IPhO 2022 · Theory problem 1

Permanent magnets

Strong permanent magnets are made from NdFeB alloy which obeys a very wide hysteresis loop so that the magnetization \(J\) can be assumed to be constant over a wide range of applications; in what follows, we assume that \(J\equiv 1.5\,\mathrm{T}/\mu_0\), where \(\mu_0=4\pi\times10^{-7}\,\mathrm{N/A^2}\), and the magnetization of all the permanent magnets is homogeneous. Magnetization is defined as the volume density of the magnetic dipole moment of the matter.

Hint 1. The following equality might be useful:

\[ \sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{\pi^4}{90}. \]

Hint 2. The magnetic field created by a spherical magnet is identical to that of a point dipole. The magnetic fields created by magnets of other shapes become equivalent to a point dipole fields only at distances much larger than their diameter.

Hint 3. Electric and magnetic fields of electric and magnetic point dipoles as functions of coordinates and of the corresponding dipole moment are similar, i.e. one can be obtained from the other by multiplying it by a constant factor.

Hint 4. The induced field due to a boundary condition can always be replaced by some configuration of field sources outside the given boundaries.

Part A. Interaction of magnets (4.5 points)

When the distance to a magnet is much larger than its size, the magnetic field created by it can be approximated with the magnetic field of its dipole moment \(\vec m\),

\[ \vec B=\frac{\mu_0}{4\pi r^3}(2\vec m_{\parallel}-\vec m_{\perp}). \]

Here \(r=|\vec r|\), and we have decomposed the dipole moment into components parallel and perpendicular to the radius vector \(\vec r\) drawn from the dipole to the observation point, \(\vec m=\vec m_{\perp}+\vec m_{\parallel}\).

A.1 — 0.6pt

Find the magnitude of the interaction force between two coaxially placed cylindrical magnets of diameter \(d=20\,\mathrm{mm}\) and thickness \(h=2\,\mathrm{mm}\), magnetized parallel to their axis, if the distance between the centres of the magnets is \(L=20\,\mathrm{cm}\). You may assume that \(L\gg d,h\).

Figure 1

A.2 — 0.4pt

At distances much larger than \(\frac h2\), the field created by the magnet from task A.1 is the same as that created by a circular current \(I\). Find \(I\).

A.3 — 1.0pt

Find the interaction force between the magnets for the setup of task A.1 if instead \(L=5\,\mathrm{mm}\). You may assume that \(d\gg L\gg h\).

A.4 — 1.0pt

Identical spherical magnets of diameter \(\delta=5\,\mathrm{mm}\), bound together by magnetic attraction, form a chain. What is the maximal length \(l\) for such a chain which does not break under its own weight when hanging beneath the topmost magnet? The density of NdFeB magnets \(\rho=7500\,\mathrm{kg/m^3}\).

A.5 — 1.5pt

Consider the chain from part A.4. Obtain an expression for the magnitude of the magnetic \(B\)-field at such a point \(P\) which is at distance \(r\) from one of the chain's endpoint \(O\), and the angle between the chain and the line \(OP\) is \(\theta\) (cf. figure below), assuming that \(l\gg r\) and \(r\sin\theta\gg\delta\)

Figure 2

Part B. Interaction with ferromagnets (3.5 points)

Now we assume that in addition to the permanent magnets we have also plates made from a ferromagnetic material, similar to what is used in transformer cores. In the situations we're concerned with, it can be considered to have a constant but very large relative permeability \(\mu_r\sim10^5\).

Hint 5. Large permeability means that magnetic field lines near the outside surface of an object made of the material are nearly perpendicular to the surface. This is similar to the behavior of electric field lines near the outside surface of a conductor.

B.1 — 1.0pt

A spherical magnet from part A.4 is at a distance \(s=\delta\) from a thick infinite ferromagnetic plate (see the answer sheet). The magnetization of the sphere is oriented perpendicular to the plate. Sketch the field lines in the cross-section shown in the answer sheet. In that figure, three points (denoted as 1, 2, and 3) are marked; you need to show field lines passing through each of these points in their full length, i.e. as much as fits into the figure.

