The following travelling electromagnetic wave $E_x=0$,$E_y=E_0 \sin (kx + \omega t)$,$E_z=-2E_0 \sin (kx - \omega t)$ is:

$A$ plane electromagnetic wave travels in free space along the $x$-direction. The electric field component of the wave at a particular point of space and time is $E = 6 \; Vm^{-1}$ along the $y$-direction. Its corresponding magnetic field component, $B$, would be:

An electromagnetic wave of frequency $1 \times 10^{14} \,Hz$ is propagating along the $z$-axis. The amplitude of the electric field is $4 \,Vm^{-1}$. What is the energy density of the electric field? (Permittivity of free space $\varepsilon_0 = 8.8 \times 10^{-12} \,C^2 \,N^{-1} \,m^{-2}$)

The electric field of a plane electromagnetic wave of wave number $k$ and angular frequency $\omega$ is given by $\vec{E} = E_0(\hat{i} + \hat{j}) \sin(kz - \omega t)$. Which of the following gives the direction of the associated magnetic field $\vec{B}$?

In the propagation of electromagnetic waves,the angle between the direction of propagation and the plane of polarisation is (in $^{\circ}$)

The amplitude of the electric field associated with a light beam of intensity $\frac{15}{\pi} \text{ W m}^{-2}$ is (in $\text{ N C}^{-1}$)