If an electromagnetic wave propagating through vacuum is described by $E_y=E_0 \sin (k x-\omega t)$; $B_z=B_0 \sin (k x-\omega t)$, then
$E_0 k=B_0 \omega$
$E_0 B_0=\omega k$
$E_0 \omega=B_0 k$
$E_0 B_0=\frac{\omega}{k}$
In the $EM$ wave the amplitude of magnetic field $H_0$ and the amplitude of electric field $E_o$ at any place are related as
Nearly $10 \%$ of the power of a $110\,W$ light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of $1\, m$ from the bulb to a distance of $5\,m$ is $a \times 10^{-2}\,W / m ^{2}$. The value of ' $a$ ' will be.
A mathematical representation of electromagnetic wave is given by the two equations $E = E_{max}\,\, cos (kx -\omega\,t)$ and $B = B_{max} cos\, (kx -\omega\,t),$ where $E_{max}$ is the amplitude of the electric field and $B_{max}$ is the amplitude of the magnetic field. What is the intensity in terms of $E_{max}$ and universal constants $μ_0, \in_0.$
If $\vec{E}$ and $\vec{K}$ represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by : $(\omega-$ angular frequency) :
Magnetic field in a plane electromagnetic wave is given by
$\vec B = {B_0}\,\sin \,\left( {kx + \omega t} \right)\hat jT$
Expression for corresponding electric field will be Where $c$ is speed of light