An electromagnetic (EM) wave is propagating i | 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 $\vec{S} \propto \vec{E} 	imes

Vedclass · Accepted Answer

$+z$ direction

Vedclass · Answer

(B) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is $\vec{S} \propto \vec{E} 	imes \vec{B}$.Given that the velocity $\vec{v}$ is along the $+x$ axis $(\hat{i})$ and the electric field $\vec{E}$ is along the $+y$ axis $(\hat{j})$.We know that $\vec{v} \parallel (\vec{E} 	imes \vec{B})$.Substituting the known directions: $\hat{i} = \hat{j} 	imes \vec{B}_{dir}$.Using the properties of unit vectors in a Cartesian coordinate system,we know that $\hat{j} 	imes \hat{k} = \hat{i}$.Therefore,the direction of the magnetic field $\vec{B}$ must be along the $+z$ axis $(\hat{k})$.Thus,the oscillating magnetic field is along the $+z$ direction.

An electromagnetic $(EM)$ wave is propagating in a medium with a velocity $\vec{v} = v\hat{i}$. The instantaneous oscillating electric field of this $EM$ wave is along the $+y$ axis. Then the direction of the oscillating magnetic field of the $EM$ wave will be along:

Similar Questions

$A$ sky wave with a frequency $55 \, MHz$ is incident on the $D$-region of the Earth's atmosphere at $45^\circ$. The angle of refraction is ......... $^\circ$ (electron density for the $D$-region is $400 \, \text{electrons}/cm^3$).

Suppose that the intensity of a laser is $\left(\frac{315}{\pi}\right) \ W/m^2$. The $rms$ electric field,in units of $V/m$,associated with this source is close to the nearest integer. Given: $\epsilon_0 = 8.86 \times 10^{-12} \ C^2 N^{-1} m^{-2}$ and $c = 3 \times 10^8 \ m/s$.

Which one of the following is the property of a monochromatic,plane electromagnetic wave in free space?

Electromagnetic waves are transverse in nature is evident by

Vedclass Test Series

Exam Paper Generator

Online Exam Module