The magnitude of the magnetic dipole moment of Earth is $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{T}$. (a) If the origin of this magnetism were a magnetized iron sphere at the center of Earth, what would be its radius? (b) What fraction of the volume of Earth would such a sphere occupy? Assume complete alignment of the dipoles. The density of Earth’s inner core isrole="math" localid="1662967671929" $\mathbf{14}\mathbf{}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$ .The magnetic dipole moment of an iron atom is $\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{T}$. (Note:Earth’s inner core is in fact thought to be in both liquid and solid forms and partly iron, but a permanent magnet as the source of Earth’s magnetism has been ruled out by several considerations. For one, the temperature is certainly above the Curie point.)

${\mathrm{\mu }}_{\mathrm{total}}=8.0\times {10}^{22}\mathrm{J}/\mathrm{T}$

$\text{μ}=2.1\times {10}^{-3}\text{J}/\text{T}$

Mass of iron atom,

$\begin{array}{c}\mathrm{m}=56\mathrm{u}\\ =56\times 1.66\times {10}^{-27}\mathrm{kg}\end{array}$

Here, we need to use the equation of mass related with the dipole moment. Using the volume equation of sphere, we can make the equation for radius and solve it. For the fraction of volume of earth, we can take the ratio of volume.

Formulae:

Total dipole moment is expressed as follows:${\mathrm{\mu }}_{\mathrm{total}}=\mathrm{N\mu }$

Total mass, $\mathrm{mass}=\mathrm{Nm}$

To find the radius of sphere$\left(\text{R}\right)$:

We have

$\text{mass}=\text{density}\times \text{volume}$

$\mathrm{Nm}=\left(\frac{4}{3}\right){\mathrm{\pi R}}^3\times \mathrm{\rho }$

role="math" localid="1662968332816" $\frac{{\text{μ}}_{\text{total}}}{\mathrm{\mu }}\times \mathrm{m}=\left(\frac{4}{3}\right){\mathrm{\pi R}}^3\times \mathrm{\rho }$

Rearranging for$\mathrm{R}$:

$\begin{array}{c}={\left(\frac{3{\mathrm{m\mu }}_{\mathrm{total}}}{4\mathrm{\pi \rho \mu }}\right)}^{\frac{1}{3}}\\ ={\left(\frac{3\times (56\times 1.66\times {10}^{27})(8.0\times {10}^{22})}{4\mathrm{\pi }(14\times {10}^3)\times (2.1\times {10}^{-23})}\right)}^{\frac{1}{3}}\\ =1.8\times {10}^5\mathrm{m}\end{array}$

The radius of sphere is, $\text{R}=1.8\times {10}^5\text{m}$

To find the fraction of volume occupied by the sphere:

$\begin{array}{c}\frac{{\text{V}}_{\text{s}}}{{\mathrm{V}}_{\mathrm{e}}}=\frac{\frac{4}{3}{\mathrm{\pi R}}_{\mathrm{s}}^3}{\frac{4}{3}{\mathrm{\pi R}}_{\mathrm{e}}^3}\\ =\frac{{\mathrm{R}}_{\mathrm{s}}^3}{{\mathrm{R}}_{\mathrm{e}}^3}\\ ={\left(\frac{1.8\times {10}^5}{6.37\times {10}^6}\right)}^3\\ =2.3\times {10}^{-5}\end{array}$

The fraction of volume is, $\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{e}}}=2.3\times {10}^{-5}$