Question

The electric field of an electromagnetic wave in a vacuum is ${\mathrm{E}}_{\mathrm{y}}=(20.0\mathrm{V}/\mathrm{m})\cos \left[\left(6.28\times {10}^8\right)\mathrm{x}-\mathrm{\omega t}\right]$, where is inlocalid="1648922287898" $\mathrm{m}$andlocalid="1648922295234" $\mathrm{t}$is in s. What are the wave's (a) wavelength, (b) frequency, and (c) magnetic field amplitude?

Short answer

a. Wavelength of EM wave is${10}^{-8}\mathrm{m}$.

b.Frequency of EM wave is $3\times {10}^{16}\mathrm{Hz}$.

c.Magnetic field amplitude EM wave is$6.67\times {10}^{-8}\mathrm{T}$.

Step-by-step solution

Calculation for wavelength of EM waves (part a)

(a).

As it stands,

${\mathrm{E}}_{\mathrm{y}}=(20.0\mathrm{V}/\mathrm{m})\cos \left[\left(6.28\times {10}^8\right)\mathrm{x}-\mathrm{\omega t}\right]$

${\mathrm{E}}_{\mathrm{y}}=\mathrm{Ecos}\left[\left(6.28\times {10}^8\right)\mathrm{x}-\mathrm{\omega t}\right]$

Wherelocalid="1648922353481" $\mathrm{E}=20\mathrm{V}/\mathrm{m}$

$\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$

$\mathrm{\lambda }=\frac{2\mathrm{\pi }}{6·28\times {10}^8}\mathrm{m}$

$\mathrm{\lambda }={10}^{-8}\mathrm{m}$

Calculation for frequency of EM waves (part b)

(b).

$\mathrm{c}=\mathrm{f}·\mathrm{\lambda }$

Both should be divided bylocalid="1648922366586" $\mathrm{\lambda }$,

$\frac{\mathrm{c}}{\mathrm{\lambda }}=\frac{\mathrm{f}·\mathrm{\lambda \lambda }}{\mathrm{\lambda }}$

$\mathrm{f}=\frac{\mathrm{c}}{\mathrm{\lambda }}$

$\mathrm{f}=\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{{10}^{-8}\mathrm{m}}$

$\mathrm{f}=3\times {10}^{16}\mathrm{Hz}$

Calculation for magnetic field amplitude of EM waves (part c)

(c).

As it stands,

$\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$

$\left|\mathrm{B}\right|=\frac{\left|\mathrm{E}\right|}{\mathrm{c}}$

$\left|\mathrm{B}\right|=\frac{20\mathrm{V}/\mathrm{m}}{3\times {10}^8\mathrm{m}/\mathrm{s}}$

$\left|\mathrm{B}\right|=6.67\times {10}^{-8}\mathrm{T}$