Use of Amperian loop to calculate the $B$-field inside a co-axial wire 

The problem asks to calculate the magnetic field at point $a$ of a co-axial cable with a wire at the center surrounded by a rubber layer, and then another hollowed cylindrical wire followed by a second rubber layer. The currents are as indicated with each going into and out of the page. 

The solution the professor gave me was using an Amperian loop surrounding the centered wire with a radius of $d$. The resulting magnetic field at $a$ is identical to that generated by a single infinitely long current-carrying wire. The professor argues that because of symmetry, the magnetic field of the cylindrical wire does not contribute to the total magnetic field experienced at point $a$. 

My question is, because point $a$ is not at the center of the circle, how does symmetry come into play in canceling out the magnetic field caused by the outer wire at point $a$?

 A: Ampere's law in magnetostatics is
$$ \oint \vec{B} \cdot d\vec{l} = \mu_0 \mu_r I_c ,$$
where  the left hand side is a closed line integral around a loop and the right hand side contains $I_c$, the total current passing through the same loop.
The symmetry that your Professor is talking about is the cylindrical symmetry which allows you to write down the left hand side of this equation in a simpler way. This symmetry means that the magnitude of the B-field must only depend on $R$ and not on $z$ or $\phi$. It also means that the B-field must be in be in the $\phi$ direction: it cannot be in the $R$ direction because it would have a non-zero divergence at the centre and the Biot-Savart law tells us it must be perpendicular to the current.
If this is the case, and we choose to evaluate the line integral in a loop around the z-axis, such that $d\vec{l} = dl \hat{\phi}$, where $\hat{\phi}$ is a unit vector, then Ampere's law becomes
$$ \oint \vec{B} \cdot dl\ \hat{\phi} = 2\pi R B_{\phi} = \mu_0 \mu_r I_c$$
Because the right hand side features only the current enclosed by the loop then for point $a$ the current in the outer conductor can be ignored as your Prof suggests.
Why symmetry is required to solve this problem easily
If the outer conductor did not have cylindrical symmetry than you could not necessarily assume that the B-field was entirely in the $\phi$ direction.
One can imagine two forms of symmetry breaking. (i) The cable is not co-axial. This wouldn't matter - because of symmetry around a new axis, you could still assume that there was no B-field in the region between the two conductors that was due to the current in the outer conductor. The B-field at $a$ would still just depend on the distance from $a$ to the centre of the inner wire. (ii) The outer cable had a non-uniform current density, so that the current running through it depended on $\phi$ or it had a uniform current density, but its thickness varied with $\phi$. This does matter and means the B-field due to the outer conductor at point $a$ would not be zero.
To demonstrate the latter consider a hollow cylindrical conductor with a uniform current density, but where the central hole is hollowed out using a central axis displaced by a small amount from the axis of the outer perimeter, so that the thickness of the conductor varies with $\phi$. To calculate the B-field at $a$ due to this arrangement one could consider the field at $a$ imagining the conductor was solid and with a central axis defined by the outer perimeter. Then, subtract from this the magnetic field that would be due to a solid cylindrical conductor with radius equal to the inner boundary but with a displaced
central axis. The result would be a non-zero field perpendicular to a line joining the centres of the respective inner and outer cylinder axes. You would then have to add this non-zero field to the field due to an inner wire.
A: This system enjoys the cylindrical symmetry. It allows you to conclude that the magnetic field does not depend on $\phi$ or $z$, just $r$. Then you can use the integrals over loops and 4-th Maxwell equation (as your professor suggested) to show that the result indeed does not depend on $I_2$.
