Question

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 160 V/m and the magnetic field is 3.18×10^?2 T . The ions next enter a uniform magnetic field of magnitude 1.73×10^?2 T that is oriented perpendicular to their velocity.

(A) If the radius of the path of the ions in the second magnetic field is 17.1 cm , what is their mass?

Answer 1

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)

                 = (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  

here , centripetal acceleration a = v²/r

      force F = mv²/r   ................ (2)  

compare eq (1) & eq (2) , we get

         Bvq = mv²/r  

         Bq = mv/r  

mass m = Bqr/v

             = (1.75×10^?2 T)(1.6*10^-19 C)(0.175 m) /(5031.44 m/s)

             = 9.738*10^-26 kg