I've read similar questions and answers given thereto but find them unsatisfactory. So please don't mark my question as "duplicate". The question may as well be a duplicate, but it's still waiting for an answer.
I mean: an answer is requested understandable by a 12th grade student, who knows the integral form but lacks the mathematical instruments to see equivalence with the differential form and to use the latter to derive a wave equation.
$\let\eps=\varepsilon \let\sig=\sigma \def\rA{{\rm A}} \def\rB{{\rm B}} \def\rC{{\rm C}} \def\rD{{\rm D}} \def\rE{{\rm E}} \def\rF{{\rm F}} \def\rG{{\rm G}} \def\rH{{\rm H}} \def\cR{{\cal R}} \def\bB{{\bf B}} \def\bE{{\bf E}} \def\bdl{{\bf dl}} \def\bdS{{\bf dS}} \def\D#1#2{{d#1 \over d#2}}$ I'm afraid I don't know of a real answer. Here you will find something that could - I hope - help to understand what happens when an e.m. wave propagates and why its speed in vacuum has the value it has.
A figure would help my argument but I have no time to draw it accurately. I'll give exact geometrical specifications so that a reader will be able to draw it by himself.
I'm going to show how an e.m. wave propagates between two parallel conducting plates. So this isn't a real e.m. wave in vacuum - it's sort of what is usually named a "guided wave". But it shows some relevant facts about how electric and magnetic field are created, maintained and propagated at a precisely determined speed.
The experiment
You're given a metal plate, width $b$, length $\gg b$. It's placed in the $x=0$ plane, between $y=0$ and $y=b$. Its lower end is at $z=0$, the upper end at an undetermined, positive $z$. The plate is grounded.
A second parallel plate (same width) is placed at $x=a\ll b$. Initially this plate is insulated and uncharged. Therefore no fields are present, neither electric nor magnetic.
At time $t=0$ a battery (emf = $V$) is connected to the plates, positive to the rear ($x=0$) plate, negative to the front ($x=a$) one, in points $y=z=0$. The battery is left there for the rest of experiment.
Question: what do we expect to observe, according to Maxwell's equations?
Preliminary assumptions and predictions
We expect the capacitor formed by both plates will be charged by the battery, but the charging won't be instantaneous. Some time is required for charge to flow towards increasing $z$. (Some time would also occur in order charge moves in $y$ direction, but I'll neglect that, since plates are long and thin.)
More exactly, I'll assume that at any time $t>0$ the capacitor is charged at its final amount in all points having $z<v\,t$, where $v$ (a velocity) is our main unknown. Points at $z>v\,t$ remain in the uncharged state. In other words, the "charged" state propagates in the $z$-direction, at a speed $v$.
The charged state has a potential difference $V$ between plates and an electric field $E=V/a$ in between, directed as $+x$. Then plates will bring a charge of surface density $$\sig = \eps_0 E = \eps_0\,{V \over a}$$ for positive (rear) plate, the opposite for negative one. At time $t$ the total charged area is $b\,v\,t$ and the total charge is $$Q(t) = \sig\,b\,v\,t = \eps_0\,b\,v\,t\,{V \over a}.$$
Current and magnetic field
As charged state propagates to increasing $z$ a charge must flow over plates at a rate $$I = \D Qt = \eps_0\,b\,v\,{V \over a}.$$ The current has opposite directions on both plates: positive (towards increasing $z$) on the rear positive plate, negative on the other. At time $t$ the current is present only for $z<v\,t$. The upper parts are unaffected.
A current entails a magnetic field $\bB$. It's directed along $+y$ and is uniform in the space between plates for $0<z<v\,t$. It's zero elsewhere. $B$ is computed as usual, choosing a convenient integration path. I'll take a rectangle $\cR=\rm ABCD$, whose vertices are $$\rA = (a/2, 0, z) \qquad \rB = (a/2, b, z) \qquad \rC = (a', b, z) \qquad \rD = (a', 0, z)$$ where $z$ is arbitrary but $<vt$ and $a'>a$. Then $$\oint_\cR \bB\cdot\bdl = bB$$ and from Ampère's theorem $$b\,B = \mu_0\,I = \mu_0\,\eps_0\,b\,v\,{V \over a}$$ $$B = \mu_0\,\eps_0\,v\,{V \over a}.\tag1$$
E is not conservative!
It's easy to verify the above statement. Choose some $z$ and stay put there. For all $t<z/v$ you'll see $E=0$, $B=0$. For $t>z/v$ both fields have become non null (with an abrupt variation at $t=z/v$ in our model, but don't care).
Now consider another loop: a rectangle $\cR'=\rm EFGH$, defined by $$\rE = (0, b/2, z-h) \qquad \rF = (a, b/2, z-h) \qquad \rG = (a, b/2, z+h) \qquad \rH = (0, b/2, z+h)$$ with $h$ a small positive value. At all times $(z-h)/v<t<(z+h)/v$ we have $$\oint_{\cR'} \bE \cdot \bdl = V \tag2$$ whereas the integral vanishes at all other times, before and after.
We should have expected that, since there is a variation of $\bB$ in time and very likely also its flux over $\cR'$ will be varying. Indeed ve have, for every $(z-h)/v<t<(z+h)/v$: $$\Phi_B = \int_{\cR'} \bB \cdot \bdS = -B\,a\,(vt - z + h)$$ $$\D{\Phi_B}t = -B\,a\,v.\tag3$$ Inserting (1) into (3): $$\D{\Phi_B}t = -\mu_0\,\eps_0\,v^2\,V.\tag4$$
And we are here: inserting (2) and (4) into Faraday's law we see it's satisfied if and only if $$\mu_0\,\eps_0\,v^2 = 1$$ $$v = {1 \over \sqrt{\eps_0\,\mu_0}}.$$