The electric field component of an electromag | vedclass

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Question

(B) The general equation for a plane electromagnetic wave is $\vec E = E_0 \cos(ky + \omega t)\hat i$ or $\vec E = E_0 \cos(ky - \omega t)\hat i$.Comp

Vedclass · Accepted Answer

$-\hat j, 3.5 \, m$

Vedclass · Answer

(B) The general equation for a plane electromagnetic wave is $\vec E = E_0 \cos(ky + \omega t)\hat i$ or $\vec E = E_0 \cos(ky - \omega t)\hat i$.Comparing the given equation $\vec E = 3\cos(1.8y + 5.4 	imes 10^8 t)\hat i$ with the standard form $\vec E = E_0 \cos(ky + \omega t)\hat i$, we identify the wave vector $k = 1.8 \, 	ext{rad/m}$.The positive sign between $ky$ and $\omega t$ indicates that the wave is propagating in the negative $y$-direction, which is represented by the unit vector $-\hat j$.The wavelength $\lambda$ is calculated using the relation $\lambda = \frac{2\pi}{k}$.Substituting $k = 1.8$, we get $\lambda = \frac{2 	imes 3.14159}{1.8} \approx 3.49 \, 	ext{m}$, which rounds to $3.5 \, 	ext{m}$.Thus, the direction of propagation is $-\hat j$ and the wavelength is $3.5 \, 	ext{m}$.

The electric field component of an electromagnetic wave in vacuum is given as $\vec E = 3\cos(1.8y + 5.4 \times 10^8 t)\hat i$. Its direction of propagation and wavelength are:

Similar Questions

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