10.02.05 · em-sr / magnetostatics

Magnetic scalar potential and the demagnetization factor

shipped3 tiersLean: none

Anchor (Master): Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 5.9-5.12; Osborn, Demagnetizing Factors of the General Ellipsoid (1945)

Intuition Beginner

In free space with no currents, you can write the magnetic field as the gradient of a magnetic scalar potential, just as you write the electric field as the gradient of an electrostatic potential. This works because Ampere's law says the H-field has zero curl wherever there is no free current.

Inside a magnetised material, things get more interesting. The H-field "wants" to oppose the magnetisation — this is the demagnetisation field. The shape of the sample determines how strong this opposition is. A flat disc magnetised perpendicular to its face has a huge demagnetisation field (almost fully opposing M). A long thin rod magnetised along its axis has almost no demagnetisation field.

The demagnetisation factor $N_{d}$ quantifies this shape effect. For a uniformly magnetised ellipsoid, $H_{in} = - N_{d} M$ . A sphere has $N_{d} = 1/3$ . A long cylinder has $N_{d} \approx 0$ . A thin slab has $N_{d} \approx 1$ . These numbers control the internal field of every permanent magnet and every magnetic recording medium.

Visual Beginner

Shape	Magnetisation direction	$N_{d}$
Sphere	Any	1/3
Long cylinder	Along axis	0
Long cylinder	Transverse	1/2
Thin slab	Perpendicular	1
Thin slab	In-plane	0

Worked example Beginner

A uniformly magnetised sphere has $N_{d} = 1/3$ . If the magnetisation is $M = 1 0^{6}$ A/m (a strong permanent magnet material), the demagnetisation field is:

$H_{d} = - N_{d} M = - (1/3) (1 0^{6}) = - 3.33 \times 1 0^{5}$ A/m.

The B-field inside is $B = μ_{0} (H + M) = μ_{0} (1 - 1/3) M = (2/3) μ_{0} M \approx 0.84$ T.

For a long cylinder of the same material magnetised along its axis ( $N_{d} = 0$ ): $H_{d} = 0$ and $B = μ_{0} M \approx 1.26$ T. The sphere loses one-third of its potential field to the demagnetisation effect; the cylinder loses none.

Check your understanding Beginner

Formal definition Intermediate+

Magnetic scalar potential. In a region with no free current ( $J_{f} = 0$ ), Ampere's law gives $\nabla \times H = 0$ , so we can write:

H = - \nabla Φ_{m}

where $Φ_{m}$ is the magnetic scalar potential. Using $\nabla \cdot B = 0$ and $B = μ_{0} (H + M)$ :

\nabla \cdot H = - \nabla \cdot M

so $Φ_{m}$ satisfies the Poisson equation with source $- \nabla \cdot M$ :

\nabla^{2} Φ_{m} = \nabla \cdot M

The quantity $ρ_{m} = - \nabla \cdot M$ is the bound magnetic charge density (a mathematical convenience, not a physical entity). The surface bound magnetic charge is $σ_{m} = M \cdot \hat{n}$ .

For uniform magnetisation ( $\nabla \cdot M = 0$ inside), the only source of $Φ_{m}$ is the surface charge $σ_{m} = M \cdot \hat{n}$ , giving:

Φ_{m} (r) = \frac{1}{4 π} \oint_{S} \frac{σ _{m} ( r ^{'} )}{∣ r - r ^{'} ∣} d a^{'} = \frac{1}{4 π} \oint_{S} \frac{M \cdot n ^ ^{'}}{∣ r - r ^{'} ∣} d a^{'}

Demagnetisation tensor. For a uniformly magnetised ellipsoid with semi-axes $a, b, c$ , the H-field inside is uniform and related to $M$ by:

H_{in} = - N \cdot M

where $N$ is the demagnetisation tensor, diagonal in the principal-axis frame of the ellipsoid with $N_{x} + N_{y} + N_{z} = 1$ . For a sphere: $N_{x} = N_{y} = N_{z} = 1/3$ . For a general ellipsoid, the $N_{i}$ are expressed in terms of elliptic integrals.

Boundary conditions on $Φ_{m}$ . At the surface of a magnetic material:

$Φ_{m}$ is continuous (from $H_{t}$ continuous).
The normal derivative jumps: $\partial Φ_{m} / \partial n_{2} - \partial Φ_{m} / \partial n_{1} = - (M_{2} - M_{1}) \cdot \hat{n}$ (from $B_{n}$ continuous).

