Several metallic objects like Iron, Copper etc can provide electrostatic shielding (one should remain inside the car during thunderstorm) and several superconductors like HTS (High Temperature Superconductors) provide magnetic shielding against the magnetic fields.
So how can such fields be repelled by these types of materials? I think it might be because of their atomic arrangement but I am still not sure that why only some of the elements from the periodic table (transition metals) provide these shielding. I mean every element in periodic table have some electrons, protons and neutrons. How these atomic arrangement influence them?
Also I want to know that why magnetic field is so specific in nature (it can be shielded only the zero resistance material). Is there any relation between the zero resistance material and it's magnetic shielding.
Answer
I think so. Electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. Electromagnetic radiation consists of coupled electric and magnetic fields. The electric field produces forces on the charge carriers (i.e., electrons) within the conductor. As soon as an electric field is applied to the surface of an ideal conductor, it induces a current that causes displacement of charge inside the conductor that cancels the applied field inside, at which point the current stops.
Mathematical model
Suppose that we have a spherical shell of a (linear and isotropic) diamagnetic material with permeability $\mu$, with inner radius a and outer radius b. We then put this object in a constant magnetic field:
$\vec{H}_{0}=H_{0}\hat{z}=H_{0}\cos\theta \hat{r}$
Since there are no currents in this problem except for possible bound currents on the boundaries of the diamagnetic material, i.e. $$\nabla\times \vec B=0$$ in a simply connected region, there exists a scalar function we call magnetic scalar potential such that $$\vec{H}=-\nabla \Phi_{M}$$ where $\vec{B}=:\mu \vec{H}$. Since $$\nabla\cdot\vec B=0$$ we have Laplace's equation: $$\nabla^{2}\Phi_{M}=0.$$
In this particular problem there is azimuthal symmetry so we can write down that the solution to Laplace's equation in spherical coordinates is:
$\Phi_{M}=\sum_{l=0}^\infty \left( A_{l}r^{l}+\frac{B_{l}}{r^{l+1}}\right)P_{l}(\cos\theta)$
After matching the boundary conditions \begin{align} (\vec{H_{2}}-\vec{H_{1}})\times \hat{n}&=0 \\ (\vec{B_{2}}-\vec{B_{1}})\cdot \hat{n}&=0 \end{align} at the boundaries (where $\hat{n}$ is a unit vector that is normal to the surface pointing from side 1 to side 2), then we find that the magnetic field inside the cavity in the spherical shell is:
$\vec{H_{in}}=\eta \vec{H_{0}}$
where $\eta$ is an attenuation coefficient that depends on the thickness of the diamagnetic material and the magnetic permeability of the material:
$\eta=\frac{9\mu}{(2\mu+1)(\mu+2)-2\left(\frac{a}{b}\right)^{3}(\mu-1)^2}$
This coefficient describes the effectiveness of this material in shielding the external magnetic field from the cavity that it surrounds. Notice that this coefficient appropriately goes to 1 (no shielding) in the limit that $\mu\rightarrow1$. In the limit that $\mu \rightarrow 0$, $\infty$ this coefficient goes to $0$ (perfect shielding), then the attenuation coefficient takes on the simpler form:
$\eta=\frac{9}{2} \frac{1}{(1-\frac{a^{3}}{b^{3}}) \mu}$
which shows that the magnetic field decreases like $\mu^{-1}$ Well I can tell u about electromagnetic shielding but not magnetic shielding. Sorry for that.
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