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posted by  207217273 on 7/25/2008 8:22:45 AM  |  status: Live  

life saver for the right answer

Course Textbook Chapter Problem
Calculus Based Physics University Physics with Modern Physics (12th) by Young, Freedman 27 26
Question Details:
A singly charged ion of ^{7}{\rm Li} (an isotope of lithium) has a mass of 1.16×10−26 kg. It is accelerated through a potential difference of 200 V and then enters a magnetic field with magnitude 0.701 T perpendicular to the path of the ion.
 
What is the radius of the ion's path in the magnetic field?
Use 1.60×10−19 C for the magnitude of the charge on an electron.
  R =
 {\rm m}
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posted by Varmonds on 7/28/2008 4:15:11 AM  |  status: Live
Asker's Rating: Lifesaver   
207217273's comment:
"thanks"
Response Details:
given
mass of lithium ion m = 1.16*10-26kg
accelerating voltage V= 200v
 magnitude of magnetic field B = 0.701T
the kinetic energy of ion is
         K = eV
        1/2(mv2) = eV
           or v = √(2eV/m)
caculate for v we get the speed of the ion
the radius of circular orbit is given by
              R = mv/Bq
     here q = 1.6*10-19c
calculate for R
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