In Jackson 4.4, boundary value problems in dielectrics page 159, a simple relationship is shown between the electric potentials and hence between the electric fields of a spherical dielectric and spherical cavity in an infinite dielectric when a uniform field $E_0$ is applied. The expressions for the electric fields inside the sphere and cavity are:

$\vec{E_s} = \frac{3\epsilon_0}{\epsilon+2\epsilon_0}\vec{E_0}, \qquad \vec{E_c} = \frac{3\epsilon}{\epsilon_0+2\epsilon}\vec{E_0}$

in which $E_s$ and $E_c$ are the electric fields of the dielectric sphere and spherical cavity in a dielectric medium respectively. The reason, as explained by Jackson, is due to the change in boundary conditions when turning to the "reciprocal problem". This is pretty intuitive because what we need to do is only to invert the relative permittivity in order to get the result for the reciprocal problem.

Upon reading this I'm wondering if we can do the same for the magnetized sphere. Now imagine a similar scenario in which we have a uniformly magnetized sphere with magnetization $\vec{M}=M_0\hat{z}$ and a spherical cavity in a uniformly magnetized infinite medium with same magnetization and no external magnetic field or free currents. The vector potential and magnetic field for the magnetized sphere (only inside) are given by:

$\vec{A_s} = \frac{\mu_0}{3}M_0 r sin(\theta)\hat{\phi}, \qquad \vec{B_s} = \frac{2\mu_0}{3}\vec{M}$

I want to find a similar trick in order to get the results for the vector potential and magnetic field of the cavity. We know that in a uniform magnetized matter we have:

$\vec{B}_m=B_0\hat{z},\qquad \vec{H}_m=(\frac{B_0}{\mu_0}-M_0)\hat{z}$

For a uniform magnetic field $\vec{B}_m=B_0\hat{z}$ we can express the vector potential as (Griffiths problem 5.25):

$\vec{A}_m=\frac{-B_0}{2}\vec{r}\times \hat{z}+\nabla f$

in which $f$ is an arbitrary scalar field and upon a proper gauge fixing it can be discarded. Using superposition we can say that the magnetic vector potential due to a spherical cavity in a magnetized medium is equal to the superposition of the vector potentials for a uniformly magnetized sphere with the opposite magnetization and the vector potential of the uniformly magnetized matter. Hence, for the vector potential inside the cavity we can write:


I don't know if this is true or not because upon substituting the expressions in the equation above, it doesn't look as symmetric as the example I described for the cavity in the polarized medium in the first paragraph. What is the correct expression for the vector potential?

Moving on, we can find the magnetic field inside the cavity using superposition of the magnetic fields of a uniform magnetized material and a magnetized sphere with the opposite magnetization. The magnetic field inside a uniformly magnetized sphere is


The magnetic field inside the cavity becomes


We can express $M_0$ in terms of $B_0$ as

$\vec{H}_m=\frac{M_0}{\chi_m}\hat{z}=(\frac{B_0}{\mu_0}-M_0)\hat{z} \implies B_0=M_0(\frac{\mu_0 \mu}{\mu - \mu_0})$

in which I have used $\chi_m=\frac{\mu}{\mu_0}-1$. Substituting the expression above in the expression for $\vec{B}_c$ and simplifying the terms gives


However according to Greiner's Classical Electrodynamics book page 234, this is not correct (apparently the multiplicative factor should be inverted). What am I doing wrong here?


1 Answer 1


The easy way to do the magnetic problem once you have done the electric one is to notice that your static dielectric problem is described by Maxwell's equations \begin{eqnarray} \vec \nabla \cdot \vec D = 0 \nonumber\\ \vec \nabla \times \vec E = 0 \end{eqnarray} with $\vec D = \epsilon E$ or $\vec D = \epsilon_0 E$ in the two regions. Similarly the magnetic problem is described by Maxwell's equations \begin{eqnarray} \vec \nabla \cdot \vec B = 0 \nonumber\\ \vec \nabla \times \vec H = 0 \end{eqnarray} with $\vec B = \mu H$ or $\vec B = \mu_0 H$ in the two regions. Therefore you can just substitute $\vec E \rightarrow \vec H$, and $\epsilon \rightarrow \mu$. That is \begin{equation} \vec H_s = \frac{3\mu_0}{\mu+2\mu_0} \vec H_0 \, \ \ \ \ \ \vec H_c = \frac{3\mu}{\mu_0+2\mu} \vec H_0 \,. \end{equation}

If you want the result for $\vec B$, for the cavity \begin{equation} \vec B_0 = \mu \vec H_0 \, \ \ \ \ \ \vec B_c = \mu_0 \vec H_c \end{equation} and for the sphere \begin{equation} \vec B_0 = \mu_0 \vec H_0 \, \ \ \ \ \ \vec B_s = \mu \vec H_s \end{equation} giving \begin{equation} \vec B_s = \frac{3\mu}{\mu+2\mu_0} \vec B_0 \, \ \ \ \ \ \vec B_c = \frac{3\mu_0}{\mu_0+2\mu} \vec B_0 \end{equation}

Your mistake is that you are not superposing the correct solutions. If you want to superpose a negative magnetization, it must be surrounded by a material with permeability $\mu$ not $\mu_0$. The correct internal field for a uniform $M_0\hat z$ sphere surrounded by material $\mu$ is \begin{equation} \vec B_s = \frac{2\mu}{2\mu+\mu_0} \mu_0 M_0 \end{equation} which reduces to your result when $\mu=\mu_0$. Applying your other equations \begin{equation} \vec B_c = \hat z(B_0 -\frac{2\mu}{2\mu+\mu_0} \mu_0 M_0) = \hat zB_0\left (1 -\frac{2\mu}{2\mu+\mu_0} \frac{\mu-\mu_0}{\mu}\right) = \hat z \frac{3\mu_0}{2\mu+\mu_0}B_0 \end{equation} which is the correct cavity result.

I would recommend just solving the differential equation for either the vector potential or the magnetic scalar potential (or both) for all of these cases to check your results and to see how things fit together if you find an inconsistency in your solutions or thinking.


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