Question 1.5.5: Given E= 10e^-j(4x - kt) - V/m in free space. (a) Write all ...

Given E=10 e^{-j(4 x-k t)} - V/m in free space.

(a) Write all the four Maxwell’s equations in free space.

(b) Find ∆ × E.

(c) Find H.

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Given, E=10 e^{-4 j(4 x-k t)} a_{y} V / m  represents uniform plane wave travelling in x direction with velocity v k/4 and E_{x}=0, E_{2}=0

So,

\frac{\partial}{\partial y} E_{y}=-j 40 e^{-j(4 x-k t)}

 

\nabla \times \vec{E}=\left|\begin{array}{ccc}a_{x} & a_{y} & a_{z} \\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\E_{x} & E_{y} & E_{z}\end{array}\right|=\left|\begin{array}{ccc}a_{x} & a_{y} & a_{z} \\\frac{\partial}{\partial y} & 0 & 0 \\0 & E_{y} & 0\end{array}\right|

 

\begin{aligned}\nabla \times \vec{E} &=\vec{a}_{x}(0)-a_{y}(0)+a_{2} \cdot \frac{\partial}{\partial y} E_{y} \\&=j 40 e-^{j(4 x-k t)}=40 e^{j(4 x-k t+90)}\end{aligned}

(c) For uniform plane wave, \eta=\frac{E}{H}, \quad E=E_{y} \hat{a}_{y} Since, wave is travelling in x direction so, the direction of \vec{H} is in z-direction. 

\begin{aligned}&\hat{y}+\hat{z}=\hat{x} \\&\eta=\eta_{0}=120 \pi=\frac{E_{y}}{A z} \\&H_{2}=\frac{E_{y}}{120 \pi}=\frac{10}{120 \pi} e^{-j(4 x-k t)} \\&\vec{H}=H_{2} \vec{a}_{z}=\frac{1}{12 \pi} e^{-j(4 x-k t)} \vec{a}_{2}\end{aligned}

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