The electric field of an electromagnetic wave travelling through a medium is given by $\overline{E}(x,t) = 25 \sin(2.0 \times 10^{15}t - 10^{7}x)\hat{n}$. The refractive index of the medium is . . . . . . . (All given measurements are in $SI$ units)

An electromagnetic $(EM)$ wave is propagating in a medium with a velocity $\vec{v} = v\hat{i}$. The instantaneous oscillating electric field of this $EM$ wave is along the $+y$ axis. Then the direction of the oscillating magnetic field of the $EM$ wave will be along:

The electric field for a plane electromagnetic wave travelling in the $+y$ direction is shown. Consider a point where $\vec E$ is in the $+z$ direction. The $\vec B$ field is

In free space,an electromagnetic wave of $3 \; GHz$ frequency strikes an object of size $\frac{\lambda}{100}$,where $\lambda$ is the wavelength of the wave in free space. The phenomenon that occurs is .....

$A$ plane $EM$ wave travelling along $z-$ direction is described by $\vec E = E_0 \sin(kz - \omega t)\hat i$ and $\vec B = B_0 \sin(kz - \omega t)\hat j$. Show that:
$(i)$ The average energy density of the wave is given by $U_{av} = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} \frac{B_0^2}{\mu_0}$.
$(ii)$ The time-averaged intensity of the wave is given by $I_{av} = \frac{1}{2} c \epsilon_0 E_0^2$.

$A$ 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?