Amperes Law:
$\oint\vec{H}.d\vec{l} = I_{enc}$
Is the current, $I_{enc}$, equal to 0 outside of the conductor ($\rho > a$) because the current density, $\vec{J}$, outside is 0?
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1$\begingroup$ Since $I=\int \textbf{J}\cdot d\textbf{S}$, trivially $I=0$ whenever $\textbf{J}=\textbf{0}$ everywhere. I your case, $I_{enc}\neq 0$ because the current density is not zero everywhere. $\endgroup$– SRSJul 10, 2017 at 15:00
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2 Answers
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The current $I$ at the location of the Amperian loop is $0$ but the enclosed current is not $0$ since the enclosed current $I_{encl}$ accounts for all the current enclosed by the loop, not just the current at the location of the loop. Thus, $I_{encl}$ by the loop is $\ne 0$ even if $I=0$ where the loop is located.
answered Jul 10, 2017 at 15:27
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The left hand side of your equation defines a loop of which one element is $d\vec l$.
The current $I_{\rm enc}$ is the current which passes though a surface whose perimeter is that loop.
So outside the wire it will be the total current passing through the conductor.
