The electric field of a plane electromagnetic | vedclass

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Question

(C) For an electromagnetic wave propagating in the $x$-direction,the electric field is in the $y$-direction $(\hat{j})$,so the magnetic field must be

Vedclass · Accepted Answer

$\overrightarrow{B} = E_{0} \sqrt{\mu_{0} \epsilon_{0}} \cos(kx) \hat{k}$

Vedclass · Answer

(C) For an electromagnetic wave propagating in the $x$-direction,the electric field is in the $y$-direction $(\hat{j})$,so the magnetic field must be in the $z$-direction $(\hat{k})$ because the direction of propagation is given by $\overrightarrow{E} 	imes \overrightarrow{B}$.Given $\overrightarrow{E} = E_{0} \cos(\omega t - kx) \hat{j}$.The amplitude of the magnetic field is $B_{0} = \frac{E_{0}}{c}$,where $c = \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}$.Thus,$B_{0} = E_{0} \sqrt{\mu_{0} \epsilon_{0}}$.The magnetic field wave equation is $\overrightarrow{B} = B_{0} \cos(\omega t - kx) \hat{k}$.At $t = 0$,$\overrightarrow{B} = B_{0} \cos(-kx) \hat{k} = B_{0} \cos(kx) \hat{k}$.Substituting $B_{0}$,we get $\overrightarrow{B} = E_{0} \sqrt{\mu_{0} \epsilon_{0}} \cos(kx) \hat{k}$.

The electric field of a plane electromagnetic wave propagating along the $x$-direction in vacuum is $\overrightarrow{E} = E_{0} \hat{j} \cos(\omega t - kx)$. The magnetic field $\overrightarrow{B}$ at the moment $t = 0$ is:

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