Given E = 160 V/m
         B
=3.18×10?2    T
Now the velocity of the ion when it emerges from the velocity
selector is
              v
= E / B................(1)
               Â
= (160 V/m) / (3.18×10?2    T)
                =
5031.44 m/s
(a)
magnetic field B =1.75×10?2 T
radius of the path of the ions r = 17.5 cm = 0.175 m
magnetic force F = Bvq  ................... (1)
from Newton's second law of motion ,
     force F = ma Â
here , centripetal acceleration a = v2/r
      force F =
mv2/r  ................ (2) Â
compare eq (1) & eq (2) , we get
        Bvq =
mv2/r Â
        Bq =
mv/r Â
mass m = Bqr/v
           Â
= (1.75×10?2 T)(1.6*10-19 C)(0.175 m)
/(5031.44 m/s)
             =
9.738*10-26 kg
