# The magnetic field of an electromagnetic wave is given by:

Question:

The magnetic field of an electromagnetic wave is given by:

$\overrightarrow{\mathrm{B}}=1.6 \times 10^{-6} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(2 \hat{i}+\hat{j}) \frac{\mathrm{Wb}}{\mathrm{m}^{2}}$

The associated electric field will be :

1. (1) $\overrightarrow{\mathrm{E}}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z-6 \times 10^{15} t\right)(2 \hat{i}+\hat{j}) \frac{V}{m}$

2. (2) $\overrightarrow{\mathrm{E}}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z-6 \times 10^{15} t\right)(-2 \hat{j}+\hat{i}) \frac{V}{m}$

3. (3) $\overrightarrow{\mathrm{E}}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(-\hat{i}+2 \hat{j}) \frac{V}{m}$

4. (4) $\overrightarrow{\mathrm{E}}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(\hat{i}-2 \hat{j}) \frac{V}{m}$

Correct Option: , 3

Solution:

(3) $\mathrm{E}_{0}=\mathrm{cB}_{0}=3 \times 10^{8} \times 1.6 \times 10^{-6}=4.8 \times 10^{2} \mathrm{~V} / \mathrm{m}$

Also $\vec{S} \Rightarrow \vec{E} \times \vec{B}$

or $-\vec{K} \Rightarrow \vec{E} \times(2 \hat{i}+\hat{j})$

Therefore direction of $\vec{E} \rightarrow(-\hat{i}+2 \hat{j})$