NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/
May 26, 2020, 06:19:51 am

Discovery consists of seeing what everybody has seen and thinking what nobody has thought. ..."Albert von Szent-Gyorgyi(1893-1986, 1937 Nobel Prize for Medicine, Lived to 93)"

 Pages: [1]   Go Down
 Author Topic: Electron in magnetic field  (Read 53946 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
Hero Member

Offline

Posts: 3085

 « Embed this message on: May 16, 2005, 03:09:01 pm »

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

It is assumed there are uniform magnetic field in the simulation region except white background area (there is no magnetic field).
An electron enter magnetic field is not trapped in the magnetic field. It will come out the region eventually.

You can change charge/ mass /magnetic field and the width of the white region with sliders. Enjoy it!

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
 Logged
d211275
Newbie

Offline

Posts: 1

 « Embed this message Reply #1 on: May 09, 2007, 09:51:15 am » posted from:Suzhou,Jiangsu,China

Thanks alot.It's very useful for me.
 Logged
Fu-Kwun Hwang
Hero Member

Offline

Posts: 3085

 « Embed this message Reply #2 on: May 18, 2007, 11:51:40 pm »

You are welcomed!
 Logged
Gizmo0105
Newbie

Offline

Posts: 1

 « Embed this message Reply #3 on: September 26, 2007, 04:39:22 pm » posted from:ABERDEEN,SCOTLAND,UNITED KINGDOM

Hi, thanks a lot for the simulation its very helpful.

Im currently doing an investigation for school on electrons in magentic and electrostatic fields. However I cannot perform the most common investigation (Charge to mass ratio) as this is already in the course.
Any suggestions would be greatly appreciated
 Logged
Fu-Kwun Hwang
Hero Member

Offline

Posts: 3085

 « Embed this message Reply #4 on: September 27, 2007, 12:51:09 am »

For electron with charge q move in E and B field.
The force will be q (E+vXB)
If the acceleration is a, and mass is m, then

ma= q (E+vXB) ;
so a=(q/m) (E+vXB)

If you can measure acceleration(in any way). You will know the q/m ratio.
 Logged
Ganeshchandra
Newbie

Offline

Posts: 1

 « Embed this message Reply #5 on: October 29, 2007, 06:33:22 pm »

thanks
 Logged
basit_fastian
Newbie

Offline

Posts: 1

 « Embed this message Reply #6 on: November 16, 2007, 11:41:47 am »

hi
thanks for this beautiful site and knowledge of physics keeping on the java aplet and this aplet give me a lot of knowledge to me about eletron in magnetic field.
 Logged
Alfcon
Newbie

Offline

Posts: 2

 « Embed this message Reply #7 on: September 11, 2009, 06:51:26 am » posted from:Melbourne,Victoria,Australia

I am new to all this and have just started to teach my self some of the basic aspects of Electrons and magnetic fields.  I have looked at your java program in the Topic: Electron in magnetic field, and I have a few questions.

Q = charge
m = mass
d = Magnetic field
b = ?

What does b represent ?
 Logged
Fu-Kwun Hwang
Hero Member

Offline

Posts: 3085

 « Embed this message Reply #8 on: September 11, 2009, 08:32:26 am » posted from:Taipei,T\'ai-pei,Taiwan

B stands for magnetic field : there is a uniform magnetic field in most of the simulation region except region with white background.
d: is the width of the white back ground region where there is no magnetic field (B=0).

Try to change slider value and watch what is changed in the simulation.
 Logged
Alfcon
Newbie

Offline

Posts: 2

 « Embed this message Reply #9 on: September 11, 2009, 10:33:49 am » posted from:Melbourne,Victoria,Australia

ok got it now, one more question.

If b is the magnetic field and i can be set from 1 to 10.

Is b set to 1 =  1 Tesla, b set to 10 = 10 Tesla?

 Logged
Fu-Kwun Hwang
Hero Member

Offline

Posts: 3085

 « Embed this message Reply #10 on: September 11, 2009, 02:40:41 pm » posted from:Taipei,T\'ai-pei,Taiwan

This is a good question. Because you might reach a wrong interpreatation if you was not careful when chose unit for the ststem.

$\vec{F}=q \vec{v}\times \vec{B}$ is used to calculate the motion in the simulation.
For a simulation, the same length in the simulation can be 1cm or 1m or 1km.
It depends on how you interprete the system.

In the above system, you can select the unit of B as Tesla, then the unit in space (x,y coordinate) will be meter, charge will be Coulomb. Unit for velocity will be m/s.

If you want to interprete it as a real electron:
qe=1.6 10-19 C, me=9.1 10-31kg.
m_e/q_e=5.69 10-12 kg/C,

If you want to interprete B as Tesla, then the unit of 1 second in the simulation represent 5.69 10-12s in real case. It means that the electron will move very fast. However, it can not be faster then speed of light c=3 108m/s (less than 0.01c will be fine to ignore relativistic effect), so the unit of length can not be meter(m). However, $\mu$m (10-6m) can be used.

If the unit of B is selected as gauss (1 gauss=10-4 Tesla), then unit of length can be selected as mm (10-3m).
 Logged
 Pages: [1]   Go Up
Discovery consists of seeing what everybody has seen and thinking what nobody has thought. ..."Albert von Szent-Gyorgyi(1893-1986, 1937 Nobel Prize for Medicine, Lived to 93)"