- Home
- Standard 12
- Physics
A carbon dioxide laser emits sinusoidal electro-magnetic wave that travels in vacuum in the negative $x-$ direction. The wavelength is $10.6\,\mu $ and $\vec E$ fields is parallel to $z-$ axis, with $E_{max} = 1.5 \times 10^6\, M\, v/m$. Then vector equations for $\vec E$ and $\vec B$ as a function of time and position are
$\vec E = \hat k\, [1.5×10^6cos(8.93×10^5x+3.78×10^{14}t)]\,v/m$
$\vec B=\hat j\, [5.0×10^{-3}cos(8.93×10^5x+3.78×10^{14}t)]\,T$
$\vec E = \hat k\, [1.5×10^6cos(8.93×10^5x+3.78×10^{14}t)]\,v/m$
$\vec B=-\hat j\, [5.0×10^{-3}cos(8.93×10^5x+3.78×10^{14}t)]\,T$
$\vec E = \hat k\, [1.5×10^6cos(5.93×10^5x+1.78×10^{14}t)]\,v/m$
$\vec B=-\hat j\, [5.0×10^{-3}cos(5.93×10^5x+1.78×10^{14}t)]\,T$
$\vec E = \hat k\, [1.5×10^6cos(5.93×10^5x+1.78×10^{14}t)]\,v/m$
$\vec B=\hat j\, [5.0×10^{-3}cos(5.93×10^5x+1.78×10^{14}t)]$
Solution
$\mathrm{B}_{\max }=\frac{\mathrm{E}_{\max }}{\mathrm{C}} \Rightarrow \mathrm{B}_{\max }=\frac{1.5 \times 10^{6}}{3.0 \times 10^{8}}=5.0 \times 10^{-3} \mathrm{\,T}$
$\mathrm{K}=\frac{2 \pi}{\lambda} \Rightarrow \mathrm{K}=\frac{2 \pi}{10.6 \times 10^{-6}} \Rightarrow \mathrm{K}=5.93 \times 10^{5} \mathrm{\,rad} / \mathrm{m}$
$\omega=\mathrm{cK} \Rightarrow \omega=3 \times 10^{8}\left(5.93 \times 10^{5}\right)$
$=1.78 \times 10^{14} \mathrm{\,rad} / \mathrm{sec}$
Direction of propagation $=\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$
$\Rightarrow-\hat{\mathrm{i}}=\overrightarrow{\mathrm{E}} \times \hat{\mathrm{A}} \Rightarrow \overrightarrow{\mathrm{A}}=\hat{\mathrm{j}}$
