If the radius of the path of the ions in the second magnetic field is 17.9 cm. w

Question

If the radius of the path of the ions in the second magnetic field is 17.9 cm. what is their mass?How fast are the ions moving when they emerge from the velocity selector?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 3.12times10^-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.

Explanation / Answer

part A

In the velocity selector only ions with a particular velocity pass undeflected through the combined electric and magnetic fields of the selector..

ie ..the magnetic and electric field forces on the ions are equal and opposite

Bqv = Eq .. .. selected velocity v = E/B

v = 158V/m / 3.12*10^-2T = v = 5064.102m/s

part B

The magnetic force (Bqv) provides the centripetal force (mv²/r) for the ions ..
Bqv = mv²/r

m = Bqr / v

=(3.12*10^-2T)(1.60*10^-19C)(0.179m) / (5064.102m/s) .. .. m = 1.73*10^-25 kg

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