These are the same mathematical structure as the electrostatic boundary conditions with $σ_{m}$ playing the role of surface charge.

Caveats

$Φ_{m}$ is not single-valued if the region encloses a current loop. If a free current $I$ is enclosed by a path, $\oint H \cdot d l = I \neq = 0$ , and $Φ_{m}$ changes by $- I$ around the loop. The scalar potential is only useful in simply-connected current-free regions.
The "magnetic charge" $ρ_{m} = - \nabla \cdot M$ is a mathematical fiction. There are no magnetic monopoles. The formalism works because it reproduces the correct bound currents via the equivalence $\nabla \times M$ and $M \times \hat{n}$ .

Key derivation Intermediate+

Derivation (Uniform internal field of a magnetised ellipsoid).

Theorem. A uniformly magnetised ellipsoid produces a uniform H-field inside: $H_{in} = - N \cdot M$ .

Proof. The magnetic scalar potential satisfies $\nabla^{2} Φ_{m} = 0$ inside the ellipsoid (since $\nabla \cdot M = 0$ for uniform $M$ ) with surface source $σ_{m} = M \cdot \hat{n}$ .

For an ellipsoid with semi-axes $a, b, c$ and $M = M_{0} \hat{x}$ , the surface charge is $σ_{m} = M_{0} n_{x}$ where $n_{x}$ is the $x$ -component of the outward normal. By the linearity of the Laplace equation, $Φ_{m}$ inside is a linear function of $M$ :

Φ_{m}^{in} = - \frac{1}{2} i, j \sum N_{ij} M_{j} x_{i}

The gradient gives $H_{i}^{in} = \sum_{j} N_{ij} M_{j}$ . By the symmetry of the ellipsoid, $N_{ij}$ is diagonal in the principal frame. The constraint $\nabla \cdot B = 0$ applied to the uniform field gives $B_{x} + B_{y} + B_{z} = 0$ in the appropriate combination, which translates to $Tr (N) = N_{x} + N_{y} + N_{z} = 1$ .

The specific values involve elliptic integrals. For a prolate spheroid ( $a > b = c$ ):

N_{a} = \frac{1 - e ^{2}}{e ^{2}} (\frac{1}{2 e} ln \frac{1 + e}{1 - e} - 1), e^{2} = 1 - b^{2} / a^{2}

For a sphere ( $a = b = c$ ): $N_{a} = N_{b} = N_{c} = 1/3$ . $□$

Bridge. The demagnetisation tensor builds toward a complete description of the internal fields in magnetic samples of any shape. The foundational insight is that only ellipsoids (and their degenerate forms: spheres, cylinders, slabs) produce uniform internal demagnetisation fields. The central message is that the demagnetisation factor controls the B-field inside a permanent magnet and therefore its useful external field. This is exactly the reason why permanent magnets are shaped as long bars (low $N_{d}$ along the axis) rather than flat discs (high $N_{d}$ perpendicular to the face). Putting these together, the magnetic scalar potential generalises the electrostatic potential formalism 10.01.02 to magnetostatics in matter, and the demagnetisation factor appears again in the magnetic energy 10.03.05 where it determines the self-energy of a finite magnetised sample.

Exercises Intermediate+

Exercise 1 (easy, symbolic).

Show that the magnetic scalar potential for a uniformly magnetised sphere of radius $R$ with $M = M_{0} \hat{z}$ is $Φ_{m}^{in} = (M_{0} /3) r cos θ$ inside and $Φ_{m}^{out} = (M_{0} R^{3} /3) cos θ / r^{2}$ outside.

Hint

Inside: $\nabla^{2} Φ_{m} = 0$ , with $Φ_{m}$ linear in $z = r cos θ$ . Outside: dipole potential $\propto cos θ / r^{2}$ . Match boundary conditions at $r = R$ .

Answer

Inside, $Φ_{m}^{in} = (M_{0} /3) r cos θ$ is harmonic ( $\nabla^{2} Φ_{m} = 0$ , verified by direct calculation in spherical coordinates for a function proportional to $r cos θ$ ). The H-field is $H^{in} = - \nabla Φ_{m}^{in} = - (M_{0} /3) \hat{z}$ , confirming $N_{d} = 1/3$ .

Outside, the surface charge $σ_{m} = M_{0} cos θ$ produces a dipole potential. The solution of $\nabla^{2} Φ_{m} = 0$ outside with the correct angular dependence ( $cos θ$ ) is $Φ_{m}^{out} = C cos θ / r^{2}$ . Continuity of $Φ_{m}$ at $r = R$ : $(M_{0} /3) R = C / R^{2}$ , giving $C = M_{0} R^{3} /3$ .

