A singly charged positive ion has a mass of 5.2x10^-27 kg. After being accelerated from rest through a potential difference of 750V, the ion enters a magnetic field of 0.65T. Calculate the radius of the path of the ion in the field.

Answer 1

eU=mv²/2
v=sqrt{2eU/m} = sqrt{2•1.6•10⁻¹⁹ •750/5.2•10⁻²⁷) =6.7•10⁴ m/s
mv²/R=evB
R=mv/eB =5.2•10⁻²⁷•6.7•10⁴/1.6•10⁻¹⁹•0.65 = 3.4•10⁻³ m