The radius of the orbit in which ions moving is determined by the relation as given below
$$
\frac{m v^{2}}{R}=q v B
$$
where $m$ is the mass, $v$ is velocity, $q$ is charge of ion and $B$ is the flux density of the magnetic field, so that $q v B$ is the magnetic force acting on the ion, and $\frac{m v^{2}}{R}$ is the centripetal force on the ion moving in a curved path of radius $R$.

The angular frequency of rotation of the ions about the vertical field $B$ is given by
$\omega=\frac{v}{R}=\frac{q B}{m}=2 \pi v$
where $v$ is frequency. Energy of ion is given by
or
$$
\begin{aligned}
E &=\frac{1}{2} m v^{2}=\frac{1}{2} m(R \omega)^{2} \\
&=\frac{1}{2} m R^{2} B^{2} \frac{q^{2}}{m^{2}} \\
E &=\frac{1}{2} \frac{R^{2} B^{2} q^{2}}{m}
\end{aligned}
$$ ... (i)
If ions are accelerated by electric potential $V$, then energy attained by ions
$$
E=q V ...(ii)
$$
From Eqs. (i) and (ii), we get
$$
q V=\frac{1}{2} \frac{R^{2} B^{2} q^{2}}{m}
$$
or
$$
\frac{q}{m}=\frac{2 V}{R^{2} B^{2}}
$$
If $V$ and $B$ are kept constant, then
$$
\frac{q}{m} \propto \frac{1}{R^{2}}
$$