Question

The magnetic field of an electromagnetic wave is given by:
$\overrightarrow{B}=1.6\times {10}^{-6}\cos (2\times {10}^{7}z+6\times {10}^{15}t)(2\hat{i}+\hat{j})\frac{\text{Wb}}{{\text{m}}^2}$

The associated electric field will be:

• A. $\overrightarrow{E}=4.8\times {10}^2\cos (2\times {10}^{7}z+6\times {10}^{15}t)(\hat{i}-2\hat{j})\frac{\text{V}}{\text{m}}$
• B. $\overrightarrow{E}=4.8\times {10}^2\cos (2\times {10}^{7}z+6\times {10}^{15}t)(-\hat{i}+2\hat{j})\frac{\text{V}}{\text{m}}$
• C. $\overrightarrow{E}=4.8\times {10}^2\cos (2\times {10}^{7}z-6\times {10}^{15}t)(2\hat{i}+\hat{j})\frac{\text{V}}{\text{m}}$
• D. $\overrightarrow{E}=4.8\times {10}^2\cos (2\times {10}^{7}z-6\times {10}^{15}t)(-2\hat{j}+\hat{i})\frac{\text{V}}{\text{m}}$

Answer 1

The correct option is B $\overrightarrow{E}=4.8\times {10}^2\cos (2\times {10}^{7}z+6\times {10}^{15}t)(-\hat{i}+2\hat{j})\frac{\text{V}}{\text{m}}$

For an electromagnetic wave,

$E_0=c{B}_0=3\times {10}^8\times 1.6\times {10}^{-6}=4.8\times {10}^2\text{V/m}$

Also, $\overrightarrow{S}=\overrightarrow{E}\times \overrightarrow{B}$

and $-\hat{k}\parallel [\overrightarrow{E}\times (2\hat{i}+\hat{j}\left)\right]$

Therefore, direction of $\overrightarrow{E}\text{is along}(-\hat{i}+2\hat{j})$

Hence, $\left(C\right)$ is the correct answer.