Question 3.6.4: Determine the H-field due to a uniformly magnetized bipolar ...

Use spherical coordinates with the sphere oriented with its magnetization along the z-axis as shown in Fig. 3.42a,

M = M_{s}\hat{z}.

The H-field can be expressed in terms of a scaar potential

H = – ∇ φ_{m},            (3.290)

Because ρ_{m} =- ∇ ⋅ M = 0, we have

∇^{2}φ_{m} = 0            (3.291)

(Section 3.4). Given the orientation of the sphere, φ_{m} will have no Φ dependence. Let φ_{m}^{(1)}(r,θ) and Φ_{m}^{(2)}(r,θ) denote the solutions to Eq. (3.291) inside and outside the magnet, respectively. These functions must be well behaved at r = 0, and r = ∞, respectively,

φ_{m}^{(1)} (0,θ) < ∞,            (3.292)

and

Φ_{m}^{(2)} (∞,θ) < ∞.            (3.293)

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

φ_{m}^{(1)} (r,θ) = \sum\limits_{n=0}^{∞}A_{n}r^{n}P_{n} (cos(θ))        (r < a),

and

φ_{m}^{(2)} (r,θ) = \sum\limits_{n=0}^{∞}C_{n}r^{-(n+1)}P_{n} (cos(θ))        (r ≥ a),

We apply the boundary conditions

H_{θ}^{(1)}(a,θ) =  H_{θ}^{(2)}(a,θ)  (H_{t}^{(1)} = H_{t}^{(2)}),

and

B_{r}^{(1)}(a,θ) =  B_{r}^{(2)}(a,θ)  (B_{n}^{(1)} = B_{n}^{(2)}),

at the surface of the sphere. This gives

\sum\limits_{n=0}^{∞}\frac{(A_{n}a^{n} – C_{n}a^{-(n+1)})}{a} \frac{d}{dθ}P_{n} (cos(θ)) = 0,      (3.294)

and

μ_{0}(-∇ φ_{m}^{(1)} + M_{s}\hat{z}) ⋅ \hat{r} = – μ_{0} ∇φ_{m}^{(2)} ⋅ \hat{r},            (3.295)

respectively. The second condition (3.295) reduces to

0 = μ_{0}M_{s}cos (θ) – μ_{0}C_{0}a^{-2}
– μ_{0}\sum\limits_{n=0}^{∞}((A_{n}na^{n-1} + C_{n}(n+1)a^{-(n+2)})) P_{n} (cos(θ)).        (3.296)

The Legendre polynomials P_{n}(cos (θ)) are linearly independent relative to one another. Therefore, each of the terms containing P_{n}(cos (θ)) or d/(dθ)P_{n}(cos (θ)) in Eqs. (3.294) and (3.296) must separately be zero for each value of n. In particular, when n = 0 we have

\frac{d}{dθ}P_{0}(cos (θ)) = 0,

and

μ_{0}C_{0}a^{-2} = 0.

Therefore, C_{0} = 0. The coefficient A_{0} is undetermined from these conditions, but it represents an additive constant that may be set to zero without affecting the fields. For n = 1 we have

A_{1} – C_{1}a^{-3} = 0

and

M_{s} – A_{1} – 2C_{1}a^{-3} = 0,

which we solve simultaneously and obtain

C_{1} = \frac{1}{3}M_{s}a^{3}       (3.297)

and

A_{1} = \frac{1}{3}M_{s}.        (3.298)

Continuing in this fashion we find that the remaining terms A_{n}= C_{n} = 0 for n ≥ 2. Thus,

φ_{m}^{(1)}= \frac{1}{3}M_{s} r cos (θ),

and

φ_{m}^{(2)}= \frac{1}{3}M_{s} (\frac{a^{3}}{r^{2}})  cos (θ).

The corresponding fields are

H^{(1)} = – \frac{1}{3}M_{s}\hat{z}          (r < a),       (3.299)

and

H^{(1)} = – \frac{1}{3}M_{s}(\frac{a^{3}}{r^{3}}) [2cos(θ)\hat{r}+sin(θ)\hat{θ}].             (3.300)

Notice that inside the sphere, H is in opposite direction to the magnetization. This is the demagnetization field (Section 1.8). Finally, we determine B^{(1)} from Eqs. (3.289) and (3.299):

B^{(1)} = μ_{0}(H^{(1)}+M_{s}\hat{z})
= μ_{0}\frac{2}{3}M_{s}\hat{z}.

The B- and H-fields inside and outside the sphere are shown in Fig. 3.42b and c, respectively. The orientations of B, H, and M inside the sphere are shown in Fig. 3.42d.