If the amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $270 \ nT$,the amplitude of the electric field part of the wave is: (in $NC^{-1}$)

If $\vec{E}$ and $\vec{B}$ are the electric field and magnetic field vectors of an electromagnetic wave respectively,then what is the direction of propagation of the electromagnetic wave?

The magnetic field of an $\text{E.M.}$ wave is given by $\overrightarrow{ B }=\left(\frac{\sqrt{3}}{2} \hat{ i }+\frac{1}{2} \hat{ j }\right) 30 \sin \left[\omega\left( t -\frac{ z }{ c }\right)\right]$ ($\text{S.I.}$ units). The corresponding electric field in $\text{S.I.}$ units is:

The magnetic field in a plane electromagnetic wave is given as $B = (3 \times 10^{-7} \text{ T}) \sin (3 \times 10^4 x + 9 \times 10^{12} t) \hat{j}$. The electric field of this wave is given as:

If a plane electromagnetic wave has electric field oscillations of frequency $3 \text{ GHz}$, then the wavelength of the wave is (speed of light in vacuum $= 3 \times 10^8 \text{ m/s}$) (in $\text{ m}$)

The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by $\vec{E} = 30(2 \hat{x} + \hat{y}) \sin \left[2 \pi \left(5 \times 10^{14} t - \frac{10^7}{3} z\right)\right] \text{V m}^{-1}$. Which of the following option$(s)$ is(are) correct?
[Given: The speed of light in vacuum,$c = 3 \times 10^8 \text{ m s}^{-1}$]
$(A)$ $B_x = -2 \times 10^{-7} \sin \left[2 \pi \left(5 \times 10^{14} t - \frac{10^7}{3} z\right)\right] \text{Wb m}^{-2}$.
$(B)$ $B_y = 2 \times 10^{-7} \sin \left[2 \pi \left(5 \times 10^{14} t - \frac{10^7}{3} z\right)\right] \text{Wb m}^{-2}$.
$(C)$ The wave is polarized in the $xy$-plane with a polarization angle $\theta = \tan^{-1}(0.5)$ with respect to the $x$-axis.
$(D)$ The refractive index of the medium is $2$.