The electric field of a plane electromagnetic wave is given by $\overrightarrow{E} = E_{0}(\hat{x} + \hat{y}) \sin(kz - \omega t)$. Its magnetic field will be given by

$A$ plane electromagnetic wave of frequency $20\, MHz$ propagates in free space along the $x$-direction. At a particular space and time, $\overrightarrow{E} = 6.6 \hat{j} \, V/m$. What is $\overrightarrow{B}$ at this point?

In fog, photographs of the objects taken with infrared radiations are more clear than those obtained during visible light because

Statement-$I$ :- During propagation of electromagnetic wave,$\vec{E}$,$\vec{B}$ and direction of propagation are perpendicular to each other.
Statement-$II$ :- During propagation of electromagnetic wave,the energy density due to electric and magnetic fields are equal.

In a plane electromagnetic wave travelling in free space,the electric field component oscillates sinusoidally at a frequency of $2.0 \times 10^{10} \text{ Hz}$ and amplitude $48 \text{ V m}^{-1}$. Then the amplitude of the oscillating magnetic field is: (Speed of light in free space $= 3 \times 10^8 \text{ m s}^{-1}$)

The electric field in a plane electromagnetic wave is given by $\overrightarrow{E} = 200 \cos \left[ (0.5 \times 10^{3} \text{ m}^{-1}) x - (1.5 \times 10^{11} \text{ rad/s}) t \right] \hat{j} \text{ V/m}$. If this wave falls normally on a perfectly reflecting surface having an area of $100 \text{ cm}^{2}$,and the radiation pressure exerted by the electromagnetic wave on the surface is $\frac{x}{10^{9}} \text{ N/m}^{2}$,find the value of $x$.