Item 8 Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 158 V/m and the magnetic field is 3.13 Times 10^2 T. The ions next enter a uniform magnetic field of magnitude 1.73 Times 10^-2 T that is oriented perpendicular to their velocity. How fast are the ions moving when they emerge from the velocity selector? If the radius of the path of the ions in the second magnetic field is 17.4 cm, what is their mass? Solution A) In Velocity selector Electric field & Magnetic field forces on ions are equal & opposite ,this is because it allows ions with a particular velocity only passes undeflected. so we can say, Bqv = Eq B:magnetic field density E:Electric field v:velocity at which the ions remains undeflected. v=E/B v = 158V/m / 3.13*10-2 T => v = 4.78*103 m/s B) as it enteres next in uniform magnetic field of value B=1.73*10-2 T Work done in uniform magnetic field is zero.It only changes the direction.So B provides centripetal force for the ions while changing direction in circular path of radius r=17.4cm. Bqv = mv²/r m = Bqr / v = (1.73*10-2T)(1.6*10-19C)(0.174m) / (4.78*103m/s) =>m = 1*10-25 kg.