The Traveling Electromagnetic Wave, Qualitatively — Question 6

Section 33-2 · Physics pages 5402–5403

Question 6 A charge moving with the wave sees partially opposing electric and magnetic forces At an event a +x vacuum wave has E = E∗ˆy and B = (E∗/c)ˆz. A positive test charge q has instantaneous velocity v = βcˆx, with 0 < β < 1. Use the laboratory Lorentz force. +x Ey Bz out v Schematic; labels, not drawing lengths, specify the data. (a) Find the total force and compare its electric and magnetic contributions. (b) Repeat for velocity -beta c along x and find the ratio of force magnitudes.
Show solutionHide solution
Question 6 - Solution A charge moving with the wave sees partially opposing electric and magnetic forces +x Ey Bz out v (a) Method and calculation FE = qE∗ˆy, FB = qβcˆx × (E∗ˆz/c) = −qβE∗ˆy. F+ = qE∗(1 −β)ˆy . (b) Interpretation and further calculation v = −βcˆx ⇒F−= qE∗(1 + β)ˆy . |F−|/|F+| = (1 + β)/(1 −β) . Physical check These are instantaneous laboratory forces; the test charge is not assumed to maintain a straight trajectory under the transverse force.

Question 6

A charge moving with the wave sees partially opposing electric and magnetic forces

At an event a +x vacuum wave has 𝑬=E*ŷ\mathbf E=E_*\hat y and 𝑩=(E*/c)ẑ\mathbf B=(E_*/c)\hat z. A positive test charge q has instantaneous velocity 𝒗=βcx̂\mathbf v=\beta c\hat x, with 0<β<10<\beta<1. Use the laboratory Lorentz force.

Diagram shown in the original SVG worksheet.

  1. Find the total force and compare its electric and magnetic contributions.

  2. Repeat for velocity -beta c along x and find the ratio of force magnitudes.

Question 6 - Solution

A charge moving with the wave sees partially opposing electric and magnetic forces

Diagram shown in the original SVG worksheet.

(a) Method and calculation

𝑭E=qE*ŷ,𝑭B=qβcx̂×(E*ẑ/c)=−qβE*ŷ.\mathbf F_E=qE_*\hat y,\quad\mathbf F_B=q\beta c\hat x\times(E_*\hat z/c)=-q\beta E_*\hat y.𝑭+=qE*(1−β)ŷ.\boxed{\mathbf F_+=qE_*(1-\beta)\hat y}.

(b) Interpretation and further calculation

𝒗=−βcx̂⇒𝑭−=qE*(1+β)ŷ.\mathbf v=-\beta c\hat x\Rightarrow\boxed{\mathbf F_-=qE_*(1+\beta)\hat y}.|F−|/|F+|=(1+β)/(1−β).\boxed{|F_-|/|F_+|=(1+\beta)/(1-\beta)}.

Physical check

These are instantaneous laboratory forces; the test charge is not assumed to maintain a straight trajectory under the transverse force.