Halbach array

Cancellation of magnetic components resulting in a one-sided flux

Although this magnetic flux distribution seems somewhat counter-intuitive to those familiar with simple bar magnets or solenoids, the reason for this flux distribution can be intuitively visualised using Mallinson's original diagram (note that it uses the negative y component, unlike the diagram in Mallinson's article). The diagram shows the field from a strip of ferromagnetic material with alternating magnetization in the y direction (top left) and in the x direction (top right). Note that the field above the plane is in the same direction for both structures, but the field below the plane is in opposite directions. The effect of superimposing both of these structures is shown in the figure.

The crucial point is that the flux will cancel below the plane and reinforce itself above the plane. In fact, any magnetization pattern where the components of magnetization are ${\displaystyle \pi /2}$ out of phase with each other will result in a one-sided flux. The mathematical transform that shifts the phase of all components of some function by ${\displaystyle \pi /2}$ is called a Hilbert transform; the components of the magnetization vector can therefore be any Hilbert-transform pair (the simplest of which is simply ${\displaystyle \sin(x)\cos(y)}$, as shown in the diagram above).

The magnetic field around an infinite Halbach array of cube magnets. The field does not cancel perfectly due to the discrete magnets used.

The field on the non-cancelling side of the ideal, continuously varying, infinite array is of the form[6]

${\displaystyle F(x,y)=F_{0}e^{ikx}e^{-ky},}$

where

${\displaystyle F(x,y)}$ is the field in the form ${\displaystyle F_{x}+iF_{y}}$,

${\displaystyle F_{0}}$ is the magnitude of the field at the surface of the array,

${\displaystyle k}$ is the wavenumber (i.e., the spatial frequency) ${\displaystyle 2\pi /\lambda .}$

The advantages of one-sided flux distributions are twofold:

• The field is twice as large on the side on which the flux is confined (in the idealized case).
• There is no stray field produced (in the ideal case) on the opposite side. This helps with field confinement, usually a problem in the design of magnetic structures.

Although one-sided flux distributions may seem somewhat abstract, they have a surprising number of applications ranging from the refrigerator magnet through industrial applications such as the brushless DC motor, voice coils,[7] magnetic drug targeting[8] to high-tech applications such as wiggler magnets used in particle accelerators and free-electron lasers.

This device is also a key component of the Inductrack Maglev train[9] and Inductrack rocket-launch system,[10] wherein the Halbach array repels loops of wire that form the track after the train has been accelerated to a speed able to lift.

Flux distribution for a refrigerator magnet

The simplest example of a one-sided flux magnet is a refrigerator magnet. These are usually composed of powdered ferrite in a binder such as plastic or rubber. The extruded magnet is exposed to a rotating field giving the ferrite particles in the magnetic compound a magnetization resulting in a one-sided flux distribution. This distribution increases the holding force of the magnet when placed on a permeable surface, compared to the holding force from, say, a uniform magnetization of the magnetic compound.

Schematic diagram of a free-electron laser

Scaling up this design and adding a top sheet gives a wiggler magnet, used in synchrotrons and free-electron lasers. Wiggler magnets wiggle, or oscillate, an electron beam perpendicular to the magnetic field. As the electrons are undergoing acceleration, they radiate electromagnetic energy in their flight direction, and as they interact with the light already emitted, photons along its line are emitted in phase, resulting in a "laser-like" monochromatic and coherent beam.

The design shown above is usually known as a Halbach wiggler. The magnetization vectors in the magnetized sheets rotate in the opposite senses to each other; above, the top sheet's magnetization vector rotates clockwise, and the bottom sheet's magnetization vector rotates counter-clockwise. This design is chosen so that the x components of the magnetic fields from the sheets cancel, and the y components reinforce, so that the field is given by

${\displaystyle H_{y}\approx \cos(kx),}$

where k is the wavenumber of the magnetic sheet given by the spacing between magnetic blocks with the same magnetization vector.

Schematic of a Halbach array consisting of a series magnetized rods

Equal-gearing arrangement for a variable Halbach array

A series of magnetic rods, magnetized perpendicular to their axes, can be arranged into a Halbach array. If each rod is then rotated alternately through 90°, the resultant field moves from one side of plane of the rods to the other, as shown schematically in the figure.

