The magnetic field of an electromagnetic wave is given by: B→=1.6×10^(-6)cos⁡ 2×10^7z+6×10^15t 2i^+j^Wb/(m^2) The associated electric field will be:

Q69 / 90 · JEE Main 2019 (08 Apr Shift 2) · Physics

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

A
$\vec{E} = 4.8 \times 10^{2} c o s (2 \times 10^{7} z - 6 \times 10^{15} t) (- 2 \hat{j} + \hat{i}) \frac{V}{m}$
B
$\vec{E} = 4.8 \times 10^{2} c o s (2 \times 10^{7} z + 6 \times 10^{15} t) (\hat{i} - 2 \hat{j}) \frac{V}{m}$
Correct answer
C
$\vec{E} = 4.8 \times 10^{2} c o s (2 \times 10^{7} z + 6 \times 10^{15} t) (- \hat{i} + 2 \hat{j}) \frac{V}{m}$
D
$\vec{E} = 4.8 \times 10^{2} c o s (2 \times 10^{7} z - 6 \times 10^{15} t) (2 \hat{i} + \hat{j}) \frac{V}{m}$

Given magnetic field,

\vec{B} = 1.6 \times 10^{- 6} c o s (2 \times 10^{7} z + 6 \times 10^{15} t) (2 \hat{i} + \hat{j}) \frac{W b}{m^{2}}

The relation between magnetic field and electric field is

E = c B = (3 \times 10^{8}) \times (1.6 \times 10^{- 6} \times \sqrt{5}) = 4.8 \times 10^{2} \times \sqrt{5}

The direction of light propagation will be along

\vec{E} \times \vec{B}

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

Hence, the correct answer is option: 'C'.