Electromagnetic Waves

Concept:An electromagnetic wave's electric field is expressed as $\vec{E} = \vec{E}_0 \sin(\vec{k} \cdot \vec{r} + \omega t)$. The phase term gives the wave vector $\vec{k}$ and propagation direction. In vacuum, $c = \omega / k$. The magnetic field is $\vec{B} = \frac{1}{\omega} (\vec{k} \times \vec{E})$.Explanation:The given electric field is $E_0 \sin(3y + 4z + \omega t) \hat{i}$.Compare with $\vec{k} \cdot \vec{r} + \omega t = k_x x + k_y y + k_z z + \omega t$.Thus $k_x = 0$, $k_y = 3$, $k_z = 4$, so $\vec{k} = 3\hat{j} + 4\hat{k}$.The wave travels opposite to $\vec{k}$ (since phase has $+\omega t$), so direction is $-\hat{k}/|\vec{k}| = -\frac{1}{5}(3\hat{j} + 4\hat{k})$. Option A is correct.Magnitude of wave vector: $k = \sqrt{0^2 + 3^2 + 4^2} = 5\ \mathrm{m}^{-1}$. Option B is incorrect.Using $c = \omega / k$, $\omega = c k = (3 \times 10^8)(5) = 1.5 \times 10^9\ \mathrm{rad\ s^{-1}}$. Option C is correct.For magnetic field: $\vec{B} = \frac{1}{\omega}(\vec{k} \times \vec{E})$. Compute $\vec{k} \times \vec{E} = (3\hat{j}+4\hat{k}) \times (E_0\sin(\ldots)\hat{i}) = E_0\sin(\ldots)[3(\hat{j}\times\hat{i}) + 4(\hat{k}\times\hat{i})] = E_0\sin(\ldots)(-3\hat{k} + 4\hat{j}) = E_0\sin(\ldots)(4\hat{j}-3\hat{k})$.Thus $\vec{B} = \frac{E_0}{\omega}\sin(\ldots)(4\hat{j}-3\hat{k}) = \frac{E_0}{5c}\sin(\ldots)(4\hat{j}-3\hat{k})$, not $\frac{E_0}{c}\sin(\ldots)(4\hat{j}-3\hat{k})$. Option D is incorrect.Answer:The correct statements are (A) and (C).