The electric field component of an electromag | vedclass

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(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

$-\hat j, 3.5 \, m$

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(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

$A$ sky wave with a frequency $55 \, MHz$ is incident on the $D$-region of the Earth's atmosphere at $45^\circ$. The angle of refraction is ......... $^\circ$ (electron density for the $D$-region is $400 \, \text{electrons}/cm^3$).

The unit of $\sqrt{\frac{2 I}{\varepsilon_0 c}}$ is
($I$: intensity of an electromagnetic wave,$c$: speed of light)

Pick out the statement which is not true.

If the direction of propagation of an electromagnetic wave is along the $z$-axis,then the directions of the electric field $(E)$ and magnetic field $(B)$ are:

For plane electromagnetic waves propagating in the $z$-direction,which one of the following combinations gives the correct possible direction for $\vec{E}$ and $\vec{B}$ fields respectively?

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