An electromagnetic wave has an electric field | vedclass

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(D) The given electric field is $\vec E = E_z \hat z = 6.0\,\cos(kx - \omega t)\hat z$. Comparing this with the standard form $\vec E = E_0 \cos(kx -

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$+y$

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(D) The given electric field is $\vec E = E_z \hat z = 6.0\,\cos(kx - \omega t)\hat z$. Comparing this with the standard form $\vec E = E_0 \cos(kx - \omega t)\hat z$,we see that the wave is propagating in the $+x$ direction (since the argument is $kx - \omega t$). Thus,the direction of propagation $\hat k$ is $+x$ (or $\hat i$). The relationship between the direction of the electric field $\vec E$,magnetic field $\vec B$,and the direction of propagation $\hat k$ is given by $\hat k = \hat E 	imes \hat B$. Here,$\hat E = \hat z$ (or $\hat k_{unit}$) and $\hat k = \hat i$. Substituting these,we get $\hat i = \hat k_{unit} 	imes \hat B$. Using the cross product rules ($\hat k 	imes \hat i = \hat j$,$\hat k 	imes \hat j = -\hat i$,etc.),we find that $\hat B = \hat j$ (or $+y$). Therefore,the direction of the magnetic field is $+y$.

An electromagnetic wave has an electric field given by the expression (in Cartesian coordinates) $\vec E(x,t) = 6.0\,\cos(1 \times 10^7x - 3 \times 10^{15}t)\hat z$. What is the direction of the magnetic field at time $t = 0$ and position $x = 0$?

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