The electric fields of two plane electromagne | vedclass

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(C) The electric fields are $\overrightarrow{E}_{1}=E_{0} \hat{j} \cos (\omega t-kx)$ and $\overrightarrow{E}_{2}=E_{0} \hat{k} \cos (\omega t-ky)$.At

Vedclass · Accepted Answer

$E_{0} q(0.8 \hat{i}+\hat{j}+0.2 \hat{k})$

Vedclass · Answer

(C) The electric fields are $\overrightarrow{E}_{1}=E_{0} \hat{j} \cos (\omega t-kx)$ and $\overrightarrow{E}_{2}=E_{0} \hat{k} \cos (\omega t-ky)$.At $t=0$ and origin $(0,0,0)$,$\overrightarrow{E}_{1} = E_{0} \hat{j}$ and $\overrightarrow{E}_{2} = E_{0} \hat{k}$.The corresponding magnetic fields are given by $\overrightarrow{B} = \frac{1}{c} (\hat{n} 	imes \overrightarrow{E})$,where $\hat{n}$ is the direction of propagation.For $\overrightarrow{E}_{1}$,$\hat{n} = \hat{i}$,so $\overrightarrow{B}_{1} = \frac{1}{c} (\hat{i} 	imes E_{0} \hat{j}) = \frac{E_{0}}{c} \hat{k}$.For $\overrightarrow{E}_{2}$,$\hat{n} = \hat{j}$,so $\overrightarrow{B}_{2} = \frac{1}{c} (\hat{j} 	imes E_{0} \hat{k}) = \frac{E_{0}}{c} \hat{i}$.The Lorentz force is $\overrightarrow{F} = q(\overrightarrow{E}_{1} + \overrightarrow{E}_{2}) + q(\overrightarrow{v} 	imes (\overrightarrow{B}_{1} + \overrightarrow{B}_{2}))$.Substituting values: $\overrightarrow{F} = q(E_{0} \hat{j} + E_{0} \hat{k}) + q(0.8 c \hat{j} 	imes (\frac{E_{0}}{c} \hat{k} + \frac{E_{0}}{c} \hat{i}))$.$\overrightarrow{F} = q E_{0} \hat{j} + q E_{0} \hat{k} + 0.8 q E_{0} (\hat{j} 	imes \hat{k}) + 0.8 q E_{0} (\hat{j} 	imes \hat{i})$.Using cross products $\hat{j} 	imes \hat{k} = \hat{i}$ and $\hat{j} 	imes \hat{i} = -\hat{k}$:$\overrightarrow{F} = q E_{0} \hat{j} + q E_{0} \hat{k} + 0.8 q E_{0} \hat{i} - 0.8 q E_{0} \hat{k} = q E_{0} (0.8 \hat{i} + \hat{j} + 0.2 \hat{k})$.

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