The electric field of a plane polarized elect | vedclass

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(B) The given electric field is $\vec E(x,z,t) = 10 \hat j \cos(6x + 8z - \omega t)$.Since the wave propagates in free space,the wave speed is $c = \o

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$\frac{1}{c}(6\hat k - 8\hat i) \cos(6x + 8z - 10ct)$

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(B) The given electric field is $\vec E(x,z,t) = 10 \hat j \cos(6x + 8z - \omega t)$.Since the wave propagates in free space,the wave speed is $c = \omega / k$. The wave vector is $\vec k = 6\hat i + 8\hat k$. The magnitude is $k = \sqrt{6^2 + 8^2} = 10$.Thus,$\omega = ck = 10c$.The electric field is $\vec E = 10 \hat j \cos(6x + 8z - 10ct)$.The amplitude of the magnetic field is $B_0 = E_0 / c = 10/c$.The direction of propagation is $\hat n = \vec k / k = (6\hat i + 8\hat k) / 10 = 0.6\hat i + 0.8\hat k$.The magnetic field is given by $\vec B = \frac{1}{c} (\hat n 	imes \vec E)$.$\vec B = \frac{1}{c} [(0.6\hat i + 0.8\hat k) 	imes 10\hat j] \cos(6x + 8z - 10ct)$.$\vec B = \frac{1}{c} [6(\hat i 	imes \hat j) + 8(\hat k 	imes \hat j)] \cos(6x + 8z - 10ct)$.Since $\hat i 	imes \hat j = \hat k$ and $\hat k 	imes \hat j = -\hat i$,we get:$\vec B = \frac{1}{c} (6\hat k - 8\hat i) \cos(6x + 8z - 10ct)$.

The electric field of a plane polarized electromagnetic wave in free space at time $t = 0$ is given by the expression $\vec E(x,z) = 10\hat j \cos(6x + 8z)$. The magnetic field $\vec B(x,z,t)$ is given by: ($c$ is the velocity of light)

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