Figure 3

B.2 — 1.0pt

Now the spherical magnet is brought into direct contact with the plate. Which direction is taken by the magnetization vector of the spherical magnet at a stable equilibrium and what is the normal force between the plate and the magnet? Mark the correct direction(s) with a tick in the corresponding box in the answer sheet. Incorrect ticks will reduce your score.

Figure 4

B.3 — 1.5pt

Now a magnet from part A.1 is placed between two thick circular ferromagnetic plates of diameter \(D=2d\) so that the flat faces of the magnet are pressed against the plates and all three discs are coaxial. Find the magnetic force \(F\) acting on each plate. Hint: You may neglect the magnetic field both outside the ferromagnetic plates and outside the gap between them.

Part C. (Anti)ferromagnetic order (2 points)

The magnetic properties of materials are due to the magnetic dipole moments of electrons and atomic nuclei. If the dipole moments orient themselves parallel to each other, the field created by them is magnified — these are ferromagnetic materials. On the other hand, if for each dipole moment there is another antiparallel dipole moment nearby, the fields cancel out — these are anti-ferromagnetic materials.

In what follows, we consider a very large number of spherical magnets of part A.4, arranged at the nodes of a two-dimensional lattice; see real photos of stable equilibrium configurations below. Assume that all the magnetization vectors lie in the plane of the figure. Consider in your calculations only nearest-neighbour interactions (on the figure of C.1, each magnet has four nearest neighbours, and on the figure of C.2 — six).

C.1 — 0.8pt

Show the magnetization directions of the magnets in the figure below. You are not required to prove that the configuration you suggested is the only possibility. You still need to justify that the configuration you suggested is indeed stable. Find the energy needed to pull one magnet out of this lattice from somewhere in the middle of the lattice, assuming the other magnets are kept stationary. Does this configuration correspond to the order of ferromagnetic or antiferromagnetic materials?

Figure 5
Figure 6

C.2 — 1.2pt

Answer the same questions as in task C.1 for the configuration shown in the figure below.

Figure 7
Figure 8

Physical Constants

Below is the list of physical constants you might use in your solution in addition to the quantities given in the problem text.

Physical constant Notation Numerical value
Speed of light in vacuum \(c\) \(299\,792\,458\,\mathrm{m/s}\)
Vacuum permeability (magnetic constant) \(\mu_0\) \(4\pi\cdot10^{-7}\,\mathrm{N/A^2}\); \([1\,\mathrm{N/A^2}=1\,\mathrm{Vs/Am}]\)
Vacuum permittivity (electric constant) \(\varepsilon_0\) \(8.8541878128(13)\cdot10^{-12}\,\mathrm{F/m}\); \([1\,\mathrm{F/m}=1\,\mathrm{As/Vm}]\)
Elementary charge \(e\) \(1.602176634\cdot10^{-19}\,\mathrm{C}\); \([1\,\mathrm{C}=1\,\mathrm{A\cdot s}]\)
Electron rest mass \(m_{\mathrm e}\) \(9.1093837015(28)\cdot10^{-31}\,\mathrm{kg}\)
Proton rest mass \(m_{\mathrm p}\) \(1.67262192369(51)\cdot10^{-27}\,\mathrm{kg}\)
Neutron rest mass \(m_{\mathrm n}\) \(1.67492749804(95)\cdot10^{-27}\,\mathrm{kg}\)
Atomic mass unit \(m_{\mathrm{amu}}\) \(1.66053906660(50)\cdot10^{-27}\,\mathrm{kg}\)
Gravitational constant \(G\) \(6.67430(15)\cdot10^{-11}\,\mathrm{m^3/(kg\cdot s^2)}\)
Avogadro constant \(N_{\mathrm A}\) \(6.02214076\cdot10^{23}\,\mathrm{mol^{-1}}\)
Molar gas constant \(R\) \(8.31446261815324\ldots\,\mathrm{J/(K\cdot mol)}\)
Boltzmann constant \(k_{\mathrm B}\) \(1.380649\cdot10^{-23}\,\mathrm{J/K}\)
Stefan–Boltzmann constant \(\sigma\) \(5.670374419\ldots\cdot10^{-8}\,\mathrm{W/(m^2\cdot K^4)}\)
Planck constant \(h\) \(6.62607015\cdot10^{-34}\,\mathrm{J\cdot s}\)
Reduced Planck constant \(\hbar=h/(2\pi)\) \(1.0545718176\ldots\cdot10^{-34}\,\mathrm{J\cdot s}\)

Official English text of the problem.

