For an electromagnetic wave propagating through vacuum, k , E and ω...

For an electromagnetic wave, the direction of wave propagation $\hat{k}$ is along the direction of $\vec{E} \times \vec{B}$.

Since $\vec{E}$, $\vec{B}$, and $\vec{k}$ form a mutually perpendicular right-handed system, the direction of the magnetic field is given by $\hat{B} = \hat{k} \times \hat{E}$.

The magnitudes of the electric and magnetic fields are related by $B = \dfrac{E}{c}$.

Using the relation for the speed of light $c = \dfrac{\omega}{k}$, we get $B = \dfrac{kE}{\omega}$.

Combining the magnitude and direction gives:

$\vec{B} = B\hat{B} = \dfrac{kE}{\omega} (\hat{k} \times \hat{E})$

Since $\vec{k} = k\hat{k}$ and $\vec{E} = E\hat{E}$, this simplifies to:

$\vec{B} = \dfrac{\vec{k} \times \vec{E}}{\omega}$

Answer: $\dfrac{\vec{k} \times \vec{E}}{\omega}$