Equating mass and charge ratio of cathode rays

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SUMMARY

The discussion centers on deriving the mass-to-charge ratio (q/m) of cathode rays using the equation q/m = (2V)/(B^2r^2). The user begins with the equality of electric and magnetic forces, leading to the expression for velocity (v = E/B). By equating kinetic energy (E[kinetic] = (mv^2)/2) with the work done by the electric field (qV), the user arrives at the conclusion that q/m can be expressed as E^2/(2VB^2). The derivation is confirmed to be correct through the application of centripetal force principles.

PREREQUISITES
  • Understanding of electric and magnetic forces
  • Familiarity with kinetic energy equations
  • Knowledge of centripetal force concepts
  • Basic principles of cathode ray behavior
NEXT STEPS
  • Study the derivation of the Lorentz force law in electromagnetism
  • Learn about the motion of charged particles in magnetic fields
  • Explore the principles of electric potential energy and its relation to kinetic energy
  • Investigate applications of mass-to-charge ratio in particle physics
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism and particle dynamics, as well as researchers interested in the properties of cathode rays.

redruM
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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?
 
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F[magnetic] = qvB

Elecrons will be deflected in a circular path, so

F[centripetal] = 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.
 
thanks a lot..

works out perfectly
 

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