Does Ampere's law not hold for superconductors?

Question

I saw this question, and the answer says that

though it's possible there could be specific conditions involving superconductors where it's not safe to use Maxwell's laws, perhaps someone else can comment on this.

I thought Ampere's law should always hold if there is a steady current, because it follows from the time independent form of one of the Maxwell's equations, $\nabla \times \vec{B} = \mu_o \vec{J}$. If Ampere's law does not hold, Maxwell's equations would be violated (unless there is an unaccounted current, which changes the magnetic field). Recently I found a question in the exercises of chapter 34 in Solid State Physics by Ashcroft and Mermin,

A current of $I$ Amperes flows in a cylindrical superconducting wire of radius $r$ cm. Show that when the field produced by the current immediately outside the wire is $H_c$ (in Gauss), then $$I = 5 r H_c$$

Now, if we naively use Ampere's law on a loop just outside the surface of a cylindrical wire of infinite length carrying current $I$ in the $\hat{z}$ direction, we would get $\vec{B} = \frac{4 \pi I}{2 \pi r c} \hat{\phi} = \frac{2 I}{ r c} \hat{\phi}$ (follows from the equation $\nabla \times \vec{B} = \frac{4 \pi}{c} \vec{J}$, this is just Ampere's circuital law).

Then $\frac{I}{c} = 0.5 r B$.(The question says $5$, not $0.5$)

Also, if the current is only along the cylindrical surface, then Ampere's law does predict that the magnetic field inside the conductor is $0$, so this magnetic field distribution agrees with Meissner's effect .

It may just as well be that there is a misprint, $0.5 \rightarrow 5$. I am curious to know if something exotic goes on inside a superconductor, which causes an apparent violation of Ampere's law.

Answer 1

It turns out that the missing factor of $10$ was due to the use of mixed units in the question (the current is in SI units, while length and magnetic field are in Gaussian units). Ampere's law does hold, as expected , because otherwise Maxwell's equations won't remain valid.

Instead of using Maxwell's equations in Gaussian units, let us use SI units, and later convert the units of distance and magnetic field to CGS. We know, from Ampere's law, $B_{\text{in Tesla}} = \frac{\mu_0 I}{2\pi r} = 2 \times 10^{-7} \frac{I_\text{in Amp}}{r_{\text{in m}}}$.

Now we can substitute $B_{\text{in Tesla}} = 10^{-4} B_{\text{in Gauss}}$, and $r_\text{in m} = 10^{-2} r_\text{in cm}$, and we do get $I_\text{in Amp} =5 r_\text{in cm} B_{\text{in Gauss}}$