What help us determine the polarization of electromagnetic wave . Does perpendicular electric and magnetic field determine it or does the direction of propagation ?
[Physics] How the polarization of electromagnetic wave is determined
electromagnetic-radiationpolarization
Related Solutions
I am still not sure I understand your question but from what I gathered you are confused about the behavior of an EM wave (let's say monochromatic) so I'll discuss the properties of waves of increasing difficulty (eventually returning to EM).
EM waves are not easy to imagine because they are vectorial in nature. That is, in every point of the space there is an arrow describing the magnitude and direction of the $E$ field. Let's simplify this a imagine that at every point there is just a single value, say pressure, so we are be talking about sound waves. It's useful to consider the notion of a wave front. This is a collection of points that have the same phase (think of the circles that form if you throw a stone into the water). For a monochromatic wave $$ p = p_0 \exp(i\omega t - i{\mathbf k \cdot r})$$ (characterized by the wave-vector $\mathbf k$ and frequency $\omega$), it will be just a plane. You can imagine that plane moving at the speed of propagation $c = {\omega \over |{\mathbf k}|}$ into the direction given by $\mathbf k$. Let's emphasize that all points of that plane will oscillate in the same manner; which simplifies the picture by letting us consider just one point from that plane. We are left with the following image
Imagine that the direction of the propagation of the wave is the horizontal axis, the phase of the wave is carried out on the vertical axes and the plane is represented by any point of the curve (e.g. pick the top of the wave) that travels to the right.
Now, let's consider vector fields. These can be written as ${\mathbf E} = (E_x, E_y, E_z)$ and the previous discussion applies for every component of the field. So there are now three independent oscillations each with its own phase. Except that EM field is massless and so cannot oscillate in the direction of propagation (so called longitudinal polarization); this fact can be derived from Maxwell's equations. So this leaves us with two transversal polarizations $E_x$ and $E_y$ (taken as complex numbers because we also need to take account of their phases) by identifying the $\mathbf z$ axis with the direction of propagation $\mathbf k$. As in the previous case, for monochromatic wave the wavefront will be a plane. But now we have two independent polarization and instead of simple oscillation we will in general obtain an elliptical polarization
Again, single point on the curve represents a whole wavefront plane. Note that special cases include circular (when $E_y$ lags by $\pi /2$ behind $E_x$ but has equal magnitude) and linear polarizations (when $E_y$ and $E_x$ are in phase).
Notes
We haven't discussed the magnetic part of EM wave $\mathbf B$ because it's determined by just knowing the wave vector $\mathbf k$ and the $\mathbf E$ field. The important point being that it will be perpendicular to both. So the above discussion about the travel of wavefront applies also to $\mathbf B$ field.
The monochromatic waves are not realistic as you might imagine. They are infinite in extent (whole wavefront plane oscillates and it sweeps entire universe during its travel). In reality the waves travel as concentrated wave-packets. But the above discussion is still useful because we can use Fourier analysis to pass to the frequency image of the wave and decompose the packet into monochromatic waves.
If we are not in the vacuum then the picture is much more complex. The $\mathbf E$ and $\mathbf B$ fields need not be perpendicular to the $\mathbf k$ vector (because the photons gain effective mass in the medium thanks to refractive index), the energy travels in different direction than the wave itself and so on.
You write the integral formulation of Faraday's law, but there is also the equivalent differential formalism: $$ \nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}\tag{1} $$ which can be proven in a straight-forward manner (and ought to be done in your standard E&M textbooks).
Using standard planar waves equations, $$ E_y=E_0\sin\left(kx-\omega t\right) \\ B_z=B_0\sin\left(kx-\omega t\right), \\ $$ then Equation (1) gives us that $$ kE_0\cos(kx-\omega t)=\omega B_0\cos(kx-\omega t) $$ Which enforces the well-known relation that $E_0=cB_0$. So clearly plane waves do satisfy Faraday's law.
A similar situation exists with Ampere's law, usually written as $$ \oint\mathbf B\cdot d\mathbf l=\frac{1}{c^2}\frac{d\phi_E}{dt} $$ which leads to $$ \nabla\times\mathbf B=\frac{1}{c^2}\frac{\partial\mathbf E}{\partial t} $$ (I'm ignoring the current density here).
Best Answer
You can uniquely define the polarisation of a plane wave from any of the following:
The electric field vector as a function of time $\vec{E}(t)$ and the magnetic field (or induction) $\vec{H}(t)$ (or $\vec{B}(t)$;
The wavevector $\vec{k}$ and two scalar functions of time, the latter being the transverse components (in the plane at right angles to $\vec{k}$) of either the electric or magnetic field (or magnetic induction);
In the case of a nearly monochromatic wave, the vector functions of time in 1. and the two scalars in 2. can be reduced to complex scalars, which define the amplitude and phase of sinusoidally varying quantities. The alternative 2. together with an implicit knowledge of the wavevector is what we are using when we represent a write down a pure polarisation state as a $2\times 1$ vector $\psi$ of complex scalars, called the Jones vector. The scalars define magnitude and phase of the two transverse components of $\vec{E}$ (or $\vec{H}$, $\vec{B}$, as appropriate).
If only the relative phase of the two complex scalars is important, we can represent a pure polarisation state by an implicit definition of the wavevector and three real scalars: the Stokes parameters $s_j = \psi^\dagger \sigma_j \psi$, where the $\sigma_j$ are the Pauli spin matrices.
A partially polarised state is most readily thought of in quantum terms: we consider a general partially-polarised state to be a classical probabilistic mixture of pure polarisation states, defined by the $2\times2$, Hermitian complex density matrix (as well as an implicit definition of the wavevector direction). An equivalent characteristation is through the Mueller calculus, as discussed in my answer here. The classical description, in terms of random processes, is much fiddlier, messier and subtler than the quantum, and takes a full chapter in Born and Wolf, Principles of Optics" to describe (Emil Wolf was one of the pioneers in the rigorous description of partially polarised and partially coherent light).