The electric field associated with an electromagnetic wave...

The given electric field is $\vec{E} = E_0 \sin(3y + 4z + \omega t)\,\hat{i}$.

Comparing the phase $\phi = 3y + 4z + \omega t$ with the standard wave equation phase $\vec{k} \cdot \vec{r} - \omega t$, we can rewrite it as $-(-3y - 4z - \omega t)$.

Thus, the wave vector is $\vec{k} = -3\hat{j} - 4\hat{k}$.

The direction of wave propagation is given by the unit vector $\hat{n}$:

$\hat{n} = \dfrac{\vec{k}}{|\vec{k}|} = \dfrac{-3\hat{j} - 4\hat{k}}{\sqrt{(-3)^2 + (-4)^2}} = -\dfrac{1}{5}(3\hat{j} + 4\hat{k})$

The magnitude of the wave vector is $|\vec{k}| = 5$ m$^{-1}$.

The angular frequency $\omega$ is:

$\omega = c|\vec{k}| = (3 \times 10^8) \times 5 = 1.5 \times 10^9$ rad s$^{-1}$

The magnetic field $\vec{B}$ is given by:

$\vec{B} = \dfrac{1}{c}(\hat{n} \times \vec{E}) = \dfrac{1}{c} \left[ -\dfrac{1}{5}(3\hat{j} + 4\hat{k}) \right] \times [E_0 \sin(3y + 4z + \omega t)\,\hat{i}]$

$\vec{B} = -\dfrac{E_0}{5c} \sin(3y + 4z + \omega t) [3(\hat{j} \times \hat{i}) + 4(\hat{k} \times \hat{i})]$

$\vec{B} = -\dfrac{E_0}{5c} \sin(3y + 4z + \omega t) (-3\hat{k} + 4\hat{j}) = \dfrac{E_0}{5c} \sin(3y + 4z + \omega t) (-4\hat{j} + 3\hat{k})$

Answer: The wave is travelling in $-\dfrac{1}{5}(3\hat{j} + 4\hat{k})$ direction.; The value of $\omega$ is $1.5 \times 10^9$ rad s$^{-1}$.