The Traveling Electromagnetic Wave, Qualitatively — Question 5

Section 33-2 · Physics pages 5400–5401

Question 5 A dipole antenna radiates only its transverse acceleration component Use a supplied far-zone dipole model E = C[¨p −(¨p · ˆn)ˆn] at equal distance, with constant C and ¨p = aˆz. The observation direction makes angle theta with +z. ¨p ˆn θ Schematic; labels, not drawing lengths, specify the data. (a) Show that the field is transverse and find its magnitude. (b) Compare the field amplitudes on the axis, at 30 degrees, and in the equatorial plane.
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Question 5 - Solution A dipole antenna radiates only its transverse acceleration component ¨p ˆn θ (a) Method and calculation ˆn · E = C[(¨p · ˆn) −(¨p · ˆn)] = 0. |E|2 = C2a2(1 −cos2 θ) ⇒|E| = |Ca| sin θ . (b) Interpretation and further calculation |E(0)| = 0, |E(30◦)| = |Ca|/2, |E(90◦)| = |Ca| . |E(30◦)|/|E(90◦)| = 1/2 . Physical check The angular null follows from projecting out the longitudinal component; it is not a slower propagation speed along the axis.

Question 5

A dipole antenna radiates only its transverse acceleration component

Use a supplied far-zone dipole model 𝑬=C[𝒑̈−(𝒑̈⋅n̂)n̂]\mathbf E=C[\ddot{\mathbf p}-(\ddot{\mathbf p}\cdot\hat n)\hat n] at equal distance, with constant C and 𝒑̈=aẑ\ddot{\mathbf p}=a\hat z. The observation direction makes angle theta with +z.

Diagram shown in the original SVG worksheet.

  1. Show that the field is transverse and find its magnitude.

  2. Compare the field amplitudes on the axis, at 30 degrees, and in the equatorial plane.

Question 5 - Solution

A dipole antenna radiates only its transverse acceleration component

Diagram shown in the original SVG worksheet.

(a) Method and calculation

n̂⋅𝑬=C[(𝒑̈⋅n̂)−(𝒑̈⋅n̂)]=0.\hat n\cdot\mathbf E=C[(\ddot{\mathbf p}\cdot\hat n)-(\ddot{\mathbf p}\cdot\hat n)]=0.|𝑬|2=C2a2(1−cos⁡2θ)⇒|E|=|Ca|sin⁡θ.|\mathbf E|^2=C^2a^2(1-\cos^2\theta)\Rightarrow\boxed{|E|=|Ca|\sin\theta}.

(b) Interpretation and further calculation

|E(0)|=0,|E(30∘)|=|Ca|/2,|E(90∘)|=|Ca|.\boxed{|E(0)|=0,\quad|E(30^\circ)|=|Ca|/2,\quad|E(90^\circ)|=|Ca|}.|E(30∘)|/|E(90∘)|=1/2.\boxed{|E(30^\circ)|/|E(90^\circ)|=1/2}.

Physical check

The angular null follows from projecting out the longitudinal component; it is not a slower propagation speed along the axis.