Magnetic field produced by a finite conductive wire vs. superconducting wire As I learn more about the amazing properties and possibilities of superconductors, I wanted to understand and briefly compare the differences of when a copper wire of finite length($L_f$) creates a magnetic field($B_w$) to a superconductor.
1) Does the Biot-Savart Law apply equally to both? Or would the formula change for a superconductor? In calculating the magnitude of $B_w$ at any point away from the conductor.
2) The distribution of the magnetic field($B_w$) produced by a copper conductor would somewhat look like this:

There is a magnetic field also within the wire, in contrast, a superconductor would expel $B_w$ within, would the result be a stronger/denser field lines at certain regions?
3) Would Type I and Type II superconductors greatly differ with respect to the two questions above?
 A: The Biot-Savart law applies.   The 'perfect diamagnetism' of
a superconductor means that only the outer skin of the superconductor
carries current.   This may, or may not, be a significant difference; there is a similar effect in metals,
which is frequency-dependent: at 2 MHz, for instance, only the outer
50 microns of a copper wire is conducting (this is called the  'skin effect').
At low frequency, though, a copper wire has nearly uniform current 
distribution (no concentration at the surface), so some B field is
inside the bulk of the wire, and the inductance value for a copper
wire is thus slightly higher than that of a superconductor (even
though the magnetic field OUTSIDE the wire surface will be the same).
The absence of B field inside a superconductor means all the current
must be on the surface (to a depth of usually a fraction of one micron).
Type I superconductors, if current gets high enough, will allow some
internal magnetic field and become non-superconductors at the same time.
Type 2 superconductors, will allow some internal magnetic field while
remaining superconducting (but at higher field they also become non-superconducting).   So, a type 2 superconductor may, in a limited current range,
have some internal B field while remaining a superconductor.  As long as the
conductor shape is cylindrical, though, field outside the wire is the
same for copper, type 1 superconductor, and type 2 superconductor.
