The electric field for an electromangetic wave in free space is $\mathbf{E}=\mathbf{i} 30 \cos \left(k z-5…

The electric field for an electromangetic wave in free space is $\mathbf{E}=\mathbf{i} 30 \cos \left(k z-5 \times 10^8 t\right)$, where magnitude of $E$ is in $\mathrm{V} / \mathrm{m}$. The magnitude of wave vector, $k$ is (velocity of em wave in free space $=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ )
  1. $0.46 \mathrm{rad} \mathrm{m}^{-1}$
  2. $3 \mathrm{rad} \mathrm{m}{ }^{-1}$
  3. $1.66 \mathrm{rad} \mathrm{m}^{-1}$
  4. $0.83 \mathrm{rad} \mathrm{m} \mathrm{m}^{-1}$

Solution

The given
$\mathbf{E}=\mathbf{i} 30 \cos \left(k z-5 \times 10^8 t\right)$
We know that,
$\mathbf{E}=\mathrm{iV} \cos (k z-\omega t)$
Comparing the both equations,
$\omega=5 \times 10^8 \mathrm{rad} / \mathrm{s}$
But we know that also
$K=\frac{2 \pi}{\lambda} \text { and } \omega=2 \pi \nu$
where, $\lambda$ is wavelength and $v$ is frequency of the wave
$\begin{array}{ll}
\therefore & \frac{\omega}{k}=\frac{2 \pi v}{2 \pi / \lambda}=v \lambda=C \\
\Rightarrow & C=\frac{\omega}{k} \\
\text {or } & k=\frac{\omega}{C}=\frac{5 \times 10^8}{3 \times 10^8} \\
\Rightarrow & k=\frac{5}{3}=1.66 \mathrm{rad} / \mathrm{m}
\end{array}$

Asked in: AP EAMCET 2014

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