The electric field in an electromagnetic wave is given as $\vec{E}=20 \sin \omega\left(t-\frac{x}{c}\right) \vec{j} NC ^{-1}$ Where $\omega$ and $c$ are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of $5 \times 10^{-4}\, m ^3$ will be $…..\times 10^{-13}\,J$

(Given $\varepsilon_0=8.85 \times 10^{-12}\,C ^2 / Nm ^2$ )

The electric field of a plane electromagnetic wave propagating along the $x$ direction in vacuum is $\overrightarrow{ E }= E _{0} \hat{ j } \cos (\omega t – kx )$. The magnetic field $\overrightarrow{ B },$ at the moment $t =0$ is :

A plane electromagnetic wave, has 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$….nT$

$\left(\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{\circ} \frac{ Nm ^{2}}{ C ^{2}}\right.$ and speed of $1 ight$ $\left.=3 \times 10^{8}\, ms ^{-1}\right)$

Light wave is travelling along y-direction. If the corresponding $\vec E$ vector at any time is along the $x-$axis, the direction of $\vec B$ vector at that time is along

What happens to the intensity of light from a bulb if the distance from the bulb is doubled? As a laser beam travels across the length of a room, its intensity essentially remains constant. What geometrical characteristic of $LASER$ beam is responsible for the constant intensity which is missing in the case of light from the bulb ?