The electric field for an electromagnetic wav | vedclass

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Question

(C) The given electric field equation is $E = \hat{i} 30 \cos (kz - 5 	imes 10^8 t)$.The standard form of a traveling electromagnetic wave is $E = E_

Vedclass · Accepted Answer

$1.66 \ rad \ m^{-1}$

Vedclass · Answer

(C) The given electric field equation is $E = \hat{i} 30 \cos (kz - 5 	imes 10^8 t)$.The standard form of a traveling electromagnetic wave is $E = E_0 \cos (kz - \omega t)$.Comparing the given equation with the standard form,we get the angular frequency $\omega = 5 	imes 10^8 \ rad/s$.The relationship between the speed of light $c$,angular frequency $\omega$,and wave vector $k$ is given by $c = \frac{\omega}{k}$.Rearranging for $k$,we get $k = \frac{\omega}{c}$.Substituting the given values,$k = \frac{5 	imes 10^8 \ rad/s}{3 	imes 10^8 \ m/s} = \frac{5}{3} \ rad/m$.Therefore,$k \approx 1.66 \ rad/m$.

The electric field for an electromagnetic wave in free space is $E = \hat{i} 30 \cos (kz - 5 \times 10^8 t)$,where the magnitude of $E$ is in $V/m$. The magnitude of the wave vector $k$ is (velocity of $EM$ wave in free space $= 3 \times 10^8 \ m/s$).

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