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:
Answer & explanation
Correct answer: option 1
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
$∴f=\frac{6×10^8}{2\pi}, c=3×10^8m/s$
$λ=\frac{c}{f}=\frac{3×10^8×2\pi}{6×10^8}=\pi$
$∴k=\frac{2\pi}{λ}=\frac{2\pi}{\pi}=2$