A plane electromagnetic wave propagates along | vedclass

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

Question

(B) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is $\overrightarrow{S} = \frac{1}{\

Vedclass · Accepted Answer

$E_y, B_z$

Vedclass · Answer

(B) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is $\overrightarrow{S} = \frac{1}{\mu_0} (\overrightarrow{E} 	imes \overrightarrow{B})$.Thus,the direction of propagation is along $\overrightarrow{E} 	imes \overrightarrow{B}$.Given that the wave propagates along the $+x$ direction,we have $\hat{i} = \hat{E} 	imes \hat{B}$.If $\overrightarrow{E}$ is along the $y$-axis $(\hat{j})$ and $\overrightarrow{B}$ is along the $z$-axis $(\hat{k})$,then $\hat{j} 	imes \hat{k} = \hat{i}$.Therefore,the components of the electric field and magnetic field are $E_y$ and $B_z$ respectively.

$A$ plane electromagnetic wave propagates along the $+x$ direction in free space. The components of the electric field,$\overrightarrow{E}$ and magnetic field,$\overrightarrow{B}$ vectors associated with the wave in a Cartesian frame are:

Similar Questions

If the frequency of an electromagnetic wave is $60 \text{ MHz}$ and it travels in air along the $z$-direction,then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other. The wavelength of the wave (in $\text{m}$) is:

$A$ plane electromagnetic wave of frequency $35 \ MHz$ travels in free space along the $X$-direction. At a particular point (in space and time) $\overrightarrow{E} = 9.6 \ \hat{j} \ V/m$. The value of the magnetic field at this point is:

An electromagnetic wave,going through vacuum,is described by $E = E_0 \sin(kx - \omega t)$. Which of the following is independent of wavelength?

The electric field part of an electromagnetic wave in a medium is represented by $E_x = 0$; $E_y = 2.5 \frac{N}{C} \cos \left[ (2\pi \times 10^6 \frac{rad}{s})t - (\pi \times 10^{-2} \frac{rad}{m})x \right]$; $E_z = 0$. The wave is:

To radiate $EM$ signal of wavelength $\lambda$ with high efficiency,the antennas should have a minimum size equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module