Magnetic field created by a long wire

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The magnetic field created by an electric current in a long straight wire is not conservative due to its non-zero circulation around an amperian loop that encloses the wire. While the field is locally conservative everywhere except at the wire's center (r=0), it fails to meet the criteria for global conservativeness because the line integral depends on the path taken. The path integral for a conservative vector field should be independent of the path, but in this case, it varies for different closed paths. Thus, despite some characteristics of conservativeness, the overall conclusion is that the magnetic field is not conservative. Understanding this distinction involves complex concepts such as cohomology groups and the nature of vector fields in non-simply connected spaces.
Caio Graco
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The magnetic field created by an electric current in a long straight wire is conservative or not conservative?

A field is conservative when its circulation closed path is zero.

For amperiana curve that surrounds the wire circulation is non-zero, but to a curve which does not involve the wire circulation is zero. And now?
 
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Caio Graco said:
For amperiana curve that surrounds the wire circulation is non-zero, but to a curve which does not involve the wire circulation is zero. And now?
Well, the magnetic field created by an electric current in a long straight wire will always surround/include the wire.
 
Hesch said:
Well, the magnetic field created by an electric current in a long straight wire will always surround/include the wire.

But I can choose amperiana curve so as not to move the wire and thus the movement is not null unlike the curve surrounding the wire. Then the magnetic field is conservative or not?
 
Caio Graco said:
The magnetic field created by an electric current in a long straight wire is conservative or not conservative?

A field is conservative when its circulation closed path is zero.

For amperiana curve that surrounds the wire circulation is non-zero, but to a curve which does not involve the wire circulation is zero. And now?
The path integral of a conservative vector field is independent of the path. It just depends on the end points. So if it is zero for some closed paths and not for others then it is not conservative.
 
DaleSpam said:
The path integral of a conservative vector field is independent of the path. It just depends on the end points. So if it is zero for some closed paths and not for others then it is not conservative.

The line integral is not over a simply connected space. The z-axis is not included. Things get a little tricky. This can get into some very abstract things involving cohomology groups that I would like to know more of besides the buzz-word.
 
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The scalar field about the wire is ##f= \frac{\mu_0 I}{2 \pi} [ log(r) + c ] ##.

##B = df = \frac{\mu_0 I}{2 \pi r}d\phi##. B first appears to be a conservative field.

##B = B_\phi d\phi ##

But ##\oint B_\phi d\phi = \frac{\mu_0 I}{2 \pi r} 2 \pi n ##, where n is the winding number.

##B## is locally conservative everywhere but at ##r=0##, though not globally conservative.
 
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It may be shown from the equations of electromagnetism, by James Clerk Maxwell in the 1860’s, that the speed of light in the vacuum of free space is related to electric permittivity (ϵ) and magnetic permeability (μ) by the equation: c=1/√( μ ϵ ) . This value is a constant for the vacuum of free space and is independent of the motion of the observer. It was this fact, in part, that led Albert Einstein to Special Relativity.
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