Answers

All force values below are magnitudes.

  • A.1:
    \[ \boxed{F=\frac{3\pi\mu_0J^2d^4h^2}{32L^4} \simeq2.1\times10^{-4}\,\mathrm N} \] The force is attractive for the directions shown.

  • A.2: A circular current of diameter \(d\), with \[ \boxed{I=Jh\simeq2.4\times10^3\,\mathrm A}. \]

  • A.3:
    \[ \boxed{F\simeq\frac{\mu_0J^2h^2d}{2L}\simeq14\,\mathrm N} \] The force is attractive.

  • A.4: In the long-chain approximation, \[ \boxed{l_{\max}\simeq\frac{\pi^4\mu_0J^2}{360\rho g} \simeq6.6\,\mathrm m}. \]

  • A.5:
    \[ \boxed{B\simeq\frac{\mu_0J\delta^2}{24r^2}} \] This magnitude is independent of \(\theta\) within the stated approximations.

  • B.1: The exterior field is that of the real dipole together with an equally strong mirror dipole pointing in the same direction. Field lines meet the plate normally. The line through 1 forms a short return loop in air; the lines through 2 and 3 belong to larger loops carrying flux through the plate. Their detailed directions are described in the Solution.

  • B.2: Tick the first drawing in each row: magnetization perpendicular to the plate, either toward it or away from it. \[ \boxed{F=\frac{\pi\mu_0J^2\delta^2}{24}\simeq5.9\,\mathrm N} \] The magnetic force is attractive.

  • B.3: Each plate is pulled toward the magnet and the other plate, with \[ \boxed{F=\frac{3\pi\mu_0J^2d^2}{32} \simeq2.1\times10^2\,\mathrm N}. \]

  • C.1: All moments in a horizontal row point in the same direction; successive rows point oppositely. This is antiferromagnetic order. \[ \boxed{W=\frac{\pi\mu_0J^2\delta^3}{24} \simeq2.9\times10^{-2}\,\mathrm J}. \]

  • C.2: All moments point in the same horizontal direction. This is ferromagnetic order, passing the single-magnet rotational stability test described below. \[ \boxed{W=\frac{\pi\mu_0J^2\delta^3}{48} \simeq1.5\times10^{-2}\,\mathrm J}. \] This shows stability against rotating a single magnet only; in the nearest-neighbour model, some collective rotations of many magnets lower the energy.

Solution

The magnetization and the two relevant dipole moments are

\[ J=\frac{1.5}{\mu_0} =1.19366\times10^6\,\mathrm{A/m}, \]

\[ m_{\mathrm c}=J\frac{\pi d^2h}{4} =0.750\,\mathrm{A\,m^2}, \qquad m_{\mathrm s}=J\frac{\pi\delta^3}{6} =0.078125\,\mathrm{A\,m^2}. \]

For two permanent dipoles, the interaction energy is

\[ U_{12}= \frac{\mu_0}{4\pi r^3} \left[ \vec m_1\cdot\vec m_2 -3(\vec m_1\cdot\hat r)(\vec m_2\cdot\hat r) \right]. \]

Equivalently, \(U_{12}=-\vec m_2\cdot\vec B_1\). Forces follow by differentiating this energy with respect to position, holding the permanent moments fixed.

A.1

Since \(L\gg d,h\), both cylinders can be replaced by point dipoles. Their moments are parallel to each other and to their common axis, so

\[ U(L)=-\frac{\mu_0m_{\mathrm c}^2}{2\pi L^3}. \]

The force tends to decrease \(L\), and its magnitude is

\[ F=\frac{dU}{dL} =\frac{3\mu_0m_{\mathrm c}^2}{2\pi L^4}. \]

Substituting \(m_{\mathrm c}=J\pi d^2h/4\),

\[ \boxed{F=\frac{3\pi\mu_0J^2d^4h^2}{32L^4}}. \]

With \(L=0.20\,\mathrm m\),

\[ F=2.109375\times10^{-4}\,\mathrm N \simeq\boxed{2.1\times10^{-4}\,\mathrm N}. \]

A.2

A uniformly magnetized body can be represented by bound currents. Its bound surface-current density is

\[ \vec K_{\mathrm b}=\vec J\times\hat n. \]

For an axially magnetized cylinder, this current flows around the curved side, with magnitude \(K_{\mathrm b}=J\). Integrating over the cylinder thickness gives an equivalent circular current

\[ \boxed{I=Jh}. \]

At distances large compared with the thickness, the thin cylindrical current sheet can be replaced by a single loop of diameter \(d\).

