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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 goal 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 mu-metal, and is an alloy of 5 percent copper, 2 percent chromium, 77 percent nickel, and 16 percent iron. The maximum permeability of mu-metal is about $ 10^5\,\mu_0$ . This material also possesses a particularly low retentivity and coercivity. Unfortunately, mu-metal 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 $ {\bf B}_0 = B_0 \,{\bf e}_z$ . Suppose that the inner radius of the shell is $ a$ , and the outer radius is $ b$ . Because there are no free currents in the problem, we can write $ {\bf H} = -\nabla\phi_m$ . Furthermore, because $ {\bf B} = \mu \,{\bf H}$ and $ \nabla\cdot{\bf 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 $ {\bf H}$ and $ {\bf B}$ , respectively, must be continuous at $ r=a$ and $ r=b$ . The boundary conditions on $ {\bf H}$ merely imply that the scalar potential $ \phi_m$ must be continuous at $ r=a$ and $ r=b$ . The boundary conditions on $ {\bf B}$ yield

$\displaystyle \mu_0\,\frac{\partial\phi_m(r=a-,\theta)}{\partial r}$ $\displaystyle = \mu \,\frac{\partial\phi_m(r=a+,\theta)}{\partial r},$ (743)
$\displaystyle \mu_0\,\frac{\partial\phi_m(r=b+,\theta)}{\partial r}$ $\displaystyle = \mu \,\frac{\partial\phi_m(r=b-,\theta)}{\partial r}.$ (744)

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

$\displaystyle \phi_m = - \frac{B_0}{\mu_0}\,r\cos\theta + \frac{\alpha}{r^{\,2}} \cos\theta$ (745)

for $ r>b$ ,

$\displaystyle \phi_m = \left(\beta\, r + \frac{\gamma}{r^{\,2}}\right) \cos\theta$ (746)

for $ b\geq r\geq a$ , and

$\displaystyle \phi_m = \delta\,r\cos\theta$ (747)

for $ r<a$ . This potential is certainly a solution of Laplace's equation throughout space. It yields the uniform magnetic field $ {\bf B}_0$ as $ r\rightarrow \infty$ , and satisfies physical boundary conditions at $ r=0$ and infinity. Because there is a uniqueness theorem associated with Poisson's equation (see Section 2.3), we can be certain that this potential is the correct solution to the problem provided that the arbitrary constants $ \alpha$ , $ \beta$ , et cetera, 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

$\displaystyle \beta \,a + \frac{\gamma}{a^{\,2}} = \delta\, a,$ (748)


$\displaystyle \beta \,b + \frac{\gamma}{b^{\,2}} = - \frac{B_0}{\mu_0}\, b + \frac{\alpha}{b^{\,2}}.$ (749)

The boundary conditions (744) and (745) yield

$\displaystyle \mu_0 \,\delta = \mu\left(\beta - \frac{2\,\gamma}{a^{\,3}}\right),$ (750)


$\displaystyle \mu_0\left(-\frac{B_0}{\mu_0} -\frac{2\,\alpha}{b^{\,3}}\right) = \mu \left(\beta - \frac{2\,\gamma} {b^{\,3}}\right).$ (751)

It follows that

$\displaystyle \mu_0\,\alpha$ $\displaystyle = \left[ \frac{(2\,\mu+\mu_0)\,(\mu-\mu_0) } {(2\,\mu+\mu_0)\,(\m...
...mu_0) - 2\,(a^{\,3}/b^{\,3})\,(\mu-\mu_0)^{\,2}}\right](b^{\,3}-a^{\,3}) \,B_0,$ (752)
$\displaystyle \mu_0\,\beta$ $\displaystyle =-\left[ \frac{3\,(2\,\mu+\mu_0)\,\mu_0} {(2\,\mu+\mu_0)\,(\mu+2\,\mu_0) - 2\,(a^{\,3}/b^{\,3})\,(\mu-\mu_0)^{\,2}}\right] B_0,$ (753)
$\displaystyle \mu_0\,\gamma$ $\displaystyle = - \left[ \frac{3\,(\mu-\mu_0)\,\mu_0} {(2\,\mu+\mu_0)\,(\mu+2\,\mu_0) - 2\,(a^{\,3}/b^{\,3})\,(\mu-\mu_0)^{\,2}}\right] a^{\,3} B_0,$ (754)
$\displaystyle \mu_0\,\delta$ $\displaystyle = -\left[ \frac{9\,\mu\,\mu_0}{(2\,\mu+\mu_0)\,(\mu+2\,\mu_0) - 2\,(a^{\,3}/b^{\,3})\,(\mu-\mu_0)^{\,2}}\right] B_0.$ (755)

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

$\displaystyle {\bf B} \simeq \frac{3}{2}\, \frac{\mu_0}{\mu}\, \frac{a}{d}\, {\bf B}_0.$ (756)

Thus, given that $ \mu\simeq 10^5 \mu_0$ for mu-metal, 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. Note, however, that as the external field-strength, $ B_0$ , is increased, the mu-metal 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 mu-metal, or similar magnetic materials.

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Next: Magnetic Energy Up: Magnetostatics in Magnetic Media Previous: Soft Iron Sphere in
Richard Fitzpatrick 2014-06-27