Ferromagnetism — Question 8

Question 8

An air gap moves a permanent magnet away from its zero-field remanence

A closed permanent-magnet circuit has magnet length ell, a gap g, equal cross-sectional area throughout, and no free current. Neglect leakage and fringing. On its recoil line the magnet obeys $B=B_r+\mu_0\mu_{\mathrm{rec}}H_m$; the gap has $H_g=B/\mu_0$.

1. Use Ampere law to find B in the gap and Hm in the magnet.

2. Find g/ell needed to reduce B to Br/2, and explain why B generally differs from Br even with no current.

Question 8 - Solution

An air gap moves a permanent magnet away from its zero-field remanence

(a) Method and calculation

$$H_m\ell+H_gg=0\Rightarrow H_m=-\frac{Bg}{\mu_0\ell}.$$$$B=B_r-\mu_{\mathrm{rec}}Bg/\ell\Rightarrow\boxed{B=\frac{B_r}{1+\mu_{\mathrm{rec}}g/\ell}}.$$

(b) Interpretation and further calculation

$$B=B_r/2\Rightarrow\boxed{g/\ell=1/\mu_{\mathrm{rec}}}.$$$$\boxed{H_m=-\frac{B_rg}{\mu_0(\ell+\mu_{\mathrm{rec}}g)}}\ne0\quad(g>0).$$

Physical check

Zero free current fixes the total circulation of H, not H at every point; the gap creates a demagnetizing operating field.