The same result follows by equating the magnetic moments:

\[ I\frac{\pi d^2}{4} =J\frac{\pi d^2h}{4}. \]

Numerically,

\[ I=(1.19366\times10^6)(2.0\times10^{-3}) =2.3873\times10^3\,\mathrm A \simeq\boxed{2.4\,\mathrm{kA}}. \]

Its circulation is chosen by the right-hand rule to produce the specified magnetization direction.

A.3

The condition \(L\gg h\) permits the current-loop representation from A.2. Because \(L\ll d\), corresponding nearby portions of the two loops can be treated as parallel straight wires separated by \(L\).

The attractive force per unit length between parallel currents is

\[ \frac{dF}{d\ell}=\frac{\mu_0I^2}{2\pi L}. \]

To leading order, this force acts along the common axis. Multiplying by the circumference \(\pi d\) gives

\[ F\simeq\frac{\mu_0I^2}{2\pi L}\,\pi d =\frac{\mu_0I^2d}{2L}. \]

Thus

\[ \boxed{F\simeq\frac{\mu_0J^2h^2d}{2L}}. \]

For \(L=5.0\times10^{-3}\,\mathrm m\),

\[ F\simeq14.3239\,\mathrm N \simeq\boxed{14\,\mathrm N}. \]

This is the leading result under the stated hierarchy \(d\gg L\gg h\).

A.4

The spheres align with their moments along the chain, all pointing in the same direction. Each sphere has mass

\[ M=\rho\frac{\pi\delta^3}{6}. \]

Two spheres whose centres are \(n\delta\) apart attract with force

\[ F_n=\frac{3\mu_0m_{\mathrm s}^2}{2\pi(n\delta)^4} =\frac{F_1}{n^4}, \]

where

\[ F_1=\frac{3\mu_0m_{\mathrm s}^2}{2\pi\delta^4} =\frac{\pi\mu_0J^2\delta^2}{24}. \]

For a chain containing \(N\) spheres, the held top sphere must support the other \(N-1\) spheres. The largest upward magnetic force it can exert on them, while all spheres remain touching, is

\[ F_{\mathrm{top}} =F_1\sum_{n=1}^{N-1}\frac1{n^4}. \]

Consequently,

\[ (N-1)Mg \leq F_1\sum_{n=1}^{N-1}\frac1{n^4}. \]

The uppermost junction is limiting: a cut farther down supports less weight, while its magnetic attraction includes at least one pair at every separation present across the uppermost junction.

For a long chain,

\[ F_{\mathrm{top}} \simeq F_1\frac{\pi^4}{90}. \]

Also, to leading order in \(1/N\), the suspended weight is

\[ Mg\frac l\delta =\frac{\rho\pi\delta^2g}{6}\,l. \]

Equating this to the limiting magnetic force,

\[ \frac{\rho\pi\delta^2g}{6}\,l_{\max} = \frac{\pi\mu_0J^2\delta^2}{24}\frac{\pi^4}{90}. \]

Therefore,

\[ \boxed{l_{\max}\simeq \frac{\pi^4\mu_0J^2}{360\rho g}}. \]

The sphere diameter cancels: the magnetic holding force and the chain’s mass per unit length both scale as \(\delta^2\).

Taking \(g=9.81\,\mathrm{m/s^2}\),

\[ F_1=5.8594\,\mathrm N,\qquad F_{\mathrm{top}}\simeq6.3417\,\mathrm N, \]

\[ l_{\max}\simeq6.5848\,\mathrm m \simeq\boxed{6.6\,\mathrm m}. \]

Counting the supported top sphere and rounding to a whole number of spheres changes the length only by an amount of order \(\delta\).

