Refocusing of paraxial electron from electron gun

Click For Summary
SUMMARY

The discussion focuses on the refocusing of paraxial electrons emitted from an electron gun when subjected to a uniform magnetic field. The key equation derived is d = sqrt(8π²mV/eB²), which describes the distance at which these electrons converge on the X-axis. The participants clarify the derivation process, correcting an initial miscalculation regarding the factor of 2 in the equation. The cyclotron frequency and the relationship between work, potential difference, and kinetic energy are also emphasized as critical components in this analysis.

PREREQUISITES
  • Understanding of electron dynamics in magnetic fields
  • Familiarity with cyclotron motion and frequency
  • Knowledge of kinetic energy and potential energy relationships
  • Basic proficiency in algebra and physics equations
NEXT STEPS
  • Study the principles of cyclotron motion in detail
  • Explore the derivation of Lorentz force in magnetic fields
  • Learn about the applications of electron guns in experimental physics
  • Investigate the effects of varying magnetic field strengths on electron trajectories
USEFUL FOR

Physics students, educators, and researchers interested in electromagnetism and particle dynamics, particularly those focusing on electron behavior in magnetic fields.

mmainak
Messages
8
Reaction score
0

Homework Statement


Electrons emitted with negligible speed from an electron gun are accelerated through a potential difference v along the X-axis. These electrons emerge from a narrow hole into a uniform magnetic field B directed along this axis. However, some of the electrons emerging from the hole make slightly divergent angles as shown in figure. Show that these paraxial electrons are refocused on the X-axis at a distance

sqrt:(8.pi^2.m.V/e.B^2).

Homework Equations

The Attempt at a Solution



m.w^2.r =B.e.V

w= B.e/m ... cyclotron frequency

T= 2.pi.m/B.e

e.V/m = f(acc).d => f= e.V/m.d

d= 1/2.f.T^2 = 1/2. e.V/m.d . 4.pi^2.m^2/B^2.e^2

=> d=sqrt:(2.pi^2.m.V/B^2.e)

2 should be 8 in the actual answer.

Where did I go wrong?
 

Attachments

  • diagram.jpg
    diagram.jpg
    8.8 KB · Views: 634
Last edited:
Physics news on Phys.org
W = qV (W=work, V=potential difference)
(mv^2)/2 = qV (v=velocity)
v = sqrt.(2qV/m)

now, time period, T = 2.pi.m/Be
so, d = v.T
d = sqrt.(8.pi^2.m.V/B^2.e)
:)
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 9 ·
Replies
9
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K