Question

A carbon dioxide laser emits sinusoidal electro-magnetic wave that travels in vacuum in the negative $x$ -direction. The wavelength is $10.6\mu m$ and $\overset{ \rightarrow }{E}$ fields is parallel to z-axis, with $E_{max}=1.5\times 10^{6}Mvm^{- 1}$ Then vector equations for $\overset{ \rightarrow }{E}$ and $\overset{ \rightarrow }{B}$ as a function of time and position are:-

• A. $\overset{ \rightarrow }{E}=\hat{k}\left[1 . 5 \times \left(10\right)^{6} cos \left(8 . 93 \times \left(10\right)^{5} x + 3 . 78 \times \left(10\right)^{14} t\right)\right]vm^{- 1}$
  $\overset{ \rightarrow }{B}=\hat{j}\left[5 . 0 \times \left(10\right)^{- 3} cos \left(8 . 93 \times \left(10\right)^{5} x + 3 . 78 \times \left(10\right)^{14} t\right)\right]T$
• B. $\overset{ \rightarrow }{E}=\hat{k}\left[1 . 5 \times \left(10\right)^{6} cos \left(8 . 93 \times \left(10\right)^{5} x + 3 . 78 \times \left(10\right)^{14} t\right)\right]vm^{- 1}$
  $\overset{ \rightarrow }{B}=-\hat{j}\left[5 . 0 \times \left(10\right)^{- 3} cos \left(8 . 93 \times \left(10\right)^{5} x + 3 . 78 \times \left(10\right)^{14} t\right)\right]T$
• C. $\overset{ \rightarrow }{E}=\hat{k}\left[1 . 5 \times \left(10\right)^{6} cos \left(5 . 93 \times \left(10\right)^{5} x + 1 . 78 \times \left(10\right)^{14} t\right)\right]vm^{- 1}$
  $\overset{ \rightarrow }{B}=-\hat{j}\left[5 . 0 \times \left(10\right)^{- 3} cos \left(5 . 93 \times \left(10\right)^{5} x + 1 . 78 \times \left(10\right)^{14} t\right)\right]T$
• D. $\overset{ \rightarrow }{E}=\hat{k}\left[1 . 5 \times \left(10\right)^{6} cos \left(5 . 93 \times \left(10\right)^{5} x + 1 . 78 \times \left(10\right)^{14} t\right)\right]vm^{- 1}$
  $\overset{ \rightarrow }{B}=\hat{j}\left[5 . 0 \times \left(10\right)^{- 3} cos \left(5 . 93 \times \left(10\right)^{5} x + 1 . 78 \times \left(10\right)^{14} t\right)\right]$

Answer 1

Answer: (D)

Let $I=\lim _{n \rightarrow \infty} \sum_{r=1}^{4 n} \frac{\sqrt{n}}{\sqrt{r}(3 \sqrt{r}+4 \sqrt{n}}$
$=\lim _{x \rightarrow \infty} \sum_{r=1}^{4 m} \frac{1}{n} \cdot \frac{1}{\sqrt{\frac{\Gamma}{n}}} \cdot \frac{1}{\frac{(3 \sqrt{r}+4 \sqrt{n}}{n}}$
$=\lim _{n \rightarrow \infty} \sum_{r=1}^{4 n} \frac{1}{\sqrt{\frac{\Gamma}{n}}} \cdot \frac{1}{\left(3 \sqrt{\frac{\Gamma}{n}}+4\right.} \cdot \frac{1}{n}$
$=\int_{0}^{4} \frac{1}{\sqrt{x}(3 \sqrt{x}+4} d x$
Put $3 \sqrt{x}+4=t$
$\Rightarrow 3 \cdot \frac{1}{2 \sqrt{x}} d x=d t$
$\Rightarrow \frac{1}{\sqrt{x}} d x=\frac{2}{3} d t$
$\Rightarrow I=\int_{4}^{10} \frac{1}{t^{2}} \cdot \frac{2}{3} d t=-\frac{2}{3}\left[\frac{1}{t}\right]_{t}^{10}$
$\Rightarrow a=\frac{2}{3}\left(\frac{1}{10}-\frac{1}{4}\right)=\frac{1}{10}$
$\Rightarrow a=10$