Current producing electromagnetic force

[jtbell]

Do you know about the right-hand rule for magnetic fields?

Magnetic fields of currents

 

[stmartin]

Do you know about the right-hand rule for magnetic fields?

Magnetic fields of currents

I know the right-hand rule very well. But why it is like that? IT must be logical staments. What would you understand if somebody tell you: " the current is going into one direction and the magnetic field in other? "

 

[jtbell]

I suspected that you might ask that question!

It comes from Ampere's Law, one of Maxwell's four fundamental equations for electric and magnetic fields. In differential form:

$$\nabla \times \vec B = \frac {\vec J} {\epsilon_0 c^2} + \frac {1}{c^2} \frac {\partial \vec E} {\partial t}$$

The "$\nabla \times$" is the "curl" operator. It has a specific "handedness" which leads to the right-hand rule for the magnetic field $\vec B$.

More stuff about Maxwell's Equations

Now, I'm going to guess that your next question will be, "Why is Ampere's Law like that?" which is equivalent to asking "Why are Maxwell's equations like that?"

My answer to that question is basically the same one that I've already given twice before, to similar questions:

https://www.physicsforums.com/showpost.php?p=1370588&postcount=4

You probably won't be happy with that answer, but it's the best I can do.

 

[stmartin]

Can somebody explain better and fundamental the creating of magnetic field around the electrons and its direction? Waht I see you are familiar with this, please help. Thank you very much, all.

 

[stmartin]

help pleaseeee.

 

[Meir Achuz]

For an electron at rest, there is a Coulomb electric field and a magnetic dipole magnetic field, each given by standard textbook equations.
If the electron is moving with constant velocity v, a Lorentz transformation will give the E and B fields of the moving electron. This is done in advanced textbooks. It is a bit complicated because E and B are part of a second rank tensor, and the r coordinate also has to be Lorentz transformed.

 

[Mohammed0]

Why you thing that the electric field lines (electric force) will be
compressed?

If electric field lines are homogeneous space (seam) bonded to the
heterogenous seam of the 'particle',
{Or each the 'boundary condition of the other}
then homogeneous space, unlike
heterogeneous space, permits "two, or more, points to be in the same
place at the same time
(xref: Einstein - Botzmann vs. Fermi - Dirac statistics).

Therefore when [exterior sound] the
heterogeneous seam (boundary condition, fermion, 1/2 spin,
only one point in anyone place at one time) is
'displaced', it compresses the homogeneous space against itself. Since
homogenous space permits two points to be in the same place at the same
time, to what degree (xref: elasticity), or since homogeneous space
can be actively homogenous by definition seeking for all points in that
direction, or seam, or space, to actually be in the same place at the
same time, or how close to that ideal symmetry, THEN you would expect a
compression of the 'field' upon motion of the 'charge'.

I hadn't thought to be able to express it that way until your
question. Thanks.

But E-M charge, unlike gravity, is 'reverse' homogeneous, or heterogeneous, to itself, except 'opposite' charges are actively homogeneous to one another while remaining passively homogeneous to all other E-M neutral points . So besides the 'local' infinity of Euclid's zero dimensional point, is there a 'distant' infinity where [^] all the horizons meet? {But they are not parallel}. Is that the 'other side' of the local, zero dimensional, infinity? Is that what
E-M charge 'converges upon' regarding 'like' charge?

Rigid vs. elastic

Elastic in which direction on which side of the 'curve'? Deviation
from perfect symmetry expressed as a 'curve'.