The Traveling Electromagnetic Wave, Qualitatively — Question 4

Section 33-2 · Physics pages 5398–5399

Question 4 Oblique propagation fixes the transverse magnetic-field components A vacuum plane wave propagates along ˆn = (3ˆx + 4ˆy)/5. At an event its electric field is E∗ˆz. Use the right-handed Cartesian convention. x y ˆn B ⊙Ez Schematic; labels, not drawing lengths, specify the data. (a) Find the vector magnetic field and verify that it is transverse. (b) Find the direction of E cross B and check the magnitude relation.
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Question 4 - Solution Oblique propagation fixes the transverse magnetic-field components x y ˆn B ⊙Ez (a) Method and calculation B = ˆn × E c = E∗ 5c (4ˆx −3ˆy) . ˆn · B = E∗ 25c(12 −12) = 0. (b) Interpretation and further calculation E × B = E2 ∗ 5c (3ˆx + 4ˆy) = E2 ∗ c ˆn. |B| = |E∗|/c . Physical check Transversality refers to the propagation direction, which need not coincide with a coordinate axis.

Question 4

Oblique propagation fixes the transverse magnetic-field components

A vacuum plane wave propagates along n̂=(3x̂+4ŷ)/5\hat n=(3\hat x+4\hat y)/5. At an event its electric field is E*ẑE_*\hat z. Use the right-handed Cartesian convention.

Diagram shown in the original SVG worksheet.

  1. Find the vector magnetic field and verify that it is transverse.

  2. Find the direction of E cross B and check the magnitude relation.

Question 4 - Solution

Oblique propagation fixes the transverse magnetic-field components

Diagram shown in the original SVG worksheet.

(a) Method and calculation

𝑩=n̂×𝑬c=E*5c(4x̂−3ŷ).\mathbf B=\frac{\hat n\times\mathbf E}{c}=\boxed{\frac{E_*}{5c}(4\hat x-3\hat y)}.n̂⋅𝑩=E*25c(12−12)=0.\hat n\cdot\mathbf B=\frac{E_*}{25c}(12-12)=0.

(b) Interpretation and further calculation

𝑬×𝑩=E*25c(3x̂+4ŷ)=E*2cn̂.\mathbf E\times\mathbf B=\frac{E_*^2}{5c}(3\hat x+4\hat y)=\frac{E_*^2}{c}\hat n.|𝑩|=|E*|/c.\boxed{|\mathbf B|=|E_*|/c}.

Physical check

Transversality refers to the propagation direction, which need not coincide with a coordinate axis.