Magnetic Materials — Question 1

Question 1

Internal field measurements determine susceptibility without an applied-field shortcut

Inside a linear isotropic sample, $H=800\,\mathrm{A/m}$ and $B/\mu_0=1000\,\mathrm{A/m}$. Use $B=\mu_0(H+M)$ and $M=\chi H$.

1. Find M, susceptibility, and relative permeability.

2. If the sample is an ellipsoid with demagnetizing factor N=0.20 and $H=H_a-NM$, find the applied field and apparent ratio M/Ha.

Question 1 - Solution

Internal field measurements determine susceptibility without an applied-field shortcut

(a) Method and calculation

$$M=B/\mu_0-H=\boxed{200\,\mathrm{A/m}}.$$$$\boxed{\chi=M/H=0.25,\quad\mu_r=1+\chi=1.25}.$$

(b) Interpretation and further calculation

$$H_a=H+NM=800+0.20(200)=\boxed{840\,\mathrm{A/m}}.$$$$\boxed{M/H_a=5/21\simeq0.238095\ne\chi}.$$

Physical check

Intrinsic susceptibility uses the internal H; the applied field differs when the sample has a demagnetizing field.