NEET Properties of EM Waves MCQs & PYQs

Electromagnetic waves are transverse in nature, and all EM waves travel with the same speed in a vacuum. The average electric energy density is equal to the average magnetic energy density, so only statement c is correct.

The velocity of light in a medium is $v = \frac{1}{\sqrt{\mu \epsilon}} = \frac{1}{\sqrt{\mu_0 \mu_r \epsilon_0 \epsilon_r}}$.
Since the velocity of light in vacuum is $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, substituting this gives $v = \frac{c}{\sqrt{\epsilon_r \mu_r}}$.

The direction of propagation of an EM wave is given by $\vec{E} \times \vec{B}$. It is propagating in $+x$ direction (i.e., $\hat{i}$).
Checking option a: $(-\hat{j} + \hat{k}) \times (-\hat{j} - \hat{k}) = \hat{j} \times \hat{k} - \hat{k} \times \hat{j} = \hat{i} - (-\hat{i}) = 2\hat{i}$, which is parallel to $+x$.
Therefore, option a provides a valid combination.

Total power received by the surface is $P = Flux \times Area = 20 W/cm^2 \times 20 cm^2 = 400 W$.
The energy received in 1 minute ($60 s$) is $E = P \times t = 400 W \times 60 s$.
$E = 24000 J = 24 \times 10^3 J$.

In an electromagnetic wave, the energy is equally divided between the electric and magnetic fields.
The average energy density of the electric field equals that of the magnetic field ($u_E = u_B$).
Therefore, their ratio of contributions to the intensity is $1 : 1$.

Comparing the given equation with the standard wave equation $B = B_0 \sin(kx + \omega t)$, we get the wave number $k = \pi \times 10^3 m^{-1}$.
The wavelength is related to the wave number by $\lambda = \frac{2\pi}{k}$.
Substituting $k$, we get $\lambda = \frac{2\pi}{\pi \times 10^3} = 2 \times 10^{-3} m$.

The direction of propagation of an EM wave is given by the cross product $\vec{E} \times \vec{B}$.
Here, velocity is along $+x$ ($\hat{i}$) and electric field is along $+y$ ($\hat{j}$). We need $\hat{j} \times \vec{B} = \hat{i}$.
Since $\hat{j} \times \hat{k} = \hat{i}$, the magnetic field must be along the $+z$ direction ($\hat{k}$).

The peak value of the electric field is $E_0 = \sqrt{2} E_{rms} = 6\sqrt{2} V/m$.
The peak value of the magnetic field is $B_0 = \frac{E_0}{c} = \frac{6\sqrt{2}}{3 \times 10^8}$.
$B_0 = 2\sqrt{2} \times 10^{-8} \approx 2.828 \times 10^{-8} T \approx 2.83 \times 10^{-8} T$.

A stationary charge produces only a static electric field, while a charge moving at constant velocity produces a steady magnetic field along with it.
Only an accelerating (or oscillating) charge produces continuously changing electric and magnetic fields that sustain each other.
Therefore, an accelerating charge produces a propagating electromagnetic wave.

The momentum of the incident radiation is $p = \frac{E}{C}$.
Since the surface is perfectly reflecting, the radiation reflects back with momentum $-p$.
The momentum transferred to the surface is the change in momentum: $\Delta p = p - (-p) = 2p = \frac{2E}{C}$.