Uniformly Magnetized Sphere

Consider a sphere of radius $a$ , with a uniform permanent magnetization ${\bf M}= M_0 \,{\bf e}_z$ , surrounded by a vacuum region. The simplest way of solving this problem is in terms of the scalar magnetic potential introduced in Equation (701). It follows from Equations (703) and (704) that $\phi_m$ satisfies Laplace's equation,

$$\nabla^{\,2}\phi_m = 0,$$ (717)

because there is zero volume magnetic charge density in a vacuum, or a uniformly magnetized magnetic medium. However, according to Equation (711), there is a magnetic surface charge density,

$$\sigma_m = {\bf e}_r\cdot{\bf M} = M_0 \cos\theta,$$ (718)

on the surface of the sphere. Here, $r$ and $\theta$ are spherical coordinates. One of the matching conditions at the surface of the sphere is that the tangential component of ${\bf H}$ must be continuous. It follows from Equation (701) that the scalar magnetic potential must be continuous at $r=a$ , so that

$$\phi_m(r=a_+,\theta) = \phi_m(r=a_-,\theta).$$ (719)

Integrating Equation (703) over a Gaussian pill-box straddling the surface of the sphere yields

$$\left[\frac{\partial\phi_m}{\partial r}\right]_{r=a-}^{r=a+} = -\sigma_m = -M_0 \cos\theta.$$ (720)

In other words, the magnetic charge sheet on the surface of the sphere gives rise to a discontinuity in the radial gradient of the magnetic scalar potential at $r=a$ .

The most general axisymmetric solution to Equation (718) that satisfies physical boundary conditions at $r=a$ and $r=\infty$ is

$$\phi_m(r,\theta) =\sum_{l=0,\infty}A_l \,r^{\,l} P_l(\cos\theta)$$ (721)

for $r<a$ , and

$$\phi_m(r,\theta) =\sum_{l=0,\infty} B_l\, r^{-(l+1)} P_l(\cos\theta)$$ (722)

for $r\geq a$ . The boundary condition (720) yields

$$B_l = A_l \,a^{\,2\,l+1}$$ (723)

for all $l$ . The boundary condition (721) gives

$$- \frac{(l+1)\,B_l}{a^{\,l+2}} - l \,A_l\, a^{\,l-1} = - M_0\, \delta_{l1}$$ (724)

for all $l$ , because $P_l(\cos\theta)=\cos\theta$ . It follows that

$$A_l = B_l = 0$$ (725)

for $l\neq 1$ , and

$$A_1 = \frac{M_0}{3},$$ (726)

$$B_1 = \frac{M_0\, a^{\,3}}{3}.$$ (727)

Thus,

$$\phi_m(r,\theta) = \frac{M_0\, a^{\,2}}{3} \frac{r}{a^{\,2}} \cos\theta$$ (728)

for $r<a$ , and

$$\phi_m(r, \theta) = \frac{M_0 \,a^{\,2}}{3} \frac{a}{r^{\,2}} \cos\theta$$ (729)

for $r\geq a$ . Because there is a uniqueness theorem associated with Poisson's equation (see Section 2.3), we can be sure that this axisymmetric potential is the only solution to the problem that satisfies physical boundary conditions at $r=0$ and infinity.

In the vacuum region outside the sphere,

$${\bf B} = \mu_0\, {\bf H} = -\mu_0\, \nabla\phi_m.$$ (730)

It is easily demonstrated from Equation (730) that

$${\bf B}(r>a) = \frac{\mu_0}{4\pi}\left[ -\frac{{\bf m}}{r^{\,3}} + \frac{3\,({\bf m}\cdot{\bf r})\,{\bf r}}{r^{\,5}} \right],$$ (731)

where

$${\bf m} = \frac{4}{3}\,\pi\, a^{\,3}\, {\bf M}.$$ (732)

This, of course, is the magnetic field of a magnetic dipole of moment ${\bf m}$ . [See Section 5.5.] Not surprisingly, the net dipole moment of the sphere is equal to the integral of the magnetization ${\bf M}$ (which is the dipole moment per unit volume) over the volume of the sphere.

Figure 4: Schematic demagnetization curve for a permanent magnet.

Inside the sphere, we have ${\bf H} = -\nabla\phi_m$ and ${\bf B} =\mu_0\,({\bf H} + {\bf M})$ , giving

$${\bf H} = -\frac{{\bf M}}{3},$$ (733)

and

$${\bf B} = \frac{2}{3} \,\mu_0\, {\bf M}.$$ (734)

Thus, both the ${\bf H}$ and ${\bf B}$ fields are uniform inside the sphere. Note that the magnetic intensity is oppositely directed to the magnetization. In other words, the ${\bf H}$ field acts to demagnetize the sphere. How successful it is at achieving this depends on the shape of the hysteresis curve in the negative $H$ and positive $B$ quadrant. This curve is sometimes called the demagnetization curve of the magnetic material that makes up the sphere. Figure 4 shows a schematic demagnetization curve. The curve is characterized by two quantities: the retentivity $B_R$ (i.e., the residual magnetic field strength at zero magnetic intensity) and the coercivity $\mu_0 \,H_c$ (i.e., the negative magnetic intensity required to demagnetize the material. The latter quantity is conventionally multiplied by $\mu_0$ to give it the units of magnetic field-strength). The operating point (i.e., the values of $B$ and $\mu_0 \,H$ inside the sphere) is obtained from the intersection of the demagnetization curve and the curve $B = \mu \,H$ . It is clear from Equations (734) and (735) that

$$\mu = -2 \,\mu_0$$ (735)

for a uniformly magnetized sphere in the absence of external fields. The magnetization inside the sphere is easily calculated once the operating point has been determined. In fact, $M_0 = B - \mu_0\, H$ . It is clear from Figure 4 that for a magnetic material to be a good permanent magnet it must possess both a large retentivity and a large coercivity. A material with a large retentivity but a small coercivity is unable to retain a significant magnetization in the absence of a strong external magnetizing field.