The Traveling Electromagnetic Wave, Quantitatively — Question 1

Section 33-3 · Physics pages 5412–5413

Question 1 A fully specified field reveals its carrier scales and first maximum In SI units a vacuum wave is E = 4.00 cos(20πx −6.00π × 109t + π/3)ˆy V/m. Use c = 3.00 × 108 m/s. x E c Schematic; labels, not drawing lengths, specify the data. (a) Find frequency, wavelength, propagation direction, and magnetic-field expression. (b) At x=0 find the first positive time at which Ey reaches its positive maximum.
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Question 1 - Solution A fully specified field reveals its carrier scales and first maximum x E c (a) Method and calculation k = 20π ⇒λ = 2π/k = 0.100 m, f = ω/(2π) = 3.00 GHz. B = (1.3333 × 10−8 T) cos(20πx −6.00π × 109t + π/3)ˆz ; ˆn = +ˆx. (b) Interpretation and further calculation x = 0 : −ωt + π/3 = 0 ⇒t = π/(3ω). tmax = 5.5556 × 10−11 s = 55.56 ps . Physical check The phase coefficient is angular frequency, not cyclic frequency; omitting 2pi would give the wrong period.

Question 1

A fully specified field reveals its carrier scales and first maximum

In SI units a vacuum wave is 𝑬=4.00cos⁡(20πx−6.00π×109t+π/3)ŷ\mathbf E=4.00\cos(20\pi x-6.00\pi\times10^9t+\pi/3)\hat y V/m. Use c=3.00×108c=3.00\times10^8 m/s.

Diagram shown in the original SVG worksheet.

  1. Find frequency, wavelength, propagation direction, and magnetic-field expression.

  2. At x=0 find the first positive time at which Ey reaches its positive maximum.

Question 1 - Solution

A fully specified field reveals its carrier scales and first maximum

Diagram shown in the original SVG worksheet.

(a) Method and calculation

k=20π⇒λ=2π/k=0.100m,f=ω/(2π)=3.00GHz.k=20\pi\Rightarrow\lambda=2\pi/k=0.100\,\mathrm m,\quad f=\omega/(2\pi)=3.00\,\mathrm{GHz}.𝑩=(1.3333×10−8T)cos⁡(20πx−6.00π×109t+π/3)ẑ;n̂=+x̂.\boxed{\mathbf B=(1.3333\times10^{-8}\,\mathrm T)\cos(20\pi x-6.00\pi\times10^9t+\pi/3)\hat z};\quad\hat n=+\hat x.

(b) Interpretation and further calculation

x=0:−ωt+π/3=0⇒t=π/(3ω).x=0:\quad-\omega t+\pi/3=0\Rightarrow t=\pi/(3\omega).tmax=5.5556×10−11s=55.56ps.\boxed{t_{\max}=5.5556\times10^{-11}\,\mathrm s=55.56\,\mathrm{ps}}.

Physical check

The phase coefficient is angular frequency, not cyclic frequency; omitting 2pi would give the wrong period.