Magnetic field in a wire with constant current 
I assume I have a wire parellel to the $z$ axis and with radius $R$. A constant current $I$ flows through it in the $z$ direction. I want to know the magnetic field inside the wire at distance $r<R$.
In the figure, the pink dots represent the flow of electrons in the z direction for $r<R$. The red dots represent the the flow of electrons in the z direction for $r>R$. I added the green magnetic field $B_o$, so that one might be wrong.
The formula I came across is:
$$  B=\frac{\mu_{0} I r}{2 \pi R^2} = \frac{\mu_{0} I_{enc} }{2 \pi r}  \text{ for } r<R$$
Why does the magnetic field (blue in figure) at $r$ only depend on the magnetic field caused by the enclosed current $I_{enc}$ (the pink dots) ? 
Why isn't there an influence by the magnetic field caused by the remainder of the current [$ I-I_{enc}$] (red dots)? For example, from $r \angle 40$ radially towards $R \angle 40$ there is current flowing in the $z$ direction (3 consecutive red dots in figure) which at $r \angle 40$ causes a magnetic field (green $B_o$ in figure) that is opposite to the one caused by the enclosed current (blue in figure), or am I wrong? 
 A: The relationship you propose only applies if the current has symmetry with respect to the axis of the wire.
Amperes law is that the line integral of the B-field around a closed loop equals the enclosed current times $\mu$.
$$\oint \vec{B}\cdot d\vec{l} = \mu I$$
To get your relationship you need to assume that on any circular path around the axis, that (I) the B-field is parallel to the path and (II) is of constant magnitude. These conditions are satisfied for circular symmetry of the current distribution.
Effectively what happens is that the B -field due to the current in the annulus around the circle exactly cancels to zero inside the circle. But I stress again, this only happens when everything is nicely symmetric.
In your diagram, I count 7 dots in the outer part of the wire in the top left quadrant and 5 in the bottom right. Therefore the current distribution does not have circular symmetry, the B-field in the wire will not have circular symmetry and you cannot easily use Ampere's law to calculate it! I suspect this is either careless drawing, or done by someone who isn't (yet) completely understanding when and how you can apply Ampere's law. So, assuming the drawing is accurate then you are not (completely) wrong. Though of course one would have to integrate the contributions from all parts of the wire.
A: The magnetic fields for the currents outside exactly cancel each other... That's just how the law of magnetism works.
A: Ampere's Law. That magnetic field that you calculated depends on $r$ and the enclosed current is $j*ds$ where $j$ is the current density and $ds$ is surface differential. So if you want to know the magnetic field outside of a wire you should take a $r>R$. 