Check: continuity of $B_{n} = μ_{0} (H_{n} + M_{n})$ at $r = R$ . Inside: $H_{r}^{in} = - (M_{0} /3) cos θ$ , $B_{r}^{in} = μ_{0} (- M_{0} /3 + M_{0}) cos θ = (2/3) μ_{0} M_{0} cos θ$ . Outside: $H_{r}^{out} = 2 M_{0} R^{3} cos θ / (3 r^{3})$ , $B_{r}^{out} = (2/3) μ_{0} M_{0} cos θ$ at $r = R$ . Continuous, as required.

Exercise 3 (medium, symbolic).

Compute the demagnetisation factor of a long cylinder magnetised perpendicular to its axis (transverse magnetisation).

Hint

A cylinder is an ellipsoid with $a \to \infty$ (infinitely long). Use $N_{x} + N_{y} + N_{z} = 1$ and the known values.

Answer

For an infinitely long cylinder: $N_{z} = 0$ along the axis (from the limit $a \to \infty$ of the prolate spheroid formula). By symmetry of the transverse plane, $N_{x} = N_{y}$ . The sum rule gives $N_{x} + N_{y} + N_{z} = 2 N_{x} + 0 = 1$ , so $N_{x} = N_{y} = 1/2$ .

A cylinder magnetised transversely has $N_{d} = 1/2$ , meaning the demagnetisation field is half the magnetisation: $H_{d} = - (1/2) M_{⊥}$ . The B-field inside is $B = μ_{0} (H + M) = (1/2) μ_{0} M_{⊥}$ , half what it would be without demagnetisation.

This is why anisotropic ferrite magnets are magnetised along their easy axis, which is designed to be the long axis of the cylinder, minimising $N_{d}$ .

Exercise 4 (medium, symbolic).

A permanent magnet in the shape of a sphere has a hysteresis curve with $M_{r} = 8 \times 1 0^{5}$ A/m (remanent magnetisation) and $H_{c} = 5 \times 1 0^{4}$ A/m (coercive field). Find the B-field at the operating point (the intersection of the demagnetisation line with the hysteresis curve in the second quadrant).

Hint

The operating point satisfies $H = - N_{d} M = - (1/3) M$ (sphere). On the hysteresis curve, $B = μ_{0} (H + M)$ .

Answer

The operating point satisfies $H = - M /3$ (demagnetisation line for a sphere). On the second-quadrant hysteresis curve, $H < 0$ and $M > 0$ . The intersection gives:

$H = - M /3$ , $B = μ_{0} (H + M) = μ_{0} (M - M /3) = (2/3) μ_{0} M$ .

For the linear part of the hysteresis curve (approximation), $M \approx M_{r} + (H / H_{c}) (M_{r} - 0)$ . At $H = - M_{r} /3$ : $M \approx M_{r} (1 - M_{r} / (3 H_{c}))$ . With $M_{r} = 8 \times 1 0^{5}$ and $H_{c} = 5 \times 1 0^{4}$ : $M_{r} / (3 H_{c}) = 800/ (150) = 5.33 > 1$ , so the operating point is below the coercive field — the sphere cannot sustain this magnetisation.

This shows that spherical magnets are impractical for most applications: the demagnetisation factor of 1/3 is too large. Real permanent magnets are long cylinders or horseshoe shapes to minimise $N_{d}$ .

Exercise 5 (medium, symbolic).

The magnetic scalar potential satisfies $\nabla^{2} Φ_{m} = \nabla \cdot M$ with boundary condition $\partial Φ_{m} / \partial n ∣_{outside} - \partial Φ_{m} / \partial n ∣_{inside} = - σ_{m}$ at the surface. Show that this is the same mathematical problem as the electrostatic Poisson equation with charge density $ρ_{e} = - \nabla \cdot P$ and surface charge $σ_{e} = P \cdot \hat{n}$ .

Hint

Replace $M$ with $P$ and $Φ_{m}$ with $Φ_{e}$ (with a sign). The equations are identical in form.

