Physics MCQs for NEET — Practice Questions with Answers

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.

according to the question 
The dimensions of permittivity ${\epsilon }_0=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$ 
Dimensions of electric field, E = [MLT^-3 A ^-1] 
Therefore the Dimensions of $\frac{1}{2}{\epsilon }_0{E}^2=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$ X [M²L²T^-6 A^-2] = [ML^-1T^-2]

In a longitudinal wave, the particles of the medium oscillate about their mean or equilibrium position along the direction of propagation of the wave itself. Sound waves are longitudinal in nature. In transverse waves, the particles of the medium oscillate about their mean or equilibrium position at right angles to the direction of propagation of the wave itself. Light waves being electromagnetic are transverse waves.