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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,

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

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

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

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

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

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

for $ r<a$ , and

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

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

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

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

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

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

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

for $ l\neq 1$ , and

$\displaystyle A_1$ $\displaystyle = \frac{M_0}{3},$ (726)
$\displaystyle B_1$ $\displaystyle = \frac{M_0\, a^{\,3}}{3}.$ (727)


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

for $ r<a$ , and

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

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

It is easily demonstrated from Equation (730) that

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


$\displaystyle {\bf m} = \frac{4}{3}\,\pi\, a^{\,3}\, {\bf M}.$ (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.
\epsfysize =3in

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

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


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

$\displaystyle \mu = -2 \,\mu_0$ (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.

next up previous
Next: Soft Iron Sphere in Up: Magnetostatics in Magnetic Media Previous: Permanent Ferromagnets
Richard Fitzpatrick 2014-06-27