A plane electromagnetic wave of frequency $$35$$ MHz travels in free space along the $$X$$-direction. At a particular point (in space and time) $$\vec{E} = 9.6\hat{j}$$ V m$$^{-1}$$. The value of magnetic field at this point is:

For a plane electromagnetic wave travelling in free space, the electric field and magnetic field are related by: $$B = \frac{E}{c}$$, where $$c = 3 \times 10^8$$ m/s is the speed of light.

Given: $$\vec{E} = 9.6\hat{j}$$ V/m. So $$E = 9.6$$ V/m.

Substituting into the formula:

$$B = \frac{9.6}{3 \times 10^8} = 3.2 \times 10^{-8}$$ T

Now we need to find the direction of $$\vec{B}$$. For an EM wave, the direction of propagation is along $$\vec{E} \times \vec{B}$$.

The wave travels along the $$x$$-direction, so $$\vec{E} \times \vec{B}$$ must be along $$\hat{i}$$.

Since $$\vec{E}$$ is along $$\hat{j}$$, we need $$\vec{B}$$ along $$\hat{k}$$, because $$\hat{j} \times \hat{k} = \hat{i}$$.

Therefore, $$\vec{B} = 3.2 \times 10^{-8}\hat{k}$$ T, which is Option (1).