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Magnetic shielding

There are many situations, particularly in experimental physics, where it is desirable to shield a certain region from magnetic fields. This can be achieved by surrounding the region in question by a material of high permeability. It is vitally important that a material used as a magnetic shield does not develop a permanent magnetization in the presence of external fields, otherwise the material itself may become a source of magnetic fields. The most effective commercially available magnetic shielding material is called Mumetal, and is an alloy of 5% Copper, 2% Chromium, 77% Nickel, and 16% Iron. The maximum permeability of Mumetal is about $10^5\,\mu_0$. This material also possesses a particularly low retentivity and coercivity. Unfortunately, Mumetal is extremely expensive. Let us investigate how much of this material is actually required to shield a given region from an external magnetic field.

Consider a spherical shell of magnetic shielding, made up of material of permeability $\mu$, placed in a formerly uniform magnetic field ${\bfm B}_0 = B_0 \,\hat{\bfm z}$. Suppose that the inner radius of the shell is $a$ and the outer radius is $b$. Since there are no free currents in the problem, we can write ${\bfm H} = -\nabla\phi_m$. Furthermore, since ${\bfm B} = \mu {\bfm H}$ and $\nabla\!\cdot\!{\bfm B} = 0$, it is clear that the magnetic scalar potential satisfies Laplace's equation, $\nabla^2\phi_m=0$, throughout all space. The boundary conditions are that the potential must be well behaved at $r=0$ and $r\rightarrow\infty$, and also that the tangential and the normal components of ${\bfm H}$ and ${\bfm B}$, respectively, must be continuous at $r=a$ and $r=b$. The boundary conditions on ${\bfm H}$ merely imply that the scalar potential $\phi_m$ must be continuous at $r=a$ and $r=b$. The boundary conditions on ${\bfm B}$ yield

$\displaystyle \mu_0\,\frac{\partial\phi_m(r=a-)}{\partial r}$ $\textstyle =$ $\displaystyle \mu \,\frac{\partial\phi_m(r=a+)}{\partial r},$ (598)
$\displaystyle \mu_0\,\frac{\partial\phi_m(r=b+)}{\partial r}$ $\textstyle =$ $\displaystyle \mu \,\frac{\partial\phi_m(r=b-)}{\partial r}.$ (599)

Let us try the following test solution for the magnetic potential:

\begin{displaymath}
\phi_m = - \frac{B_0}{\mu_0}\,r\cos\theta + \frac{\alpha}{r^2} \cos\theta
\end{displaymath} (600)

for $r>b$,
\begin{displaymath}
\phi_m = \left(\beta\, r + \frac{\gamma}{r^2}\right) \cos\theta
\end{displaymath} (601)

for $b\geq r\geq a$, and
\begin{displaymath}
\phi_m = \delta\,r\cos\theta
\end{displaymath} (602)

for $r<a$. This potential is certainly a solution of Laplace's equation throughout space. It yields the uniform magnetic field ${\bfm B}_0$ as $r\rightarrow\infty$, and satisfies physical boundary conditions at $r=0$ and infinity. Since there is a uniqueness theorem associated with Poisson's equation, we can be certain that this potential is the correct solution to the problem provided that the arbitrary constants $\alpha$, $\beta$, etc. can be adjusted in such a manner that the boundary conditions at $r=a$ and $r=b$ are also satisfied.

The continuity of $\phi_m$ at $r=a$ and $r=b$ requires that

\begin{displaymath}
\beta \,a + \frac{\gamma}{a^2} = \delta\, a,
\end{displaymath} (603)

and
\begin{displaymath}
\beta \,b + \frac{\gamma}{b^2} = - \frac{B_0}{\mu_0}\, b + \frac{\alpha}{b^2}.
\end{displaymath} (604)

The boundary conditions (3.175) yield
\begin{displaymath}
\mu_0 \,\delta = \mu\left(\beta - \frac{2\gamma}{a^3}\right),
\end{displaymath} (605)

and
\begin{displaymath}
\mu_0\left(-\frac{B_0}{\mu_0} -\frac{2\alpha}{b^3}\right) = \mu \left(\beta - \frac{2\gamma}
{b^3}\right).
\end{displaymath} (606)

It follows that
$\displaystyle \mu_0\,\alpha$ $\textstyle =$ $\displaystyle \left[ \frac{(2\mu+\mu_0)(\mu-\mu_0) }
{(2\mu+\mu_0)(\mu+2\mu_0) - 2(a^3/b^3)(\mu-\mu_0)^2}\right](b^3-a^3) B_0,$  
      (607)
$\displaystyle \mu_0\,\beta$ $\textstyle =$ $\displaystyle -\left[ \frac{3(2\mu+\mu_0)\mu_0}
{(2\mu+\mu_0)(\mu+2\mu_0) - 2(a^3/b^3)(\mu-\mu_0)^2}\right] B_0,$ (608)
$\displaystyle \mu_0\,\gamma$ $\textstyle =$ $\displaystyle - \left[ \frac{3(\mu-\mu_0)\mu_0}
{(2\mu+\mu_0)(\mu+2\mu_0) - 2(a^3/b^3)(\mu-\mu_0)^2}\right] a^3 B_0,$ (609)
$\displaystyle \mu_0\,\delta$ $\textstyle =$ $\displaystyle -\left[ \frac{9\mu\mu_0}{(2\mu+\mu_0)(\mu+2\mu_0) - 2(a^3/b^3)(\mu-\mu_0)^2}\right] B_0.$ (610)

Consider the limit of a thin, high permeability shell for which $b=a+d$, $d/a\ll1$, and $\mu/\mu_0\gg 1$. In this limit, the field inside the shell is given by

\begin{displaymath}
{\bfm B} \simeq \frac{3}{2} \frac{\mu_0}{\mu} \frac{a}{d}\, {\bfm B}_0.
\end{displaymath} (611)

Thus, if $\mu\simeq 10^5 \mu_0$ for Mumetal, then we can reduce the magnetic field strength inside the shell by almost a factor of 1000 using a shell whose thickness is only 1/100th of its radius. Clearly, a little Mumetal goes a long way! Note, however, that as the external field strength, $B_0$, is increased, the Mumetal shell eventually saturates, and $\mu/\mu_0$ gradually falls to unity. Thus, extremely strong magnetic fields (typically, $B_0\stackrel {_{\normalsize >}}{_{\normalsize\sim}}1$ tesla) are hardly shielded at all by Mumetal, or similar magnetic materials.


next up previous
Next: Magnetic energy Up: The effect of dielectric Previous: A soft iron sphere
Richard Fitzpatrick 2002-05-18