Ferromagnetism — Question 5

Section 32-10 · Physics pages 5360–5361

Question 5 A saturated core keeps the vacuum differential response In a narrow reversible region deep in positive saturation, a core obeys M(H) = Ms −a/H2 with a positive and H sufficiently large. Take B = µ0(H + M). B ℓ N turns; cross section A Schematic; labels, not drawing lengths, specify the data. (a) Find differential permeability and its large-H limit. (b) For a long tightly wound solenoid of length ell, area A and N turns, derive the small-signal inductance about a bias field Hb. Compare it with the empty-coil inductance.
Show solutionHide solution
Question 5 - Solution A saturated core keeps the vacuum differential response B ℓ N turns; cross section A (a) Method and calculation dM dH = 2a H3 ⇒µdiff= µ0(1 + 2a/H3) . H →∞⇒µdiff→µ0 . (b) Interpretation and further calculation H = NI/ℓ, Λ = NAB ⇒Lsmall = dΛ dI Hb = N2A ℓ µdiff(Hb). Lsmall/L0 = 1 + 2a/H3 b →1 . Physical check Saturation removes most incremental magnetization response, but it does not make B or incremental inductance vanish.

Question 5

A saturated core keeps the vacuum differential response

In a narrow reversible region deep in positive saturation, a core obeys M(H)=Ms−a/H2M(H)=M_s-a/H^2 with a positive and H sufficiently large. Take B=μ0(H+M)B=\mu_0(H+M).

Diagram shown in the original SVG worksheet.

  1. Find differential permeability and its large-H limit.

  2. For a long tightly wound solenoid of length ell, area A and N turns, derive the small-signal inductance about a bias field Hb. Compare it with the empty-coil inductance.

Question 5 - Solution

A saturated core keeps the vacuum differential response

Diagram shown in the original SVG worksheet.

(a) Method and calculation

dMdH=2aH3⇒μdiff=μ0(1+2a/H3).\frac{dM}{dH}=\frac{2a}{H^3}\Rightarrow\boxed{\mu_{\mathrm{diff}}=\mu_0(1+2a/H^3)}.H→∞⇒μdiff→μ0.H\to\infty\Rightarrow\boxed{\mu_{\mathrm{diff}}\to\mu_0}.

(b) Interpretation and further calculation

H=NI/ℓ,Λ=NAB⇒Lsmall=dΛdI|Hb=N2Aℓμdiff(Hb).H=NI/\ell,\quad\Lambda=NAB\Rightarrow L_{\mathrm{small}}=\left.\frac{d\Lambda}{dI}\right|_{H_b}=\frac{N^2A}{\ell}\mu_{\mathrm{diff}}(H_b).Lsmall/L0=1+2a/Hb3→1.\boxed{L_{\mathrm{small}}/L_0=1+2a/H_b^3\to1}.

Physical check

Saturation removes most incremental magnetization response, but it does not make B or incremental inductance vanish.