A.5

Because \(r\sin\theta\gg\delta\), the discrete chain can be replaced by a continuous line of dipoles. Its dipole moment per unit length is

\[ \lambda_m=\frac{m_{\mathrm s}}{\delta} =\frac{\pi J\delta^2}{6}. \]

Place \(O\) at the origin and let the chain extend along \(+z\), from \(0\) to \(l\). Choose all moments along \(+z\). In the exterior region, define a magnetic scalar potential by

\[ \vec B=-\mu_0\nabla\Phi. \]

The potential of the line is

\[ \Phi(\vec r) =\frac{\lambda_m}{4\pi} \int_0^l \frac{z-z'}{|\vec r-z'\hat z|^3}\,dz'. \]

Since

\[ \frac{\partial}{\partial z'} \frac1{|\vec r-z'\hat z|} = \frac{z-z'}{|\vec r-z'\hat z|^3}, \]

integration gives

\[ \Phi(\vec r) =\frac{\lambda_m}{4\pi} \left( \frac1{|\vec r-l\hat z|} -\frac1r \right). \]

Thus the exterior field is equivalent to that of two opposite effective magnetic poles at the endpoints:

\[ \vec B= \frac{\mu_0\lambda_m}{4\pi} \left[ \frac{\vec r-l\hat z}{|\vec r-l\hat z|^3} -\frac{\vec r}{r^3} \right]. \]

For \(l\gg r\), the distant endpoint contributes negligibly, leaving

\[ \vec B\simeq -\frac{\mu_0\lambda_m}{4\pi r^2}\hat r. \]

Hence

\[ \boxed{B\simeq\frac{\mu_0J\delta^2}{24r^2}}. \]

The field is radial about the nearby endpoint, so its magnitude is independent of \(\theta\). Reversing all moments reverses the field direction.

B.1

Let the plate occupy \(x<0\), with its surface at \(x=0\), and let the sphere’s centre be at \(x=a>0\). In the answer-sheet drawing,

\[ \vec m=-m_{\mathrm s}\hat x. \]

For a highly permeable plate, the exterior tangential field must nearly vanish. This boundary condition is produced by placing a mirror dipole at \(x=-a\), with the same normal orientation:

\[ \vec m_{\mathrm{image}}\simeq-m_{\mathrm s}\hat x. \]

More precisely, its magnitude is multiplied by

\[ \frac{\mu_r-1}{\mu_r+1}\simeq1. \]

At the surface, the tangential components of the two dipole fields cancel. If \(y\) denotes height above the sphere’s centre, the remaining normal component is

\[ B_x(0,y) = \frac{\mu_0m_{\mathrm s}}{2\pi} \frac{y^2-2a^2}{(a^2+y^2)^{5/2}}. \]

Thus flux enters the plate near the sphere’s axis, where \(|y|<\sqrt2\,a\), and emerges farther away, where \(|y|>\sqrt2\,a\).

The requested sketch is described as follows:

  • The sphere’s left side is its north side, and its right side is its south side. Exterior arrows run from the left side toward the right side along curved return paths. Inside the sphere, arrows run approximately right to left.
  • Through point 1: draw a short upper return loop entirely in the air outside the plate. It leaves the sphere’s upper-left surface, rises through point 1, bends over toward the right, and enters the sphere’s upper-right surface.
  • Through point 2: draw a larger loop carrying flux upward through the plate. At point 2 it points mainly upward and somewhat toward the surface. It bends rightward and emerges from the surface above point 3, perpendicular to the surface, before curving back to the sphere’s south side. Traced backward, its segment inside the plate descends and connects to flux entering the plate nearer the magnet.
  • Through point 3: draw a line emerging approximately horizontally to the right from the plate. It then curves downward and rightward toward the sphere’s south side. Continue it backward through the plate as a curved path returning to a lower entry point, supplied by flux from the sphere’s north side.
  • The lower-half pattern is the reflection of the upper-half pattern about the horizontal line through the sphere’s centre. Every line continues through the material and magnet to form a closed magnetic-field line.

For the ideal half-space limit, the field inside the plate is approximately twice the field that the real dipole alone would create there. This fixes the curved continuations through point 2 as well as their directions.

