Magnetic Materials — Question 7

Section 32-7 · Physics pages 5304–5305

Question 7 A dilute magnetic mixture can have zero net susceptibility A dilute composite has volume fraction f of species A with susceptibility +4a and fraction 1-f of species B with susceptibility -a, where a is small and positive. Neglect interactions and local-field corrections, so both species experience the same H. f 1 −f Schematic; labels, not drawing lengths, specify the data. (a) Find the effective susceptibility and the fraction giving zero total M. (b) At that fraction, find the separate contributions to magnetization per unit total volume and explain why zero net susceptibility does not imply zero microscopic response.
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Question 7 - Solution A dilute magnetic mixture can have zero net susceptibility f 1 −f (a) Method and calculation M = f(4aH) + (1 −f)(−aH) = a(5f −1)H. χeff= a(5f −1), f∗= 1/5 . (b) Interpretation and further calculation Mtotal A = f∗4aH = 4aH/5, M total B = −(1 −f∗)aH = −4aH/5. Mtotal A + Mtotal B = 0 with each contribution nonzero . Physical check The cancellation is a volume-average result in the supplied dilute model, not the absence of atomic magnetic moments.

Question 7

A dilute magnetic mixture can have zero net susceptibility

A dilute composite has volume fraction f of species A with susceptibility +4a and fraction 1-f of species B with susceptibility -a, where a is small and positive. Neglect interactions and local-field corrections, so both species experience the same H.

Diagram shown in the original SVG worksheet.

  1. Find the effective susceptibility and the fraction giving zero total M.

  2. At that fraction, find the separate contributions to magnetization per unit total volume and explain why zero net susceptibility does not imply zero microscopic response.

Question 7 - Solution

A dilute magnetic mixture can have zero net susceptibility

Diagram shown in the original SVG worksheet.

(a) Method and calculation

M=f(4aH)+(1−f)(−aH)=a(5f−1)H.M=f(4aH)+(1-f)(-aH)=a(5f-1)H.χeff=a(5f−1),f*=1/5.\boxed{\chi_{\mathrm{eff}}=a(5f-1),\quad f_*=1/5}.

(b) Interpretation and further calculation

MAtotal=f*4aH=4aH/5,MBtotal=−(1−f*)aH=−4aH/5.M_A^{\mathrm{total}}=f_*4aH=4aH/5,\quad M_B^{\mathrm{total}}=-(1-f_*)aH=-4aH/5.MAtotal+MBtotal=0with each contribution nonzero.\boxed{M_A^{\mathrm{total}}+M_B^{\mathrm{total}}=0\quad\text{with each contribution nonzero}}.

Physical check

The cancellation is a volume-average result in the supplied dilute model, not the absence of atomic magnetic moments.