The electric field for an electromagnetic wav | vedclass

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Question

(C) The given equation for the electric field is $E = i 30 \cos (k z - 5 	imes 10^8 t)$.Comparing this with the standard wave equation $E = E_0 \cos

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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