A plane electromagnetic wave of frequency $f=50\text{ MHz}$ travels in free space along the positive $x$-direction. At a particular point in space and time, the electric field is $\vec{E}=6.3\hat{j}\text{ V/m}$. What is the corresponding magnetic field vector $\vec{B}$ at this point?

Solution

1. Given the electric field $\vec{E}=6.3\hat{j}\text{ V/m}$ and the speed of light $c=3.0\times 10^8\text{ m/s}$, the magnitude of the magnetic field is $$\begin{array}{rl}B& =E/c\\ & =6.3/(3.0\times 10^8)\\ & =2.1\times 10^{-8}\text{ T}\end{array}$$.
2. The wave propagates in the positive $x$-direction ($\hat{i}$), and since the direction of propagation is given by the cross product $\vec{E}\times \vec{B}$, we have $\hat{i}=\hat{j}\times \hat{k}$.
3. Therefore, the magnetic field vector must be in the $\hat{k}$ direction, resulting in $\vec{B}=2.1\times 10^{-8}\hat{k}\text{ T}$.
4. Hence the answer is (A).