An electromagnetic wave travelling in the x- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The amplitude of the magnetic field is given by $B_0 = \frac{E_0}{c}$.Given $E_0 = 27 \, Vm^{-1}$ and $c = 3 	imes 10^8 \, ms^{-1}$,we have $B_0

Vedclass · Accepted Answer

$\vec{B}(x, t) = (9 	imes 10^{-8} \, T) \hat{k} \sin [2\pi (1.5 	imes 10^{-6} \, x - 2 	imes 10^{14} \, t)]$

Vedclass · Answer

(D) The amplitude of the magnetic field is given by $B_0 = \frac{E_0}{c}$.Given $E_0 = 27 \, Vm^{-1}$ and $c = 3 	imes 10^8 \, ms^{-1}$,we have $B_0 = \frac{27}{3 	imes 10^8} = 9 	imes 10^{-8} \, T$.Since the wave travels in the $x-$ direction,the magnetic field must oscillate in the $y-z$ plane. Thus,the direction is $\hat{j}$ or $\hat{k}$.The wave number $k$ is given by $k = \frac{\omega}{c} = \frac{2\pi f}{c} = \frac{2\pi 	imes 2 	imes 10^{14}}{3 	imes 10^8} = \frac{4\pi}{3} 	imes 10^6 \approx 4.19 	imes 10^6 \, m^{-1}$.Specifically,$k = \frac{2\pi}{\lambda} = \frac{2\pi f}{c} = 2\pi 	imes \frac{2 	imes 10^{14}}{3 	imes 10^8} = 2\pi 	imes (0.666 	imes 10^6) \approx 2\pi 	imes (1.5 	imes 10^{-6})^{-1}$ is not quite right; let's calculate $k = \frac{2\pi f}{c} = \frac{2\pi 	imes 2 	imes 10^{14}}{3 	imes 10^8} = \frac{4\pi}{3} 	imes 10^6 \approx 4.18 	imes 10^6 \, m^{-1}$.Comparing with the options,the term inside the sine function is $2\pi (\frac{x}{\lambda} - ft)$. Here $\frac{1}{\lambda} = \frac{f}{c} = \frac{2 	imes 10^{14}}{3 	imes 10^8} = 0.666 	imes 10^6 \approx 1.5 	imes 10^{-6} \, m^{-1}$.Thus,the correct form is $\vec{B}(x, t) = (9 	imes 10^{-8} \, T) \hat{k} \sin [2\pi (1.5 	imes 10^{-6} \, x - 2 	imes 10^{14} \, t)]$.

An electromagnetic wave travelling in the $x-$ direction has a frequency of $2 \times 10^{14} \, Hz$ and an electric field amplitude of $27 \, Vm^{-1}$. From the options given below,which one describes the magnetic field for this wave?

Similar Questions

Which one of the following statements is not the property of light?

The electric field component of a monochromatic radiation is given by
$\vec E = 2E_0 \hat i \cos kz \cos \omega t$
Its magnetic field $\vec B$ is then given by

For a plane electromagnetic wave,$E_x = 0$,$E_y = 2.5 \frac{N}{C} \cos \left[ (2\pi \times 10^6 \frac{rad}{s})t - (\pi \times 10^{-2} \frac{rad}{m})x \right]$,and $E_z = 0$. Which of the following is true?

$A$ plane electromagnetic $(EM)$ wave is propagating along the $x$-direction. It has a wavelength of $4 \text{ mm}$. If the electric field is in the $y$-direction with a maximum magnitude of $60 \text{ Vm}^{-1}$,the equation for the magnetic field is:

$A$ plane electromagnetic wave of frequency $28 \, MHz$ travels in free space along the positive $x$-direction. At a particular point in space and time, the electric field is $9.3 \, V/m$ along the positive $y$-direction. The magnetic field (in $T$) at that point is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module