Answer

The electrostatic potential satisfies $\nabla^{2} Φ_{e} = - ρ_{e} / ϵ_{0} = \nabla \cdot P / ϵ_{0}$ (from $\nabla \cdot D = ρ_{f}$ and $D = ϵ_{0} E + P$ , with $ρ_{f} = 0$ giving $\nabla \cdot E = - \nabla \cdot P / ϵ_{0}$ ). The boundary condition on $Φ_{e}$ at the surface: $ϵ_{0} (E_{2 n} - E_{1 n}) = σ_{e} = P \cdot \hat{n}$ , which gives $\partial Φ_{e} / \partial n ∣_{1} - \partial Φ_{e} / \partial n ∣_{2} = σ_{e} / ϵ_{0}$ .

The magnetic problem has $\nabla^{2} Φ_{m} = \nabla \cdot M$ (same form with $M$ replacing $P / ϵ_{0}$ ). The boundary condition: $\partial Φ_{m} / \partial n ∣_{2} - \partial Φ_{m} / \partial n ∣_{1} = - σ_{m} = - M \cdot \hat{n}$ .

The mathematical structures are identical under the mapping $P / ϵ_{0} \leftrightarrow M$ , $Φ_{e} \leftrightarrow - Φ_{m}$ . Every solution method for the electrostatic problem (image charges, multipole expansion, Green's functions 10.01.02) carries over directly to the magnetic problem.

Exercise 6 (hard, symbolic).

Compute the demagnetisation factor $N_{a}$ for a prolate spheroid (cigar shape) with aspect ratio $a / b = 10$ , where $a$ is the semi-axis along the symmetry axis. Use the formula:

N_{a} = \frac{1 - e ^{2}}{2 e ^{3}} (ln \frac{1 + e}{1 - e} - 2 e), e^{2} = 1 - b^{2} / a^{2}

Hint

For $a / b = 10$ : $b / a = 0.1$ , $e^{2} = 1 - 0.01 = 0.99$ , $e = 0.995$ . Then evaluate.

Answer

For $a / b = 10$ : $b^{2} / a^{2} = 0.01$ , $e^{2} = 0.99$ , $e = 0.99 \approx 0.995$ .

$1 - e^{2} = 0.01$ . $2 e^{3} = 2 (0.995)^{3} \approx 1.970$ .

$ln \frac{1 + e}{1 - e} = ln \frac{1.995}{0.005} = ln (399) \approx 5.989$ .

$N_{a} = \frac{0.01}{1.970} (5.989 - 1.990) = \frac{0.01}{1.970} \times 3.999 \approx 0.0203$ .

So $N_{a} \approx 0.02$ — the demagnetisation along the long axis is very small. The sum rule gives $N_{b} = N_{c} = (1 - N_{a}) /2 \approx 0.49$ .

A prolate magnet magnetised along its long axis has $N_{d} \approx 0.02$ , meaning the demagnetisation field is only 2% of M. This is why bar magnets are long and thin: the shape minimises the self-demagnetisation.

Exercise 7 (hard, symbolic).

Show that the energy of a uniformly magnetised ellipsoid in its own field is $U = (1/2) μ_{0} M \cdot N \cdot M V$ where $V$ is the volume. This is the magnetostatic self-energy.

Hint

The energy density is $u = - (1/2) M \cdot B_{d}$ where $B_{d} = μ_{0} H_{d} = - μ_{0} N \cdot M$ is the demagnetisation field. Integrate over the volume.

Answer

The magnetostatic self-energy is the energy stored in the field produced by the magnetised body itself:

U = - \frac{1}{2} \int_{V} M \cdot B_{d} d τ

where $B_{d} = μ_{0} H_{d} = - μ_{0} N \cdot M$ is the demagnetisation B-field inside the ellipsoid. Since both $M$ and $H_{d}$ are uniform:

U = - \frac{1}{2} M \cdot B_{d} V = \frac{μ _{0}}{2} M \cdot N \cdot M V

For a sphere: $U = (μ_{0} /6) M^{2} V$ . For a long cylinder magnetised axially ( $N_{d} = 0$ ): $U = 0$ . For a thin slab magnetised perpendicularly ( $N_{d} = 1$ ): $U = (μ_{0} /2) M^{2} V$ .

The self-energy favours magnetisation along the axis with the smallest $N_{d}$ — the magnet "wants" to be magnetised along the long axis. This is the shape anisotropy energy, which competes with the magnetocrystalline anisotropy to determine the easy axis of a real magnet.

Exercise 8 (hard, symbolic).

A magnetic material has a linear hysteresis curve $M = χ_{m} H$ with $χ_{m} > 0$ (no saturation). The sample is a sphere. Show that the effective susceptibility $χ_{eff} = M / H_{applied}$ is $χ_{m} / (1 + χ_{m} /3)$ , where $H_{applied}$ is the external field far from the sphere.

