Editing Halbach array (section)

==Cylinder==
[[File:Halbach cylinder.png|thumb|A ferromagnetic cylinder showing various magnetization patterns and magnetic field|220x220px]]
[[File:Halbach cylinder2.png|thumb|right|Cylinder magnetization]]
A '''Halbach cylinder''' is a magnetized cylinder composed of [[ferromagnetic]] material producing (in the idealized case) an intense magnetic field confined entirely within the cylinder, with zero field outside. The cylinders can also be magnetized such that the magnetic field is entirely outside the cylinder, with zero field inside. Several magnetization distributions are shown in the figures.

The direction of magnetization within the ferromagnetic material, in plane perpendicular to the axis of the cylinder, is given by

: <math>M = M_r\left[\cos\left((k - 1)\left(\varphi - \frac{\pi}{2}\right)\right) \widehat{\rho} + \sin\left((k - 1)\left(\varphi - \frac{\pi}{2}\right)\right) \widehat{\varphi}\right],</math>

where ''M<sub>r</sub>'' is the ferromagnetic [[remanence]] (A/m). A positive value of ''k''&nbsp;−&nbsp;1 gives an internal magnetic field, and a negative one gives an external magnetic field.

Ideally, these structures would be created from an infinite-length cylinder of magnetic material with the direction of magnetization continuously varying. The magnetic flux produced by this ideal design would be perfectly uniform and be entirely confined to either the bore of the cylinder or the outside of the cylinder. Of course, the ideal case of infinite length is not realizable, and in practice the finite length of the cylinders produces ''end effects'', which introduce non-uniformities in the field.<ref name="Mhiochain1999"/><ref name="Bjoerk2011"/> The difficulty of manufacturing a cylinder with a continuously varying magnetization also usually leads to the design being broken into segments.

===Applications===
These cylindrical structures are used in devices such as brushless AC motors, magnetic couplings and high-field cylinders. Both brushless motors and coupling devices use multipole field arrangements:
* Brushless motors or alternators typically use cylindrical designs in which all the flux is confined to the centre of the bore (such as ''k''&nbsp;=&nbsp;4 above, a 6-pole rotor) with the AC coils also contained within the bore. Such self-shielding motor or alternator designs are more efficient and produce higher torque or output than conventional motor or alternator designs.
* Magnetic-coupling devices transmit torque through magnetically transparent barriers (that is, the barrier is non-magnetic or is magnetic but not affected by an applied magnetic field), for instance, between sealed containers or pressurised vessels. The optimal torque couplings consists of a pair of coaxially nested cylinders with opposite +''k'' and −''k'' flux magnetization patterns, as this configuration is the only system for infinitely long cylinders that produces a torque.<ref name="Bjoerk2010"/> In the lowest-energy state, the outer flux of the inner cylinder exactly matches the internal flux of the outer cylinder. Rotating one cylinder relative to the other from this state results in a restoring torque.
*Cylindrical Halbach Arrays are used in Portable [[MRI]] Scanners.<ref>{{cite web |url=https://www.stanfordmagnets.com/halbach-magnets-history-types-and-uses.html |title=Halbach Magnets: History, Types, and Uses |last=Marchio |first=Cathy |date=May 23, 2024 |website=Stanford Magnets |access-date=July 31, 2024}}</ref> They offer the potential for a relatively lightweight, low to mid-field system with no [[cryogenics]], a small fringe field, and no electrical power requirements or heat dissipation needs.<ref>{{cite journal |last1=Cooley |first1=C. Z. |last2=Haskell |first2=M. W. |year=2018 |title=Design of Sparse Halbach Magnet Arrays for Portable MRI Using a Genetic Algorithm |journal=IEEE Transactions on Magnetics |volume=54 |issue=1 |pages=1-12 |doi=10.1109/TMAG.2017.2751001|pmc=5937527 }}</ref> The reduced stray fields also enhance safety and minimize interference with surrounding electronic devices.<ref>{{cite journal |last1=Sarwar |first1=A. |last2=Nemirovski |first2=A. |year=2012 |title=Optimal Halbach permanent magnet designs for maximally pulling and pushing nanoparticles |journal=Journal of Magnetism and Magnetic Materials |volume=234 |issue=5 |pages=742-754 |doi=10.1016/j.jmmm.2011.09.008|pmc=3547684 }}</ref>

