Electromagnetic Waves MCQs for NEET — Physics Questions with Answers

As E=cB

so required ratio$$\begin{gathered} \frac{\mathrm{E}}{\mathrm{B}}=\mathrm{c} \\ \frac{\mathrm{B}}{\mathrm{E}}=\frac{1}{\mathrm{c}} \end{gathered}$$

Therefore, the ratio of amplitude of magnetic field

to the amplitude of electric field for an electromagnetic

wave propagating in vacuum is equal to reciprocal

of speed of light.

Decreasing order of wavelength of various rays Microwave$>$Infrafed$>$Ultraviolet$>$Gamma rays

The time-varying electric and magnetic fields are mutually perpendicular to each other and also perpendicular to the direction of propagation of the wave.

The electric field of electromagnetic wave : 

                    $\overrightarrow{\mathbf{E}}=10 \cos ({10}^{7}t\mathit{ }\mathit{\pm }\mathit{ }kx)\hat{\mathbf{j}}$

     Amplitude=10 V/m

$$\begin{gathered} \because \mathrm{c}=\frac{\mathrm{\omega }}{\mathrm{k}} \\ \therefore 3\times {10}^8=\frac{{10}^7}{\mathrm{k}} \\ \mathrm{or} \mathrm{k}=\frac{1}{30} \\ \mathrm{or} \frac{2\mathrm{\pi }}{\mathrm{\lambda }}=\frac{1}{30} \\ \mathrm{or} \mathrm{\lambda }=188.4 \mathrm{m} \end{gathered}$$  

So, (1) and (3) statements are correct.           

Comparing the given equation :

     $$\begin{gathered} {\mathrm{E}}_{\mathrm{y}}=2.5\frac{\mathrm{N}}{\mathrm{C}}\cos \left[\left(2\mathrm{\pi }\times {10}^6\frac{\mathrm{rad}}{\mathrm{s}}\right)\mathrm{t} -\left(\mathrm{\pi }\times {10}^{-2}\frac{\mathrm{rad}}{\mathrm{m}}\right)\mathrm{x}\right] \\ \mathrm{With} \mathrm{the} \mathrm{standard} \mathrm{equation} \\ {\mathrm{E}}_{\mathrm{y}}={\mathrm{E}}_{\mathrm{o}}\cos (\mathrm{\omega t}-\mathrm{kx}) \\ \mathrm{we} \mathrm{get} \\ \mathrm{\omega }=2\mathrm{\pi f}=2\mathrm{\pi }\times {10}^6 \\ \therefore \mathrm{f}={10}^6 \mathrm{Hz} \\ \mathrm{Moreover}, \mathrm{we} \mathrm{know} \mathrm{that} \\ \frac{2\mathrm{\pi }}{\mathrm{\lambda }}=\mathrm{k}=\mathrm{\pi }\times {10}^{-2} {\mathrm{m}}^{-1} \\ \Rightarrow \mathrm{\lambda }=200 \mathrm{m} \end{gathered}$$

Hence, the wave is moving along positive x-direction with frequency ${10}^6 \mathrm{Hz}$ and wavelength of 200 m.

The velocity of electromagnetic radiation is the velocity of light (c), i.e, 

     c=$\frac{1}{\sqrt{{\mu }_o{\epsilon }_o}}$

where ${\mu }_o$ is the permeability and ${\epsilon }_o$ is the permittivity of free space.

Wave impedance $Z=\sqrt{\frac{{\mu }_r}{{\epsilon }_r}}\times \sqrt{\frac{{\mu }_0}{{\epsilon }_0}}$

$=\sqrt{\frac{50}{2}}\times 376.6=1883\Omega$

(b) 

$\beta -rays$ consist of charged particles. EM waves are chargeless and massless

Let λ be the electromagnetic wave, 
To find the direction of λ, place the right-hand fingers in the direction electric field then curl them towards the magnetic field, now the direction of λ is represented by the thumb. 
$$\begin{gathered} \lambda =\overrightarrow{E}\times \overrightarrow{B} \\ =E_0\hat{i}\times B_0\hat{j} \end{gathered}$$ 
$\overrightarrow{E}\times \overrightarrow{B}$ points in the direction of wave propagation.