electromagnetism - Calculate electric field induced by a changing magnetic field

I want to plot the electric field (as a vector field plot) which is induced by a changing magnetic field for some simple cases.

Suppose for example that the magnetic field changes linearly (or quatratically) with t. Then you may calculate the curl $\nabla \times \vec{E}$ of $\vec{E}$ via:

$$ \nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} $$

as in this question. However this doesn't give you a unique solution for $\vec{E}$.

I guess that the solution will become unique if I add some boundary conditions. But I don't know how to do this in detail.

Suppose you have the experimental setup, that the magnetic field is (homogenous and) perpendicular to a given area $A$ in the $x_1$,$x_2$ plane (for example $A$ being a circular or rectangular area) and zero for every point $P$ which $x_1$ and $x_2$ coordinates are outside of $A$.

What is needed for the experimental setup to make the solution unique. How to calculate the solution in detail, what will it look like?

Edit: Since there are no charges in my example you have also the equation $\nabla\cdot \vec{E} = 0$. But I don't see how this helps.

Edit2

Here is what I tried so far:

If you suppose that the electric field is symmetric around the $x_3$ axis (but why can I assume this?) one could procceed as follows:

Let $\gamma$ be a circular path in the $x_1$-$x_2$-plane with radius $r$ and center $(0,0,0)$:

$$ 2\pi r |\vec{E}| = \int_{\gamma} \vec{E} d\vec{s} = -\frac{d\Phi}{dt} = -\pi r^2 \frac{d |\vec{B}(t)|}{d t} $$

which implies:

$$ |\vec{E}| = - \frac{r}{2} \frac{d |\vec{B}(t)|}{d t} $$

But this gives only the absolute value of $\vec{E}$ and not the direction. Furthermore it depends linearly on $r$ which would means that it goes to $\infty$ if $r$ goes to $\infty$. But that seems to be very unintuitive to me.

Also there is no distinguished point, so I should perhaps better assume translation invariance...

Edit 3

Bonus question: How to solve it via differential equations (not using the integral form)?

Answer

You're almost there. For the symmetry argument: first notice that Faraday's law, $\oint\textbf{E} \cdot d\textbf{l}=-\frac{d\Phi}{dt}$, looks the same as Ampère's law from electrostatics: $\oint\textbf{B}\cdot d\textbf{l}=\mu_0 I$.

Now consider a current (or a homogeneous current density) pointing in the positive $x_3$-direction. What is the direction of the magnetic field such a current would produce? Indeed, using the right-hand rule (or any of your favorite symmetry arguments) it readily follows that the $\textbf{B}$-field encircles the current, i.e. it is symmetric around the $x_3$-axis and points anticlockwise. I trust that you are familiar with symmetries of magnetic fields produced by steady currents.

Now compare the form of Faraday's and Ampère's laws. Because the laws look exactly the same, it's easy to see that the electric field due to a flux decrease in the $x_3$-direction will have the same symmetry as a magnetic field due to a current (density) in the $x_3$ -direction. Hence here the $\bf{E}$-field will also be symmetric around the $x_3$-axis and will point in the azimuthal direction! (It'll point clockwise if the flux increases, but this will follow from the calculation.)

Therefore, we can do the calculation in just the way you did, yielding

$\textbf{E}=-\frac{r}{2}\frac{dB}{dt}\hat{\phi}$ as you noted. (Here $\textbf{B}(t)=B(t) \hat{x}_3$.)

Note that in your calculation you had already assumed that $\textbf{E}$ is in the $\hat{\phi}$ direction when you said that $2\pi r |\vec{E}| = \int_{\gamma} \vec{E} d\vec{s}$, since this assumes that $\bf{E}$ and $d\bf{s}$ are parallel.

The last thing to notice is that your calculation holds when your contour $\gamma$ is in the circle $A$ where the flux changes. Let $R$ be the radius of $A$. If $\gamma$ is outside $A$, then it encloses all of the flux, hence $\frac{d\Phi}{dt}=\pi R^2\frac{dB}{dt}$ so that for $r>R$ we get

$\textbf{E}=-\frac{R^2}{2r}\frac{dB}{dt}\hat{\phi}$,

which nicely vanishes as $r\rightarrow \infty$.

Finally, notice that if we would calculate the magnetic field produced by a volume charge density $\textbf{J}=J_0\hat{x}_3$ with $J_0$ constant, and replaced in our answer $J_0$ by $-\frac{1}{\mu_0}\frac{dB}{dt}$, we'd get exactly the electric field above.