The electric field vector associated with electromagnetic wave in free space is given by $E=40 \cos \left(6 \times 10^8 t-k x\right) Vm^{-1}$. The value of k in SI unit is:

2

The correct answer is Option (1) → 2

$E=E_0\cos(wt-kx)$ [general form]

given,

$E=40\cos(6×10^8t-kx)$

∴ $E_0$, Amplitude of electric intensity = 40 N/C

$w$, Angular frequency = $dπf=6×10^8$

k, Wave number

$∴f=\frac{6×10^8}{2\pi}, c=3×10^8m/s$

$λ=\frac{c}{f}=\frac{3×10^8×2\pi}{6×10^8}=\pi$

$∴k=\frac{2\pi}{λ}=\frac{2\pi}{\pi}=2$