What Produces a Magnetic Field? — Question 7

Section 28-1 · Physics pages 4484–4485

Question 7 Random microscopic moments have a nonzero root-mean-square sum even when their mean vanishes N independent moments have isotropic random orientations and fixed magnitude mu. Assume ⟨⃗µi⟩= 0 and ⟨⃗µi ·⃗µj⟩= 0 for distinct i and j. Let⃗M = ∑ i⃗µi. Schematic; labels, not drawing lengths, specify the data. (a) Find the mean vector and root-mean-square magnitude of M. (b) Compare the rms magnitude with the fully aligned moment for N = 1020. State why a zero average does not mean each microscopic source disappears.
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Question 7 - Solution Random microscopic moments have a nonzero root-mean-square sum even when their mean vanishes (a) Method and calculation ⟨⃗M⟩= ∑ i ⟨⃗µi⟩= 0. ⟨M2⟩= ∑ i ⟨µ2 i ⟩+ 2 ∑ i<j ⟨⃗µi ·⃗µj⟩= Nµ2. (b) Interpretation and further calculation Mrms = √ Nµ, Maligned = Nµ. Mrms/Maligned = N−1/2 = 10−10. Individual moments remain present; their directions largely cancel. Physical check The square-root scaling relies on independent, unbiased orientations; correlated domains need a different model.

Question 7

Random microscopic moments have a nonzero root-mean-square sum even when their mean vanishes

N independent moments have isotropic random orientations and fixed magnitude mu. Assume ⟨μ→i⟩=0\langle\vec\mu_i\rangle=0 and ⟨μ→i⋅μ→j⟩=0\langle\vec\mu_i\cdot\vec\mu_j\rangle=0 for distinct i and j. Let M→=∑iμ→i\vec M=\sum_i\vec\mu_i.

Diagram shown in the original SVG worksheet.

  1. Find the mean vector and root-mean-square magnitude of M.

  2. Compare the rms magnitude with the fully aligned moment for N=1020N=10^{20}. State why a zero average does not mean each microscopic source disappears.

Question 7 - Solution

Random microscopic moments have a nonzero root-mean-square sum even when their mean vanishes

Diagram shown in the original SVG worksheet.

(a) Method and calculation

⟨M→⟩=∑i⟨μ→i⟩=0.\langle\vec M\rangle=\sum_i\langle\vec\mu_i\rangle=0.⟨M2⟩=∑i⟨μi2⟩+2∑i<j⟨μ→i⋅μ→j⟩=Nμ2.\langle M^2\rangle=\sum_i\langle\mu_i^2\rangle+2\sum_{i<j}\langle\vec\mu_i\cdot\vec\mu_j\rangle=N\mu^2.

(b) Interpretation and further calculation

Mrms=Nμ,Maligned=Nμ.M_{\mathrm{rms}}=\sqrt N\mu,\quad M_{\mathrm{aligned}}=N\mu.Mrms/Maligned=N−1/2=10−10.M_{\mathrm{rms}}/M_{\mathrm{aligned}}=N^{-1/2}=10^{-10}.Individual moments remain present; their directions largely cancel.

Physical check

The square-root scaling relies on independent, unbiased orientations; correlated domains need a different model.