Curl H outside a long, thin wire of constant current

rude man
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Homework Statement


Per Maxwell's equation and my H. H. Skilling EM textbook, curl H = 0 in the absence of current density.

But consider a long, thin wire along the z axis carrying time-invariant current I. By Ampere's law, at a point (x,0) outside the wire, H = I/2πx j . But curl H computed by Cramer's rule from the curl determinant is not zero.

2. Relevant equations

curl H = J + ∂D /∂t
2πxH = I

The Attempt at a Solution



Since I is time-invariant there is no time-varying electric field so ∂D/∂t = 0.

Since H = H j = I/2πx j at (x,0),
curl H = ∂H/∂x k - ∂H/∂z i
but for a long wire, ∂H/∂z = 0 so we wind up with
curl H = dH/dx k
= -I/2π x2 k ≠ 0.

So how is the paradox resolved, please? If you're familiar with the concept of little "paddle wheels" (Skilling uses) indicating curl, even there it seems curl H ≠ 0 outside the wire.

All glory and thanks to the resolver!
 
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If you want to work in Cartesian coordinates, then first work out the general expression for the components of H in terms of x and y. Since you are going to take derivatives, you don't want to let y = 0 until after taking the derivatives.
 

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Thanks, T. Will explore! I had a faint idea I shouldn't ignore the x component of H but never paid it enough attention.
 
I switched to cylindrical coordinates, and lo & behold ... :smile:

I won't bother with the cartesian but obviously am now a Believer! Thanks again, will get to sleep a bit faster now ...
 
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The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?

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