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 $\vec{E} = E_0 \hat{n} e^{i k_0[(x+y+z) - ct]}$.Comparing this with the standard form $\vec{E} = E_0 \hat{n} e^{i(\vec

Vedclass · Accepted Answer

$\hat{n} = \frac{\hat{i} - \hat{k}}{\sqrt{2}}; v = \frac{c}{\sqrt{3}}$

Vedclass · Answer

(B) The given electric field is $\vec{E} = E_0 \hat{n} e^{i k_0[(x+y+z) - ct]}$.Comparing this with the standard form $\vec{E} = E_0 \hat{n} e^{i(\vec{k} \cdot \vec{r} - \omega t)}$, we identify the wave vector $\vec{k} = k_0(\hat{i} + \hat{j} + \hat{k})$.The magnitude of the wave vector is $k = |k_0(\hat{i} + \hat{j} + \hat{k})| = k_0 \sqrt{1^2 + 1^2 + 1^2} = k_0 \sqrt{3}$.The speed of the wave in the medium is $v = \frac{\omega}{k} = \frac{k_0 c}{k_0 \sqrt{3}} = \frac{c}{\sqrt{3}}$.The refractive index $n = \frac{c}{v} = \sqrt{3}$.Since the wave is transverse, $\vec{E} \cdot \vec{k} = 0$, which implies $\hat{n} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$.Given that $\vec{E}$ is polarized in the $x-z$ plane, $\hat{n}$ must be in the $x-z$ plane, so $\hat{n} = a\hat{i} + b\hat{k}$.From $\hat{n} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$, we get $a + b = 0$, so $a = -b$.Normalizing $\hat{n}$, we get $\hat{n} = \frac{\hat{i} - \hat{k}}{\sqrt{2}}$.Thus, both options $B$ and $C$ are correct.

The electric field of a plane electromagnetic wave in a medium is given by $\vec{E}(x, y, z, t) = E_0 \hat{n} e^{i k_0[(x+y+z)-ct]}$, where $c$ is the speed of light in free space. The $\vec{E}$ field is polarized in the $x-z$ plane. If the speed of the wave in the medium is $v$, then:

Similar Questions

The Poynting vector $\vec S$ is defined as a vector whose magnitude is equal to the wave intensity and whose direction is along the direction of wave propagation. Mathematically,it is given by $\vec S = \frac{1}{{\mu _0}}(\vec E \times \vec B)$. Show the nature of the $\vec S$ vs $t$ graph.

What are electromagnetic waves?

The amplitude of the electric field in a parallel beam of plane electromagnetic waves of intensity $53.1 \ W \ m^{-2}$ is (Permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12} \ C^2 \ N^{-1} \ m^{-2}$) (in $N \ C^{-1}$)

The frequency of an electromagnetic wave is $3 \, MHz$. If the refractive index of the medium is $n = 4$,what happens to the wavelength and frequency when it passes through the medium?

The energy density of an electromagnetic wave in a vacuum is given by the relation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module