Question

The electric field associated with an electromagnetic wave in vacuum is given by $ E = \hat{i}\,40\,\cos\,\left(kz-6\times10^{8}\,t\right) $ , where $ E $ , $ z $ and $ t $ are $ Volt/m $ , meter and second respectively. The value of wave vector $ k $ is

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