Electric Field Vector from Magnetic Field in EM Wave

An electromagnetic wave traveling in free space has its magnitude of electric field $E$ and magnetic field $B$ related by the speed of light $c$: $E = c B$

Given:

• Speed of light in free space, $c = 3 \times 10^8 \text{ m/s}$
• Magnetic field vector, $\vec{B} = 2 \times 10^{-7} \hat{j} \text{ T}$

Calculating the magnitude of the electric field $E$: $E = (3 \times 10^8 \text{ m/s}) \times (2 \times 10^{-7} \text{ T}) = 60 \text{ V/m}$

The direction of wave propagation $\hat{v}$ is along the positive $x$-direction ($\hat{i}$), and it is given by the cross product of the unit vectors of the electric field ($\hat{E}$) and magnetic field ($\hat{B}$): $\hat{v} = \hat{E} \times \hat{B}$

Substituting $\hat{v} = \hat{i}$ and $\hat{B} = \hat{j}$: $\hat{i} = \hat{E} \times \hat{j}$

Since $(-\hat{k}) \times \hat{j} = \hat{i}$, the unit vector along the electric field must be: $\hat{E} = -\hat{k}$

Combining the magnitude and direction, the corresponding electric field vector is: $\vec{E} = -60 \hat{k} \text{ V/m}$

Hence, the correct option is B.