Defaults are singly-charged Ar (z=1, m=40 u, V=2000 V, B=0.50 T). Launch an isotope pair from a preset and the lighter isotope lands on the inner spot, the heavier one farther out, separating them. Elementary charge e = 1.602×10⁻¹⁹ C, atomic mass unit u = 1.6605×10⁻²⁷ kg, relativistic corrections neglected.
Left = ion source and accelerator (voltage V) / centre = magnetic-field region B (x marks point into the page) / each ion flies along a semicircle leaving a trail / right edge = detector plane. Lighter isotopes land on the inner spot, heavier ones farther out.
X = detector position 2r (mm, heavier = farther right) / Y = relative intensity / detector hits accumulate as peaks, each labelled by m/q. Blue curve = theoretical radius r versus m/z.
An ion of charge q = ze accelerated through voltage V gains kinetic energy qV, which fixes its speed.
Ion speed after acceleration:
$$v = \sqrt{\frac{2qV}{m}}$$In the magnetic field the Lorentz force $\mathbf{F} = q\,\mathbf{v}\times\mathbf{B}$ is always perpendicular to the velocity and acts as the centripetal force. Solving $qvB = \dfrac{m v^{2}}{r}$ for $r$ gives the orbit radius:
$$r = \frac{m v}{q B} = \frac{1}{B}\sqrt{\frac{2 m V}{q}}$$Validation: singly-charged ²⁰Ne (m=20 u, q=e), V=1000 V, B=0.1 T → v ≈ 9.82×10⁴ m/s, r ≈ 0.204 m. The ²²Ne radius ratio is √(22/20) ≈ 1.0488, matching the ²⁰/²² separation produced by this tool.
Mass-to-charge ratio (in u/e):
$$\frac{m}{q} = \frac{r^{2} B^{2}}{2 V}$$Kinetic energy:
$$KE = q V = \tfrac{1}{2} m v^{2}$$Here $q = ze$ ($e = 1.602\times10^{-19}$ C), $m$ is the ion mass [kg] (1 u = $1.6605\times10^{-27}$ kg), $V$ the acceleration voltage [V], $B$ the magnetic flux density [T] and $r$ the radius [m]. Relativistic corrections are negligible for v < 0.1c.