A ferromagnet is one for which the magnetization **M**(**r**)
is given and the free current density **j**_{f} = 0.

(a) For this case, write down the relevant Maxwell
equations for the magnetic induction **B** and the magnetic field
**H** in
the absence of any time-varying electric field.

(b) For this ferromagnet relate **B**,
**H**, and
**M**.

(c) Define the magnetic scalar potential
Φ_{M} and state why this is a
valid concept in this case.

(d) Show that
∇^{2}Φ_{M}
can be expressed in terms of **M** by a Poisson equation and give its formal
solution in the absence of boundaries by comparing with the solution of the electrostatic Poisson equation.

Solution:

- Concepts:

Magnetostatics, magnetic materials - Reasoning:

Starting from Maxwell's equations we derive an expression for the magnetic scalar potential for permanent magnets. - Details of the calculation:

(a)**∇**∙**B**= 0,**∇**×**B**= μ_{0}**j**,**j**=**j**_{f }+**j**_{m},**j**=_{m}**∇**×**M**.

**H**=**B**/μ_{0}-**M**,**∇**×**H**=**j**_{f}.

Since**j**_{f}= 0 we have**∇**∙**B**= 0,**∇**×**H**= 0.

(b)**H**=**B**/μ_{0}-**M**.

(c) The curl of**H**is zero everywhere, therefore we may write**H**= -**∇**Φ_{M}.(d)

**∇**∙**H**=**∇**∙**B**/μ_{0}-**∇**∙**M**.

Therefore ∇^{2}Φ_{M}=**∇∙M**.

In electrostatics, the solution to the Poisson equation ∇^{2}Φ = -ρ/ε_{0}in free space is

Φ(**r**) = [1/(4πε_{0})]∫ρ(**r**') dV'/|**r**-**r**'|.

The same equations have the same solutions.

Φ_{M}(**r**) = [-1/(4π)]∫**∇∙M**(**r**') dV'/|**r**-**r**'|.

A sphere of linear magnetic
material is placed in an originally uniform magnetic field magnetic field B_{0}.
Find the new field inside the sphere.

Solution:

- Concepts:

The uniqueness theorem, boundary conditions,**B**= μ**H**. - Reasoning:

The external magnetic field will magnetize the sphere. The magnetic field of a uniformly magnetized sphere is uniform inside the sphere (**B**= 2**M**μ_{0}/3) and a pure dipole field outside the sphere. Assume that the total field inside the sphere is the superposition of the field of a uniform magnetized sphere and the constant external field**B**_{0}. The superposition of these two fields fulfills all the boundary conditions for**B**,

(**B**_{2}-**B**_{1})·**n**_{2}= 0, (**B**_{2}-**B**_{1})·**t**_{2}= μ_{0}**k**·**n**,

on the surface of the sphere.

We have a unique solution as long as inside**B**= μ**H**= μ_{0}(1 + χ_{m})**H**inside the sphere.

**B**= μ**H**determines the magnetization surface current density. - Details of the calculation:

**B**_{in}=**B**_{0}+ 2**M**μ_{0}/3,**H**_{in}=**B**_{in},**M**= χ_{m}**H**_{in}= (χ_{m}/μ)**B**_{in}.

(1 - 2μ_{0}χ_{m}/(3μ))**B**_{in}=**B**_{0}.**B**_{in}=**B**_{0}(3 + 3χ_{m})/(3 + χ_{m}).

The magnetic moment of the sphere is**m**= (4πR^{3}/3)**M**= (4πR^{3}/3)(χ_{m}/μ)**B**_{in}, where R is the radius of the sphere. For r > R we have

**B**_{out}=**B**_{0}+ (μ_{0}/4π)[3(**m**×**r**)**r**/r^{5}-**m**/r^{3}].

A small spherical cavity of radius a is made in a permanent magnet of
uniform magnetization **M**. Find
**B** and
**H** at the center of the
cavity.

Solution:

- Concepts:

The principle of superposition - Reasoning:

Maxwell's equations are linear equations. The principle of superposition holds. We may write

**B**_{cavity}=**B**_{material}-**B**_{uniformly magnetized sphere}. - Details of the calculation:

For a uniformly magnetized sphere we have:

**B**_{out}= pure dipole field,**B**_{in}= uniform, in the direction of**M**.

Let**M**= M**k**,**m**= (4/3)πa^{3}**M**.**B**_{out}= (μ_{0}/4π)[3(**m∙r**)**r**/r^{5}-**m**/r^{3}].

B_{out_θ}= (μ_{0}/4π)(m/r^{3})sinθ, B_{out_r}= (μ_{0}/4π)(2m/r^{3})cosθ.

**B**_{in}= B_{in}**k**.**B**is continuous at**r**= a**k**.

**B**_{in}= (μ_{0}/4π)(2m/a^{3})**k**= (μ_{0}/4π)(8/3)π**M**.

We therefore have**B**_{cavity}=**B**_{material}- (μ_{0}/4π)(8/3)π**M**.

