[quote author=Fu-Kwun Hwang link=topic=2564.msg9723#msg9723 date=1355581761]
You have learned how to draw electric field line from http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=539.0

You can use the same method to draw magnetic field line.

I tried to open ejs model from your jar file. However, I was not able to understand what you were doing in the model (evolution page: lambdax, lambday, lambdaz???)
[/quote]
it is very difficult for me to implement it is 3D as i only have very limited knowledge, the codes you invented http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=539.0 are too advanced for me redesign in 3D.
i can remix :)

[quote author=Fu-Kwun Hwang link=topic=2564.msg9723#msg9723 date=1355581761]
I tried to open ejs model from your jar file. However, I was not able to understand what you were doing in the model (evolution page: lambdax, lambday, lambdaz???)
[/quote]
according to prof mike gallis http://www.compadre.org/osp/bulletinboard/TDetails.cfm?ViewType=2&TID=2226&CID=51612&#PID51613

For a parameterized path in space ?(t) tangent in space which "follows" a field E(r) the rate of change of the path (the "velocity" of the trajectory) is in the direction of the field. The first try:

d?(t)/dt = E(?(t))

this method works in principle, but in numerical approaches, the computed path ?(t) will have large step sizes where the field is strong, which is probably undesireable. I prefer:

d?(t)/dt = E(?(t))/|E(?(t))|

which gives the path a "constant velocity".

For a dipole field (from Jackson):

E(r) = (3n(p?n)-p)/|r-r0|3
r = field point
r0 = location of dipole
p = dipole moment
n = normal vector pointing from source to field point = (r-r0)/|r-r0|

For the example coded here, the source is at the origin (r0 = 0) and the dipole is oriented along the z axis. With the coordinates of the field point given as x,y,z (and r= x2+y2+z2)1/2 ) the field components are
Ex=3xzp/r5
Ey=3yzp/r5
Ez=3z2p/r5 -p/r3