91Ó°ÊÓ

${\mathrm{μ}}_{\mathrm{total}}=8.0×{10}^{22}\mathrm{J}/\mathrm{T}$

$\text{μ}=2.1×{10}^{−3}\text{J}/\text{T}$

Mass of iron atom,

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

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:${\mathrm{μ}}_{\mathrm{total}}=\mathrm{\pm ·Î¼}$

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

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

We have

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

$\mathrm{Nm}=\left(\frac{4}{3}\right){\mathrm{Ï€¸é}}^3×\mathrm{ÒÏ}$

role="math" localid="1662968332816" $\frac{{\text{μ}}_{\text{total}}}{\mathrm{μ}}×\mathrm{m}=\left(\frac{4}{3}\right){\mathrm{Ï€¸é}}^3×\mathrm{ÒÏ}$

Rearranging for$\mathrm{R}$:

$\begin{array}{c}={\left(\frac{3{\mathrm{³¾Î¼}}_{\mathrm{total}}}{4\mathrm{Ï€ÒÏμ}}\right)}^{\frac{1}{3}}\\ ={\left(\frac{3×(56×1.66×{10}^{27})(8.0×{10}^{22})}{4\mathrm{Ï€}(14×{10}^3)×(2.1×{10}^{−23})}\right)}^{\frac{1}{3}}\\ =1.8×{10}^5\mathrm{m}\end{array}$

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

To find the fraction of volume occupied by the sphere:

$\begin{array}{c}\frac{{\text{V}}_{\text{s}}}{{\mathrm{V}}_{\mathrm{e}}}=\frac{\frac{4}{3}{\mathrm{Ï€¸é}}_{\mathrm{s}}^3}{\frac{4}{3}{\mathrm{Ï€¸é}}_{\mathrm{e}}^3}\\ =\frac{{\mathrm{R}}_{\mathrm{s}}^3}{{\mathrm{R}}_{\mathrm{e}}^3}\\ ={\left(\frac{1.8×{10}^5}{6.37×{10}^6}\right)}^3\\ =2.3×{10}^{−5}\end{array}$

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