Equating mass and charge ratio of cathode rays

  • Thread starter redruM
  • Start date
  • #1
15
0
hi:smile:

i was approached by this problem,

prove q/m = (2V)/(B^2r^2) [2V over B-squared r-squared]

so far i have,

F[electric] = qE

F[magnetic] = Qvb

F[electric] = F[magnetic]

therefore, v = E/B

-----
now i am a bit confused/wrong
-----

E[kinetic] = (mv^2)/2

therefore, q x V = (mv^2)/2

therefore, q/m = E^2/(2VB^2)

jus by looking at the result, i have to get E= 2V/r.

any insights will be greatly appreciated....
if my method even correct?
 

Answers and Replies

  • #2
80
0
F[magnetic] = qvB

Elecrons will be deflected in a circular path, so

F[centripital] = mv^2/r

Equate these, and get v = Bqr/m

Electrons accelerate through a potential, so

E[kinetic] = mv^2/2 = qV

Substitute v = Bqr/m into this, and solve for q/m.
 
  • #3
15
0
thanks a lot..

works out perfectly
 

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