**B**_{material}depends on the shape of the magnetized material.

**H**_{cavity}=**B**_{cavity}/μ_{0}.

A short cylinder (length l and radius a) of iron is magnetized
along the axis of the cylinder. Calculate **H** and
**B** on the axis,
both inside and outside.

Solution:

- Concepts:

Magnetization M, magnetization current densities, the field of a solenoid

**j**=_{m}**∇**×**M**,**k**=_{m}**M**×**n**. - Reasoning:

Magnetization always results in magnetization current densities. Uniform magnetization results in surface current densities. The uniform magnetization of a cylinder results in a surface current density like that of a solenoid. - Details of the calculation:

For a solenoid we have B_{z}= (μ_{0}In/2)(cosθ_{1}- cosθ_{2}).

(We choose the symmetry axis to be the z-axis,)

(I*n) is the current per unit length, i.e. the surface current density.

For the magnetized cylinder we therefore have on the axis B_{z}= (μ_{0}k_{m}/2)(cosθ_{1}- cosθ_{2}).

**k**_{m}= M (**z**/z) ×(**ρ**/ρ) = M (**φ**/φ). B_{z}= (μ_{0}M/2)(cosθ_{1}- cosθ_{2}).

If the center of the cylinder is at the origin then

cosθ_{1}= (z + l/2)/(a^{2}+ (z + l/2)^{2})^{½}, cosθ_{2}= (z - l/2)/(a^{2}+ (z - l/2)^{2})^{½}.

**H**=**B**/μ_{0}outside.**H**=**B**/μ_{0}-**M**inside.**H**=**k**[(M/2)(cosθ_{1}- cosθ_{2}) - M] inside.

At z = l/2 + e (e --> 0) we have**H**=**k**(M/2)cosθ_{1},**H**points into the positive z-direction.

At z = l/2 - e (e --> 0) we have**H**=**k**[(M/2)cosθ_{1 }- M],**H**points into the negative z-direction.

**H**reverses direction at the boundary

Remember: The same current distributions produce the same fields.

The normal component of**H**is not continuous,**∇∙H**≠ 0 in general.

A coil of N turns is wrapped around an iron ring of radius d
and cross section A (d >> A^{½}). Assume a
constant permeability μ >> 1 for the iron.

(a) What is the magnetic flux
Φ = ∫**B∙**d**A** as a function of
current I?

(b) If a gap of width b, (b^{2 }<< A) is cut in the
ring, what is the flux for the same current I?

(c) What is the field energy in the iron and in the gap?

(d) With such a gap in the ring, what is the self inductance?

Solution:

- Concepts:

Ampere's law for**H**, ∮_{Γ}**H∙**d**l**= I_{free through Γ},**H**=**B**/μ_{0}-**M**, magnetic field energy, self inductance - Reasoning:

The problem has enough symmetry to let us calculate H from Ampere's law of H alone.

Iron is not a lih material, but we assume that for the field strength B given in the problem we may write**B**= μ**H**. After a small gap has been cut into the ring we use the boundary conditions for**B**to find**B**and**H**in the gap. We use the energy density u = ½**B∙H**to find the stored energy. - Details of the calculation:

(a)**H**and**B**point into the φ direction, H = NI/(2πr), B = μNI/(2πr).

If d >> A^{½}, then B inside the coil is nearly constant, B = μNI/(2πd).

Flux Φ = ∫B_{n}dA, Φ = μN^{2}IA/(2πd), Φ is proportional to I.

(b) If b^{2 }<< A, then**B**_{iron}≈**B**_{gap}= μ_{0}**H**_{gap}, because the normal component of**B**is continuous.

H_{iron}(2πd - b) + H_{gap}b = NI, (B/μ)(2πd - b) + (B/μ_{0})b = NI, from Ampere's law for**H**.

B = μμ_{0}NI/[μ_{0}(2πd - b) + μb], Φ = μμ_{0}N^{2}IA/[μ_{0}(2πd - b) + μb].

(c) The energy density u = ½**B∙H**= (1/(2μ))B^{2}.

The field energy in the iron is

U_{iron}= A(2πd - b)(μ/2)(μ_{0}NI/[μ_{0}(2πd - b) + μb])^{2}.

The field energy in the gap is

U_{gap}= Ab(μ_{0}/2)(μNI/[μ_{0}(2πd - b) + μb])^{2}.

U_{iron}+ U_{gap}= ½μμ_{0}N^{2}I^{2}A/[μ_{0}(2πd - b) + μb].

(d) Φ = LI or U = ½LI^{2}yield L = μμ_{0}N^{2}A/[μ_{0}(2πd - b) + μb].

Comment:

When you are confronted with a small gap between opposite magnetic poles, you make the same approximation that you make when you are confronted with a small gap between large oppositely charged plates (parallel plate capacitor). You assume that the field in the gap is uniform and neglect edge effects.

In this problem you again use Ampere's law for**H**. Remember that the normal component of**B**is continuous, but the normal component of**H**is not.

An infinitely long straight wire carrying a steady current
I, lies along the axis of a linear paramagnetic cylinder of radius R and
permeability μ.

