Author Topic: Charged particle motion in E/B Field  (Read 325753 times)

blackjackchik

  • Newbie
  • *
  • Posts: 1
Re: Charged particle motion in E/M Field
« Reply #15 on: May 12, 2008, 03:35:37 pm »
Please send me source code too.

Fu-Kwun Hwang

  • Administrator
  • Hero Member
  • *****
  • Posts: 3062
    • Enjoy the fun of physics with simulations
Re: Charged particle motion in E/B Field
« Reply #16 on: October 24, 2008, 11:13:54 am »
Quote
My students and I ran your  "particle in a crossed E and B field" simulation  today.  It appears that the axes shown in the applet are not the axes the program reads. For example, if I click and drag the velocity vector to a point along the +x axis, the program does not show (+vx, 0, 0) as expected. Please could you check into this and contact me?
A physics teacher sent me the above email message.
I am sorry that I did not write it clearly about how to change the initial velocity in the above applet.
Because I want user to be able to change initial velocity in 3D. However, the screen is 2D.
So user can use mouse to change the initial velocity either in the X-Y(Vz=0), Y-Z(Vx=0) or Z-X(Vy=0) plane.
Please see the attached file to find out how to switch between the above 3 different planes (Click in different region to switch between plane).

ruby amit bhatt

  • Newbie
  • *
  • Posts: 1
Re: topic36
« Reply #17 on: February 12, 2010, 01:07:17 pm »
I woud like to see your code for cyclotron simulation in java[

Fu-Kwun Hwang

  • Administrator
  • Hero Member
  • *****
  • Posts: 3062
    • Enjoy the fun of physics with simulations
Re: Charged particle motion in E/B Field
« Reply #18 on: February 12, 2010, 01:16:39 pm »
Please check out emField.java (attached file under the first message) by yourself.

hawraa

  • Newbie
  • *
  • Posts: 4
Re: Charged particle motion in E/B Field
« Reply #19 on: January 07, 2016, 12:57:30 am »
plz sir, can i have a java applet for an electron between 2 charged parallel plate(2-D | path of electron in an electric field where we use this equation Electron path:  y = [eV/2dmv2]x2)...simple one...thanks