A plane electromagnetic wave is moving in free space with velocity $$c=3\times 10^{8} m/s$$ and its eleclric field is given as $$\overrightarrow{E}=54\sin (kz-\omega t)\widehat{j} V/m$$, where $$\widehat{j}$$ is the unit vector along y-axis. The magnetic field vector $$\overrightarrow{B}$$ of the wave is :

Given $$\vec{E} = 54\sin(kz - \omega t)\hat{j}$$ V/m. Find $$\vec{B}$$.

From the argument $$(kz - \omega t)$$, the wave propagates in the $$+z$$ direction: $$\hat{k}$$.

For an EM wave, $$\vec{E}$$, $$\vec{B}$$, and the direction of propagation are mutually perpendicular, and $$\hat{E} \times \hat{B} = \hat{k}_{\text{propagation}}$$.

$$\hat{j} \times \hat{B} = \hat{k}$$. Since $$\hat{j} \times (-\hat{i}) = \hat{k}$$, we get $$\hat{B} = -\hat{i}$$.

$$B_0 = \frac{E_0}{c} = \frac{54}{3 \times 10^8} = 1.8 \times 10^{-7} \text{ T}$$

$$\vec{B} = -1.8 \times 10^{-7}\sin(kz - \omega t)\hat{i} \text{ T}$$

The correct answer is Option (3): $$-1.8 \times 10^{-7}\sin(kz - \omega t)\hat{i}$$ T.