$A$ point source of electromagnetic radiation has an average power output of $800 \ W$. Find the maximum value of the magnetic field at a distance of $3.5 \ m$ from the source.

The velocity of electromagnetic radiation in a medium of permittivity $\varepsilon_0$ and permeability $\mu_0$ is given by:

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:

If the frequency of an electromagnetic wave is $60 \text{ MHz}$ and it travels in air along the $z$-direction,then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other. The wavelength of the wave (in $\text{m}$) is:

An electromagnetic wave,going through vacuum,is described by $E = E_0 \sin(kx - \omega t)$. Which of the following is independent of wavelength?

$A$ plane electromagnetic wave has a frequency of $2.0 \times 10^{10} \ Hz$ and its energy density is $1.02 \times 10^{-8} \ J/m^3$ in vacuum. The amplitude of the magnetic field of the wave is close to $....nT$. (Given: $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ Nm^2/C^2$ and speed of light $c = 3 \times 10^8 \ m/s$)