It is necessary to recall the proofs of (Gauss's and Stokes's) theorems.
They use a static field or fields, a speed of which is infinite.
That's an issue if you use only part of Maxwell's equations. With all of Maxwell's equations, a time-varying charge requires the existance of currents which are the source of magnetic fields. And the net effect of electromagnetism on physical measurements is the sum of the effects of electric and magnetic fields.
Another way to see this is that electromagnetism has never predicted that signals travel with inifinite velocity, so your claim that E and B respond instantaneously to the charge and current distribution ignores that the charge and current distribution are physically constrained to vary in limited manner.
A third way to see this is to use the 4-dimensional version of the divergence theorem, which gives us:
$$\iiint _{\partial \Omega} \partial_k A^m d\omega^k = \iiint \int_{\Omega} \partial^k \partial_k A^m dV dt = \iiint \int_{\Omega} \partial^k \partial_k A^m dV dt = \iiint \int_{\Omega} \partial^2 A^m dV dt = \mu_0 \iiint \int _{\Omega} J^m dV dt $$
But a college level education in electromagnetism is usually only gained by people going on to actually get a degree in physics or an advanced degree in electrical engineering. So your ignorance is understandable. Your motivations to claim expertise in this field is not understood.
Spend a thought experiment:
1. Electric current in a conductor changed.
2. Do L-contour need time to learn about this event, if L-contour located at some distance from the conductor?
Electric current flows in circuits. So lets make this concrete.
Imagine an eternal loop of charge of radius R, which is large, and linear charge density of $$\ell = \frac{Q}{2 \pi R}$$.
Before time t=0, the loop is motionless. After time t=0, the loop rigidly rotates with linear velocity v, which gives a current of $$I = v \ell = \frac{Q v}{2 \pi R}$$. Or using the step function, $$I(t) = \frac{Q v}{2 \pi R} \theta(t)$$
Thus Ampère's circuital law suggests that for a loop of constant radius r just outside the charges wire (r << R), that the magnetic field is $$B = \frac{\mu_0}{2 \pi r} I(t) = \frac{\mu_0 Q v}{4 \pi^2 R r} \theta(t)$$ But this is not Maxwell's equation, but an equation of magnetostatics misapplied to a case where the current changes.
Maxwell realized that the Ampère's circuital law was wrong, and added a term he called the displacement current to it. The displacement current is not really a current, but it's the standard name for this term.
So since we have a lot of heavy lifting to do, lets review:
$$\vec{E}(\vec{r},t) = -\vec{\nabla}\varphi(\vec{r},t) - \frac{\partial}{\partial t}\vec{A}(\vec{r},t) \, ; \; \vec{B}(\vec{r},t) = \vec{\nabla} \times \vec{A}(\vec{r},t) \, ; \; \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \varphi(\vec{r},t) - \nabla^2 \varphi(\vec{r},t) \varphi(\vec{r},t) = \frac{\rho(\vec{r},t)}{\varepsilon_0} \, ; \; \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \vec{A}(\vec{r},t) - \nabla^2 \varphi(\vec{r},t) \vec{A}(\vec{r},t) = \mu_0 \vec{j}(\vec{r},t)$$
Our plan to solve $$\left( \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla^2 \varphi(\vec{r},t) \right) Y(\vec{r},t) = X(\vec{r},t)$$ is to express X in terms of sums of point functions and then have Y in terms of Green's functions. So we need to solve: $$\left( \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla^2 \varphi(\vec{r},t) \right) G(\vec{r},t,\vec{r}',t') = \delta(\vec{r}-\vec{r}') \delta(t - t')$$. Please verify that $$G(\vec{r},t,\vec{r}',t') = \frac{\delta \left( t - t' - \frac{\left| \vec{r}-\vec{r}' \right| }{c} \right) }{4 \pi \left| \vec{r}-\vec{r}' \right|}$$ is a solution.
So for the neighborhood near the wire at r'=0, we have a source term which is $$\mu_0 \vec{j}(\vec{r},t) = \frac{\mu_0 Q v}{2 \pi R} \theta(t) \delta(r_x)\delta(r_y)$$ (in the z-direction) which gives rise to vector potential $$\vec{A}(\vec{r},t) = \frac{\mu_0 Q v}{4 \pi^2 R} \frac{\theta \left(t - \left| \vec{r} \right| /c \right)}{\left| \vec{r} \right|}$$ (in the direction of the current) which gives us $$\vec{B}(\vec{r},t) = \vec{\nabla} \times \vec{A}(\vec{r},t) = \frac{\mu_0 Q v}{4 \pi^2 R \left| r \right|} \theta \left(ct - \left| \vec{r} \right| \right)$$ (in the direction expected ).
So the expectation is that the B field "turns on" at a later time depending on how far away you are from the current.
For further reading, I direct you to
any college textbook for a year-long course in electromagnetism or
http://farside.ph.utexas.edu/teaching/em/lectures/lectures.html (especially the section on the Time-Dependent Maxwell's equations).
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