This arrangement allows the field to effectively be turned on and off above or below the plane of the rods, depending on the rotation of the rods. Such a device makes an efficient mechanical magnetic latch requiring no power. A detailed study of this arrangement has shown that each rod is subject to a strong torque from its neighboring rods, and therefore requires mechanical stabilization.[11] However, a simple and efficient solution, providing both stabilization and the ability to rotate each rod alternately, is simply to provide an equal-gearing arrangement on each rod, as shown in the figure.

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 = 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.[14] 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.

Uniform field inside Halbach cylinder

For the special case of k = 2, the field inside the bore is uniform and is given by

${\displaystyle H=M_{r}\ln \left({\frac {R_{\text{o}}}{R_{\text{i}}}}\right){\widehat {y}},}$

where the inner and outer cylinder radii are R_i and R_o 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, 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.[15][16]

Three designs (A) (B) (C) producing uniform magnetic fields within their central air gap

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,[17][18] 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 = 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.[19] 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.

The method used to find the field created by the cylinder is mathematically very similar to that used to investigate a uniformly magnetised sphere.[20]

Because of the symmetry of the arrangement along the cylinder's axis, the problem can be treated as two-dimensional. Work in plane-polar coordinates ${\displaystyle (r,\theta )}$ with associated unit vectors ${\displaystyle {\hat {\mathbf {r} }}}$ and ${\displaystyle {\hat {\boldsymbol {\theta }}}}$, and let the cylinder have radial extent ${\displaystyle r_{\mathrm {i} }<r<r_{\mathrm {o} }}$. Then the magnetisation in the cylinder walls, which has magnitude ${\displaystyle M_{0}}$, rotates smoothly as

${\displaystyle \mathbf {M} _{\mathrm {cyl} }=M_{0}(\cos \theta \,{\hat {\mathbf {r} }}+\sin \theta \,{\hat {\boldsymbol {\theta }}}),}$

while the magnetisation vanishes outside the walls, that is for the bore ${\displaystyle r<r_{\mathrm {i} }}$ and surroundings ${\displaystyle r>r_{\mathrm {o} }}$.

By definition, the auxiliary magnetic field strength ${\displaystyle \mathbf {H} }$ is related to the magnetisation and magnetic flux density ${\displaystyle \mathbf {B} }$ by ${\displaystyle \mathbf {B} =\mu _{0}(\mathbf {H} +\mathbf {M} )}$. Using Gauss' law ${\displaystyle \nabla \cdot \mathbf {B} =0}$, this is equivalently

${\displaystyle \nabla \cdot \mathbf {H} =-\nabla \cdot \mathbf {M} .}$

(1)

Since the problem is static there are no free currents and all time derivatives vanish, so Ampère's law additionally requires ${\displaystyle \nabla \times \mathbf {H} =0\implies \mathbf {H} =\nabla \varphi }$, where ${\displaystyle \varphi }$ is the magnetic scalar potential (up to a sign under some definitions). Substituting this back into the previous Equation 1 governing ${\displaystyle \mathbf {H} }$ and ${\displaystyle \mathbf {M} }$, we find that we need to solve

${\displaystyle \nabla ^{2}\varphi =-\nabla \cdot \mathbf {M} ,}$

(2)

which has the form of Poisson's equation.

Consider now the boundary conditions at the cylinder-air interfaces ${\displaystyle r=r_{\mathrm {i} }}$ and ${\displaystyle r=r_{\mathrm {o} }}$. Integrating ${\displaystyle \nabla \times \mathbf {H} =0}$ over a small loop straddling the boundary and applying Stokes' theorem requires that the parallel component of ${\displaystyle \mathbf {H} }$ is continuous. This in turn requires that ${\displaystyle \varphi }$ is continuous across the boundary. (More properly this implies that ${\displaystyle \varphi }$ 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 1 across a small volume straddling the boundary and apply the divergence theorem to find

${\displaystyle \left[{\frac {\partial \varphi }{\partial r}}\right]=\pm \mathbf {M} \cdot {\hat {\mathbf {r} }},}$

where the notation ${\displaystyle [f]}$ denotes a jump in the quantity ${\displaystyle f}$ across the boundary, and in our case the sign is negative at ${\displaystyle r=r_{\mathrm {i} }}$ and positive at ${\displaystyle r=r_{\mathrm {o} }}$. 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 ${\displaystyle \mathbf {F} =F_{r}{\hat {\mathbf {r} }}+F_{\theta }{\hat {\boldsymbol {\theta }}}}$ is given by

