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How can I find the magnetic field created by an infinite straight wire of radius $R$ and current density $\vec{j}$ using only the local form of Ampere's law : $\vec{\nabla}\times \vec{B} = \mu_0 \vec{j}$ ?

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Outside of the wire where $r>R$, the curl of the B-field is zero (no current density) and inside, it is non-zero. Using symmetry, we know that if $\vec{j}$ only depends on r, which is often the case, then the B-field must also only depend on r. $\vec{B}=\vec{B}(r)$. If we look at the curl in cylindrical coordinates, we find that $\nabla \times \vec{B}=-\frac{dB_z(r)}{dr}\hat{\phi}+\frac{1}{r}\frac{d(rB_{\phi}(r))}{dr}\hat{z}$. Inside, we have $\vec{j}=j(r)\hat{z}$ for a current running along the length of the wire. So that means $\frac{1}{r}\frac{d(rB_{\phi}(r))}{dr}=\mu_0j(r)$. Integrate this to get your answer. Outside of the wire, we do the same thing but with $\vec{j}=0$, and impose the boundary condition at $r=R$ to fix the integration constant. It's definitely easier to use the integral form for this type of problem though.

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  • $\begingroup$ I actually just found the solution seconds ago and was about to edit my post, but thanks anyway :) . And I agree it's more natural with the integral form. $\endgroup$ – mwa1 May 8 '15 at 17:14

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