(a) Find **H**, **B **and
**M **inside and outside the cylinder.

(b) Compute all bound currents flowing in the cylinder.

Solution:

- Concepts:

Ampere's law for**H,**∮_{Γ}**H∙**d**l**= I_{free through Γ},**H**=**B**/μ_{0}-**M** - Reasoning:

The problem has cylindrical symmetry. We can find H from Ampere's law alone.

For linear, isotropic, homogeneous (lih) magnetic materials we have

**M**= Χ_{m}**H**,**B**= μ_{0}(**H**+**M**) = μ_{0}(1 + Χ_{m})**H**= μ_{0}κ_{m}**H**= μ**H**. - Details of the calculation:

(a) Let the current flow along the z-axis in the +z direction.**H**= I/(2πr) (**φ**/φ) .

For 0 < r < R,**M**= (μ/μ_{0}- 1)I/(2πr) (**φ**/φ) ,**B**= μI/(2πr) (**φ**/φ) ,

For r > R,**M**= 0,**B**= μ_{0}I/(2πr) (**φ**/φ)

(b)**j**=_{m}**∇**×**M**,**k**=_{m}**M**×**n**.

**M**= M_{φ}(**φ**/φ),**∇**×**M**=**k**(1/r)∂rM_{φ}/∂r = 0.**∇**×**M**is undefined at r = 0. I_{m}= ∫_{A}(**∇**×**M**)**∙n**dA = ∫_{loop}**M∙**d**s**= (μ/μ_{0}- 1)I.

In addition to the current I in the wire, a magnetization current (μ/μ_{0}- 1)I is flowing in the**k**direction in the center.

Along the surface at r = R we have a surface current**k**_{m}= (μ/μ_{0}- 1)I/(2πR) (-**k**).

A magnetization current (μ/μ_{0}- 1)I is flowing in the**-k**direction along the surface of the magnetic material.

An infinitely long cylinder of
radius R carries a "frozen in" magnetization parallel to the axis, M = kr, where
k is constant and r is the distance from the axis.

(a) Find **B** and **H** inside and outside the cylinder.

(b) Find the magnetic vector potential inside and outside the cylinder.

Solution:

- Concepts:

Magnetization currents, Ampere's law for**H** - Reasoning:

Let the symmetry axis of the cylinder be the z-axis. Then the volume magnetization current density**j**_{m}=**∇**×**M**has only a φ component,

j_{mφ}= -∂M/∂r = -k. The surface magnetization current density**k**_{m}=**M**×**n**also has only a φ component, k_{mφ}= kR.

- Details of the calculation:

(a) The problem is equivalent to that of a set of infinitely-long, nested coils. We therefore know that for r > R**B**=**H**= 0. For r < R**B**and**H**can only have z-components, and we can solve the problem using Amperes law for**H**. The contour integral of the tangential component of**H**along the close loop Γ must be zero,**H**= 0 everywhere.

Inside the magnetized material**B**= μ_{0}(**H**+**M**),**B**= μ>_{0}**M**.(b)

**B**=**∇**×**A**. Symmetry suggests that A can be chosen to only have a φ component.

Stoke's theorem: ∫_{area S }(**∇**×**A) ·n**dS = ∫**B·n**dS = ∫_{Γ}**A·**d**r**= ∫_{Γ}A_{φ}r d φ = 2πrA_{φ}.Inside: ∫

**B·n**dS = ∫_{0}^{r}μ_{0}kr 2πr dr = μ_{0}2πk r^{3}/3.

A_{φ}= μ_{0}k r^{2}/3.

Outside: ∫**B·n**dS = ∫_{0}^{R}μ_{0}kr 2πr dr = μ_{0}2πk R^{3}/3.

A_{φ}= μ_{0}k R^{3}/(3r).

At saturation, when nearly all of the atoms have their magnetic moments
aligned, the magnetic field in a sample of iron can be 2 T. If each
electron contributes a magnetic moment of one Bohr magneton, how many electrons
per atom contribute to the saturated field of the iron?

(Iron: ~8.5*10^{28} atoms/m^{3})

Solution:

- Concepts:

The saturation magnetization**M**_{sat},_{ }**B**= μ_{0}(**H**+**M**). - Reasoning:

**M**_{sat }= Nm**k**, for a magnetic field pointing in the z direction. Here m is the magnitude of the magnetic moment of each atom and N is the number of atoms per unit volume.

At saturation**B**= μ_{0}**M**_{sat}. - Details of the calculation:

The magnetization**M**is defined as the magnetic dipole moment per unit volume.

M = 8.5*10^{28}atoms/m^{3}*n*9.27*10^{-24}Am^{2}.

Here n is the number of electrons per atom that contribute to the saturated field of the iron and 9.27*10^{-24}Am^{2}is the Bohr magneton in SI units.

2 T = 4π*10^{-7}(Tm/A)*8.5*10^{28}m^{-3}*n*9.27*10^{-24}Am^{2},

n = 2.