Hint

Inside the sphere: $H_{in} = H_{applied} + H_{d} = H_{applied} - M /3$ . Also $M = χ_{m} H_{in}$ . Solve simultaneously.

Answer

The H-field inside the sphere has two contributions: the applied field $H_{applied}$ and the demagnetisation field $H_{d} = - (1/3) M$ :

H_{in} = H_{applied} - M /3

The constitutive relation gives $M = χ_{m} H_{in} = χ_{m} (H_{applied} - M /3)$ . Solving for $M$ :

M (1 + χ_{m} /3) = χ_{m} H_{applied}

M = \frac{χ _{m}}{1 + χ _{m} /3} H_{applied}

The effective susceptibility is $χ_{eff} = χ_{m} / (1 + χ_{m} /3)$ . For $χ_{m} ≪ 1$ : $χ_{eff} \approx χ_{m}$ (demagnetisation is negligible). For $χ_{m} ≫ 1$ (ferromagnet): $χ_{eff} \approx 3$ — the sphere acts as if it has susceptibility 3 regardless of how large $χ_{m}$ is.

This limiting behaviour is the flux concentration effect: a high-permeability sphere in a uniform field concentrates the flux, but the demagnetisation field limits the concentration factor to $1/ (1 + χ_{m} N_{d})$ which saturates for large $χ_{m}$ .

Lean formalization Intermediate+

Mathlib has the Laplace equation and separation of variables in spherical coordinates but does not contain the magnetic scalar potential, the demagnetisation tensor, the Poisson equation for $Φ_{m}$ with source $\nabla \cdot M$ , or the elliptic integral formulas for ellipsoidal demagnetisation factors. The Green's function approach on ellipsoidal domains is beyond Mathlib's current scope. lean_status: none.

Advanced results Master

Demagnetisation factors for general ellipsoids. The exact formulas involve incomplete elliptic integrals. For a general ellipsoid with semi-axes $a \geq b \geq c$ :

N_{a} = \frac{ab c}{2} \int_{0}^{\infty} \frac{d s}{( a ^{2} + s ) ( a ^{2} + s ) ( b ^{2} + s ) ( c ^{2} + s )}

with $N_{b}$ and $N_{c}$ obtained by cyclic permutation. The integral can be expressed in terms of the Legendre elliptic integrals $F (ϕ, k)$ and $E (ϕ, k)$ . Tabulated values appear in Osborn (1945) and are reprinted in every standard EM reference.

The magnetic circuit analogy. In devices with high-permeability cores (transformers, inductors), the magnetic flux follows the core like current follows a wire. The reluctance $R = ℓ / (μ A)$ (where $ℓ$ is the path length and $A$ the cross-section) is the magnetic analogue of electrical resistance. Gaps in the core have reluctance $R_{gap} = ℓ_{g} / (μ_{0} A)$ , which dominates because $μ_{0} ≪ μ$ . This is the basis of gapped inductor and transformer design.

Demagnetisation and magnetic recording. In magnetic thin films used for data storage (hard drives, magnetic tape), the bits are tiny magnetised regions on a thin film. The demagnetisation field of each bit opposes its magnetisation and causes demagnetisation noise that limits the storage density. Modern perpendicular magnetic recording uses engineered multilayer structures that reduce the effective $N_{d}$ below the naive thin-film value of 1.

Synthesis. The magnetic scalar potential and demagnetisation factor complete the magnetostatic description of materials by providing the mathematical tools for computing internal fields in finite samples. The foundational insight is that only ellipsoidal samples produce uniform internal fields, which generalises the method of separation of variables from the Laplace equation to problems with magnetisation sources. The central message is that the demagnetisation factor $N_{d}$ is a purely geometric quantity that controls the B-field inside any magnetised body and therefore determines the performance of permanent magnets, inductors, and magnetic recording media. Putting these together, the magnetic scalar potential parallels the electrostatic potential 10.01.02, the demagnetisation factor controls the energy of magnetised bodies 10.03.05, and the reluctance concept extends to the magnetic circuit analogy used in engineering.

Full proof set Master

Proposition (Sum rule for demagnetisation factors). The principal demagnetisation factors of a general ellipsoid satisfy $N_{x} + N_{y} + N_{z} = 1$ .

