Paramagnetism — Question 7

Section 32-9 · Physics pages 5344–5345

Question 7 A paramagnetic force thermometer has an inverse-temperature response A small fixed-volume weak paramagnet has susceptibility χ = C/T with positive C. At its location the applied field has fixed positive gradient G = ∂x(B2). Neglect demagnetization and use Fx = χV G/(2µ0). Fx ∂xB2 > 0 x Schematic; labels, not drawing lengths, specify the data. (a) Find the force and its temperature derivative at fixed field profile. (b) A measured force drops to 0.80 of its initial value. Find the final-to-initial temperature ratio and the force change for a small fractional temperature increase epsilon.
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Question 7 - Solution A paramagnetic force thermometer has an inverse-temperature response Fx ∂xB2 > 0 x (a) Method and calculation Fx = CV G 2µ0T > 0 . dFx dT = −CV G 2µ0T 2 = −Fx/T . (b) Interpretation and further calculation Ff/Fi = Ti/Tf = 0.80 ⇒Tf/Ti = 1.25 . Tf = Ti(1 + ε) ⇒∆F/Fi ≃−ε (|ε| ≪1). Physical check The material is pulled toward stronger field; this inverse-temperature scaling assumes fixed density and an unchanged field gradient.

Question 7

A paramagnetic force thermometer has an inverse-temperature response

A small fixed-volume weak paramagnet has susceptibility χ=C/T\chi=C/T with positive C. At its location the applied field has fixed positive gradient G=∂x(B2)G=\partial_x(B^2). Neglect demagnetization and use Fx=χVG/(2μ0)F_x=\chi VG/(2\mu_0).

Diagram shown in the original SVG worksheet.

  1. Find the force and its temperature derivative at fixed field profile.

  2. A measured force drops to 0.80 of its initial value. Find the final-to-initial temperature ratio and the force change for a small fractional temperature increase epsilon.

Question 7 - Solution

A paramagnetic force thermometer has an inverse-temperature response

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Fx=CVG2μ0T>0.\boxed{F_x=\frac{CVG}{2\mu_0T}>0}.dFxdT=−CVG2μ0T2=−Fx/T.\boxed{\frac{dF_x}{dT}=-\frac{CVG}{2\mu_0T^2}=-F_x/T}.

(b) Interpretation and further calculation

Ff/Fi=Ti/Tf=0.80⇒Tf/Ti=1.25.F_f/F_i=T_i/T_f=0.80\Rightarrow\boxed{T_f/T_i=1.25}.Tf=Ti(1+ε)⇒ΔF/Fi≃−ε(|ε|≪1).T_f=T_i(1+\varepsilon)\Rightarrow\boxed{\Delta F/F_i\simeq-\varepsilon}\quad(|\varepsilon|\ll1).

Physical check

The material is pulled toward stronger field; this inverse-temperature scaling assumes fixed density and an unchanged field gradient.