Charged particle motion in E/B Field

[Fu-Kwun Hwang]

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There are 12 translations, select to view translated versionDeutsch - German(by fpf: 23)English(by austenite: 25)Ελληνικά - Greek(by panmar56: 122)English(by sathrakk.mca: 14)Magyar - Hungarian(by nap61: 14)English(by maskun: 19)Deutsch - German(by sheizenreder: 14)فارسی - Persian(by sevin.mostafa: 3)English(by NAMANSAR: 1)Polski - Polish(by rychu: 1)Shqip - Albanian(by erjon: 4)Nederlands - Dutch(by R.H.M. Willems: 134)
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or

The magnetic field is always in the y direction, However, you can enter Ex,Ey,Ez to add electric field to the system.
The force for charge particle in the electric and magnetic field is
If the initial velocity is proportional to the magnetic field and there is no electric field presented, then the force is always perpendicular to the velocity and magnetic fiel d, which means that it is a circular motion , so , or which is a independent on the velocity of the particle. it only depends on and magnetic field B.





This java applet tries to show :

The motion of a charged particle in a uniform and constant electric/magnetic field

1. Particle starts at the origin of the coordinate system


2. Blue arrow starts from the origin shows the magnetic field (always in the Y direction)


3. Red arrow starts from the origin shows the electric field.


LEFT Click near the tip of the arrow, and drag the mouse to change the E field (Both direction and magnitude),
The black arrow on X-Z plane shows the drift velocity V_d


4. You can also key in values in the textFields to change E / B fields.


Do not forget to hit the RETURN key after enter the value into the textfield.


5. Change the coordinate system:


1. Translation : LEFT Click near the origin, and drag the mouse


2. Rotation: RIGHT Click within the window and drag the mouse


6. Change the initial Velocity V:


Left click the mouse button within the window and drag the mouse



7. The above changes depend on where you press the mouse button.


(X-Y, Y-Z or Z-X plane, watch the color of the axis)

Press start button to start the animation

1. The position, initial velocity and the period of the motion are shown at the top left region.


2. During the animation


1. If the trajectory is not on the X-Z plane :


1. Its color is GREEN


2. The black curve shows projection of the trajectory on the X-Z plane.


2. RED arrow reprenents velocity of the charge


3. BLUE arrow represents the force acting on the charge.


4. Press the LEFT mouse button will suspend the animation, press it again to resume.


5. Click the checkbox on the right to save all the traces


6. Drag the RIGHT mouse button to change the viewing angle.


1. Animation resumes when you release the mouse button -- with the same initial condition


2. If the trace is saved, you can view the trace from various angle.


3. Press Reset button to reset the condition

You might want to check out EJS version of Charged particle motion in E/B Field


Physics Law:

The Lorentz frce acting on a point charge q is given by F = m a = q ( E + V X B )

1. If E=0. ¡GF = q V X B is perpendicular to both V and B
   (example 1)
   


1. If initial velocity V is perpendicular to magnetic field B, the change will move in a circular orbit with


P = m V = q R X B


(R is the radius of the circular motion)


2. The period of the motion is not depend on the velocity


the angular frequency (also known as cyclotron frequency)


w = q B / m


1. Try with larger q/m value (let q/m=0.1) ! charge moves along B field


2. If EXB exists, the charge will drift in that direction with drift velocity V_drift = EXB/|B|².


called "E cross B drift"
( example 2)
 
( example 3)
 

Try it with a larger q/m value ( let q/m=0.1 => smaller radius R)!

Electrostatic Field	The work done by the field will increase the kinetci energy of the change particle K = q E¡ER (R is the displacement vector¡^ Particles with different velocities will spread out in E field
Magnetostatic Field	Magnetostatic field can not change the kinetic energy of the particle ( only change the direction of its velocity, F is always perpendicular to its velocity V¡^ The cyclotron frequency does not depend on the speed of the particle or the radius of the orbit. Particles with different velocities will not spread out in a uniform magnetostatic field.

