An electromagnetic wave with angular frequenc | vedclass

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Question

(D) The direction of propagation of an electromagnetic wave is given by the direction of the cross product of the electric field vector $\vec E$ and t

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 cross product of the electric field vector $\vec E$ and the magnetic field vector $\vec B$,i.e.,direction $\propto \vec E 	imes \vec B$.Given that the wave propagates in the $+y$ direction (unit vector $\hat j$) and the magnetic field $\vec B$ is in the $-x$ direction (unit vector $-\hat i$).Let the electric field be in the direction $\hat n$. Then,$\hat n 	imes (-\hat i) = \hat j$.Using the properties of cross products,we know that $\hat k 	imes (-\hat i) = -(\hat k 	imes \hat i) = -\hat j$. This does not match. However,$(-\hat k) 	imes (-\hat i) = \hat k 	imes \hat i = \hat j$.Thus,the electric field must be in the $+z$ direction $(\hat k)$.The wave equation for a wave traveling in the $+y$ direction is given by $\cos(\omega t - ky)$,where $k = \frac{2\pi}{\lambda}$.Therefore,the electric field vector is $\vec E = E_0 \cos \left( \omega t - \frac{2\pi}{\lambda} y ight) \hat z$.

An electromagnetic wave with angular frequency $\omega$ and wavelength $\lambda$ propagates in the $+y$ direction. Its magnetic field is directed along the $-x$ direction. The vector expression for the associated electric field (with amplitude $E_0$) is ...........

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The electric field part of an electromagnetic wave in a medium is represented by:
$E_x = 0;$
$E_y = 2.5 \, \text{N/C} \cos \left[ \left( 2\pi \times 10^6 \, \text{rad/s} \right) t - \left( \pi \times 10^{-2} \, \text{rad/m} \right) x \right];$
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