Question

A singly-charged ion enters a velocity selector that has a 0.18-T magnetic field perpendicular to an electric field of 1.6 kV/m, with both fields perpendicular to the velocity of the ion. The same magnetic field is then used to deflect the ion into a circular path of radius 12.5 cm. What velocity was selected by the velocity selector?

Answer 1

Answer: a) 8.9km/s b) 4.05 x 10^-25 kg

Explanation:

a)

To find the velocity, 

V = E/B plug in the numbers we get

V = 1.6kV/m / 0.18 T = 8.9 km/s 



b)

To find the ion mass, 

m = erB^2/E = 0.125 m x 1.6x10^-19 x (0.18)^2 / ( 1.6 x 10 ^ 3 V/m ) = 4.05 x 10 ^ -25 kg