Question 3.6.2: Figure 3.37a shows a partial cross-sectional view of a magne...

This problem constitutes a 2D boundary-value problem. As there is no applied current, we use the magnetic scalar potential formulation

H = –  ∇φ_{m},        (3.214)

and

∇^{2}φ_{m} = ∇ ⋅ M               (3.215)

(Section 3.4). The 2D geometry can be reduced by exploiting the symmetry that results from the repeating magnetic structure. Specifically, the field is symmetric about the vertical center lines of each pole. We consider a reduced rectangular region extending from the center of one pole to the center of the neighboring pole as shown in Fig. 3.37b. The reduced geometry consists of three regions with

M = \left\{\begin{matrix}M_{s}\hat{y} \qquad (region  1)\\ – M_{s}\hat{y} \qquad (region  2)\\  0 \qquad  (region  3).\end{matrix} \right.          (3.216)

We find that ρ_{m} = – ∇ ⋅ M = 0 in all three regions, and Eq. (3.215) reduces to

\frac{∂^{2}φ_{m}}{∂x^{2}} + \frac{∂^{2}φ_{m}}{∂y^{2}} = 0.

The general solution to this equation is given by Eq. (3.199). We rule out the k_{x} = k_{y} = 0 solution because of the periodicity of the geometry. Thus, we obtain

Eq. (3.199): φ(x,y)= \underbrace{(a_{1}x+a_{2}) (b_{1} y+ b_{2})}_{k_{x}= k_{y}=0} +\underbrace{\sum\limits_{n=1}^{∞}(c_{1n}sin(k_{n}x)+c_{2n}cos(k_{n}x))(d_{1n}e^{k_{n}y}+ d_{2n}e^{-k_{n}y}) }_{k_{n}= k_{x}>0}

φ_{m}^{(1)}(x,y) = (S_{1 1}e^{k_{1}y}+S_{1 2}e^{-k_{1}y}) sin(k_{1}x)
+ + (C_{1 1}e^{k_{1}y}+C_{1 2}e^{-k_{1}y}) cos(k_{1}x)    (region 1),        (3.217)
φ_{m}^{(2)}(x,y) = (S_{2 1}e^{k_{2}y}+S_{2 2}e^{-k_{2}y}) sin(k_{2}x)
+ (C_{2  1}e^{k_{2}y}+C_{2  2}e^{-k_{2}y}) cos(k_{2}x)     (region 2),        (3.218)

and

φ_{m}^{(3)}(x,y) = (S_{3  1}e^{k_{3}y}+S_{3  2}e^{-k_{3}y}) sin(k_{3}x)
+ (C_{3  1}e^{k_{3}y}+C_{3  2}e^{-k_{3}y}) cos(k_{3}x)     (region 3),        (3.219)

respectively. The coefficients in Eqs. (3.217) to (3.219) are determined by imposing the boundary conditions. Notice that the three regions in Fig. 3.37b are collectively bounded by four lines that form a rectangular area. Two sides of the rectangle, that is, the y-axis and the line x = l,  represent lines of symmetry along which the normal component H_{n} if zero. The boundaries that form the top and bottom of the rectangle are bordered by a ferromagnetic material with infinite permeability (μ = ∞). Consequently, the tangential component H_{t} vanishes along these boundaries. We evaluate the boundary conditions region by region.

Region 1: One of the boundaries that border region 1 is the y-axis along which the normal component of the field vanishes:

H_{n}^{(1)}(0,y)=- \frac{∂φ_{m}^{(1)} (x,y)}{∂x} |_{x=0} = 0        (g < y < h).

This condition yields

S_{1  1}e^{k_{1}y} =  – S_{1  2}e^{-k_{1}y}.                      (3.220)

Region 1 is also bounded by a flux plate (μ = ∞) at y = h. The tangential
component of the field is zero along this boundary, that is,

H_{t}^{(1)}(x,h) = – \frac{∂φ_{m}^{(1)} (x,y)}{∂x}|_{y=h} = 0        (0 < x< l/2).