Registed user can get files related to this applet for offline access.
Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
If java program did not show up, please download and install latest Java RUN TIME

There are 12 translations, select to view translated versionDeutsch - German(by fpf: 23)English(by austenite: 25)Ελληνικά - Greek(by panmar56: 122)English(by sathrakk.mca: 14)Magyar - Hungarian(by nap61: 14)English(by maskun: 19)Deutsch - German(by sheizenreder: 14)فارسی - Persian(by sevin.mostafa: 3)English(by NAMANSAR: 1)Polski - Polish(by rychu: 1)Shqip - Albanian(by erjon: 4)Nederlands - Dutch(by R.H.M. Willems: 134)
Higher number at the end means more translation been done.

or

Subject: applet
Date: Fri, 15 Oct 1999 14:09:41 -0400
From: Daniel Lemire <Daniel.Lemire@Videotron.ca>
To: hwang@phy03.phy.ntnu.edu.tw
Good day,

I checked your E/M applet today. Maybe you can help me out...

When I set both E and M fields to zero, I would expect the particule to
move at a constant speed in a constant direction (or possibly stay still).

Yet, no matter what I do, it doesn't seem to work.

I am just trying to show to a friend that a charged particule standing
still in a magnetic field won't feel a thing... if there is not E field.
Your applet would be perfect... but it doesn't give the desired result.
 

Daniel Lemire


--------------------------------------------------------------------------------

Subject: Re: applet
Date: Fri, 15 Oct 1999 22:11:30 -0400
From: Daniel Lemire <Daniel.Lemire@Videotron.ca>
To: Fu-Kwun Hwang <hwang@phy03.phy.ntnu.edu.tw>
Thanks for your quick reply!

Subject: applet
Date: Fri, 15 Oct 1999 14:09:41 -0400
From: Daniel Lemire <Daniel.Lemire@Videotron.ca>
To: hwang@phy03.phy.ntnu.edu.tw
Good day,

I checked your E/M applet today. Maybe you can help me out...

When I set both E and M fields to zero, I would expect the particule to
move at a constant speed in a constant direction (or possibly stay still).

Yet, no matter what I do, it doesn't seem to work.

I am just trying to show to a friend that a charged particule standing
still in a magnetic field won't feel a thing... if there is not E field.
Your applet would be perfect... but it doesn't give the desired result.
 

Daniel Lemire


--------------------------------------------------------------------------------

Subject: Re: applet
Date: Fri, 15 Oct 1999 22:11:30 -0400
From: Daniel Lemire <Daniel.Lemire@Videotron.ca>
To: Fu-Kwun Hwang <hwang@phy03.phy.ntnu.edu.tw>
Thanks for your quick reply!

[parrenas18]

Can get your codes in the particle charge in motion.... Because i have a science project about the said topic and i need to make a simulator... I need a code reference to make that project plsssssssss!!!!!!!!!!!!!!!!1

[delores.knipp]

Is it possible to download the files rather than having them e-mailed?  My firewall is stripping off the attachments to your e-mails.

Delores Knipp

[Fu-Kwun Hwang]

Please change you firewall setting or email acoount setting!

[guybrush]

Hello,
I observed that in the java applet "Charged particle motion in E/M Field" when there exits both magnetic(y-axis) and electric field (x-axis), and initial velocity has only i (x) component, charge drifts in Z direction. B causes charge to make rotational motion and E causes it to drift, this is what I assume. According to my assumption, charge should drift in the direction of electric field, in x direction. 
What is the reason of moving in Z direction? I would like to learn the nature of the motion. My second question is when we solve equation m a = q ( E + V X B ), we get (for conditons above) qE = m dVx/dt which means acceleration is constant in X axis; is it possible for a drifting motion to have constant acceleration?
Thanks for your help
Hasan Serce

[Fu-Kwun Hwang]

If the velocity of a charge particle is perpendicular to the B field,
The force q vXB will cause a particle to move in a circle with constant velocity.
From qvB=m v²/r so r=mv/(qB)
The E field will cause a particle to accelerate along the field direction.

Now, we have a charge particle in E and B field.
The circular motion radius r=mv/(qB), which is dependent on the velocity of the particle.
Because the velocity is changing due to E field, so the circular radius r also is changing,too.
The changing in radius cause the particle to drift in the EXB direction, so called EXB drift (E cross B drift).

[prendergast.david]

any chance i can get the source code for this as it would really help with a project im doing. thanks

[Fu-Kwun Hwang]

You should have received the source code now.

[prendergast.david]

Thanks very much. got it and helped loads. thanks!

[blackjackchik]

Please send me source code too.

[Fu-Kwun Hwang]

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]

I woud like to see your code for cyclotron simulation in java[

[Fu-Kwun Hwang]

Please check out emField.java (attached file under the first message) by yourself.

[hawraa]

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