You know that an electromagnetic wave is the combination of both electric and magnetic fields perpendicular to each other and both fields are perpendicular to the direction of propagation of the wave. Therefore the electromagnetic wave is a transverse wave.
Here we define the plane electromagnetic wave, satisfy it with Maxwell's equations and find the speed of light (electromagnetic wave). We apply the most basic idea to form an electromagnetic wave. The Figure 1 below shows electromagnetic waves traveling towards +x-direction. The electric field is along +y-axis and the magnetic field is along +z-axis. Both fields are perpendicular to each other and the direction of propagation is perpendicular to both fields (therefore called transverse wave).
We consider a plane (wave front) perpendicular to the direction of propagation of the wave at an instant that there is no field to the right of the plane and the fields are uniform on all other such planes (if you create them) to the left of it. Such a wave whose wave front is the plane is called plane wave. The plane at which the waves are in phase is wave front, that is they reach their maximum and minimum values at the same time.
For the electromagnetic plane wave, we've considered the fields are uniform at each plane perpendicular to the direction of propagation of the wave. You'll see the sinusoidal electromagnetic wave as the special case of plane electromagnetic wave, where the fields at each plane are in phase but the field values are different at different planes (because they very sinusoidally).
If you consider the electromagnetic waves as rays, and if all those rays are parallel to each other, they are the plane waves. On the other hand we have spherical waves emitted from a point source and they are not parallel. Spherical waves can't the strike a plane perpendicular to the direction of propagation at the same time because they have spherical wave front, and therefore they are not in phase over a plane.
But we can approximate them as plane waves if they have traveled long distance from the source. Once the waves reach far distances, they can be approximated as parallel to each other. The receiving aperture will be even much more smaller that our imaginary plane to receive this wave at the receiving end of the wireless link. The sinusoidal electromagnetic wave we produced before, can be regarded as the plane sinusoidal electromagnetic wave after it has traveled enough distance from the source. The "enough" distance means that the wave is far from the source that it can be considered as plane wave similar to the Earth's surface looks flat to you but the Earth is not flat.
Now we verify our simple plane electromagnetic wave we defined before illustrated in Figure 1 with Maxwell's equations. First we apply Gauss's laws (both electric and magnetic) to our wave. There is no electric charge in the electromagnetic wave but only the field, so we need to verify that the net flux (electric and magnetic) through a closed surface is zero.
To do that we make a rectangular box as shown in Figure 2, and note that, \(\vec E\) is perpendicular to both top and bottom areas of the box, that is \(\vec E\) is parallel or antiparallel to \(\vec A\), that is through one area it is \(EA\) and through another area it is \(-EA\). Through the left and right sides, \(\vec E\) is perpendicular to \(\vec A\), and the net flux is zero. Similar argument can be made for the Gauss's law for magnetism.
Now we apply Faraday's law to our electromagnetic wave. In Figure 3 you can see a rectangle \(abcd\) parallel to the xy-plane. At time \(t\), the wave has traveled a distance of \(c\,dt\) where \(c\) is the speed of the wave. The well known equation of Faraday's law is
\[\vec E \cdot d\vec l = -\frac{d\Phi_B}{dt} \tag{1}\label{1}\]
Let's first determine the rate of change of magnetic flux. The area swept by the magnetic field in that time if \(h\) is the height of the rectangle is \(A = hc\,dt\). The magnetic field is constant in our case, and it is parallel to the area vector (note that we have chosen the direction of \(\vec A\) towards +z-direction) so, the rate of change of magnetic flux is
\[\frac{d\Phi_B}{dt} = \frac{Bhc\,dt}{dt} = Bch\]
Now we evaluate the left hand side of the Equation \eqref{1}. We are going to integrate in anticlockwise direction with our choice of the direction of area vector. Can you guess why? If we had chosen the direction of area vector towards -z-direction, we would have integrated in clockwise direction. It's the Lenz's law we are talking about right now. We just wanted the negative sign to pop up in our expression after evaluating.
The \(\vec E\) is perpendicular with sides \(ab\) and \(ef\) and \(\vec E \cdot d\vec l \) is zero for these cases. And there is no \(\vec E\) for the side \(cd\). Therefore, the side \(da\) only contributes to the integral.
\[\oint \vec E \cdot d\vec l = -Eh\]
We evaluated both sides of the Equation \eqref{1}. Now substituting the values in that equation, we obtain
\[E = cB \tag{2} \label{2}\]
The result we obtained here is the relationship of electric and magnetic field and the ratio of electric field to magnetic field is the speed of the wave, that is \(c = E/B\). Let's quickly move to satisfy our wave with the Ampere-Maxwell's law now. The equation of this law (without the conduction current; if you do not know Maxwell's equations, you can not understand electromagnetic waves) is
\[\oint \vec B \cdot d\vec l = \mu_0\epsilon_0\frac{d\Phi_E}{dt} \tag{3} \label{3}\]
Similarly we evaluate the right hand side of the above equation especially \(d\Phi_E/dt\). The Figure 3 shows the similar rectangle \(abcd\) we used before but it's plane is parallel to xz-plane now.
The process is the same, in time \(dt\) the area swept out by the wave (electric field ) \(A = hc\,dt\) if \(h\) is the height of the rectangle shown. And the electric field is parallel to the area vector (we've chosen the direction of area vector towards +y-axis). Therefore, the rate of change of electric flux is
\[\frac{d\Phi_E}{dt} = \frac{Ehcdt}{dt} = Ehc\]
Now in Ampere's law, which direction you prefer to integrate, anticlockwise around the rectangle right? The magnetic field. Similar to the case for electric field, only the side \(da\) contributes to the integral, and therefore
\[\oint \vec B \cdot d\vec l = Bh\]
Now all you got to do is substitute the corresponding values in Equation \eqref{3} and you get
\[B = \mu_0\epsilon_0Ec \tag{4} \label{4}\]
The above equation gives \(c = B/\mu_0\epsilon_0E\), and you can eliminate \(E\) from Equation \eqref{2} and get the value of \(c\) to be
\[c = \frac{1}{\sqrt{\mu_0\epsilon_0}} \tag{5} \label{5}\]
What we did here is that we satisfied Maxwell's equations to our plane electromagnetic wave and found the expression of the speed of the wave. The speed found if you put the values of \(\mu_0\) and \(\epsilon_0\) is approximately \(3 \times 10^8 \text{m/s}\) in vacuum. The speed of light in other mediums can be found by replacing \(\mu_0\) by \(\mu\) (permeability of the dielectric) and \(\epsilon_0\) by \(\epsilon\) (permittivity of the dielectric). Since light is the electromagnetic wave, we just found the speed of light!
Why is the ratio of \(E\) to \(B\) so large? You'll see in the energy carried by the electromagnetic wave that the energy densities due to both fields are indeed equal to each other. You can not compare these fields directly as these fields are measured in different units.
