The magnetic field of a plane electromagnetic wave is $$\vec{B} = 3 \times 10^{-8} \sin\left[200\pi(y + ct)\right]\hat{i}$$ T. Where, $$c = 3 \times 10^8$$ m s$$^{-1}$$ is the speed of light. The corresponding electric field is:

We are given the magnetic field of a plane electromagnetic wave:

$$\vec{B} = 3 \times 10^{-8} \sin[200\pi(y + ct)]\,\hat{i}\,\text{T}$$

where $$c = 3 \times 10^8\,\text{m/s}$$.

We need to find the corresponding electric field $$\vec{E}$$.

The argument of the sine function is $$200\pi(y + ct)$$. Since it has the form $$(y + ct)$$, this wave is traveling in the $$-y$$ direction (negative $$y$$-axis). The unit propagation vector is $$\hat{n} = -\hat{j}$$.

The amplitude of the magnetic field is $$B_0 = 3 \times 10^{-8}\,\text{T}$$.

The amplitude of the electric field is:

$$E_0 = cB_0 = 3 \times 10^8 \times 3 \times 10^{-8} = 9\,\text{V/m}$$

Now we determine the direction of $$\vec{E}$$. For an electromagnetic wave, the direction of energy propagation (Poynting vector) is along $$\vec{E} \times \vec{B}$$.

We need: $$\vec{E} \times \vec{B}$$ to point in the propagation direction $$-\hat{j}$$.

Since $$\vec{B}$$ is along $$\hat{i}$$, let us check if $$\vec{E}$$ is along $$-\hat{k}$$:

$$(-\hat{k}) \times \hat{i} = -(\hat{k} \times \hat{i}) = -\hat{j}$$

This gives the correct propagation direction $$-\hat{j}$$.

Therefore, $$\vec{E}$$ is in the $$-\hat{k}$$ direction:

$$\vec{E} = -9\sin[200\pi(y + ct)]\,\hat{k}\,\text{V/m}$$

The correct answer is Option D.