This condition is satisfied when

C_{1  1}e^{k_{1}h} = –  C_{1  2}e^{-k_{1}h}.          (3.221)

Applying Eqs. (3.220) and (3.221) to Eq. (3.217) gives

∂φ_{m}^{(1)} (x,y) = C_{1}sinh(k_{1}(y-h))cos(k_{1}x),          (3.222)

where C_{1} is a constant

Region 2: The conditions for region 2 are similar to those of region 1. Specifically, the normal component of the field is zero along the symmetry line defined by x = l, and the tangential field is zero along the upper boundary defined by y = h:

H_{n}^{(2)}(l,y) =  -\frac{∂φ_{m}^{(2)} (x,y)}{∂x} |_{x=l}= 0        (g < y < h),              (3.223)

and

H_{t}^{(2)}(x,y) =  -\frac{∂φ_{m}^{(2)} (x,y)}{∂x} |_{y=h}= 0        (l/2 < x < l).              (3.224)

Applying the first condition to Eq. (3.218) gives

S_{1  2}e^{k_{2}y} =  – S_{2  2}e^{-k_{2}y}.                      (3.225)

and

k_{2} = \frac{nπ}{l}      (n = 1, 2, 3, . . .).            (3.226)

Then, the second condition is satisfied when

C_{2  1}e^{k_{2}h} = –  C_{2  2}e^{-k_{2}h}.          (3.227)

Relations (3.225) to (3.227) reduce Eq. (3.218) to

∂φ_{m}^{(2)} (x,y) = C_{2}sinh(\frac{nπ}{l}(y-h))cos(\frac{nπ}{l}x),          (3.228)

where C_{2} is a constant.

Magnet-magnet interface: At the interface between the two magnets the normal component of B and the tangential component of H must be continuous, that is,

B_{n}^{(1)} (l/2,y)= B_{n}^{(2)}(l/2,y)           (g < y < h),        (3.229)

and

H_{t}^{(1)} (l/2,y)= H_{t}^{(2)}(l/2,y)           (g < y < h),        (3.230)

where

B_{n}^{(i)} (l/2,y)=- μ_{0}\frac{∂φ_{m}^{(i)} (x,y)}{∂x} |_{x=l/2}        (i = 1, 2),

and

H_{t}^{(i)} (l/2,y)=  – \frac{∂φ_{m}^{(i)} (x,y)}{∂y} |_{x=l/2}        (i = 1, 2).

From these relations we obtain

C_{1}=C_{2},                 (3.231)

and

k_{1}=k_{2}.                 (3.232)

Therefore, the solutions to both regions are identical. We form a general solution φ_{mag} for these regions using superposition, that is,

φ_{mag}(x,y)= \sum\limits_{n=1}^{∞}C_{mag,n}sinh(\frac{nπ}{l}(y-h)) cos(\frac{nπ}{l}x).             (3.233)

Region 3: For region 3 there are two boundaries about which the field is symmetric and on which the normal component of H is zero. These boundaries are vertical lines defined by x = 0 and x = l. Therefore,

H_{n}^{(3)} (0,y) = 0            (0 < y < g),        (3.234)

and

H_{n}^{(3)} (l,y) = 0            (0 < y < g),          (3.235)

When these conditions are imposed, we find that

S_{3  1}e^{k_{3}y} =  – S_{3  2}e^{-k_{3}}.                      (3.236)

and

k_{3} = \frac{nπ}{l}      (n = 1, 2, 3, . . .).          (3.237)

respectively. Also, as the bottom of the region borders a flux plate, the tangential component of the field vanishes there:

H_{t}^{(3)} (x,0) = 0            (0 < x < l),          (3.238)

This implies that

C_{3  1} = – C_{3  2}.                (3.239)

We call this region the gap, and form its general solution via superposition,

φ_{gap}(x,y)= \sum\limits_{n=1}^{∞}C_{gap,n}sinh(\frac{nπ}{l}(y-h)) cos(\frac{nπ}{l}x).             (3.240)

The coefficients C_{gap,n} are determined at the magnet-gap interface.

