A carbon dioxide laser emits a sinusoidal electromagnetic wave that travels in vacuum in the negative x-direction. The wavelength is 10.6 μm (in the infrared; see Fig. 32.4) and the $\stackrel{\to }{E}$ field is parallel to the z-axis, with Write vector equations for and as functions of time and position. SOLUTION IDENTIFY and SET UP: Equations (32.19) describe a wave traveling in the negative x-direction with $\stackrel{\to }{E}$ along the y-axis—that is, a wave that is linearly polarized along the y-axis. By contrast, the wave in this example is linearly polarized along the z-axis. At points where $\stackrel{\to }{E}$is in the positive z-direction, $\stackrel{\to }{B}$must be in the positive y-direction for the vector product $\stackrel{\to }{E}$ x $\stackrel{\to }{B}$ to be in the negative x-direction (the direction of propagation). Figure 32.15 shows a wave that satisfies these requirements. EXECUTE: A possible pair of wave functions that describe the wave shown in Fig. 32.15 are The plus sign in the arguments of the cosine functions indicates that the wave is propagating in the negative x-direction, as it should. Faraday’s law requires that Emax = cBmax[Eq. (32.18)], so Bmax = $\frac{\mathrm{{E}_{\mathrm{max}}}}{c}$ = $\frac{1.5×{10}^{6}\phantom{\rule{2ex}{0ex}}\text{V/m}}{3×{10}^{8}\phantom{\rule{2ex}{0ex}}\text{V/m}}$ = $5×{10}^{-3}\phantom{\rule{2ex}{0ex}}\text{T}$ To check unit consistency, note 1 V = 1 Wb/s that and 1 Wbm2 = 1T We have so the wave number and angular frequency are
We have λ= 10.6 x 10-6so the wave number and angular frequency are $k=\frac{\mathrm{2\pi }}{\lambda }=\frac{\mathrm{2\pi }\phantom{\rule{2ex}{0ex}}\text{rad}}{10.6×{10}^{-6}\phantom{\rule{2ex}{0ex}}\text{m}}=5.93×{10}^{5}\phantom{\rule{2ex}{0ex}}\text{rad/m}$
$\omega =\mathrm{ck}=\left(3×{10}^{8}\phantom{\rule{2ex}{0ex}}\text{m/s}\right)\left(5.93×{10}^{5}\phantom{\rule{2ex}{0ex}}\text{rad/m}\right)$
$\stackrel{\to }{\mathrm{E\left(x,t\right)}}=\stackrel{^}{k}\left(1.5×{10}^{6}\phantom{\rule{2ex}{0ex}}\text{rad/m}\right)×\mathrm{cos}\left[\left(5.93×{10}^{5}\phantom{\rule{2ex}{0ex}}\text{rad/m}\right)x+\left(1.78×{10}^{14}\phantom{\rule{2ex}{0ex}}\text{rad/s}\right)t\right]$ $\stackrel{\to }{\mathrm{E\left(x,t\right)}}=\stackrel{^}{k}\left(1.5×{10}^{6}\phantom{\rule{2ex}{0ex}}\text{rad/m}\right)×\mathrm{cos}\left[\left(5.93×{10}^{5}\phantom{\rule{2ex}{0ex}}\text{rad/m}\right)x+\left(1.78×{10}^{14}\phantom{\rule{2ex}{0ex}}\text{rad/s}\right)t\right]$ +(1.78× 1014 rad/s)t ] MathML operators - − − To specify subtraction × × × To specify multiplication ÷ ÷ ÷ To specify division ≠ ≠ ≠ To specify not equals ≈ ≈ ≈ To specify approximately equals < < < To specify less than ≤ ≤ ≤ To specify less than or equals