Question 28 - 2019 (08 Apr 2019 Shift 2)

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

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

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

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

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