- #1

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*finite*length?

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- Thread starter Ja4Coltrane
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- #1

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- #2

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- #3

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The equation is pretty to look at, but in almost all cases e.g. (finite wire) no one has steps for trying to find a solution or an approximation to one...

...because we just transform it into a different equation which we can solve beatuifully: we define the MAGNETIC VECTOR POTENTIAL [tex] A [/tex] so that its CURL is the magnetic field [tex] B[/tex], and then [tex] A[/tex] satisfies LAPLACES EQUATION, which is a standard (easy) PARTIAL DIFFERENTIAL EQUATION.

If numerical integration of laplaces equation cannot proceed for some reason then you can always integrate the field of current elements according to the biot-savaart law.

- #4

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There's a good explanation by nrqed in this thread:

https://www.physicsforums.com/archive/index.php/t-118298.html

nrqed said:Ampere's law is valid (well, if you include the term added by Maxwell), no matter what. *BUT* the integral over \vec B \cdot \dl is in general difficult to do. It is only in a simple case like an (idealized) infinite wire that you can say that (with a circular closed loop centered around the wire) that vec B \cdot \dl = B dl and that, moreover, the magnitude of the B field is a constant which may be taken out of the integral.

It's the same as for Gauss' law for the electric field. It is *always* valid but the integral is easy to do only in specific cases with a lot of symmetry (spherical, cylindrical or planar symmetry). When a system is more complex and there is no obviosu symmetry, it is not that the law fails, it is rather than it is not very useful because it involves an integral very difficult to do. The reason books look at those special cases (infinte planes, infinite wires, infinite cylinders, etc) is that these are the only cases where the integrals involved in Gauss' and Ampere's laws are easy to carry out. It does not mean that the laws are not valid al the time, simply that they are not terribly useful.

- #5

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In fact, out of some elementary and high symmetric problems, you must use Biot-Savart.

- #6

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I FEEL SO GOOD!!

I USED THE AMPERE MAXWELL EQUATION TO FIND THE B FIELD FROM A FINITE WIRE!!!

- #7

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Care to post your solution? I'm interested.

- #8

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First, we must consider that the meaning of a current in a finite wire is the same as that of a staticaly charged rod moving with uniform velocity. With this in mind, one can determine the rate of change of electric flux through some surface by first finding the electric flux from a point charge a distance x from the surface (a circle). Then one can integrate over the length of the rod to find an expression from the flux of a full rod of length "L" and with distance "a" from the cross sectional area of the circular path. After doing this, one can find the derivative of this with respect to time (noting that this is an implicit differentiation). Now you have the rate of change of electric flux with respect to time. Plugg it into the Ampere-Maxwell law, and you will find that you can use it to determine the magnetic field.

- #9

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This is the electric field created by a charge q in movement:

[tex]\vec E={-q\over 4\pi \epsz}\left[{\vec e_{r'}\over r'^2}+

{r'\over c}{d\ \over dt}\left({\vec e_{r'}\over r'^2}\right) +

{1\over c^2}{d^2\ \over dt^2}\left(\vec e_{r'}\right)\right][/tex]

This formula and its user manual can be found in tome II chap. 21-1 of the Feynman Lectures on Physics.

You are ignoring the second term (the wake field). The third is zero if speed is constant.

- #10

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uh oh...............

I'm sort of in high school...get back to me in a year or two...lol

I'm sort of in high school...get back to me in a year or two...lol

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