An electromagnetic wave propagating through a | vedclass

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

Question

(A) For an electromagnetic wave in a vacuum,the relationship between the amplitudes of the electric field $(E_0)$ and the magnetic field $(B_0)$ is gi

Vedclass · Accepted Answer

$E_0 k = B_0 \omega$

Vedclass · Answer

(A) For an electromagnetic wave in a vacuum,the relationship between the amplitudes of the electric field $(E_0)$ and the magnetic field $(B_0)$ is given by $E_0 = c B_0$,where $c$ is the speed of light in a vacuum.We know that the speed of light $c$ is related to the angular frequency $\omega$ and the wave number $k$ by the equation $c = \frac{\omega}{k}$.Substituting this into the first relation,we get $E_0 = \left( \frac{\omega}{k} \right) B_0$.Rearranging this equation,we obtain $E_0 k = B_0 \omega$.Therefore,the correct option is $A$.

An electromagnetic wave propagating through a vacuum is described by $E = E_0 \sin(kx - \omega t)$ and $B = B_0 \sin(kx - \omega t)$. Which of the following equations is true?

Similar Questions

Assertion $(A):$ $A$ charge moving in a circular orbit can produce electromagnetic waves.
Reason $(R):$ The source of electromagnetic waves should be in accelerated motion.

The nature of electromagnetic wave is

If a source is transmitting electromagnetic waves of frequency $8.2 \times 10^6 \,Hz$, then the wavelength of the electromagnetic waves transmitted from the source will be (in $\,m$)

How is the average energy density in an electromagnetic wave related to the magnetic field?

The electric field associated with an electromagnetic $(EM)$ wave in vacuum is given by $\vec{E} = \hat{i} 40 \cos (kz - 6 \times 10^{8} t)$,where $E$,$z$,and $t$ are in $V/m$,$m$,and $s$ respectively. The value of the wave vector $k$ is .... $m^{-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module