Exercise 5.5.7: Potentials of a magnetic field that grows linearly

$\nabla \cdot A=\partial A_x/\partial x+\partial A_y/\partial y=0$ and $V=0$, so the pair satisfies both the Coulomb and the Lorenz gauge.

With $V=0$,

since $-y{a}_x+x{a}_y=\rho a_{\varphi }$. At $\rho =5\mathrm{cm}$, $|E_{\varphi }|=0.5\times 0.05/(2\times 0.01)=1.25V/m$.

On the circle $dl=\rho d\varphi a_{\varphi }$ and $E_{\varphi }=-1.25V/m$ is constant. Upper half ($\varphi$ from $0$ to $\pi$): $E_{\varphi }\pi \rho =-1.25\times \pi \times 0.05=-0.196\mathrm{V}$. Lower half ($\varphi$ from $0$ to $-\pi$): $+0.196\mathrm{V}$. On the diameter ($y=0$) $E_x=-\frac{B_0}{2T}(-y)=0$: $0\mathrm{V}$.

Upper half minus lower half is the integral around the whole circle traversed counter-clockwise: $-0.393\mathrm{V}=-\pi {\rho }^2{B}_0/T=-d\Phi /dt$, Faraday's law.

Yes, inside the solenoid. $E$ does not depend on time, so the displacement current ${\epsilon }_0\partial E/\partial t$ is zero, consistent with $\nabla \times B=0$ for a uniform $B$. The components of $A$ are linear in $x$, $y$ and $t$, so ${\nabla }^{2}A=0$ and ${\partial }^{2}A/\partial t^2=0$: the homogeneous wave equation holds in this source-free region.