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.