin{gathered} qvB=\hspace{0.17em}\frac{m{v}^2}{r} \\ \Rightarrow r=\frac{mv}{qB} \end{gathered}$$
Here, m is the mass of the particle and v is its velocity.
The amount of charge of all the given particles is same; hence, for a given charge, $r\propto m$. And since the electron is the lightest of all the particles, it describes the smallest circle when projected with the same velocity perpendicular to the magnetic field.