# The electric field of a plane polarized electromagnetic wave in free space at time t=0 is given by an expression

Question:

The electric field of a plane polarized electromagnetic wave in free space at time $t=0$ is given by an expression

$\overrightarrow{\mathrm{E}}(x, y)=10 \hat{\mathrm{j}} \cos [(6 x+8 z)]$

The magnetic field $\overrightarrow{\mathrm{B}}(x, z, t)$ is given by: ( $\mathrm{c}$ is the velocity of light)

1. (1) $\frac{1}{\mathrm{c}}(6 \hat{\mathrm{k}}+8 \hat{\mathrm{i}}) \cos [(6 x-8 z+10 \mathrm{c} t)]$

2. (2) $\frac{1}{\mathrm{c}}(6 \hat{\mathrm{k}}-8 \hat{\mathrm{i}}) \cos [(6 x+8 z-10 \mathrm{c} t)]$

3. (3) $\frac{1}{\mathrm{c}}(6 \hat{\mathrm{k}}+8 \hat{\mathrm{i}}) \cos [(6 x+8 z-10 \mathrm{c} t)]$

4. (4) $\frac{1}{\mathrm{c}}(6 \hat{\mathrm{k}}-8 \hat{\mathrm{i}}) \cos [(6 x+8 z+10 \mathrm{c} t)]$

Correct Option: 2,

Solution:

(2) $\overrightarrow{\mathrm{E}}=10 \hat{\mathrm{j}} \cos [(6 \hat{\mathrm{i}}+8 \hat{\mathrm{k}}) \cdot(x \hat{\mathrm{i}}+\mathrm{zk})]$

$=10 \hat{\mathrm{j}} \cos [\overrightarrow{\mathrm{K}} \cdot \overrightarrow{\mathrm{r}}]$

$\therefore \overrightarrow{\mathrm{K}}=6 \hat{\mathrm{i}}+8 \hat{\mathrm{K}} ;$ direction of waves travel

i. e. direction of ' $c$ '.

$\hat{\mathrm{C}} \times \hat{\mathrm{E}}=\frac{-4 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}}{5}$

$\overrightarrow{\mathrm{B}}=\frac{\mathrm{E}}{\mathrm{C}}=\frac{10}{\mathrm{C}}$

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

or, magnetic field $\overrightarrow{\mathrm{B}}(\mathrm{x}, \mathrm{z}, \mathrm{t})=\frac{1}{\mathrm{C}}$

$(6 \hat{k}-8 \hat{i}) \cos (6 x+8 z-10 c t)$