Ytdyklly

Let’s say we have a current wire with a current $I$ flowing. We know there is a field of $B=frac{mu_0I}{2pi r}$ by using Ampère's law, and a simple integration path which goes circularly around the wire. Now if we take the path of integration as so the surface spans doesn’t intercept the wire we trivially get a $B=0$ which is obviously incorrect.

I see that I have essentially treated it as if there is no current even present. But a similar argument is used in other situations without fault.

Take for example a conducting cylinder with a hollow, cylindrical shaped space inside. By the same argument there is no field inside.

To further illustrate my point, the derivation of the B field inside of a solenoid requires you to intercept the currents. You can’t simply do the loop inside of the air gap.

This, at least to me, seems like the same thing, and I can’t justify why one is incorrect and the other is incorrect. Please point out why I am stupid.

You aren't stupid, you are just learning things the same way we all did: the school of hard knocks. Ampere's law says that the integral around that closed path is zero, not that the field is zero at every point. What the law tells us is that the field is sometimes "positive" and sometimes "negative" on that path, and when we add up contributions from everywhere on the path, we get zero.

Your argument is incorrect. $oint vec Bcdot dvec ell$ is $0$ when no current is enclosed but this does not imply $B=0$: you cannot use $oint vec Bcdot dvec ell= Btimes 2pi r =mu_0 I_{encl}$ since $vec B$ is not constant on the loop defined by the contour: in other words, $ointvec Bcdot dvec ell$ is not $Btimes 2pi r$ unless the contour is one where $vec Bcdot dvec ell$ is constant.

You basically have to evaluate a line integral $oint_{rm C} vec B cdot dvec l$ over a closed loop which in general is difficult to do.

Here are two examples to show that even if there is a magnetic field present the line integral is zero.

In the left hand case $$oint_{rm C} vec B cdot dvec l = int_{rm WX} vec B cdot dvec l+int_{rm XY} vec B cdot dvec l+int_{rm YZ} vec B cdot dvec l+int_{rm ZX} vec B cdot dvec l = B,2R+0+(-B,2R)+0 =0$$

The right hand case is a little trickier.

$$oint_{rm C} vec B cdot dvec l = int_{0}^{2pi} B,Rdtheta ,sin theta = 0$$

and if one looks at a quadrant between $theta =0$ and $theta = frac pi 2$ the line integral is $displaystyle int_{0}^{frac{pi}{2}} B,Rdtheta ,sin theta = +BR$

Then going round in $frac pi 2$ steps the integral around the closed loop is $+BR+BR+(-BR)+(-BR) =0$ as before.