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8.Electromagnetic waves
medium
The electric field associated with an $e.m.$ wave in vacuum is given by $\vec E = \hat i\,40\,\cos \,\left( {kz - 6 \times {{10}^8}\,t} \right)$. where $E$, $z$ and $t$ are in $volt/m$, meter and seconds respectively. The value of wave factor $k$ is ....... $m^{-1}$.
A
$6$
B
$3$
C
$2$
D
$0.5$
Solution
$E = \hat i\,\,\,40\,\,\,\cos \,\,(kz\, – \,6 \times {10^8}\,\,t)$
$2 \pi f=6 \times 10^{8}$
$\mathrm{f}=\frac{6 \times 10^{8}}{2 \pi}$
$C=f \lambda$
$\lambda=\frac{3 \times 10^{8}}{6 \times 10^{8}} 2 \pi=\pi$
$\boxed{{\text{K}} = \frac{{2\pi }}{\lambda } = \frac{{2\pi }}{\pi } = 2}$
Standard 12
Physics
