Ferromagnetism — Question 2

Section 32-10 · Physics pages 5354–5355

Question 2 A model hysteresis loop distinguishes remanence from coercive field A model major loop has increasing-field branch M↑= Ms tanh[(H −Hc)/H0] and decreasing-field branch M↓= Ms tanh[(H + Hc)/H0], with positive constants. These branches are phenomenological. H M Schematic; labels, not drawing lengths, specify the data. (a) Find remanent M on each branch and the H values where M is zero. (b) At those zero-M points find B. Explain why the field that makes B zero generally differs from the field that makes M zero.
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Question 2 - Solution A model hysteresis loop distinguishes remanence from coercive field H M (a) Method and calculation M↓(0) = Ms tanh(Hc/H0), M↑(0) = −Ms tanh(Hc/H0) . M↑= 0 at H = Hc; M↓= 0 at H = −Hc . (b) Interpretation and further calculation B = µ0(H + M) ⇒B M=0 = ±µ0Hc̸ = 0 . B = 0 ⇐⇒M(H) = −H , which is a different intersection. Physical check Always specify whether a quoted coercivity refers to zero induction B or zero magnetization M.

Question 2

A model hysteresis loop distinguishes remanence from coercive field

A model major loop has increasing-field branch M↑=Mstanh⁡[(H−Hc)/H0]M_\uparrow=M_s\tanh[(H-H_c)/H_0] and decreasing-field branch M↓=Mstanh⁡[(H+Hc)/H0]M_\downarrow=M_s\tanh[(H+H_c)/H_0], with positive constants. These branches are phenomenological.

Diagram shown in the original SVG worksheet.

  1. Find remanent M on each branch and the H values where M is zero.

  2. At those zero-M points find B. Explain why the field that makes B zero generally differs from the field that makes M zero.

Question 2 - Solution

A model hysteresis loop distinguishes remanence from coercive field

Diagram shown in the original SVG worksheet.

(a) Method and calculation

M↓(0)=Mstanh⁡(Hc/H0),M↑(0)=−Mstanh⁡(Hc/H0).\boxed{M_\downarrow(0)=M_s\tanh(H_c/H_0),\quad M_\uparrow(0)=-M_s\tanh(H_c/H_0)}.M↑=0 at H=Hc;M↓=0 at H=−Hc.\boxed{M_\uparrow=0\text{ at }H=H_c;\quad M_\downarrow=0\text{ at }H=-H_c}.

(b) Interpretation and further calculation

B=μ0(H+M)⇒B|M=0=±μ0Hc≠0.B=\mu_0(H+M)\Rightarrow\boxed{B\big|_{M=0}=\pm\mu_0H_c\ne0}.B=0⇔M(H)=−H,which is a different intersection.\boxed{B=0\iff M(H)=-H},\quad\text{which is a different intersection.}

Physical check

Always specify whether a quoted coercivity refers to zero induction B or zero magnetization M.