The peak electric field produced by the radiation coming from the $80\, W$ bulb at a distance of $10\, m$ is $\frac{x}{10} \sqrt{\frac{\mu_{0} c }{\pi}} \,\frac{ V }{ m }$. The efficiency of the bulb is $10\, \%$ and it is a point source. The value of $x$ is ...... .

An electromagnetic wave in vacuum has the electric and magnetic field $\vec{E}$ and $\vec{B}$,which are always perpendicular to each other. The direction of polarization is given by $\vec{X}$ and that of wave propagation by $\vec{k}$. Then:

The electric field of a plane electromagnetic wave in a medium is given by $\vec{E}(x, y, z, t) = E_0 \hat{n} e^{i k_0[(x+y+z)-ct]}$,where $c$ is the speed of light in free space. The $\vec{E}$ field is polarized in the $x-z$ plane. If the speed of the wave in the medium is $v$,then:

What is the range of frequency of $EM$ waves that are reflected back by the ionosphere?

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 magnitude of the electric field of a plane electromagnetic wave travelling in free space is $E$. If $\mu_0$ and $\varepsilon_0$ are respectively permeability and permittivity of the free space,then the magnitude of magnetic field of the wave is