Question 3.6.3: Determine the magnetic field of a bipolar cylindrical magnet...

We solve a 2D boundary-value problem. Choose a coordinate system with the magnetization along the x-axis, which gives

M = M_{s}\hat{x}
= M_{s} [cos (Φ)\hat{r} – sin (Φ)\hat{Φ}].               (3.264)

Because there are no current sources we can represent H in terms of a magnetic scalar potential

H = – ∇ φ_{m},          (3.265)

where

∇^{2} φ_{m} = 0            (3.266)

(Section 3.4). Let φ_{m}^{(1)} (r,Φ) and φ_{m}^{(2)} (r,Φ)denote the solutions to Eq. (3.266) inside and outside the magnet, respectively. From symmetry, these solutions must be even functions of Φ,

φ_{m}^{(i)} (r,Φ)= φ_{m}^{(i)}(r, -Φ)      (i = 1, 2).        (3.267)

They must also be well behaved at r = 0 and r = ∞, respectively,

φ_{m}^{(1)}  (0,Φ) < ∞,            (3.268)
φ_{m}^{(2)}  (∞,Φ) < ∞.            (3.269)

The general forms for φ_{m}^{(1)} (r,Φ) and φ_{m}^{(2)} (r,Φ) compatible with these conditions are

φ_{m}^{(1)} (r,Φ)= \sum\limits_{n=1}^{∞} a_{n} r^{n} cos(nΦ)          (r ≤ a),           (3.270)

and

φ_{m}^{(2)} (r,Φ)= \sum\limits_{n=1}^{∞} b_{n} r^{-n} cos(nΦ)          (a ≤ r),           (3.271)

which follow from Eq. (3.263). The coefficients a_{n} and b_{n} are determined from the boundary conditions imposed at the surface of the magnet (r = a). Specifically, the potential φ_{m} (r, Φ) and the normal component of B must be continuous at this surface, that is,

Eq. (3.263): φ(r,Φ) = a_{0}+b_{0} ln(r)+\sum\limits_{n=1}^{∞}(a_{n}r^{n}+b_{n}r^{-n})(c_{n}sin(nΦ)+d_{n}cos(nΦ))

φ_{m}^{(1)} (a,Φ)= φ_{m}^{(2)}(a, Φ),          (3.272)

and

  [- ∇φ_{m}^{(1)}(a,Φ) + M] ⋅ \hat{r}  = – ∇φ_{m}^{(2)}(a,Φ) ⋅ \hat{r},                (3.273)

respectively. These conditions yield the following expression for φ_{m}^{(2)} (r,Φ):

φ_{m}^{(2)}(r,Φ) = \frac{M_{s}}{2} \frac{a^{2}}{r} cos(Φ)        (a < r).

Outside the magnet B = μ_{0} H, and therefore

B_{r}(r,Φ) =  μ_{0}\frac{M_{s}}{2} \frac{a^{2}}{r} cos(Φ)        (a < r),            (3.274)

and

B_{Φ}(r,Φ) =  μ_{0}\frac{M_{s}}{2} \frac{a^{2}}{r} sin(Φ)        (a < r),            (3.275)

Also, because B = ∇ × A we have

A(r,Φ) = \frac{μ_{0}M_{s}}{2} \frac{a^{2}}{r}sin(Φ) \hat{z} (a < r).            (3.276)

Calculations: We demonstrate the use of Eqs. (3.274) and (3.275) with some sample calculations. Consider a cylinder with a radius a = 2.5 mm, and a magnetization M_{s} = 4.3 ×10^{5} (A/m) (B_{r} = 5400 G). The fields B_{r}(r, Φ) and B_{Φ}(r, Φ) are computed for r = 3.8 mm and 0 ≤ Φ ≤ π (Fig. 3.41). Notice that B_{r}(r, Φ) peaks at Φ = 0. This shows that the radial field component peaks at the center of the north pole as expected.