$A$ light bulb emits power of $800 \, W$. What is the maximum value of the magnetic field at a distance of $3.5 \, m$ from the bulb?

If $\vec{E}$ and $\vec{K}$ represent the electric field and propagation vectors of electromagnetic $(EM)$ waves in a vacuum,respectively,then the magnetic field vector $\vec{B}$ is given by: (where $\omega$ is the angular frequency).

If the electric field intensity of a uniform plane electromagnetic wave is given as $E = -301.6 \sin (kz - \omega t) \hat{a}_{x} + 452.4 \sin (kz - \omega t) \hat{a}_{y} \text{ V/m}$. Then,the magnetic intensity $H$ of this wave in $\text{A/m}$ will be (Given: Speed of light in vacuum $c = 3 \times 10^{8} \text{ m/s}$,permeability of vacuum $\mu_{0} = 4\pi \times 10^{-7} \text{ N/A}^{2}$)

$A$ 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:

The sun delivers $10^{3} \,W m^{-2}$ of electromagnetic flux on the Earth's surface. The total power that is incident on a roof of dimensions $6 \,m \times 30 \,m$ is . . . . . . .

Show that the radiation pressure exerted by an $EM$ wave of intensity $I$ on a surface kept in vacuum is $\frac{I}{c}$.