A Circulating Charged Particle — Question 7

Section 28-5 · Physics pages 4564–4565

Question 7 Relativistic momentum enlarges an orbit beyond the classical radius estimate For perpendicular motion use the supplied relativistic relations p = γmv, r = p/(|q|B) and γ = (1 −v2/c2)−1/2. A particle travels at v=.800c in a uniform field; radiation losses are neglected. r⃗ B into the page Schematic; labels, not drawing lengths, specify the data. (a) Compare its radius and period with the classical formulas evaluated at the same speed. (b) If an observer infers |q|/m from the classical expression v/(Br), find the inferred-to-true ratio.
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Question 7 - Solution Relativistic momentum enlarges an orbit beyond the classical radius estimate r⃗ B into the page (a) Method and calculation γ = (1 −.8002)−1/2 = 5/3, r = γrclassical. T = 2πr/v = γ 2πm |q|B = 5 3Tclassical. (b) Interpretation and further calculation (|q| m ) inferred = v Br = 1 γ (|q| m ) true . ηinferred/ηtrue = 3/5 = .600. Physical check The magnetic force still does no work; the correction comes from the relation between momentum and velocity.

Question 7

Relativistic momentum enlarges an orbit beyond the classical radius estimate

For perpendicular motion use the supplied relativistic relations p=γmvp=\gamma mv, r=p/(|q|B)r=p/(|q|B) and γ=(1−v2/c2)−1/2\gamma=(1-v^2/c^2)^{-1/2}. A particle travels at v=.800c in a uniform field; radiation losses are neglected.

Diagram shown in the original SVG worksheet.

  1. Compare its radius and period with the classical formulas evaluated at the same speed.

  2. If an observer infers |q|/m from the classical expression v/(Br), find the inferred-to-true ratio.

Question 7 - Solution

Relativistic momentum enlarges an orbit beyond the classical radius estimate

Diagram shown in the original SVG worksheet.

(a) Method and calculation

γ=(1−.8002)−1/2=5/3,r=γrclassical.\gamma=(1-.800^2)^{-1/2}=5/3,\quad r=\gamma r_{\mathrm{classical}}.T=2πr/v=γ2πm|q|B=53Tclassical.T=2\pi r/v=\gamma\frac{2\pi m}{|q|B}=\tfrac53T_{\mathrm{classical}}.

(b) Interpretation and further calculation

(|q|m)inferred=vBr=1γ(|q|m)true.\left(\frac{|q|}{m}\right)_{\mathrm{inferred}}=\frac v{Br}=\frac1\gamma\left(\frac{|q|}{m}\right)_{\mathrm{true}}.ηinferred/ηtrue=3/5=.600.\eta_{\mathrm{inferred}}/\eta_{\mathrm{true}}=3/5=.600.

Physical check

The magnetic force still does no work; the correction comes from the relation between momentum and velocity.