===Uniform fields===
[[File:Halbach array by Zureks.png|thumb|Uniform field inside Halbach cylinder|220x220px]]
For the special case of ''k'' = 2, the field inside the bore is uniform and is given by

: <math>H = M_r \ln\left(\frac{R_\text{o}}{R_\text{i}}\right) \widehat{y},</math>

where the inner and outer cylinder radii are ''R''<sub>i</sub> and ''R''<sub>o</sub> respectively. ''H'' is in the ''y'' direction. This is the simplest form of the Halbach cylinder, and it can be seen that if the ratio of outer to inner radii is greater than [[e (mathematical constant)|''e'']], the flux inside the bore actually exceeds the [[remanence]] of the magnetic material used to create the cylinder. However, care has to be taken not to produce a field that exceeds the coercivity of the permanent magnets used, as this can result in demagnetization of the cylinder and the production of a much lower field than intended.<ref name="Bjoerk2015"/><ref name="Insinga2016"/>

[[File:Halbach cylinder3.png|thumb|Three designs (A) (B) (C) producing uniform magnetic fields within their central air gap|220x220px]]
This cylindrical design is only one class of designs that produce a uniform field inside a cavity within an array of permanent magnets. Other classes of design include wedge designs, proposed by Abele and Jensen, in which wedges of magnetized material are arranged to provide uniform field within cavities inside the design as shown.

The direction of magnetization of the wedges in (A) can be calculated using a set of rules given by Abele and allows for great freedom in the shape of the cavity. Another class of design is the magnetic mangle (B), proposed by Coey and Cugat,<ref name="Coey2003"/><ref name="Cugat1998"/> in which uniformly magnetized rods are arranged such that their magnetization matches that of a Halbach cylinder, as shown for a 6-rod design. This design greatly increases access to the region of uniform field, at the expense of the volume of uniform field being smaller than in the cylindrical designs (although this area can be made larger by increasing the number of component rods). Rotating the rods relative to each other results in many possibilities, including a dynamically variable field and various dipolar configurations. It can be seen that the designs shown in (A) and (B) are closely related to the ''k''&nbsp;=&nbsp;2 Halbach cylinder. Other very simple designs for a uniform field include separated magnets with soft iron return paths, as shown in figure (C).

In recent years, these Halbach dipoles have been used to conduct low-field [[NMR]] experiments.<ref>{{cite journal |author=Raich, H. |author2=Blümler, P. |title=Design and construction of a dipolar Halbach array with a homogeneous field from identical bar magnets: NMR Mandhalas |journal=Concepts in Magnetic Resonance Part B: Magnetic Resonance Engineering |date=21 October 2004 |volume=23B |pages=16–25 |doi=10.1002/cmr.b.20018|s2cid=58309210 }}</ref> Compared to commercially available ([[Bruker]] Minispec) standard plate geometries (C) of permanent magnets, they, as explained above, offer a huge bore diameter, while still having a reasonably homogeneous field.

====Derivation in the ideal case====
The method used to find the field created by the cylinder is mathematically very similar to that used to investigate a uniformly magnetised sphere.<ref>{{cite web
 | title = Uniformly Magnetized Sphere
 | url = https://farside.ph.utexas.edu/teaching/jk1/Electromagnetism/node61.html
 | last1 = Fitzpatrick
 | first1 = Richard
 | date = 2014-06-27
 | access-date =  2022-03-28
 | publisher = [[The University of Texas at Austin]]
}}</ref>

Because of the symmetry of the arrangement along the cylinder's axis, the problem can be treated as two-dimensional. Work in [[Polar coordinate system|plane-polar coordinates]] <math>(r, \theta)</math> with associated unit vectors <math>\hat\mathbf{r}</math> and <math>\hat\boldsymbol{\theta}</math>, and let the cylinder have radial extent <math>r_\mathrm{i} < r < r_\mathrm{o}</math>. Then the [[Magnetization|magnetisation]] in the cylinder walls, which has magnitude <math>M_0</math>, rotates smoothly as
:<math>\mathbf{M}_\mathrm{cyl} = M_0 (\cos\theta\, \hat\mathbf{r} + \sin\theta\, \hat\boldsymbol{\theta}),</math>
while the magnetisation vanishes outside the walls, that is for the bore <math>r < r_\mathrm{i}</math> and surroundings <math>r > r_\mathrm{o}</math>.

