NEET Practice Questions (MCQs) with Answers & Solutions

An electromagnetic wave can be created by accelerating charges; moving charges back and forth will produce oscillating electric and magnetic fields, and these travel at the speed of light.

J.C. Maxwell showed that for logical consistency, a changing electric field must also produce a magnetic field. Further, since magnetic fields have always been associated with currents, Maxwell postulated that this current was proportional to the rate of change of the electric field and called it displacement current.

$\overrightarrow{S}=\overrightarrow{E}\times \overrightarrow{B}$

The electric and magnetic field components of a linearly polarized electromagnetic wave oscillate in such a way that they peak at the same time and they become zero at the same time but they point to different directions in space, separated by an angle of $900$

(b) Given root mean square value of electric field,

                $E_{rms}= 6 V/m$

We know that, peak value of electric field, 

   $E_o=\sqrt{2}{E}_{rms}$

$\Rightarrow E_o=\sqrt{2}\times 6 V/m$

Also, we know that, $c=\frac{E_o}{B_o}$

where, c= speed of light in vacuum 

$B_o= peak value of magnetic field$

$$\begin{gathered} \Rightarrow B_o=\frac{\sqrt{2}\times 6}{3\times {10}^8} \\ \Rightarrow B_o=\frac{8.48}{3}\times {10}^{-8} \\ \Rightarrow B_o=2.83\times {10}^{-8} T \end{gathered}$$

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.