Magnet-gap interface: At the interface between the magnet and the gap the tangential component of H must be continuous, that is,

H_{mag,t}(x,g) = H_{gap,t}(x,g)            (0 < x < l).         (3.241)

This condition, along with the fact that the functions sin(nπx/l) are orthogonal on (0, l), implies that

C_{mag,n}  =C_{gap,n} \frac{sinh(nπg/l)}{sinh(nπ(g-h)/l)}                 (3.242)

In addition, the normal component of B must also be continuous, that is,

B_{mag,n}(x,g) = B_{gap,n}(x,g)      (0 < x < l).     (3.243)

Because

B_{mag,n} =\left\{\begin{matrix} μ_{0}(H_{mag,n}+M_{s}) \qquad (0  <  x  <  l/2)\\ μ_{0}(H_{mag,n}-M_{s} \qquad (l/2  <  x  <  l)\end{matrix} \right.                    (3.244)

Eq. (3.243) reduces to

\sum\limits_{n=1}^{∞} \frac{nπ}{l}C_{gap,n} K(n,h,g,l) cos(\frac{nπx}{l})= \left\{\begin{matrix} M_{s} \qquad (0  <  x  <  l/2) \\ -M_{s} \qquad (l/2  <  x  <  l) \end{matrix} \right.             (3.245)

where

K(n,h,g,l) = cosh(\frac{nπg}{l}) – sinh (\frac{nπg}{l}) coth (\frac{nπ}{l}(g-h)) .      (3.246)

Again, exploiting the orthogonality of cos(nπx/l) on (0, l) we find that

C_{gap,n}= \frac{2M_{s}}{nπK(n,h,g,l)} \{ \int_{0}^{l/2} cos(\frac{nπx}{l}) dx  – \int_{l/2}^{l} cos(\frac{nπx}{l}) dx \}  .          (3.247)

After integrating Eq. (3.247) we obtain

C_{gap,n}=- \frac{4lM_{s}(-1)^{(n-1)/2}}{(nπ)^{2}K(n,h,g,l)}       (n = 1, 3, 5, . . .),        (3.248)

which fully defines the solution for the gap region. Specifically, substitute Eq. (3.248) into Eq. (3.240) and obtain

φ_{gap}(x,y) = -4lM_{s} \sum\limits_{n=1,3,5,…..}^{∞} \frac{(-1)^{(n-1)/2}}{(nπ)^{2}K(n,h,g,l)} sinh (\frac{nπy}{l}) cos (\frac{nπx}{l}).            (3.249)

From Eq. (3.249) we find that the vertical component of the field in this region is

B_{gap,y}(x,y) =  -4μ_{0}M_{s}  \sum\limits_{n=1,3,5,…..}^{∞} \frac{(-1)^{(n-1)/2}}{(nπ)K(n,h,g,l)} cosh (\frac{nπy}{l}) cos (\frac{nπx}{l}),            (3.250)

where g = the gap, h = g + t_{m}, t_{m} = height of the magnet, and K(n, h, g, l) is as specified in Eq. (3.246).

Calculations: We demonstrate the use of Eq. (3.250) with some sample calculations. Consider a structure with l = 4.8 cm, t_{m} = 1.28 cm, and M_{s} = 3000 (A/m). We compute the gap field B_{gap,y}(x, y) at y = 0.375 mm and 0 ≤ x ≤ l, for three different gaps g = 6, 8, and 10 mm (Fig. 3.38). A similar calculation is performed with y = 2.5, 5 and 7.5 mm, for a fixed gap g = 10 mm (Fig. 3.39).