thumb|220x220px|The flux diagram of a Halbach array
thumb|220x220px|A Halbach array, showing the orientation of each piece's magnetic field. This array would give a strong field underneath, while the field above would cancel.A Halbach array () is a special arrangement of permanent magnets that augments the magnetic field on one side of the array while cancelling the field to near zero on the other side.
The effect was also discovered by John C. Mallinson in 1973, and these "one-sided flux" structures were initially described by him as a "curiosity", although at the time he recognized from this discovery the potential for significant improvements in magnetic tape technology.
Physicist Klaus Halbach, while at the Lawrence Berkeley National Laboratory during the 1980s, independently invented the Halbach array to focus particle accelerator beams.
Linear arrays
Magnetization
thumb|Cancellation of magnetic components resulting in a one-sided flux|220x220px
The magnetic flux distribution of a linear Halbach array may seem somewhat counter-intuitive to those familiar with simple magnets or solenoids. The reason for this flux distribution can be visualised using Mallinson's original diagram (note that it uses the negative y component, unlike the diagram in Mallinson's article).
: <math>F(x, y) = F_0 e^{ikx} e^{-ky},</math>
where
: <math>F(x, y)</math> is the field in the form <math>F_x + i F_y</math>,
: <math>F_0</math> is the magnitude of the field at the surface of the array,
: <math>k</math> is the wavenumber (i.e., the spatial frequency) <math>2\pi / \lambda.</math>
Applications
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, missing references.)
- 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.
Thus they have a number of applications, ranging from flat refrigerator magnets through industrial applications such as the brushless DC motor, voice coils, magnetic drug targeting and Inductrack rocket-launch system 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. The reduced stray fields also enhance safety and minimize interference with surrounding electronic devices.
Uniform fields
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, 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. 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.
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 <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 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 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' law <math>\nabla\cdot\mathbf{B} = 0</math>, this is equivalently
Since the problem is static there are no free currents and all time derivatives vanish, so 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 governing <math>\mathbf{H}</math> and <math>\mathbf{M}</math>, we find that we need to solve
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 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
Similarly, the gradient of a scalar field <math>f</math> is given by
Combining these two relations, the Laplacian <math>\nabla^2 f = \nabla\cdot(\nabla f)</math> becomes
Using Equation , 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 , which is that we want to solve, becomes by using Equation
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 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 , 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
at the inner boundary and
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 and , 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. 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.
thumb|Magnetic levitation using a planar Halbach array and concentric structure windings|220x220px
Sphere
If the two-dimensional magnetic distribution patterns of the Halbach cylinder are extended to three dimensions, the result is the Halbach sphere. These designs have an extremely uniform field within the interior of the design, as they are not affected by the "end effects" prevalent in the finite-length cylinder design. The magnitude of the uniform field for a sphere also increases to 4/3 the amount for the ideal cylindrical design with the same inner and outer radii. However, for a spherical struction, access to the region of uniform field is usually restricted to a narrow hole at the top and bottom of the design.
The equation for the field in a Halbach sphere is
: <math>B = \frac{4}{3} B_0 \ln\left(\frac{R_\text{o{R_\text{i\right).</math>
Higher fields are possible by optimising the spherical design to take account of the fact that it is composed of point dipoles (and not line dipoles). This results in the stretching of the sphere to an elliptical shape and having a non-uniform distribution of magnetization over the component parts. Using this method, as well as soft pole pieces within the design, 4.5 T in a working volume of 20 mm<sup>3</sup> was achieved, although over a smaller working volume of 0.05 mm<sup>3</sup>. As hard materials are temperature-dependent, refrigeration of the entire magnet array can increase the field within the working area further. This group also reported development of a 5.16 T Halbach dipole cylinder in 2003.
External links
- Passive levitation of a shaft
- Electric model aircraft motor
- magnet levitated vehicle switching system