By definition, the auxiliary [[Magnetic field|magnetic field strength]] <math>\mathbf{H}</math> is related to the magnetisation and [[magnetic flux density]] <math>\mathbf{B}</math> by <math>\mathbf{B} = \mu_0 (\mathbf{H} + \mathbf{M})</math>. Using [[Gauss's law for magnetism|Gauss' law]] <math>\nabla\cdot\mathbf{B} = 0</math>, this is equivalently
{{NumBlk
 | :
 | <math>\nabla\cdot\mathbf{H} = -\nabla\cdot\mathbf{M}.</math>
 | {{EquationRef|1}}
}}
Since the problem is static there are no free currents and all time derivatives vanish, so [[Ampère's circuital law|Ampère's law]] additionally requires <math>\nabla\times\mathbf{H}=0 \implies \mathbf{H} = \nabla\varphi</math>, where <math>\varphi</math> is the [[magnetic scalar potential]] (up to a sign under some definitions). Substituting this back into the previous Equation {{EquationNote|1}} governing <math>\mathbf{H}</math> and <math>\mathbf{M}</math>, we find that we need to solve
{{NumBlk
 | :
 | <math>\nabla^2\varphi = -\nabla\cdot\mathbf{M},</math>
 | {{EquationRef|2}}
}}
which has the form of [[Poisson's equation]].

Consider now the boundary conditions at the cylinder-air interfaces <math>r = r_\mathrm{i}</math> and <math>r = r_\mathrm{o}</math>. Integrating <math>\nabla \times \mathbf{H} = 0</math> over a small loop straddling the boundary and applying [[Stokes' theorem]] requires that the parallel component of <math>\mathbf{H}</math> is continuous. This in turn requires that <math>\varphi</math> is continuous across the boundary. (More properly this implies that <math>\varphi</math> must differ by a ''constant'' across the boundary, but since the physical quantities we are interested in depend on gradients of this potential, we can arbitrarily set the constant to zero for convenience.) To obtain a second set of conditions, integrate Equation {{EquationNote|1}} across a small volume straddling the boundary and apply the [[divergence theorem]] to find
:<math>\left[\frac{\partial \varphi}{\partial r}\right] = \pm \mathbf{M} \cdot \hat{\mathbf{r}},</math>
where the notation <math>[f]</math> denotes a jump in the quantity <math>f</math> across the boundary, and in our case the sign is negative at <math>r = r_\mathrm{i}</math> and positive at <math>r = r_\mathrm{o}</math>. The sign difference is due to the relative orientation of the magnetisation and the surface normal to the part of the integration volume inside the cylinder walls being opposite at the inner and outer boundaries.

