advanced32 min studyLesson 526 in Physics

Magnetic anisotropy and domain walls

Balance exchange and easy-axis anisotropy in a wall, then explain why magnetostatic energy can favor domains.

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What you’ll understand

  • Derive a wall profile length and energy per area with explicit units.
  • Distinguish a profile parameter from a conventional wall width.
  • Explain how dipolar energy and geometry compete with wall formation.

Exchange tends to keep neighboring magnetic orientations similar, but it does not by itself choose a direction relative to the crystal. Spin–orbit coupling and the crystal environment can supply that preference through magnetic anisotropy. A domain wall then balances the cost of changing orientation rapidly against the cost of spending space away from an easy direction. Whether a specimen forms many domains requires another balance: walls cost energy, while domains can reduce a long-range magnetostatic field.

From local orientation to a continuum energy

Let the magnetization have approximately fixed magnitude Ms and direction n. A continuum model is useful when n varies over distances much greater than the atomic spacing and well below scales that would obscure a wall. For a uniaxial material with easy axis z, take an anisotropy density Ksin⁡2θK\sin^2\theta with K positive, where theta is the angle to that axis. Both plus and minus z are minima, separated by orientations that cost anisotropy energy.

For a flat one-dimensional wall rotating in a fixed plane, exchange gives a density proportional to the square of the angular gradient. The exchange stiffness A has units joules per metre, so this density has joules per cubic metre. Integrating across the wall leaves energy per wall area. This simplified functional excludes magnetostatic charges, applied fields, chiral interactions, and variations of the magnetization magnitude. Those exclusions are appropriate for deriving a basic uniaxial wall, but not for predicting every thin-film texture.

σ[θ]=∫−∞∞ ⁣[A(dθdx)2+Ksin⁡2θ]dx,θ(−∞)=0,θ(+∞)=π\sigma[\theta]=\int_{-\infty}^{\infty}\!\left[A\left(\frac{d\theta}{dx}\right)^2+K\sin^2\theta\right]dx,\qquad \theta(-\infty)=0,\quad\theta(+\infty)=\pi
σ is wall energy per area in J/m², x is the coordinate normal to the wall, A is in J/m, and K is in J/m³. The boundary conditions specify a 180-degree wall between equally favored easy-axis domains.↔ Scroll wide equations horizontally.

Minimization fixes both a width scale and a surface energy

The Euler–Lagrange equation has a first integral in which exchange and anisotropy densities are equal along the minimum-energy profile. A nonzero integration constant would conflict with the uniform zero-energy domains at both infinities. Choosing the increasing-angle branch gives an angular gradient proportional to sine theta. Separating variables produces an exponential in position, centered at an arbitrary wall position because this uniform model has translation symmetry.

The profile length is the square root of A divided by K. A larger exchange stiffness favors a broader, gentler rotation, while larger anisotropy penalizes the interior and narrows the wall. A widely used conventional Bloch-wall width is pi times that length, so reported widths can differ by pi without disagreeing physically. Substituting the optimum profile into the energy gives four times the square root of the product AK. Increasing anisotropy therefore narrows the wall while increasing its surface cost.

δ=AK,θ(x)=2arctan⁡e(x−x0)/δ,σwall=4AK,wB=πδ\delta=\sqrt{\frac AK},\qquad \theta(x)=2\arctan e^{(x-x_0)/\delta},\qquad \sigma_{\mathrm{wall}}=4\sqrt{AK},\qquad w_{\mathrm B}=\pi\delta
δ is the profile length, not automatically the quoted experimental width. x0 is the wall center. The energy and width expressions hold for the stated isolated uniaxial wall without additional magnetostatic or chiral contributions.↔ Scroll wide equations horizontally.

Domains require a nonlocal energy comparison

Uniform magnetization can create a stray magnetic field whose energy extends outside the sample. For a fixed magnetization distribution, the demagnetizing field satisfies magnetostatics, and its energy can be written as μ0∫R3∣Hd∣2dV/2\mu_0\int_{\mathbb R^3}|\mathbf H_d|^2dV/2. Breaking a large specimen into appropriately arranged domains may reduce this energy enough to pay for walls. Exchange alone would prefer a uniform orientation and cannot explain why an otherwise homogeneous ferromagnet splits into many oppositely magnetized regions.