${\displaystyle \nabla \cdot \mathbf {F} ={\frac {1}{r}}{\frac {\partial }{\partial r}}(rF_{r})+{\frac {1}{r}}{\frac {\partial F_{\theta }}{\partial \theta }}.}$

(3)

Similarly, the gradient of a scalar field ${\displaystyle f}$ is given by

${\displaystyle \nabla f={\frac {\partial f}{\partial r}}{\hat {\mathbf {r} }}+{\frac {1}{r}}{\frac {\partial f}{\partial \theta }}{\hat {\boldsymbol {\theta }}}.}$

(4)

Combining these two relations, the Laplacian ${\displaystyle \nabla ^{2}f=\nabla \cdot (\nabla f)}$ becomes

${\displaystyle \nabla ^{2}f={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \theta ^{2}}}.}$

(5)

Using Equation 3, the magnetisation divergence in the cylinder walls is

${\displaystyle {\begin{aligned}\nabla \cdot \mathbf {M} _{\mathrm {cyl} }&={\frac {1}{r}}{\frac {\partial }{\partial r}}(rM_{0}\cos \theta )+{\frac {1}{r}}{\frac {\partial }{\partial \theta }}(M_{0}\sin \theta )\\[5pt]&={\frac {M_{0}\cos \theta }{r}}+{\frac {M_{0}\cos \theta }{r}}\\[5pt]&={\frac {2M_{0}\cos \theta }{r}}.\end{aligned}}}$

Hence Equation 2, which is that we want to solve, becomes by using Equation 5

${\displaystyle {\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial \varphi }{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}\varphi }{\partial \theta ^{2}}}={\begin{cases}0&r<r_{\mathrm {i} },\\-{\frac {2M_{0}\cos \theta }{r}}&r_{\mathrm {i} }<r<r_{\mathrm {o} },\\0&r>r_{\mathrm {0} }.\end{cases}}}$

(6)

Look for a particular solution of this equation in the cylinder walls. With the benefit of hindsight, consider ${\displaystyle \varphi _{\mathrm {p} }=r\ln r\cos \theta }$, because then we have

${\displaystyle {\begin{aligned}{\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial \varphi _{\mathrm {p} }}{\partial r}}\right)&={\frac {1}{r}}{\frac {\partial }{\partial r}}\left[r{\frac {\partial }{\partial r}}(r\ln r\cos \theta )\right]\\[5pt]&={\frac {\cos \theta }{r}}{\frac {\partial }{\partial r}}\left[r(\ln r+1)\right]\\[5pt]&={\frac {\cos \theta }{r}}(\ln r+1+1)\\[5pt]&={\frac {\ln r\cos \theta }{r}}+{\frac {2\cos \theta }{r}}\end{aligned}}}$

and also

${\displaystyle {\begin{aligned}{\frac {1}{r^{2}}}{\frac {\partial ^{2}\varphi _{\mathrm {p} }}{\partial \theta ^{2}}}&={\frac {1}{r^{2}}}{\frac {\partial ^{2}}{\partial \theta ^{2}}}(r\ln r\cos \theta )\\[5pt]&=-{\frac {\ln r\cos \theta }{r}}.\end{aligned}}}$

Hence ${\displaystyle \nabla ^{2}\varphi _{\mathrm {p} }={\frac {2\cos \theta }{r}}}$, and comparison with Equation 6 shows that ${\displaystyle -M_{0}\varphi _{\mathrm {p} }=-M_{0}r\ln r\cos \theta }$ is the appropriate particular solution.

Now consider the homogeneous equation for Equation 6, namely ${\displaystyle \nabla ^{2}\varphi _{\mathrm {h} }=0}$. 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 ${\displaystyle \theta }$ (such that all the physical quantities are single-valued) is given by

${\displaystyle \varphi _{\mathrm {h} }=A_{0}+B_{0}\theta +C_{0}\ln r+\sum _{n=1}^{\infty }\left(A_{n}r^{n}+B_{n}r^{-n}\right)\cos n\theta +\sum _{n=1}^{\infty }\left(C_{n}r^{n}+D_{n}r^{-n}\right)\sin n\theta ,}$

where the ${\displaystyle A_{i},\ B_{i},\ C_{i},\ D_{i}}$ 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

