- how Galilean transformations which are wrong (are approximately correct) give the correct answer for k?
The Lorentz prediction and the Galilean prediction must agree in the limit that $v \to 0$ (or in the limit that $c \to \infty$). This is because $v=0$ corresponds to no transformation at all, so they had better both agree there. So if you take the transformation and evaluate it for smaller and smaller $v$, you'll find that $k=1$ still has to be true.
- Why we should assume that there are two electric fields ,one in the lab frame and one in the other , but just one magnetic field in both frames?
That's just the Galilean Transformation of the EM field. To see how it relates to the relativistic case, the Lorentz Transformation of the EM field is:
$$\mathbf{E}' = \gamma \left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right ) - \left ({\gamma-1} \right ) ( \mathbf{E} \cdot \mathbf{\hat{v}} ) \mathbf{\hat{v}}$$
$$\mathbf{B}' = \gamma \left( \mathbf{B} - \frac {\mathbf{v} \times \mathbf{E}}{c^2} \right ) - \left ({\gamma-1} \right ) ( \mathbf{B} \cdot \mathbf{\hat{v}} ) \mathbf{\hat{v}}$$
When you take the limit that $c \to \infty$, we know that $\gamma \to 1$, so it just becomes:
$$\mathbf{E}' = \mathbf{E} + \mathbf{v} \times \mathbf{B}$$
$$\mathbf{B}' = \mathbf{B}$$
I assume what is meant by Faraday's law of induction is what Griffiths refers to as the "universal flux rule", the statement of which can be found in this question. This covers both cases 1) and 2), even though in 1) it is justified by the third Maxwell equation1 and in 2) by the Lorentz force law.
The universal flux rule is a consequence of the third Maxwell equation, the Lorentz force law, and Gauss's law for magnetism (the second Maxwell equation). To the extent that those three laws are fundamental, the universal flux rule is not.
I won't comment on whether the universal flux rule is intuitively true. But the real relationship is given by the derivation of the universal flux rule from the Maxwell equations and the Lorentz force law. You can derive it yourself, but it requires you to either:
- know the form of the Leibniz integral rule for integration over an oriented surface in three dimensions
- be able to derive #1 from the more general statement using differential geometry
- be able to come up with an intuitive sort of argument involving infinitesimal deformations of the loop, like what is shown here.
If you look at the formula for (1), and set $\mathbf{F} = \mathbf{B}$, you see that
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{a} &= \iint_{\Sigma} \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a} + \iint_{\Sigma} \mathbf{v}(\nabla \cdot \mathbf{B}) \cdot \mathrm{d}\mathbf{a} - \int_{\partial \Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell \\
&= - \iint_{\Sigma} \nabla \times \mathbf{E} \cdot \mathrm{d}\mathbf{a} - \int_{\partial\Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell \\
&= -\int_{\partial \Sigma} \mathbf{E} + \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell
\end{align*}
where we have used the third Maxwell equation, Gauss's law for magnetism, and the Kelvin--Stokes theorem. The final expression on the right hand side is of course the negative emf in the loop, and we recover the universal flux rule.
Observe that the first term, $\iint_\Sigma \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a}$, becomes the electric part of the emf, so if the loop is stationary and the magnetic field changes, then the resulting emf is entirely due to the induced electric field. In contrast, the third term, $-\int_{\partial\Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell$, becomes the magnetic part of the emf, so if the magnetic field is constant and the loop moves, then the resulting emf is entirely due to the Lorentz force. In general, when the magnetic field may change and the loop may also move simultaneously, the total emf is the sum of these two contributions.
If you are an undergrad taking a first course in electromagnetism, you should know the statement of the universal flux rule, and you should be able to justify it by working out specific cases using the third Maxwell equation, the Lorentz force law, or some combination thereof, but I can't imagine you would be asked for the proof of the general case from scratch, as given above.
The universal flux rule only applies to the case of an idealized wire, modelled as a continuous one-dimensional closed curve in which current is constrained to flow, that possibly undergoes a continuous deformation. It cannot be used for cases like the Faraday disc. In such cases you will need to go back to the first principles, that is, the third Maxwell equation and the Lorentz force law. There is no shortcut or generalization of the flux rule that you can apply. You should be able to do this on an exam.
1 This equation is also often referred to as "Faraday's law" (which I try to avoid) or the "Maxwell--Faraday equation/law" (which I will also avoid here because of the potential to cause confusion).
Best Answer
Your paradox is not quite as different as you think. I asked a very similar question early in 2020 under the title "Conductors and induced emfs: an inconsistency?".
For what it's worth, the conclusion I came to is this. We're prepared to define an electric field in terms of the force $\mathbf F =q\mathbf E$, that would be experienced by a stationary test charge, if one were present. I think that we must do the same for a magnetic field, defining it from $ \mathbf F = q\mathbf v \times \mathbf B$ in which $\mathbf F$ is the force that would be experienced by a charge $q$ moving at velocity $\mathbf v$, if one were there.
In this way we can talk about emfs of both kinds in non-material loops, because we imagine them being replaced by material loops with charge carriers in them.
I'm far from satisfied with this answer, but at least I'm showing solidarity.