An electromagnetic wave with frequency omega | vedclass

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(D) The direction of propagation of an electromagnetic wave is given by the direction of the vector $\vec{E} 	imes \vec{B}$.Given,the wave travels in

Vedclass · Accepted Answer

$\vec{E} = E_0 \cos \left( \omega t - \frac{2\pi}{\lambda} y \right) \hat{z}$

Vedclass · Answer

(D) The direction of propagation of an electromagnetic wave is given by the direction of the vector $\vec{E} 	imes \vec{B}$.Given,the wave travels in the $+y$ direction,so the direction of propagation is $\hat{j}$.The magnetic field $\vec{B}$ is along the $+x$ axis,so $\vec{B} = B_0 \cos(\omega t - ky) \hat{i}$.We know that $\vec{E} 	imes \vec{B}$ must be in the direction of $\hat{j}$.Let $\vec{E} = E_0 \cos(\omega t - ky) \hat{n}$.Then $\hat{n} 	imes \hat{i} = \hat{j}$.Using the cross product rules,$\hat{k} 	imes \hat{i} = \hat{j}$.Therefore,the electric field vector must be along the $+z$ axis,i.e.,$\hat{k}$.Thus,$\vec{E} = E_0 \cos \left( \omega t - \frac{2\pi}{\lambda} y ight) \hat{z}$.

An electromagnetic wave with frequency $\omega$ and wavelength $\lambda$ travels in the $+y$ direction. Its magnetic field is along the $+x$ axis. The vector equation for the associated electric field (of amplitude $E_0$) is:

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