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Next: A soft iron sphere Up: The effect of dielectric Previous: Permanent ferromagnets

A uniformly magnetized sphere

Consider a sphere of radius $a$, with a uniform permanent magnetization ${\bfm M}= M_0 \,\hat{\bfm z}$, surrounded by a vacuum region. The simplest way of solving this problem is in terms of the scalar magnetic potential introduced in Eq. (3.134). From Eqs. (3.136) and (3.137), it is clear that $\phi_m$ satisfies Laplace's equation,
\begin{displaymath}
\nabla^2\phi_m = 0,
\end{displaymath} (572)

since there is zero volume magnetic charge density in a vacuum or a uniformly magnetized magnetic medium. However, according to Eq. (3.144), there is a magnetic surface charge density,
\begin{displaymath}
\sigma_m = \hat{\bfm r}\!\cdot\!{\bfm M} = M_0 \cos\theta,
\end{displaymath} (573)

on the surface of the sphere. One of the matching conditions at the surface of the sphere is that the tangential component of ${\bfm H}$ must be continuous. It follows from Eq. (3.134) that the scalar magnetic potential must be continuous at $r=a$, so that
\begin{displaymath}
\phi_m(r=a_+) = \phi_m(r=a_-).
\end{displaymath} (574)

Integrating Eq. (3.136) over a Gaussian pill-box straddling the surface of the sphere yields
\begin{displaymath}
\left[\frac{\partial\phi_m}{\partial r}\right]_{r=a-}^{r=a+} = -\sigma_m
= -M_0 \cos\theta.
\end{displaymath} (575)

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 Eq. (3.151) which satisfies physical boundary conditions at $r=a$ and $r=\infty$ is

\begin{displaymath}
\phi_m(r,\theta) =\sum_{l=0}^\infty A_l \,r^l P_l(\cos\theta)
\end{displaymath} (576)

for $r<a$, and
\begin{displaymath}
\phi_m(r,\theta) =\sum_{l=0}^\infty B_l\, r^{-(l+1)} P_l(\cos\theta)
\end{displaymath} (577)

for $r\geq a$. The boundary condition (3.153) yields
\begin{displaymath}
B_l = A_l \,a^{2l+1}
\end{displaymath} (578)

for all $l$. The boundary condition (3.154) gives
\begin{displaymath}
- \frac{(l+1)\,B_l}{a^{l+2}} - l A_l\, a^{l-1} = - M_0\, \delta_{l1}
\end{displaymath} (579)

for all $l$, since $P_l(\cos\theta)=\cos\theta$. It follows that
\begin{displaymath}
A_l = B_l = 0
\end{displaymath} (580)

for $l\neq 1$, and
$\displaystyle A_1$ $\textstyle =$ $\displaystyle \frac{M_0}{3},$ (581)
$\displaystyle B_1$ $\textstyle =$ $\displaystyle \frac{M_0 a^3}{3}.$ (582)

Thus,
\begin{displaymath}
\phi_m(r,\theta) = \frac{M_0 a^2}{3} \frac{r}{a^2} \cos\theta
\end{displaymath} (583)

for $r<a$, and
\begin{displaymath}
\phi_m(r, \theta) = \frac{M_0 a^2}{3} \frac{a}{r^2} \cos\theta
\end{displaymath} (584)

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

In the vacuum region outside the sphere

\begin{displaymath}
{\bfm B} = \mu_0 {\bfm H} = -\mu_0 \nabla\phi_m.
\end{displaymath} (585)

It is easily demonstrated from Eq. (3.162) that
\begin{displaymath}
{\bfm B}(r>a) = \frac{\mu_0}{4\pi}\left[
-\frac{{\bfm m}}{r^3} + \frac{3({\bfm m}\!\cdot\!{\bfm r})\,{\bfm r}}{r^5}
\right],
\end{displaymath} (586)

where
\begin{displaymath}
{\bfm m} = \frac{4}{3}\,\pi a^3 {\bfm M}.
\end{displaymath} (587)

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

Figure 4: Schematic demagnetization curve for a permanent magnet
\begin{figure}\epsfysize =2.5in
\centerline{\epsffile{demag.eps}}\end{figure}

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

\begin{displaymath}
{\bfm H} = -\frac{{\bfm M}}{3},
\end{displaymath} (588)

and
\begin{displaymath}
{\bfm B} = \frac{2}{3} \,\mu_0 {\bfm M}.
\end{displaymath} (589)

Thus, both the ${\bfm H}$ and ${\bfm B}$ fields are uniform inside the sphere. Note that the magnetic intensity is oppositely directed to the magnetization. In other words, the ${\bfm H}$ 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 which 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: this 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 Eqs. (3.166) and (3.167) that
\begin{displaymath}
\mu = -2 \,\mu_0
\end{displaymath} (590)

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 Fig. 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: A soft iron sphere Up: The effect of dielectric Previous: Permanent ferromagnets
Richard Fitzpatrick 2002-05-18