Shape matters because surface magnetic charges depend on the component of magnetization normal to the boundary. For a uniformly magnetized ellipsoid, the SI demagnetizing tensor gives Hd=−NM\mathbf H_d=-\mathsf N\mathbf M, with its principal factors summing to one. This convention differs from systems that use factors summing to four pi. Domain patterns, pinning, and hysteresis also depend on defects and preparation; a local minimum reached experimentally need not be the globally lowest-energy pattern. A wall-length estimate is therefore not a complete domain-size prediction.

Compute the two wall width conventions

An ideal uniaxial ferromagnet has A=1.0×10−11 J/mA=1.0\times10^{-11}\,\mathrm{J/m} and K=1.0×105 J/m3K=1.0\times10^5\,\mathrm{J/m^3}. Calculate the profile length, conventional Bloch-wall width, and minimum wall energy per area. Check the dimensional factors.

  1. The ratio is A/K=10−16 m2A/K=10^{-16}\,\mathrm{m^2}. Its square root gives δ=10−8 m=10 nm\delta=10^{-8}\,\mathrm m=10\,\mathrm{nm}. A/K is a squared length rather than a length.

  2. With the stated conventional definition, wB=πδ≃31.4 nmw_B=\pi\delta\simeq31.4\,\mathrm{nm}. The factor pi is a convention for reporting the same profile, not a second independent physical wall.

  3. The product is AK=10−6 J2/m4AK=10^{-6}\,\mathrm{J^2/m^4}, giving σwall=4×10−3 J/m2=4.0 mJ/m2\sigma_{\mathrm{wall}}=4\times10^{-3}\,\mathrm{J/m^2}=4.0\,\mathrm{mJ/m^2}. The units describe a surface cost, not a volume anisotropy density.

The result

The profile length is 10 nm, the conventional width is about 31.4 nm, and the ideal wall costs 4.0 mJ per square metre.

Explore the model

Interactive lab

Interactive lab · function

The curvature of an easy-axis minimum

For each fresh comparison, Reset and set BOTH Position x=0.50 and Coefficient a. Compare a=0.50 with a=1.00. Predict y=0.125 and 0.250 and slopes 0.50 and 1.00. Near an easy axis, sin²θ≈θ², so the graph illustrates how doubling anisotropy doubles local restoring curvature. The instrument is y=ax² and does not contain exchange gradients, walls, magnetostatics, or a domain-size calculation.

The curvature of an easy-axis minimumThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA larger local curvature penalizes tilting away from an easy direction, while the wall profile additionally requires spatial exchange.

Check your understanding

In the uniaxial wall model, what happens when K increases by a factor of four at fixed A?

Try it yourself

A specimen has Ms=8.0×10⁵ A/m. In a uniformly magnetized ellipsoid, compare demagnetizing energy densities along axes with N=0.10 and N=0.80, using ed=μ0NMs2/2e_d=\mu_0NM_s^2/2 and μ0=4π×10−7 H/m\mu_0=4\pi\times10^{-7}\,\mathrm{H/m}. Which axis is favored by this contribution alone?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compute the common factor μ0Ms²/2 first. These are SI demagnetizing factors; their values should not be multiplied by an extra four pi.

Compare with the solution

The common factor is (4π×10−7)(8.0×105)2/2≃4.021×105 J/m3(4\pi\times10^{-7})(8.0\times10^5)^2/2\simeq4.021\times10^5\,\mathrm{J/m^3}. Multiplication by 0.10 gives about 4.02×104 J/m34.02\times10^4\,\mathrm{J/m^3}, while 0.80 gives 3.22×105 J/m33.22\times10^5\,\mathrm{J/m^3}. The lower-N axis is favored by magnetostatic shape anisotropy alone, with an energy-density advantage of about 2.81×105 J/m32.81\times10^5\,\mathrm{J/m^3}. Crystal anisotropy and nonuniform domain formation can alter the final state, so this uniform-shape comparison does not determine an entire hysteresis curve.

Take it with you

  • A wall balances spatial exchange against the cost of leaving an easy direction.
  • The profile length √(A/K) and conventional width π√(A/K) must be distinguished.
  • Domains can reduce nonlocal magnetostatic energy while paying wall costs.

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