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Topic 9 of 9

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

\vec M

is the magnetic dipole moment per unit volume. Bound surface current:

\vec K_b=\vec M\times\hat n

. Bound volume current:

\vec J_b=\nabla\times\vec M

H Field

\vec H=\vec B/\mu_0-\vec M

. Ampere's law for

\vec H

\oint\vec H\cdot d\vec l=I_{\rm free,enc}

. In linear magnetic material:

\vec B=\mu_0\mu_r\vec H=\mu\vec H

Magnetic Susceptibility

\vec M=\chi_m\vec H

. Paramagnets:

\chi_m>0

(small,

\sim10^{-3}

). Diamagnets:

\chi_m<0

(very small). Ferromagnets:

\chi_m\gg1

(nonlinear, history-dependent).

Hysteresis

Ferromagnets exhibit hysteresis:

M

depends on the history of

H

. The area of the

B

H

hysteresis loop equals the energy dissipated per cycle — the basis of magnetic data storage.

Key Equations

H field

\vec{H} = \frac{\vec{B}}{\mu_0} - \vec{M}

Auxiliary magnetic field; Ampere's law for H involves only free currents.

Linear magnetic medium

\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 = \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 ( $n=1000$ turns/m, $I=2.0$ A) has an iron core with $\mu_r=1200$ . Find $H$ and $B$ inside.

Solution

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

B = \mu_0\mu_r H = 4\pi\times10^{-7}\times1200\times2000 = 4\pi\times10^{-7}\times2.4\times10^6

B = 4\pi\times10^{-7}\times2.4\times10^6 = 9.6\pi\times10^{-1}\approx3.02\text{ T}

Answer

H=2000

A/m;

B\approx3.02

Practice

Exercises

7 problems

1 of 7

Both panels show identical H arrows — H depends only on free current, not the core material. For $n=1000$ turns/m and $I=2.0$ A, find H inside (in A/m).

H field — solenoid

Current I 2.0 A → H = 2000 A/m

H depends only on free currents (nI), not on the material inside. The material only amplifies B.

H field at I = 2.0 A

H = A/m

Formula: H = nI | n = 1000/m, I = 2.0 A

2 of 7

Drag $\mu_r$ and watch B explode — an iron core amplifies B by 1200× over air! Same $n=1000$, $I=2.0$ A. Find B at $\mu_r=1200$ (in T).

B field — iron core

Relative permeability μᵣ 500 → B = 1.2566 T

Iron core amplifies B by μᵣ = 1200× compared to air. Same H = nI = 2000 A/m, but B = μ₀μᵣH grows to over 3 Tesla!

B at μᵣ = 1200

B = T

Formula: B = μ₀μᵣnI | μᵣ = 1200, n = 1000/m, I = 2.0 A

3 of 7

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

Unlock Exercise 3