If $\overrightarrow E$ and $\overrightarrow B$ are the electric and magnetic field vectors of electromagnetic waves,then the direction of propagation of the electromagnetic wave is along the direction of:

$A$ light beam is described by $E = 800 \sin \omega (t - x/c)$. An electron is allowed to move normal to the propagation of the light beam with a speed of $3 \times 10^{7} \text{ m/s}$. What is the maximum magnetic force exerted on the electron?

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 $= 3 \times 10^8 \text{ m s}^{-1}$)

$A$ radio station has two channels. One is $AM$ at $1020 \ kHz$ and the other is $FM$ at $89.5 \ MHz$. For good results,you will use:

$A$ plane electromagnetic wave is moving in free space with velocity $c = 3 \times 10^8 \ m/s$ and its electric field is given as $\vec{E} = 54 \sin(kz - \omega t) \hat{j} \ V/m$, where $\hat{j}$ is the unit vector along the $y$-axis. The magnetic field vector $\vec{B}$ of the wave is:

$A$ point source of electromagnetic radiation has an average power output of $800 \, W$. The maximum value of the electric field at a distance $4.0 \, m$ from the source is .... $V/m$.