Question

The electric field of a plane polarized electromagnetic wave in free space at time $t=0$ is given by an expression $\overrightarrow{E}(x,y)=10\hat{j}\cos \left[\right(6x+8z\left)\right]$. The magnetic field $\overrightarrow{B}(x,z,t)$ is given by

($c$ is the velocity of light)

• A. $\frac{1}{c}(6\hat{k}-8\hat{i})\cos \left[\right(6x+8z-10ct\left)\right])$
• B. $\frac{1}{c}(4\hat{k}+8\hat{i})\cos \left[\right(2x-8z+20ct\left)\right])$
• C. $\frac{1}{c}(5\hat{k}+8\hat{i})\cos \left[\right(6x+8z-80ct\left)\right])$
• D. $\frac{1}{c}(4\hat{k}-8\hat{i})\cos \left[\right(6x+8z+70ct\left)\right])$

Answer 1

The correct option is A $\frac{1}{c}(6\hat{k}-8\hat{i})\cos \left[\right(6x+8z-10ct\left)\right])$

$\overrightarrow{E}=10\hat{j}\cos \left[\right(6\hat{i}+8\hat{k}).(x\hat{i}+z\hat{k}\left)\right]=10\hat{j}\cos [\overrightarrow{K}.\overrightarrow{r}]$

Direction of waves travel i. e. direction of ${}^{\prime }{c}^{\prime }$ is $\overrightarrow{K}=6\hat{i}+8\hat{k}$

Direction of magnetic field
$\hat{c}\times \hat{E}=\frac{-4\hat{i}+3\hat{k}}{5}$

$|\overrightarrow{B}|=\frac{E}{c}=\frac{10}{c}$

$\overrightarrow{B}=\frac{10}{c}\left(\frac{-4\hat{i}+3\hat{k}}{5}\right)=\left(\frac{-8\hat{i}+6\hat{k}}{c}\right)$

Magnetic field $\overrightarrow{B}(x,z,t)=\frac{1}{c}(6\hat{k}-8\hat{i})\cos (6x+8z-10ct)$.