Question

The electric field associated with an electromagnetic wave in vacuum is given by $E=\hat{i}40\cos \left(kz-6\times 10^{8}t\right)$ , where $E$ , $z$ and $t$ are $Volt/m$ , meter and second respectively. The value of wave vector $k$ is

• A. $2m^{-1}$
• B. $0.5m^{-1}$
• C. $6m^{-1}$
• D. $3m^{-1}$

Answer 1

Answer: (A)

Given, $E=\hat{i}40\cos (kz-6\times 10^{8}t)...(i)$
Comparing eq. $(i)$ with the standard equation which is
$E=\hat{i}{E}_0\cos (kz-(\omega t)$, we get
$\omega =6\times 10^{8}rad{s}^{-1}$,
$v=c=3\times 10^{8}m{s}^{-1}$
$\therefore k=\frac{2\pi }{\lambda }=\frac{2\pi v}{v}$
$=\frac{\omega }{v}=\frac{6\times 10^8}{3\times 10^8}=2m^{-1}$