hard ferromagnetic sphere

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Consider a hard ferromagnetic sphere of radius R\, that has a uniform magnetization \mathbf{M} = M_0 \hat\mathbf{z}\, "frozen in". We wish to evaluate the magnetic field \mathbf{B}\, everywhere inside and outside the sphere.

Bound current method

The magnetization can be modeled as a bound current density

\mathbf{J}_b = \boldsymbol\nabla\times\mathbf{M} = 0\,

together with a bound surface current

\mathbf{K}_b = \mathbf{M}\times \hat\mathbf{n} = M_0 \sin\theta \hat\boldsymbol\phi\,.

Magnetic potential method

The uniform magnetization leads to no effective magnetic charge density inside the sphere. There is, however, an effective surface charge density equal to

\sigma_m = M_0 \cos\theta\,.

The magnetic potential is then

\Phi_m(\mathbf{r}) = \frac{M_0}{4\pi} R^2\int\!d\Omega'\,\frac{ \cos\theta'}{|\mathbf{r}-\mathbf{r}'|}\,,

which can be evaluated using the spherical multipole expansion

\frac{1}{|\mathbf{r}-\mathbf{r}'|} = \frac{1}{r_>} \sum_{\ell=0}^\infty \sum_{m=-\ell}^{\ell} \frac{4\pi}{2\ell + 1} \left(\frac{r_<}{r_>}\right)^\ell Y^{m}_\ell(\theta, \phi) Y^{m*}_\ell(\theta^{\prime}, \phi^{\prime})\,.

Thus

\Phi_m(\mathbf{r})\, =\frac{M_0}{4\pi} \sum_{\ell=0}^\infty \sum_{m=-\ell}^{\ell} \frac{1}{r_>} \left(\frac{r_<}{r_>}\right)^\ell \frac{4\pi}{2\ell + 1} Y^{m}_\ell(\theta, \phi) R^2\int\!d\Omega'\,\cos\theta'\, Y^{m*}_\ell(\theta^{\prime}, \phi^{\prime})\,,
=\frac{M_0}{4\pi} \sum_{\ell=0}^\infty \sum_{m=-\ell}^{\ell} \frac{1}{r_>} \left(\frac{r_<}{r_>}\right)^\ell \frac{4\pi}{2\ell + 1} Y^{m}_\ell(\theta, \phi) 2\sqrt{\frac{\pi}{3}}R^2\int\!d\Omega'\,Y^0_1(\theta',\phi') Y^{m*}_\ell(\theta^{\prime}, \phi^{\prime})\,,
=\frac{M_0 R^2}{3}\frac{r_<}{r_>^2} \cos\theta\,.

Outside

\mathbf{H}\, = -\boldsymbol\nabla \Phi_m\,,
=\frac{2 M_0 R^3 \cos\theta}{3 r^3} \hat\mathbf{r} + \frac{M_0 R}{3 r^3}\sin\theta \hat\boldsymbol\theta\,,
=\frac{3 M_0 R^3 \cos\theta}{3 r^3} \hat\mathbf{r} - \frac{M_0 R}{3 r^3}\hat\mathbf{z}\,,
= \frac {1} {4\pi} \frac{ 3(\mathbf{m}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{m}}{r^3}\,,

and

\mathbf{B} = \frac {\mu_0} {4\pi} \frac{ 3(\mathbf{m}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\mathbf{m}}{r^3}\,,

where

\mathbf{m} = \frac{4\pi R^3}{3} \mathbf{M}\, is the magnetic dipole moment of the sphere.

Inside,

\mathbf{H}\, = -\boldsymbol\nabla \Phi_m\,,
= -\frac{M_0}{3} \cos\theta\hat\mathbf{r} + \frac{M_0}{3} \sin\theta \hat\boldsymbol\theta\,,
= -\frac{\mathbf{M}}{3}\,,

so that

\mathbf{B} = \mu_0 \frac{2}{3} \mathbf{M}\,.

References

[1] [2]

  1. ↑ Jackson, John D. (1998). Classical Electrodynamics (3rd ed.). Wiley. ISBN 0-471-30932-X. 
  2. ↑ Griffiths, David J. (1998). Introduction to Electrodynamics (3rd ed.). Prentice Hall. ISBN 0-13-805326-X. 
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