The electric field part of an electromagnetic | vedclass

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

Question

(D) The given equation for the electric field is:$E = 3.1 \cos [ (1.8) y + (5.4 	imes 10^8) t ] \hat{i}$  .......... $(i)$Comparing this with the sta

Vedclass · Accepted Answer

$3.5$

Vedclass · Answer

(D) The given equation for the electric field is:$E = 3.1 \cos [ (1.8) y + (5.4 	imes 10^8) t ] \hat{i}$  .......... $(i)$Comparing this with the standard wave equation:$E = E_0 \cos (ky + \omega t)$  .......... $(ii)$We identify the wave number $k$ as:$k = 1.8 \, rad \, m^{-1}$The relationship between wavelength $\lambda$ and wave number $k$ is given by:$\lambda = \frac{2 \pi}{k}$Substituting the value of $k$ and using $\pi \approx 3.14159$:$\lambda = \frac{2 	imes 3.14159}{1.8} \approx 3.49 \, m$Rounding to the nearest provided option,we get:$\lambda \approx 3.5 \, m$

The electric field part of an electromagnetic wave in vacuum is
$E = 3.1 \, N C^{-1} \cos [ (1.8 \, rad \, m^{-1}) y + (5.4 \times 10^8 \, rad \, s^{-1}) t ] \hat{i}$
The wavelength of this part of the electromagnetic wave is ...... $m$.

Similar Questions

$A$ plane electromagnetic wave having a frequency $n = 23.9\, GHz$ propagates along the positive $z-$ direction in free space. The peak value of the electric field is $60\, V/m$. Which among the following is the acceptable magnetic field component in the electromagnetic wave?

The magnetic field of a plane electromagnetic wave is given by $\overrightarrow{B} = 3 \times 10^{-8} \cos (1.6 \times 10^3 x + 48 \times 10^{10} t) \hat{j} \text{ T}$. The associated electric field is:

The direction of the Poynting vector represents:

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

When an electromagnetic wave enters the ionized layer of the ionosphere,the relative permittivity of the ionized layer is found to be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The electric field part of an electromagnetic wave in vacuum is$E = 3.1 \, N C^{-1} \cos [ (1.8 \, rad \, m^{-1}) y + (5.4 \times 10^8 \, rad \, s^{-1}) t ] \hat{i}$The wavelength of this part of the electromagnetic wave is ...... $m$.