If the curl of the magnetic field, produced by a volume of current, $J$, should be given by $ \nabla \times B = \mu_0 J$, then should the curl of a such field, produced by a surface of current, $K$, be given by $ \nabla \times B = \mu_0 K$ (or should it be instead given by a delta-function: zero every where and $\infty$ only at the surface itself)?
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it would still be $J$ but $J$ would.be 2 dimensional
If you want to keep it "3 dimensional" it could.be expressed as a delta function if you wish. perhaps if your current only exists on the $x,y$ plane and.not on the $z$ plane it would be $$ J = j(x,y)\delta(z) $$ (assuming current is located when $z =0$)
so that when $z$ isn't 0 $j=0$ and when it is zero its infinite but its integral would be defined when solving for the fields
e.g $$ \int j(x,y)\delta(z) dx dy dz $$ is just $$ \int \delta(z)dz \int j(x,y) dx dy $$ now it's just $1 \times \int j(x,y) dS$ and we're back into 2 dimensions
mike stone
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jensen paull
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It wouldn't be correct just plugging $\vec{K}$ where previously there was a $\vec{J}$ (Notice the dimensions are not the same). If you're dealing with surface currents instead of volume currents, just apply the fact that $$ \vec{J}=\frac{d\vec{K}}{dl_\perp} $$ Where $dl_\perp$ is the direction perpendicular to the surface. In a clearer way, reverting the curl into a line integral with Stoke's theorem $$ \oint{\vec{B}·d\vec{r}}=\mu_0\int{\vec{J}·d\vec{s}}=\mu_0\int{\vec{K}·d\vec{l}} $$ Here $\vec{B}$ is to be integrated along the desired contour and $\vec{K}$ along the perpendicular direction to the current (but contained in the surface)
