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

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

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

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

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

for $r>b$ ,

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

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

$$\phi_m = \delta\,r\cos\theta$$ (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

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

and

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

The boundary conditions (744) and (745) yield

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

and

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

It follows that

$$\mu_0\,\alpha = \left[ \frac{(2\,\mu+\mu_0)\,(\mu-\mu_0) } {(2\,\mu+\mu_0)\,(\m... ...mu_0) - 2\,(a^{\,3}/b^{\,3})\,(\mu-\mu_0)^{\,2}}\right](b^{\,3}-a^{\,3}) \,B_0,$$ (752)

$$\mu_0\,\beta =-\left[ \frac{3\,(2\,\mu+\mu_0)\,\mu_0} {(2\,\mu+\mu_0)\,(\mu+2\,\mu_0) - 2\,(a^{\,3}/b^{\,3})\,(\mu-\mu_0)^{\,2}}\right] B_0,$$ (753)

$$\mu_0\,\gamma = - \left[ \frac{3\,(\mu-\mu_0)\,\mu_0} {(2\,\mu+\mu_0)\,(\mu+2\,\mu_0) - 2\,(a^{\,3}/b^{\,3})\,(\mu-\mu_0)^{\,2}}\right] a^{\,3} B_0,$$ (754)

$$\mu_0\,\delta = -\left[ \frac{9\,\mu\,\mu_0}{(2\,\mu+\mu_0)\,(\mu+2\,\mu_0) - 2\,(a^{\,3}/b^{\,3})\,(\mu-\mu_0)^{\,2}}\right] B_0.$$ (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

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