In plane-polar coordinates, the [[divergence]] of a [[vector field]] <math>\mathbf{F} = F_r\hat\mathbf{r} + F_\theta \hat\boldsymbol{\theta}</math> is given by
{{NumBlk
 | :
 | <math>\nabla\cdot\mathbf{F} = \frac{1}{r}\frac{\partial}{\partial r}(r F_r) + \frac{1}{r}\frac{\partial F_\theta}{\partial\theta}.</math>
 | {{EquationRef|3}}
}}
Similarly, the [[gradient]] of a [[scalar field]] <math>f</math> is given by
{{NumBlk
 | :
 | <math>\nabla f = \frac{\partial f}{\partial r}\hat\mathbf{r} + \frac{1}{r}\frac{\partial f}{\partial\theta}\hat\boldsymbol{\theta}.</math>
 | {{EquationRef|4}}
}}
Combining these two relations, the [[Laplace operator|Laplacian]] <math>\nabla^2 f = \nabla\cdot(\nabla f)</math> becomes
{{NumBlk
 | :
 | <math>\nabla^2 f = \frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial f}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2 f}{\partial\theta^2}.</math>
 | {{EquationRef|5}}
}}
Using Equation {{EquationNote|3}}, the magnetisation divergence in the cylinder walls is
:<math>
\begin{align}
\nabla \cdot \mathbf{M}_\mathrm{cyl} &= \frac{1}{r}\frac{\partial}{\partial r}(r M_0 \cos\theta) + \frac{1}{r} \frac{\partial}{\partial\theta}(M_0 \sin\theta)\\[5pt]
&= \frac{M_0 \cos\theta}{r} + \frac{M_0 \cos\theta}{r}\\[5pt]
&= \frac{2M_0 \cos\theta}{r}.
\end{align}
</math>
Hence Equation {{EquationNote|2}}, which is that we want to solve, becomes by using Equation {{EquationNote|5}}
{{NumBlk
 | :
 | <math>
\frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial \varphi}{\partial r}\right) + \frac{1}{r^2} \frac{\partial^2 \varphi}{\partial\theta^2} =
\begin{cases}
0 & r < r_\mathrm{i}, \\
-\frac{2M_0 \cos\theta}{r} & r_\mathrm{i} < r < r_\mathrm{o}, \\
0 & r > r_\mathrm{0}.
\end{cases}
</math>
 | {{EquationRef|6}}
}}
Look for a [[particular solution]] of this equation in the cylinder walls. With the benefit of hindsight, consider <math>\varphi_\mathrm{p} = r \ln r \cos\theta</math>, because then we have
:<math>
\begin{align}
\frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial\varphi_\mathrm{p}}{\partial r}\right) &= \frac{1}{r}\frac{\partial}{\partial r}\left[r \frac{\partial}{\partial r}(r \ln r \cos\theta)\right]\\[5pt]
&= \frac{\cos\theta}{r}\frac{\partial}{\partial r}\left[r(\ln r + 1)\right]\\[5pt]
&= \frac{\cos\theta}{r}(\ln r + 1 + 1)\\[5pt]
&= \frac{\ln r\cos\theta}{r} + \frac{2 \cos\theta}{r}
\end{align}
</math>
and also
:<math>
\begin{align}
\frac{1}{r^2}\frac{\partial^2 \varphi_\mathrm{p}}{\partial\theta^2} &=  \frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}(r \ln r \cos\theta)\\[5pt]
&= -\frac{\ln r \cos\theta}{r}.
\end{align}
</math>
Hence <math>\nabla^2 \varphi_\mathrm{p} = \frac{2 \cos\theta}{r}</math>, and comparison with Equation {{EquationNote|6}} shows that <math>-M_0 \varphi_\mathrm{p} = - M_0 r \ln r \cos\theta</math> is the appropriate particular solution.

Now consider the homogeneous equation for Equation {{EquationNote|6}}, namely <math>\nabla^2 \varphi_\mathrm{h} = 0</math>. This has the form of [[Laplace's equation]]. Through the method of [[separation of variables]], it can be shown that the general homogeneous solution whose gradient is periodic in <math>\theta</math> (such that all the physical quantities are single-valued) is given by
:<math>\varphi_\mathrm{h} = A_0 + B_0 \theta + C_0 \ln r + \sum_{n=1}^\infty \left(A_n r^n + B_n r^{-n}\right)\cos n\theta + \sum_{n=1}^\infty \left(C_n r^n + D_n r^{-n}\right)\sin n\theta,</math>
where the <math>A_i,\ B_i,\ C_i,\ D_i</math> are arbitrary constants. The desired solution will be the sum of the particular and homogeneous solutions that satisfies the boundary conditions. Again with the benefit of hindsight, let us set most of the constants to zero immediately and assert that the solution is
:<math>
\varphi =
\begin{cases}
\alpha r \cos\theta & r < r_\mathrm{i}, \\
\beta r \cos \theta - M_0 r \ln r \cos\theta & r_\mathrm{i} < r < r_\mathrm{o}, \\
0 & r > r_\mathrm{o},
\end{cases}
</math>
where now <math>\alpha,\ \beta</math> are constants to be determined. If we can choose the constants such that the boundary conditions are satisfied, then by the [[uniqueness theorem for Poisson's equation]], we must have found the solution.

