Question

A uniform plane electromagnetic wave propagates in a lossless dielectric medium of infinite extent. The electric field in the wave has the instantaneous expression

E(r,t) = (ix √3 - iz) 2 sin(2π.10^8t + 2πx/3 + 2nz/√3 + 30 ), V/m.

Find:

a. iE, the unit vector in the direction of the wave electric field
b. the amplitude Eo of the wave
c. the wavelength of the wave
d. ik, the unit vector in the direction of propagation

Answer 1

Answer:

Explanation:

From the information given:

The instantaneous expression of the electric field in the wave is:

$E(r,t)= (i_x \sqrt{3} -i_z) 2 \ sin (2 \pi*10^8t + 2 \pi x/3+2 \pi z /\sqrt{3} + 30 ^0) , \ V/m$

To determine the unit vector in line with the wave electric field, we take the first term in E(r,t) for $I_E^\to$ as:

$I_E^\to = i_x \sqrt{3}-i_z \\ \\ I_E^\to = \dfrac{i_x \sqrt{3}-i_z}{\sqrt{3 +1}} \\ \\ \mathbf{ I_E = \dfrac{i_x\sqrt{3} -i_z}{2}}$

The amplitude is denoted by the numerical value after the first term, which is:

$\mathbf{E_o = 2}$

The wavelength can be determined by using the expression:

$\beta =\dfrac{2 \pi}{\lambda }$

from the given instantaneous expression:

$\beta = \dfrac{2 \pi}{3}x+\dfrac{2 \pi}{\sqrt{3}}z$

$\beta = \sqrt{\dfrac{2 \pi}{(3)^2}+\dfrac{(2 \pi}{(\sqrt{3})^2}}$

$\beta = \sqrt{\dfrac{2 \pi}{9}+\dfrac{2 \pi}{{3}}}$

Factorizing 2π

$\beta =2 \pi \sqrt{\dfrac{1}{9}+\dfrac{1}{{3}}}$

$\beta =2 \pi \sqrt{\dfrac{9+3}{9*3}}}$

$\beta =2 \pi \sqrt{\dfrac{12}{27}}}$

$\beta =2 \pi \sqrt{\dfrac{4*3}{9*3}}}$

$\beta =2 \pi \sqrt{\dfrac{4}{9}}}$

$\beta =2 \pi\times {\dfrac{2}{3}}}$

recall from the expression using in calculating wavelength:

$\beta =\dfrac{2 \pi}{\lambda }$

∴

equating both together, we have:

$\dfrac{2 \pi}{\lambda }= 2 \pi\times {\dfrac{2}{3}}}$

$\lambda = \dfrac{3}{2}$

λ = 1.5 m

In line with the wave direction; unit vector $i_k$ can be computed as follows:

$i_k = - [ \beta_1x +\beta_2z]/\beta$

where;

$\beta_1 = \dfrac{2 \pi }{3} \ ; \ \beta_2 = \dfrac{2 \pi }{\sqrt{3}} \ ; \ \beta = \dfrac{2 \pi \times 2}{3} ;$

∴

$i_k = - \Big[\dfrac{2 \pi}{3}x + \dfrac{2 \pi}{\sqrt{3}} z\Big]\times \dfrac{1}{\dfrac{2 \pi *2}{3}}$

$i_k = - \Big[\dfrac{x}{2} + \sqrt\dfrac{{3}}{4}} z\Big]$

$i_k = - \Big[\dfrac{1}{2}x + \sqrt{\dfrac{3}{4} }z\Big]$

$\mathbf{i_k = - \Big[0.5x +0.86 z\Big]}$