B.2

Let \(\hat n\) point normally out of the plate, and decompose the sphere’s moment as

\[ \vec m=m_n\hat n+\vec m_t. \]

Its image has moment

\[ \vec m'=m_n\hat n-\vec m_t. \]

If the sphere’s centre is distance \(a\) from the plate, the image is \(2a\) away. Its field at the real dipole is

\[ \vec B_{\mathrm{image}} = \frac{\mu_0}{4\pi(2a)^3} \left(2m_n\hat n+\vec m_t\right). \]

Because the plate magnetization is induced, the interaction energy is

\[ U=-\frac12\vec m\cdot\vec B_{\mathrm{image}}. \]

Writing \(\theta\) for the angle between \(\vec m\) and the plate normal,

\[ U(a,\theta) = -\frac{\mu_0m_{\mathrm s}^2} {8\pi(2a)^3} (1+\cos^2\theta). \]

This is minimized at

\[ \theta=0\quad\text{or}\quad\theta=\pi. \]

Therefore, both directions perpendicular to the plate are stable: toward the plate and away from it. These are the first drawings in the upper and lower rows.

At either stable orientation, the attractive force has magnitude

\[ F=\frac{\partial U}{\partial a} =\frac{3\mu_0m_{\mathrm s}^2}{2\pi(2a)^4}. \]

At contact, \(a=\delta/2\), so

\[ F=\frac{3\mu_0m_{\mathrm s}^2}{2\pi\delta^4} =\boxed{\frac{\pi\mu_0J^2\delta^2}{24}}. \]

Numerically,

\[ F=5.8594\,\mathrm N \simeq\boxed{5.9\,\mathrm N}. \]

B.3

Let

\[ A=\frac{\pi d^2}{4} \]

be the magnet’s face area, and

\[ A_p=\frac{\pi D^2}{4}=4A \]

the area of each plate.

The permeability of the plates is very large, so their internal \(H\)-field is negligible. Each plate is therefore approximately an equipotential surface for the magnetic scalar potential. Neglecting fringing, the axial \(H\)-field has the same value in the magnet and in the surrounding annular air gap.

Let \(B_m\) be the signed axial field in the magnet and \(B_a\) that in the annular gap. Since the magnetization is fixed,

\[ B_m=\mu_0(H+J),\qquad B_a=\mu_0H. \]

Hence

\[ B_m-B_a=\mu_0J. \]

All flux passing through the magnet must return through the annular gap, because exterior flux is neglected. Therefore,

\[ AB_m+(A_p-A)B_a=0. \]

Solving,

\[ B_m=\mu_0J\left(1-\frac A{A_p}\right), \qquad B_a=-\mu_0J\frac A{A_p}. \]

For \(A_p=4A\),

\[ B_m=\frac34\mu_0J=1.125\,\mathrm T, \qquad B_a=-\frac14\mu_0J=-0.375\,\mathrm T. \]

A magnetic field normal to a highly permeable surface produces an attractive pressure

\[ p=\frac{B^2}{2\mu_0}. \]

At the magnet–plate contact this can be evaluated by imagining an infinitesimal vacuum slit: normal \(B\) is continuous across it. Both the central region and the annular return-flux region pull the plate inward, irrespective of their opposite field directions.

Thus

\[ F= \frac{AB_m^2+(A_p-A)B_a^2}{2\mu_0}. \]

Substitution gives

\[ F= \frac{\mu_0J^2A}{2} \left(1-\frac A{A_p}\right). \]

With \(A_p=4A\),

\[ \boxed{F=\frac{3\pi\mu_0J^2d^2}{32}}. \]

Numerically,

\[ F=210.9375\,\mathrm N \simeq\boxed{2.1\times10^2\,\mathrm N} \]

on each plate.

C.1

Define the nearest-neighbour dipolar energy scale

\[ \varepsilon_d= \frac{\mu_0m_{\mathrm s}^2}{4\pi\delta^3} =\frac{\pi\mu_0J^2\delta^3}{144} =4.8828125\times10^{-3}\,\mathrm J. \]

Choose moments along horizontal rows, reversing their direction in successive rows. On the answer sheet, the top row can point right, the middle row left, and the bottom row right:

\[ \begin{matrix} \rightarrow&\rightarrow&\rightarrow&\rightarrow\\ \leftarrow&\leftarrow&\leftarrow&\leftarrow\\ \rightarrow&\rightarrow&\rightarrow&\rightarrow \end{matrix} \]

Consider a central magnet pointing right.

