Magnetic Materials — Question 3

Section 32-7 · Physics pages 5296–5297

Question 3 Ellipsoidal demagnetization limits the measured response A linear ellipsoid has intrinsic susceptibility chi and demagnetizing factor N along the applied field. Its uniform internal field satisfies H = Ha −NM and M = χH. M Ha Hd Schematic; labels, not drawing lengths, specify the data. (a) Derive the apparent susceptibility M/Ha. (b) Invert the relation to obtain intrinsic chi from measured apparent susceptibility q. State the positive-chi consistency bound when N is positive.
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Question 3 - Solution Ellipsoidal demagnetization limits the measured response M Ha Hd (a) Method and calculation M = χ(Ha −NM) ⇒M(1 + Nχ) = χHa. q ≡M/Ha = χ 1 + Nχ . (b) Interpretation and further calculation q(1 + Nχ) = χ ⇒χ = q 1 −Nq . χ > 0, N > 0 ⇒0 < q < 1/N . Physical check As q approaches 1/N, a small measurement error can produce a large error in the inferred intrinsic susceptibility.

Question 3

Ellipsoidal demagnetization limits the measured response

A linear ellipsoid has intrinsic susceptibility chi and demagnetizing factor N along the applied field. Its uniform internal field satisfies H=Ha−NMH=H_a-NM and M=χHM=\chi H.

Diagram shown in the original SVG worksheet.

  1. Derive the apparent susceptibility M/Ha.

  2. Invert the relation to obtain intrinsic chi from measured apparent susceptibility q. State the positive-chi consistency bound when N is positive.

Question 3 - Solution

Ellipsoidal demagnetization limits the measured response

Diagram shown in the original SVG worksheet.

(a) Method and calculation

M=χ(Ha−NM)⇒M(1+Nχ)=χHa.M=\chi(H_a-NM)\Rightarrow M(1+N\chi)=\chi H_a.q≡M/Ha=χ1+Nχ.\boxed{q\equiv M/H_a=\frac{\chi}{1+N\chi}}.

(b) Interpretation and further calculation

q(1+Nχ)=χ⇒χ=q1−Nq.q(1+N\chi)=\chi\Rightarrow\boxed{\chi=\frac{q}{1-Nq}}.χ>0,N>0⇒0<q<1/N.\chi>0,\ N>0\Rightarrow\boxed{0<q<1/N}.

Physical check

As q approaches 1/N, a small measurement error can produce a large error in the inferred intrinsic susceptibility.