Crossed Fields: Discovery of the Electron — Question 7

Section 28-3 · Physics pages 4524–4525

Question 7 Correlated field calibration errors can cancel in a charge-to-mass experiment An experiment infers η = e/m = E2/(2UB2) from independently specified E, U and B. Small fractional errors are epsilonE, epsilonU and epsilonB. positive plate negative plate⃗ v⃗ E Schematic; labels, not drawing lengths, specify the data. (a) Derive the first-order fractional error in eta and a worst-case bound for independent hard error limits of .5 percent, 1.0 percent and .5 percent respectively. (b) Show what happens if E and B share exactly the same multiplicative calibration factor while U is exact.
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Question 7 - Solution Correlated field calibration errors can cancel in a charge-to-mass experiment positive plate negative plate⃗ v⃗ E (a) Method and calculation δ ln η = 2δ ln E −δ ln U −2δ ln B. δη/η ≃2ϵE −ϵU −2ϵB, |δη|/η ≲.0300. (b) Interpretation and further calculation E′ = kE, B′ = kB, U′ = U. η′ = k2E2 2Uk2B2 = η. Physical check A worst-case sum and a correlated-error cancellation answer different experimental questions.

Question 7

Correlated field calibration errors can cancel in a charge-to-mass experiment

An experiment infers η=e/m=E2/(2UB2)\eta=e/m=E^2/(2UB^2) from independently specified E, U and B. Small fractional errors are epsilonE, epsilonU and epsilonB.

Diagram shown in the original SVG worksheet.

  1. Derive the first-order fractional error in eta and a worst-case bound for independent hard error limits of .5 percent, 1.0 percent and .5 percent respectively.

  2. Show what happens if E and B share exactly the same multiplicative calibration factor while U is exact.

Question 7 - Solution

Correlated field calibration errors can cancel in a charge-to-mass experiment

Diagram shown in the original SVG worksheet.

(a) Method and calculation

δln⁡η=2δln⁡E−δln⁡U−2δln⁡B.\delta\ln\eta=2\delta\ln E-\delta\ln U-2\delta\ln B.δη/η≃2ϵE−ϵU−2ϵB,|δη|/η≲.0300.\delta\eta/\eta\simeq2\epsilon_E-\epsilon_U-2\epsilon_B,\quad|\delta\eta|/\eta\lesssim.0300.

(b) Interpretation and further calculation

E′=kE,B′=kB,U′=U.E\prime=kE,\quad B\prime=kB,\quad U\prime=U.η′=k2E22Uk2B2=η.\eta\prime=\frac{k^2E^2}{2Uk^2B^2}=\eta.

Physical check

A worst-case sum and a correlated-error cancellation answer different experimental questions.