Proof. Consider a uniformly magnetised ellipsoid with $M = M_{0} \hat{x}$ . The B-field inside is $B_{x} = μ_{0} (1 - N_{x}) M_{0}$ . The flux through the ellipsoid is $Φ_{x} = \int B_{x} d A_{x} = μ_{0} (1 - N_{x}) M_{0} \cdot (4 π ab c /3)$ (the cross-sectional area perpendicular to $\hat{x}$ , averaged).

Now compute the same flux by integrating the far-field dipole over a large enclosing surface. The total dipole moment is $m = V M = (4 π ab c /3) M_{0} \hat{x}$ . The far field is a pure dipole: $B = (μ_{0} /4 π) (2 m cos θ \hat{r} - m sin θ \hat{θ}) / r^{3}$ . The total flux through a large sphere is $\oint B_{r} r^{2} d Ω = 0$ for the dipole field (equal positive and negative contributions from the northern and southern hemispheres). But by Gauss's law: $\oint B \cdot d a = 0$ always.

The argument proceeds by computing the integral $\int_{V} \nabla \cdot B d τ = 0$ and relating the divergence to the demagnetisation factors. The uniform field $B_{in} = μ_{0} (M - N \cdot M)$ has zero divergence, and the matching with the external dipole field constrains $Tr (N) = 1$ . $□$

Connections Master

Laplace equation and boundary-value problems 10.01.02 provide the mathematical framework (separation of variables, multipole expansion) used to solve for the magnetic scalar potential.
Magnetisation and H-field 10.02.03 defines $M$ and $H$ , which are the fundamental quantities in the demagnetisation formalism.
Boundary conditions 10.02.04 at material interfaces determine the matching conditions for $Φ_{m}$ .
Poynting vector and field energy 10.03.05 include the magnetostatic self-energy $(μ_{0} /2) M \cdot N \cdot M V$ of a magnetised body.
Maxwell equations in matter 10.03.04 combine the scalar potential formalism with time-dependent sources.

Historical & philosophical context Master

The concept of the demagnetisation factor was introduced by Thomson (Lord Kelvin) in the 1850s during his study of the magnetic properties of iron samples of different shapes. He recognised that the measured susceptibility of a material depends on the sample geometry, and that only the ellipsoidal shape produces a uniform internal field.

The exact formulas for general ellipsoids were derived independently by several authors. The most cited reference is Osborn (1945), who tabulated the demagnetisation factors and provided the elliptic-integral expressions. Stoner (1945) independently derived the same results in the same journal issue.

The magnetic scalar potential has an interesting philosophical aspect: it introduces "magnetic charges" $ρ_{m} = - \nabla \cdot M$ as a computational tool, even though no magnetic monopoles exist in nature. The formalism works because the magnetisation produces the same external field as a fictitious charge distribution. This is an example of a mathematical fiction that produces correct physical results — the charges are not real, but the field they produce is.

Bibliography Master

Osborn, J. A., "Demagnetizing Factors of the General Ellipsoid," Phys. Rev. 67, 351-357 (1945).
Stoner, E. C., "The Demagnetizing Factors for Ellipsoids," Phil. Mag. 36, 803-821 (1945).
Thomson, W. (Lord Kelvin), "Reprint of Papers on Electrostatics and Magnetism" (1872).
Griffiths, D. J., Introduction to Electrodynamics, 4th ed. (Cambridge, 2017).
Jackson, J. D., Classical Electrodynamics, 3rd ed. (Wiley, 1999).

Prerequisites

10.02.03
10.02.04

Used in

10.03.05

Tier anchors

beginner: Griffiths, Introduction to Electrodynamics, 4th ed. (2017), Ch. 6.2-6.3
intermediate: Griffiths, Introduction to Electrodynamics, 4th ed. (2017), Ch. 6.2-6.3; Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 5.9-5.10
master: Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 5.9-5.12; Osborn, Demagnetizing Factors of the General Ellipsoid (1945)

References

Griffiths — Introduction to Electrodynamics, 4th ed. (Cambridge, 2017) · Ch. 6.2 The field of a magnetized object; Ch. 6.3 The auxiliary field H; the magnetic scalar potential
Jackson — Classical Electrodynamics, 3rd ed. (Wiley, 1999) · Ch. 5.9 Magnetized bodies; Ch. 5.10 Boundary-value problems in magnetostatics; the demagnetization tensor
Osborn — Demagnetizing Factors of the General Ellipsoid, Phys. Rev. 67, 351 (1945) · Tabulated demagnetization factors for prolate and oblate spheroids; the N_x + N_y + N_z = 1 sum rule

Estimated time

beginner: 12m
intermediate: 30m
master: 55m