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| Questions for those who know electromagnetism better than I do |
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| cvanc:
(...which may very well be all of you :-DD) Hi all- I'm scratching my head on a couple 'what ifs' regarding the shape of the magnetic field that gets set up around a wire carrying DC. It's well known a thin round wire throws a circular field outside itself when carrying DC current. Right hand rule and all. My Googling shows it almost always worded exactly that way: 'thin round wire'. It made wonder what would happen if you played with each of those words in turn. What shape does the magnetic field have then? What happens if the wire isn't thin? What if it's say, inches in diameter? What happens if the wire isn't round? What if the wire is triangular in cross section? Or square? Or complex? What happens if the wire isn't even a wire, but is instead a pipe (hollow)? Is there a field both inside and outside? Thanks, I'm really curious about this (what software lets me play with stuff like this?) |
| rstofer:
You wind up dealing with Maxwell's Equations and NOBODY wants to go there voluntarily. Three people in history have understood Maxwell's Equations: Maxwell, Feynman and Einstein. The rest of us just think we know what is going on. In the 48 years since I graduated, I have used Maxwell's Equations exactly never. MATLAB can help! Google for "MATLAB Maxwell's Equations". There is some chance that what you find for MATLAB will also run on the free work-alike, Octave. https://www.mathworks.com/matlabcentral/fileexchange/70394-biot-savart-law https://www.mathworks.com/matlabcentral/fileexchange/48990-magnetic-field-simulator https://en.wikipedia.org/wiki/Maxwell%27s_equations |
| ejeffrey:
Thin doesn't matter at all. A thick wire will do just fine. So will a pipe, and the field inside the pipe will be zero if everything is perfectly symmetric. If the wire cross section is some other shape the result is qualitatively the same but quantitatively different. More than a few wire diameters away it will basically look the same but near the wire will be different. |
| CatalinaWOW:
You can use Maxwell's equations for this, but for a DC current the magnetic field is given simply by Ampere's law. The thin wire assumption avoids a problem that I have never seen discussed. Many texts and online resources give solutions for a cylinder were the current density is uniform in the cross section of the cylinder. But they never discuss whether the current density would be uniform. Given the uniform current density assumption the solution is simple geometry, calculating the area inside a radius which then gives the field. Presumably doing a full Maxwell's equation simulation including forces on the electrons starting from some initial distribution would give the actual current distribution. I have never done this, and haven't thought of or seen a simplifying assumption that would justify the uniform density assumption. But like the OP I am quite sure that I am not the sharpest tool in the shed. The answers are mostly of academic interest. As ejeffrey says, the answers converge to the same thing a few diameters outside of the wire. Or in the pipe case, a few thicknesses outside the outer skin. |
| T3sl4co1l:
Well, I wouldn't disregard theory out of hand... I certainly have a good grasp of it. Granted, people call me smart, but I'm no Feynman either. Regarding wire diameter, the principle is Ampere's law, in same class as Gauss's law and so on, which state that the sum flux through a boundary only depends on what's enclosed within that boundary. In the Amperian case, it's a closed loop curve, the enclosed area of which carries a current, and the flux is the magnetic field parallel to the path. Gauss's law applies to electric charge contained within a shell, the flux of which (electric charge) extends through, perpendicular to the shell; but also to gravity for another example, where the acceleration depends on the mass contained within a shell. As with the Earth: if we could stand at the same altitude (distance from center), and imagine all its mass compacted into an infinitesimal point, we would experience the same acceleration as with all that mass distributed as it is currently. All that matters is that the mass is contained within a spherical shell beneath our feet. (The exact shape of the shell, does and doesn't matter, due to more rigorous conditions that I won't go into detail about; suffice it to say, this works best when the shell shares the same symmetry as the problem, i.e. a spherical shell for a spherical mass distribution, and the shell is larger than said distribution.) So, too, the magnetic field around a wire, at a given distance from center, is independent of the wire diameter, so long as the wire radius is smaller. And as long as the wire is cylindrically symmetric, the law still applies within it; the magnetic field drops smoothly from its value at the surface of the wire, down to zero at the center. (The story is different for AC, where self-induction within the wire causes an opposing magnetic field, cancelling out the internal field: skin effect. Needless to say, this is more complicated, so I won't go into detail; just to note there's interesting things happening when wires or currents are changing over time.) Which also gives us the tool to understand the other conditions. Hollow: no field inside. In effect, all the surrounding currents cancel out here, but we would need a lengthy calculation to show it that way! Other (non-circular) cross sections: same outside the highest peak, and same inside the lowest valley. The exact field now depends on angle as well as distance from center, so we can't say what the field is, at every point, so simply. We might need to use a numerical solution in this case (e.g. Biot-Savart law, replacing the integral with a Riemann sum). Tim |
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