Discussion
In free space, an electromagnetic wave is travelling whose wavevector is \(\vec{k} = 10(\hat{x} + \sqrt{3}\hat{y})\) m−1. The electric field component of this electromagnetic wave is given by \(\vec{E}(\vec{r}, t) = \hat{z}\ 600 \cos(\vec{k} \cdot \vec{r} - \omega t)\) V.m−1. The speed of light in free space is \(c = 3.0 \times 10^{8}\) m.s−1. The corresponding magnetic field \(\vec{B}(\vec{r}, t)\) is
Answer: B
The magnetic field is \(10^{-6}(\sqrt{3}\hat{x} - \hat{y})\cos(\vec{k}\cdot\vec{r} - \omega t)\) V.m−2.s (that is, tesla).
For a plane wave in free space \(\vec{B} = \frac{1}{c}\hat{k} \times \vec{E}\). Here \(|\vec{k}| = 10\sqrt{1 + 3} = 20\), so \(\hat{k} = (\hat{x} + \sqrt{3}\hat{y})/2\):
\[ \begin{aligned} \hat{k} \times \hat{z} &= \tfrac{1}{2}(\hat{x} \times \hat{z} + \sqrt{3}\,\hat{y} \times \hat{z}) \\ &= \tfrac{1}{2}(\sqrt{3}\hat{x} - \hat{y}) \\ B_{0} &= \frac{600}{3 \times 10^{8}} = 2 \times 10^{-6} \end{aligned} \]
So the amplitude vector is \(2 \times 10^{-6} \cdot \frac{1}{2}(\sqrt{3}\hat{x} - \hat{y})\), which is \(10^{-6}(\sqrt{3}\hat{x} - \hat{y})\). The first option forgets that \((\sqrt{3}\hat{x} - \hat{y})\) has length 2, the third reverses the direction and the powers of ten in the third and fourth are too large by 10.
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