The magnetic field in a plane electromagnetic wave is given by $$\text B_\text y$$​ = 0.2 μ T sin (8π × $$10^2$$​ z + 6π × $$10^{11}$$​t). Then the electric field of the wave is:

Correct option is D

​$$\text{To find the electric field of a plane electromagnetic wave, we use the relation between electric field } \vec{E} \text{ and magnetic field } \vec{B} \text{ in a vacuum:} \\E = c \cdot B \\\\\text{Where:} \\\quad c = 3 \times 10^8\, \text{m/s (speed of light),} \\\quad B = 0.2\, \mu\text{T} = 0.2 \times 10^{-6}\, \text{T} \\\\E = 3 \times 10^8 \times 0.2 \times 10^{-6} = 60\, \text{V/m} \\\\\text{From the question:} \\B_y = 0.2\, \mu\text{T} \cdot \sin(8\pi \times 10^2 z + 6\pi \times 10^{11} t) \\\\\text{So the corresponding electric field will be:} \\\bullet\ \text{Amplitude: } 60\, \text{V/m} \\\bullet\ \text{Direction: Since } \vec{B} \text{ is along } \hat{y} \text{ and the wave is propagating in the } \hat{z}\text{-direction, the electric field must be along } \hat{x} \\\bullet\ \text{Same wavefunction (argument of sine)}$$$$\boxed{E_x = 60\, \text{V/m} \cdot \sin(8\pi \times 10^2 z + 6\pi \times 10^{11} t)}$$​​​