The electric field of a monochromatic plane wave travelling in a lossless isotropic and homogenous medium is given by $\overrightarrow{E}(z,t)=E_0\left[\hat{x}\cos \right(\omega t-kz)+\hat{y}\sin (\omega t-kz\left)\right]$ in a right-handed orthogonal co-ordinate system. Which of the following is the correct polarization of the electromagnetic wave?

The correct option is Left-handed circularly polarized.

To determine the polarization state of the wave, we look at the given electric field equation:
$\overrightarrow{E}(z,t)=E_0\left[\hat{x}\cos \right(\omega t-kz)+\hat{y}\sin (\omega t-kz\left)\right]$

First, let us observe the magnitudes of the components in the x and y directions. Both components have the same amplitude $E_0$. Because the amplitudes are equal and the phase difference between the cosine and sine functions is 90 degrees (or π/2 radians), the polarization is circular.

Next, we determine the direction of rotation (handedness) of the electric field vector. By convention, the polarization handedness is determined by looking at the wave at a fixed position, say $z=0$, as it propagates towards the observer (which is in the +z direction, since the phase term is $\omega t-kz$).

Setting $z=0$, the electric field simplifies to:
$\overrightarrow{E}(0,t)=E_0\left[\hat{x}\cos \right(\omega t)+\hat{y}\sin (\omega t\left)\right]$

Let us trace the orientation of $\overrightarrow{E}$ in the xy-plane at successive times:
1. At $t=0$: $\overrightarrow{E}=E_0\hat{x}$ (pointing along the positive x-axis).
2. At $t=\frac{\pi }{2\omega }$: $\overrightarrow{E}=E_0\hat{y}$ (pointing along the positive y-axis).
3. At $t=\frac{\pi }{\omega }$: $\overrightarrow{E}=-E_0\hat{x}$ (pointing along the negative x-axis).

As time increases, the tip of the electric field vector rotates from the positive x-axis to the positive y-axis, which is a counter-clockwise rotation when looking down at the xy-plane (with the wave propagating towards you, i.e., in the +z direction).

By standard IEEE/engineering definition, if you point your left thumb in the direction of wave propagation (+z) and curl the fingers of your left hand, they follow this counter-clockwise rotation from the positive x-axis to the positive y-axis. Therefore, the wave is left-handed circularly polarized.