1 \begin{problem*}{24.7} % EM waves
2 Figure 24.3 shows a plane electromagnetic sinosoidal wave propogating
3 in the $x$ direction. Suppose the wavelength is $50.0\U{m}$ and the
4 electric field vibrates in the $xy$ plane with an amplitude of
5 $22.0\U{V/m}$. Calculate \Part{a} the frequency of the wave
6 and \Part{b} the magnitude and direction of \vect{B} when the electric
7 field has its maximum value in the negative $y$ direction. \Part{c}
8 Write an expression for the \vect{B} with the correct unit vector,
9 with numerical values for $B_\text{max}$, $k$, and $\omega$, and with
10 its magnidude in the form
12 B = B_\text{max}\cos(kx-\omega t)
18 This is just a units conversion
20 f = \frac{c}{\lambda} = \frac{3.00\E{8}\U{m/s}}{50.0\U{m/cycle}}
21 = 6.00\E{6}\U{cycles/s} = \ans{6.00\U{MHz}}
25 The magnitude of $B$ in an electromagnetic plane wave is given by
26 $B=E/c$. The direction of the wave's motion is given by the Poynting
27 vector $\vect{S}=\vect{E}\times\vect{B}$. Using the right-hand-rule
28 for the cross product, we see that when \vect{E} is in the $-\jhat$
29 direction and \vect{S} is in the \ihat\ direction, \vect{B} must be in
30 the $-\khat$ direction. Putting this together
32 \vect{B_0} = \frac{-E_0}{c}\khat = \ans{-73.3\U{nT}\cdot\khat}
36 Because it is a sinusoidal wave moving in the \ihat\ direction, we know
37 $B$ must look something like
39 \vect{B} = \vect{B_0} \cos(kx - \omega t + \phi) \;.
41 We already found \vect{B_0} in \Part{b}, and we don't have any phase
42 information, so we can drop $\phi$. That leaves
44 k &= \frac{2\pi}{\lambda} = 0.126\U{rad/m} \\
45 \omega &= 2\pi f = 3.77\E{7}\U{rad/s}
49 \vect{B} = \ans{-73.3\U{nT} \cdot
50 \cos(0.126\U{rad/m}\cdot x - 3.77\E{7}\U{rad/s}\cdot t) \khat} \;.