Its two horizontal neighbours also point right and lie on its dipole axis. Each contributes interaction energy

\[ U_{\mathrm{horizontal}}=-2\varepsilon_d. \]

Its two vertical neighbours point left and lie perpendicular to its moment. Each contributes

\[ U_{\mathrm{vertical}}=-\varepsilon_d. \]

The central magnet’s total interaction energy with its neighbours is therefore

\[ U_{\mathrm{central}} =2(-2\varepsilon_d)+2(-\varepsilon_d) =-6\varepsilon_d. \]

All four neighbours produce fields parallel to the central moment, with total magnitude

\[ B_{\mathrm{nn}}=\frac{6\varepsilon_d}{m_{\mathrm s}}. \]

Rotating that magnet through an angle \(\alpha\), while keeping its neighbours fixed, increases the energy by

\[ \Delta U=6\varepsilon_d(1-\cos\alpha)>0. \]

Thus it experiences a restoring torque.

For fixed square-lattice sites, the pattern also minimizes the full nearest-neighbour orientational energy. To see this, write a unit moment as \((u_i,v_i)\). Then

\[ \frac{U}{\varepsilon_d} = \sum_{\langle ij\rangle_x}(-2u_iu_j+v_iv_j) + \sum_{\langle ij\rangle_y}(u_iu_j-2v_iv_j) \geq -3\sum_i(u_i^2+v_i^2) =-3N. \]

The inequality follows from \(2|ab|\leq a^2+b^2\), and the proposed stripe pattern attains the bound in the bulk.

Removing the central magnet to infinity eliminates its four bonds, so the required work is

\[ W=0-U_{\mathrm{central}}=6\varepsilon_d. \]

There is no factor of \(1/2\): each bond involving the removed magnet is counted once. Therefore,

\[ \boxed{W=\frac{\pi\mu_0J^2\delta^3}{24} \simeq2.9\times10^{-2}\,\mathrm J}. \]

Successive rows have opposite magnetization, giving antiferromagnetic order.

C.2

Choose all moments horizontal and pointing right, including every sphere in the answer-sheet diagram.

A central sphere has two horizontal neighbours and four diagonal neighbours whose displacement vectors make angles of \(60^\circ\) or \(120^\circ\) with the moments.

Each horizontal pair contributes

\[ U_{\mathrm{horizontal}}=-2\varepsilon_d. \]

For parallel moments at angle \(\beta\) to their separation,

\[ U_{\mathrm{pair}} =\varepsilon_d(1-3\cos^2\beta). \]

Thus each diagonal neighbour contributes

\[ U_{\mathrm{diagonal}} =\varepsilon_d\left(1-\frac34\right) =\frac14\varepsilon_d. \]

Summing the six bonds,

\[ U_{\mathrm{central}} =2(-2\varepsilon_d)+4\left(\frac14\varepsilon_d\right) =-3\varepsilon_d. \]

The transverse components of the neighbours’ fields cancel. The remaining field points along the central moment and has magnitude

\[ B_{\mathrm{nn}}=\frac{3\varepsilon_d}{m_{\mathrm s}}. \]

Consequently, rotating one magnet through \(\alpha\) with all its neighbours fixed costs

\[ \Delta U=3\varepsilon_d(1-\cos\alpha)>0. \]

This establishes a restoring torque against an individual rotation. It does not establish stability against collective rotations: in the nearest-neighbour model, rotating all moments together through the same angle leaves the energy unchanged, because the six bond directions are \(60^\circ\) apart, and some non-uniform collective rotations lower the energy.

The work required to remove the central magnet is

\[ W=0-U_{\mathrm{central}}=3\varepsilon_d, \]

so

\[ \boxed{W=\frac{\pi\mu_0J^2\delta^3}{48} \simeq1.5\times10^{-2}\,\mathrm J}. \]

All moments point in the same direction, so this configuration has ferromagnetic order.

The solution was checked against the official solution.