The electric field of a plane electromagnetic | vedclass

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

Question

(B) The given electric field is $\overrightarrow{E} = E_{0}(\hat{x} + \hat{y}) \sin(kz - \omega t)$.The direction of propagation of the wave is along

Vedclass · Accepted Answer

$\frac{E_{0}}{c}(-\hat{x} + \hat{y}) \sin(kz - \omega t)$

Vedclass · Answer

(B) The given electric field is $\overrightarrow{E} = E_{0}(\hat{x} + \hat{y}) \sin(kz - \omega t)$.The direction of propagation of the wave is along the $+z$-axis,so $\hat{k} = \hat{z}$.The relation between the direction of propagation,electric field,and magnetic field is given by $\hat{k} = \hat{E} 	imes \hat{B}$.The unit vector for the electric field is $\hat{E} = \frac{\hat{x} + \hat{y}}{\sqrt{2}}$.Substituting the values: $\hat{z} = \left(\frac{\hat{x} + \hat{y}}{\sqrt{2}}ight) 	imes \hat{B}$.Solving for $\hat{B}$,we find $\hat{B} = \frac{-\hat{x} + \hat{y}}{\sqrt{2}}$.Since the magnitude of the magnetic field is $B_{0} = \frac{E_{0}}{c}$,the magnetic field is $\overrightarrow{B} = \frac{E_{0}}{c}(-\hat{x} + \hat{y}) \sin(kz - \omega t)$.

The electric field of a plane electromagnetic wave is given by $\overrightarrow{E} = E_{0}(\hat{x} + \hat{y}) \sin(kz - \omega t)$. Its magnetic field will be given by

Similar Questions

$A$ radio can tune in to any station in the $7.5\; MHz$ to $12\; MHz$ band. What is the corresponding wavelength band?

The amplitude of the electric field in an electromagnetic wave is $1 \ V/m$. The frequency of the wave is $5 \times 10^{14} \ Hz$. The wave propagates along the $z$-axis. What is the average energy density of the electric field in $J/m^3$?

Write the standard equation for a plane electromagnetic wave traveling in the $x$-direction.

If the amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $270 \ nT$,the amplitude of the electric field part of the wave is: (in $NC^{-1}$)

The electric field in $NC^{-1}$ of an electromagnetic wave is given by $E = 36 \sqrt{\pi} \sin(\omega t - kx)$. The average energy density of the electromagnetic wave due to the electric field is (Given: $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ Nm^2 C^{-2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module