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Magnetic Materials

Magnetic materials respond to applied fields through orbital and spin magnetic moments. The magnetization $\vec M$ creates bound currents that modify the field. The auxiliary field $\vec H$ plays the same role for magnetostatics that $\vec D$ plays for electrostatics — it satisfies Ampere's law with only free currents.

Key Concepts

Magnetization
M\vec M is the magnetic dipole moment per unit volume. Bound surface current: Kb=M×n^\vec K_b=\vec M\times\hat n. Bound volume current: Jb=×M\vec J_b=\nabla\times\vec M.
H Field
H=B/μ0M\vec H=\vec B/\mu_0-\vec M. Ampere's law for H\vec H: Hdl=Ifree,enc\oint\vec H\cdot d\vec l=I_{\rm free,enc}. In linear magnetic material: B=μ0μrH=μH\vec B=\mu_0\mu_r\vec H=\mu\vec H.
Magnetic Susceptibility
M=χmH\vec M=\chi_m\vec H. Paramagnets: χm>0\chi_m>0 (small, 103\sim10^{-3}). Diamagnets: χm<0\chi_m<0 (very small). Ferromagnets: χm1\chi_m\gg1 (nonlinear, history-dependent).
Hysteresis
Ferromagnets exhibit hysteresis: MM depends on the history of HH. The area of the BB-HH hysteresis loop equals the energy dissipated per cycle — the basis of magnetic data storage.

Key Equations

H field
H=Bμ0M\vec{H} = \frac{\vec{B}}{\mu_0} - \vec{M}
Auxiliary magnetic field; Ampere's law for H involves only free currents.
Linear magnetic medium
B=μ0(1+χm)H=μ0μrH\vec{B} = \mu_0(1+\chi_m)\vec{H} = \mu_0\mu_r\vec{H}
μr = 1 + χm is the relative permeability.
Solenoid with core
B=μ0μrnIB = \mu_0\mu_r n I
Field inside solenoid with ferromagnetic core; can be 1000× the air value.
Worked Example

H and B in a Magnetic Material

Problem

A solenoid (n=1000n=1000 turns/m, I=2.0I=2.0 A) has an iron core with μr=1200\mu_r=1200. Find HH and BB inside.

Solution

H=nI=1000×2.0=2000H=nI=1000\times2.0=2000 A/m.

B=μ0μrH=4π×107×1200×2000=4π×107×2.4×106B = \mu_0\mu_r H = 4\pi\times10^{-7}\times1200\times2000 = 4\pi\times10^{-7}\times2.4\times10^6
B=4π×107×2.4×106=9.6π×1013.02 TB = 4\pi\times10^{-7}\times2.4\times10^6 = 9.6\pi\times10^{-1}\approx3.02\text{ T}
Answer H=2000H=2000 A/m; B3.02B\approx3.02 T.
Practice

Exercises

7 problems
1 of 7

A solenoid (n=1000n=1000 turns/m, I=2.0I=2.0 A) in air. Find HH inside (in A/m).

A/m
2 of 7

Same solenoid but with iron core (μr=1200\mu_r=1200). Find BB inside (in T).

T
3 of 7

For a paramagnetic material with χm=3.0×103\chi_m=3.0\times10^{-3}, find μr=1+χm\mu_r=1+\chi_m.

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4 of 7

A magnetic material has B=0.50B=0.50 T and μr=100\mu_r=100. Find H=B/(μ0μr)H=B/(\mu_0\mu_r) (in A/m).

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5 of 7

Same material. Find M=χmHM=\chi_m H (in A/m) if χm=99\chi_m=99 (so μr=100\mu_r=100).

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6 of 7

Energy density in a magnetic field: u=B2/(2μ0μr)u=B^2/(2\mu_0\mu_r). For B=0.50B=0.50 T, μr=100\mu_r=100 (in J/m³).

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7 of 7

Magnetic field of a diamagnetic sphere in external field H0H_0: Hin=3H0/(μr+2)H_{\rm in}=3H_0/(\mu_r+2). For μr=0.999\mu_r=0.999 (diamagnet) and H0=1000H_0=1000 A/m, find HinH_{\rm in} (in A/m).

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Key Takeaways

  • H=B/μ0M\vec H=\vec B/\mu_0-\vec M satisfies Ampere's law for free currents; B\vec B includes all (free and bound) currents.
  • Linear media: B=μ0μrH\vec B=\mu_0\mu_r\vec H, μr>1\mu_r>1 (paramagnets), μr<1\mu_r<1 (diamagnets), μr1\mu_r\gg1 (ferromagnets).
  • Bound surface and volume currents from M\vec M are the magnetic analogs of bound charges from P\vec P.
  • Ferromagnet hysteresis: irreversible magnetization — the basis of permanent magnets and magnetic memory.