Question: The magnetic field of an electromagnetic wave is given by:
$\overrightarrow{\mathrm{B}}=1.6 \times 10^{-6} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(2 \hat{i}+\hat{j}) \frac{W b}{m^{2}}$
The associated electric field will be :
(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) $\overrightarrow{\mathrm{E}}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z-6 \times 10^{15} t\right)(-2 \hat{j}+\hat{i}) \frac{V}{m}$
(3) $\overrightarrow{\mathrm{E}}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(-\hat{i}+2 \hat{j}) \frac{V}{m}$
(4) $\overrightarrow{\mathrm{E}}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(\hat{i}-2 \hat{j}) \frac{V}{m}$
Correct Option: , 3
Solution:
(3) $\mathrm{E}_{0}=\mathrm{cB}_{0}=3 \times 10^{8} \times 1.6 \times 10^{-6}=4.8 \times 10^{2} \mathrm{~V} / \mathrm{m}$
