The magnetic field strength outside a (long) wire falls of as $r^{-1}$. The gist of your question seem to be what happens as $r \rightarrow 0$?

The answer is that the $r^{-1}$ dependence is only true outside the wire. Inside the wire you would need to use Ampere's law with a finite current density to work out what current was encircled by a chosen loop. e.g. For a uniform current density the magnetic field scales as $r$ inside the wire and $B \rightarrow 0$ as $r \rightarrow 0$.

EDIT: You are asking about mathematical abstractions (1-dimensional currents) rather than physical situations; this is how to proceed.

Ampere's law (in magnetostatics) says $$\oint \vec{B}\cdot d\vec{l} = \int \vec{J}\cdot d\vec{A}$$ If we consider an infinitely long wire defined by the z-axis, then taking a circular loop around the z-axis, the enclosed current is always the same.The B-field is therefore $\propto r^{-1}$ and would become infinite when $r=0$.

However, if we say instead that we have a uniform current density $\vec{J}$ that occupies a cylinder of radius $a$, then this treatment only applied for $r>a$.

If we allow $r<a$ then Ampere's law gives $$ 2\pi r B = \pi r^2 J$$ For any finite current density, then as $r \rightarrow 0$ then the right hand side goes to zero faster than the left hand side and $B \rightarrow 0$.

If instead you allow the current density to be infinite, so that a 1-d wire can carry a current, then do not be surprised that you get an infinite B-field! (You also need an infinite E-field because $J = \sigma \vec{E}$.)

The magnetic field strength outside a (long) wire falls off as $r^{-1}$. The gist of your question seem to be what happens as $r \rightarrow 0$?

The answer is that the $r^{-1}$ dependence is only true outside the wire. Inside the wire you would need to use Ampere's law with a finite current density to work out what current was encircled by a chosen loop. e.g. For a uniform current density the magnetic field scales as $r$ inside the wire and $B \rightarrow 0$ as $r \rightarrow 0$.

EDIT: You are asking about mathematical abstractions (1-dimensional currents) rather than physical situations; this is how to proceed.

Ampere's law (in magnetostatics) says $$\oint \vec{B}\cdot d\vec{l} = \int \vec{J}\cdot d\vec{A}$$ If we consider an infinitely long wire defined by the z-axis, then taking a circular loop around the z-axis, the enclosed current is always the same.The B-field is therefore $\propto r^{-1}$ and would become infinite when $r=0$.

However, if we say instead that we have a uniform current density $\vec{J}$ that occupies a cylinder of radius $a$, then this treatment only applied for $r>a$.

If we allow $r<a$ then Ampere's law gives $$ 2\pi r B = \pi r^2 J$$ For any finite current density, then as $r \rightarrow 0$ then the right hand side goes to zero faster than the left hand side and $B \rightarrow 0$.

If instead you allow the current density to be infinite, so that a 1-d wire can carry a current, then do not be surprised that you get an infinite B-field! (You also need an infinite E-field because $J = \sigma \vec{E}$.)

The magnetic field strength outside a (long) wire falls of as $r^{-1}$. The gist of your question seem to be what happens as $r \rightarrow 0$?

The answer is that the $r^{-1}$ dependence is only true outside the wire. Inside the wire you would need to use Ampere's law with a finite current density to work out what current was encircled by a chosen loop. e.g. For a uniform current density the magnetic field scales as $r$ inside the wire and $B \rightarrow 0$ as $r \rightarrow 0$.

The magnetic field strength outside a (long) wire falls of as $r^{-1}$. The gist of your question seem to be what happens as $r \rightarrow 0$?

The answer is that the $r^{-1}$ dependence is only true outside the wire. Inside the wire you would need to use Ampere's law with a finite current density to work out what current was encircled by a chosen loop. e.g. For a uniform current density the magnetic field scales as $r$ inside the wire and $B \rightarrow 0$ as $r \rightarrow 0$.

EDIT: You are asking about mathematical abstractions (1-dimensional currents) rather than physical situations; this is how to proceed.

Ampere's law (in magnetostatics) says $$\oint \vec{B}\cdot d\vec{l} = \int \vec{J}\cdot d\vec{A}$$ If we consider an infinitely long wire defined by the z-axis, then taking a circular loop around the z-axis, the enclosed current is always the same.The B-field is therefore $\propto r^{-1}$ and would become infinite when $r=0$.

However, if we say instead that we have a uniform current density $\vec{J}$ that occupies a cylinder of radius $a$, then this treatment only applied for $r>a$.

If we allow $r<a$ then Ampere's law gives $$ 2\pi r B = \pi r^2 J$$ For any finite current density, then as $r \rightarrow 0$ then the right hand side goes to zero faster than the left hand side and $B \rightarrow 0$.

If instead you allow the current density to be infinite, so that a 1-d wire can carry a current, then do not be surprised that you get an infinite B-field! (You also need an infinite E-field because $J = \sigma \vec{E}$.)