\begin{problem*}{24.7} % EM waves
Figure 24.3 shows a plane electromagnetic sinosoidal wave propogating
in the $x$ direction.  Suppose the wavelength is $50.0\U{m}$ and the
electric field vibrates in the $xy$ plane with an amplitude of
$22.0\U{V/m}$.  Calculate \Part{a} the frequency of the wave
and \Part{b} the magnitude and direction of \vect{B} when the electric
field has its maximum value in the negative $y$ direction.  \Part{c}
Write an expression for the \vect{B} with the correct unit vector,
with numerical values for $B_\text{max}$, $k$, and $\omega$, and with
its magnidude in the form
\begin{equation}
 B = B_\text{max}\cos(kx-\omega t)
\end{equation}
\end{problem*}

\begin{solution}
\Part{a}
This is just a units conversion
\begin{equation}
  f = \frac{c}{\lambda} = \frac{3.00\E{8}\U{m/s}}{50.0\U{m/cycle}}
    = 6.00\E{6}\U{cycles/s} = \ans{6.00\U{MHz}}
\end{equation}

\Part{b}
The magnitude of $B$ in an electromagnetic plane wave is given by
$B=E/c$.  The direction of the wave's motion is given by the Poynting
vector $\vect{S}=\vect{E}\times\vect{B}$.  Using the right-hand-rule
for the cross product, we see that when \vect{E} is in the $-\jhat$
direction and \vect{S} is in the \ihat\ direction, \vect{B} must be in
the $-\khat$ direction.  Putting this together
\begin{equation}
  \vect{B_0} = \frac{-E_0}{c}\khat = \ans{-73.3\U{nT}\cdot\khat}
\end{equation}

\Part{c}
Because it is a sinusoidal wave moving in the \ihat\ direction, we know
$B$ must look something like
\begin{equation}
  \vect{B} = \vect{B_0} \cos(kx - \omega t + \phi) \;.
\end{equation}
We already found \vect{B_0} in \Part{b}, and we don't have any phase
information, so we can drop $\phi$.  That leaves
\begin{align}
  k &= \frac{2\pi}{\lambda} = 0.126\U{rad/m} \\
  \omega &= 2\pi f = 3.77\E{7}\U{rad/s}
\end{align}
so
\begin{equation}
  \vect{B} = \ans{-73.3\U{nT} \cdot
    \cos(0.126\U{rad/m}\cdot x - 3.77\E{7}\U{rad/s}\cdot t) \khat} \;.
\end{equation}
\end{solution}