${\displaystyle \varphi ={\begin{cases}\alpha r\cos \theta &r<r_{\mathrm {i} },\\\beta r\cos \theta -M_{0}r\ln r\cos \theta &r_{\mathrm {i} }<r<r_{\mathrm {o} },\\0&r>r_{\mathrm {o} },\end{cases}}}$

where now ${\displaystyle \alpha ,\ \beta }$ 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

${\displaystyle \alpha =\beta -M_{0}\ln r_{\mathrm {i} }}$

(7)

at the inner boundary and

${\displaystyle \beta -M_{0}\ln r_{\mathrm {o} }=0}$

(8)

at the outer boundary. The potential gradient has non-vanishing radial component ${\displaystyle \beta \cos \theta -M_{0}\ln r\cos \theta -M_{0}\cos \theta }$ in the cylinder walls and ${\displaystyle \alpha \cos \theta }$ in the bore, and so the conditions on the potential derivative become

${\displaystyle (\beta \cos \theta -M_{0}\ln r_{\mathrm {i} }\cos \theta -M_{0}\cos \theta )-\alpha \cos \theta =-M_{0}\cos \theta \implies \beta -M_{0}\ln r_{\mathrm {i} }-\alpha =0\implies \alpha =\beta -M_{0}\ln r_{\mathrm {i} }}$

at the inner boundary and

${\displaystyle 0-(\beta \cos \theta -M_{0}\ln r_{\mathrm {0} }\cos \theta -M_{0}\cos \theta )=M_{0}\cos \theta \implies \beta -M_{0}\ln r_{\mathrm {o} }=0}$

at the outer boundary. Note that these are identical to Equations 7 and 8, so indeed the guess was consistent. Hence we have ${\displaystyle \beta =M_{0}\ln r_{\mathrm {o} }}$ and ${\displaystyle \alpha =M_{0}\ln \left({\frac {r_{\mathrm {o} }}{r_{\mathrm {i} }}}\right)}$, giving the solution

${\displaystyle \varphi ={\begin{cases}M_{0}\ln \left({\frac {r_{\mathrm {o} }}{r_{\mathrm {i} }}}\right)r\cos \theta &r<r_{\mathrm {i} },\\M_{0}\ln \left({\frac {r_{\mathrm {o} }}{r}}\right)r\cos \theta &r_{\mathrm {i} }<r<r_{\mathrm {o} },\\0&r>r_{\mathrm {o} }.\end{cases}}}$

Consequently, the magnetic field is given by

${\displaystyle \mathbf {H} =\nabla \varphi ={\begin{cases}M_{0}\ln \left({\frac {r_{\mathrm {o} }}{r_{\mathrm {i} }}}\right)\cos \theta \,{\hat {\mathbf {r} }}-M_{0}\ln \left({\frac {r_{\mathrm {o} }}{r_{\mathrm {i} }}}\right)\sin \theta \,{\hat {\boldsymbol {\theta }}}&r<r_{\mathrm {i} },\\M_{0}\cos \theta \left[\ln \left({\frac {r_{\mathrm {o} }}{r}}\right)-1\right]\,{\hat {\mathbf {r} }}-M_{0}\ln \left({\frac {r_{\mathrm {o} }}{r}}\right)\sin \theta \,{\hat {\boldsymbol {\theta }}}&r_{\mathrm {i} }<r<r_{\mathrm {o} },\\0&r>r_{\mathrm {o} },\end{cases}}}$

while the magnetic flux density can then be found everywhere using the previous definition ${\displaystyle \mathbf {B} =\mu _{0}(\mathbf {H} +\mathbf {M} )}$. In the bore, where the magnetisation vanishes, this reduces to ${\displaystyle \mathbf {B} _{\mathrm {bore} }={\frac {1}{\mu _{0}}}\mathbf {H} _{\mathrm {bore} }}$. Hence the magnitude of the flux density there is

${\displaystyle B_{\mathrm {bore} }=\left|\mathbf {B} _{\mathrm {bore} }\right|={\sqrt {\cos ^{2}\theta +\sin ^{2}\theta }}{\frac {M_{0}}{\mu _{0}}}\ln \left({\frac {r_{\mathrm {o} }}{r_{\mathrm {i} }}}\right)={\frac {M_{0}}{\mu _{0}}}\ln \left({\frac {r_{\mathrm {o} }}{r_{\mathrm {i} }}}\right),}$

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.

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.[21] 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.

Magnetic levitation using a planar Halbach array and concentric structure windings