The continuity conditions give
{{NumBlk
 | :
 | <math>\alpha = \beta - M_0 \ln r_\mathrm{i}</math>
 | {{EquationRef|7}}
}}
at the inner boundary and
{{NumBlk
 | :
 | <math>\beta - M_0 \ln r_\mathrm{o} = 0</math>
 | {{EquationRef|8}}
}}
at the outer boundary. The potential gradient has non-vanishing radial component <math>\beta\cos\theta - M_0 \ln r \cos\theta - M_0 \cos\theta</math> in the cylinder walls and <math>\alpha\cos\theta</math> in the bore, and so the conditions on the potential derivative become
:<math>(\beta\cos\theta - M_0 \ln r_\mathrm{i} \cos\theta - M_0\cos\theta) - \alpha \cos\theta = -M_0\cos\theta \implies \beta - M_0 \ln r_\mathrm{i} - \alpha = 0 \implies \alpha = \beta - M_0 \ln r_\mathrm{i}</math>
at the inner boundary and
:<math>0 - (\beta\cos\theta - M_0 \ln r_\mathrm{0} \cos\theta - M_0\cos\theta) = M_0\cos\theta \implies \beta - M_0 \ln r_\mathrm{o} = 0</math>
at the outer boundary. Note that these are identical to Equations {{EquationNote|7}} and {{EquationNote|8}}, so indeed the guess was consistent. Hence we have <math>\beta = M_0 \ln r_\mathrm{o}</math> and <math>\alpha = M_0 \ln\left(\frac{r_\mathrm{o}}{r_\mathrm{i}}\right)</math>, giving the solution
:<math>
\varphi =
\begin{cases}
M_0 \ln\left(\frac{r_\mathrm{o}}{r_\mathrm{i}}\right) r \cos\theta & r < r_\mathrm{i},\\
M_0 \ln\left(\frac{r_\mathrm{o}}{r}\right) r \cos\theta & r_\mathrm{i} < r < r_\mathrm{o},\\
0 & r > r_\mathrm{o}.
\end{cases}
</math>
Consequently, the magnetic field is given by
:<math>
\mathbf{H} = \nabla\varphi =
\begin{cases}
M_0 \ln\left(\frac{r_\mathrm{o}}{r_\mathrm{i}}\right) \cos\theta\,\hat\mathbf{r} - M_0 \ln\left(\frac{r_\mathrm{o}}{r_\mathrm{i}}\right) \sin\theta\,\hat\boldsymbol{\theta} & r < r_\mathrm{i},\\
M_0 \cos\theta \left[\ln\left(\frac{r_\mathrm{o}}{r}\right)-1\right]\,\hat\mathbf{r} - M_0 \ln\left(\frac{r_\mathrm{o}}{r}\right) \sin\theta\,\hat\boldsymbol{\theta} & r_\mathrm{i} < r < r_\mathrm{o},\\
0 & r > r_\mathrm{o},
\end{cases}
</math>
while the magnetic flux density can then be found everywhere using the previous definition <math>\mathbf{B} = \mu_0 (\mathbf{H} + \mathbf{M})</math>. In the bore, where the magnetisation vanishes, this reduces to <math>\mathbf{B}_\mathrm{bore} = \frac{1}{\mu_0}\mathbf{H}_\mathrm{bore}</math>. Hence the magnitude of the flux density there is
:<math>B_\mathrm{bore} = \left|\mathbf{B}_\mathrm{bore}\right| = \sqrt{\cos^2\theta + \sin^2\theta}\frac{M_0}{\mu_0} \ln\left(\frac{r_\mathrm{o}}{r_\mathrm{i}}\right) = \frac{M_0}{\mu_0} \ln\left(\frac{r_\mathrm{o}}{r_\mathrm{i}}\right),</math>
which is independent of position. Similarly, outside the cylinder the magnetisation also vanishes, and since the magnetic field vanishes there, the flux density does too. So indeed the field is uniform inside and zero outside the ideal Halbach cylinder, with a magnitude depending on its physical dimensions.

===Varying the field===
Halbach cylinders give a static field. However, cylinders can be nested, and by rotating one cylinder relative to the other, cancellation of the field and adjustment of the direction can be achieved.<ref>{{Cite web |url=http://www.tipmagazine.com/tip/INPHFA/vol-4/iss-3/p34.pdf |archive-url=https://web.archive.org/web/20060328054612/http://www.tipmagazine.com/tip/INPHFA/vol-4/iss-3/p34.pdf  |title=Tip Magazine: Magnets, Markets, and Magic Cylinders The Industrial Physicist by Michael Coey and Denis Weaire |archive-date=28 March 2006}}</ref> As the outside field of a cylinder is quite low, the relative rotation does not require strong forces. In the ideal case of infinitely long cylinders, no force would be required to rotate a cylinder with respect to the other.

[[File:Halbach_planar_array.svg|thumb|Magnetic levitation using a planar Halbach array and concentric structure windings|220x220px]]