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February 26, 2020, 01:00:38 am

Like what you dislike of those things are imortant. ...Wisdom

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 Author Topic: Shape of charged beam due to deviation in angle/velocity  (Read 11738 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: February 13, 2010, 04:48:01 pm » posted from:Taipei,T\'ai-pei,Taiwan

A charge q with velocity v enter into a region with perpendular to magnetic field B, will move in a circular orbit.
The trajectory of the beam will spread out if the bean is deviated with a small angle $\theta$ (dca in the simulation) or with small deviation in velocity (dv)

The angle $\theta$ = dca * (random() -0.5); // where random() give random number between 0-1
The velocity v=V+ dv* (random() -0.5);

The trjjectory for all different possible paths are shown in the following simulation.

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macfamous
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 « Embed this message Reply #1 on: February 26, 2010, 01:04:55 am » posted from:Yogyakarta,Yogyakarta,Indonesia

may we change for the point random .. it means become positive ..
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Fu-Kwun Hwang
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 « Embed this message Reply #2 on: February 26, 2010, 10:29:26 am » posted from:Taipei,T\'ai-pei,Taiwan

The above simulation assume the beam enter into the magnetic field vertically (but there is a small angle deviation).
I think what you mean is that the beam might enter into the field with a angle.
The code has been updated, so that you can change the incident angle with slider c, or drag the red arrow to change it direction directly.

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• Post questions to be asked to help students to think, to explore.
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Like what you dislike of those things are imortant. ...Wisdom

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