Diamagnetism — Question 10

Section 32-8 · Physics pages 5330–5331

Question 10 A susceptibility measurement needs a holder-background subtraction A weak diamagnetic sample and its holder are measured in the same uniform H with negligible demagnetization. The total moment is mtot = aH + b; the empty holder gives mh = cH + b. The sample volume is V and a is less than c. H m holder total b Schematic; labels, not drawing lengths, specify the data. (a) Find the sample susceptibility and explain the role of the intercept b. (b) If the volume estimate is 5 percent too large, find the ratio of inferred to true susceptibility and the relative error in its magnitude.
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Question 10 - Solution A susceptibility measurement needs a holder-background subtraction H m holder total b (a) Method and calculation ms = mtot −mh = (a −c)H = χV H. χ = (a −c)/V < 0 ; b −b = 0. (b) Interpretation and further calculation Vest = 1.05V ⇒χest/χtrue = 1/1.05 = 20/21 . |χest| −|χtrue| |χtrue| = −1/21 ≃−4.762% . Physical check The shared offset cancels, but the holder slope must also be subtracted; otherwise even the inferred sign can be wrong.

Question 10

A susceptibility measurement needs a holder-background subtraction

A weak diamagnetic sample and its holder are measured in the same uniform H with negligible demagnetization. The total moment is mtot=aH+bm_{\mathrm{tot}}=aH+b; the empty holder gives mh=cH+bm_h=cH+b. The sample volume is V and a is less than c.

Diagram shown in the original SVG worksheet.

  1. Find the sample susceptibility and explain the role of the intercept b.

  2. If the volume estimate is 5 percent too large, find the ratio of inferred to true susceptibility and the relative error in its magnitude.

Question 10 - Solution

A susceptibility measurement needs a holder-background subtraction

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ms=mtot−mh=(a−c)H=χVH.m_s=m_{\mathrm{tot}}-m_h=(a-c)H=\chi VH.χ=(a−c)/V<0;b−b=0.\boxed{\chi=(a-c)/V<0};\qquad b-b=0.

(b) Interpretation and further calculation

Vest=1.05V⇒χest/χtrue=1/1.05=20/21.V_{\mathrm{est}}=1.05V\Rightarrow\boxed{\chi_{\mathrm{est}}/\chi_{\mathrm{true}}=1/1.05=20/21}.|χest|−|χtrue||χtrue|=−1/21≃−4.762%.\boxed{\frac{|\chi_{\mathrm{est}}|-|\chi_{\mathrm{true}}|}{|\chi_{\mathrm{true}}|}=-1/21\simeq-4.762\%}.

Physical check

The shared offset cancels, but the holder slope must also be subtracted; otherwise even the inferred sign can be wrong.