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I'm learning about the Meissner effect, but I can't really grasp an example of Bleaney's book on electricity and magnetism.

Let's assume a hollow thick cylinder made of superconductor material, with it's axis in the orientation of an applied field $B_1$. We cool it under it's superconducting transition temperature, so it becomes a superconductor, and as a good one it expells the $B$ lines of its walls interior; there is no $B$ field inside the walls of the superconductor. The field inside and outside the cylinder remains being $B_1$.

After that, we change the applied $B_1$ field to a value $B_2$. The field in the exterior of the cylinder changes from $B_1$ to $B_2$, but the field inside the cylinder remains being $B_1$.

Why the interior field does not change as the exterior field? It would be surely compatible with the expulsion of the field lines, I though that the only constrain were that the $B$ inside the superconductor material must be zero.

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The cylinder is a perfect conductor.

Lenz's law says that when there is a change of flux inside a coil, this will set up an e.m.f. to resist that change. But the moment an e.m.f. is generated, this will create a current in the coil that resists the change in flux.

When the resistance is zero, any change in the flux will immediately be canceled by the change in current around the loop. And so the flux inside the cylinder is "locked" to whatever value it was when the cylinder became superconducting.

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This is where the field-line picture really becomes very useful.

The field-lines idea

The field line picture says: let us picture space as being filled with tons of imaginary filaments which point in the direction of some field $\vec F$. Now at any given point in space create a little flat surface of area $dA$ pointed along some normal vector $\hat n$, and thereby define the little-area-vector, $$d\vec A = \hat n~dA.$$ The idea is that we will spread out these filaments so that the number of filaments which pass through this surface is equal to $\vec F \cdot d\vec A.$ In the limit, of course, the whole space is infinitely filled with these filaments, called "field lines".

Now the reason we've chosen this is that we know Gauss's law, $$\oint_{\partial V} d\vec A \cdot \vec F = \iiint_V dV~~\nabla \cdot \vec F.$$ Translated into the field line picture this says that the number of field lines that permanently exit some volume, grows proportional to the total divergence of the field in that volume. And if you imagine shrinking this volume to a point about some point-of-divergence you can see that this means "field lines come out of a point of positive divergence and disappear into a point of negative divergence, otherwise they maintain a stable identity."

How this applies here

So the laws of electromagnetism include these two, $$\begin{align} \nabla\cdot E &= \rho/\epsilon_0,\\ \nabla \cdot B &= 0.\end{align}$$ These have a field-line interpretation: "$E$-field lines come out of positive charges and disappear into negative charges; $B$-field lines can do neither but must either head out to infinity or loop back around on themselves." In other words, a positive charge like a proton generates these field lines going out in all directions, and their spread over an increasing surface area $1/(4\pi r^2)$ is the reason for the $1/r^2$ dependence of the Coulomb force. Unlike this, there are no "magnetic monopoles" known to exist: magnetic field lines always must occur in loops or else loops that are implicitly closed out at infinity.

A field line is by definition a line on which $B \ne 0$ and it cannot pass continuously through an object which has $B = 0,$ so as long as the superconductor remains superconducting it is a solid boundary through which field lines cannot pass. If it starts to form a 2D surface enclosing some region, like a tube around an open center, then that region needs to have a constant number of field lines going through it, which means the $B$-field has a constant average strength inside. To put it another way, if a $B$-field line passed through the superconductor, look at the moment that it passes through: it would have to exist as normal outside of the superconductor, but vanish inside of it. But if this happens, then it must be due to a magnetic monopole at the surface of the superconductor; that is the only thing that can make a field line disappear into a point. Yet this is the very thing that the above equations say cannot happen.

You could create new field lines without passing through the superconductor of course, but what is its topology: where is it, and what is it?

You could for example use a pair of magnetic monopoles: separating them in the inside to "grow" just a normal field line, and then you could join the monopoles on the outside of this tube or push them out to infinity, and thereby create a new field line in this tube. The only problem is the one I already mentioned: there are no magnetic monopoles seen in nature, so this procedure to create such field lines without intersecting the tube of superconductor cannot be done in practice.

Or you can create a loop inside the tube and then try to stretch it out along the axis of the tube, but then the total number of field lines going through the tube in one direction gains $+1$ for the one side of the loop, but also gains $-1$ for the returning side of the loop, so the count doesn't change on average. You'd like to expel this "backwards" side from the tube, but it cannot pass through the tube! So that won't work either.

In practice as you ramp up or down a uniform magnetic field you see all of these parallel field lines going out to infinity and they all crowd in "from the sides", sliding laterally in. Since they can't pass into the tube without going through it, they leave the magnetic field inside of the tube constant and just "crowd in" the edges that have exited the tube. Or to take it the other direction, as you bring the external magnetic field to 0 these field lines from the inside want to spread out and even join back on themselves in a sort of magnetic-dipole-field, but they can only do that outside the tube. They cannot pass through the tube to get to that outside.

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The integral form of Faraday's law says that $\oint_C {\bf E} \cdot d{\bf l} = -\frac{d\Phi}{dt}$, where $\Phi$ is the magnetic flux through an open surface bounded by the curve $C$. Let the surface be a cross-section of the cylinder parallel to the end caps. A superconductor is a perfect conductor so it cannot contain any electric field, and $C$ lies within the hollow superconducting shell, so ${\bf E}$ is zero along $C$, and the magnetic flux cannot change over time. So the magnetic field strength inside the cylinder will stay at whatever value it had when the walls went superconducting.

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