The magnitude of the magnetic dipole moment of Earth is 8.0×1022JT. a) If the origin...

Step 1
a) The total dipole moment of a saturated sphere with iron atoms is given by: ${\displaystyle {\mu }_{\to t.}=N\mu }$ (1)
where N is the number of iron atoms in the sphere. The mass the of an iron sphere can be written as: ${\displaystyle M=Nm}$ (2)
where m is the mass of a single atom, and also it can be written in terms of the density and the volume of the sphere:
${\displaystyle M=\rho V}$
${\displaystyle M=\frac{4}{3}\pi \rho R^3}$ (3)
where ${\displaystyle V=\frac{4}{3}\pi R^3}$, by equating (2) and (3) we get:
${\displaystyle Nm=\frac{4}{3}\rho \pi R^3}$
${\displaystyle N=\frac{4\pi \rho R^3}{3m}}$
substitute into (1) to get:
${\displaystyle {\mu }_{\to t.}=\frac{4\pi \rho R^3\mu }{3m}}$
solve for R to get:
${\displaystyle R={\left(\frac{3m{\mu }_{\to t.}}{4\pi \rho \mu }\right)}^{\frac{1}{3}}}$ (4)
the mass of the iron atom is:
${\displaystyle m=56u\times \frac{1.66\times {10}^{-27}kg}{1u}=9.30\times {10}^{-26}kg}$
substitute with the givens into (4) to get:
${\displaystyle R={\left(\frac{3(9.30\times {10}^{-26}kg)(8.0\times {10}^{22}\frac{J}{T})}{4\pi (14\times {10}^{3}k\frac{g}{m^3})(2.1\times {10}^{-23}\frac{J}{T})}\right)}^{\frac{1}{3}}}$
${\displaystyle =1.82\times {10}^{5}m}$
${\displaystyle R=1.82\times {10}^{5}m}$
Step 2
b) The volume of the iron sphere is given by:
${\displaystyle V_s=\frac{4}{3}\pi R^3}$
substitute from part (a) to get:
${\displaystyle V_s=\frac{4}{3}\pi {(1.82\times {10}^{5}m)}^3}$
${\displaystyle =2.53\times {10}^{16}{m}^3}$
the volume of the earth is given by:
${\displaystyle V_e=\frac{4}{3}\pi R_e^3}$
substitute with the givens to get (where ${\displaystyle R_e=6.37\times {10}^{6}m}$):
${\displaystyle V_e=\frac{4}{3}\pi {(6.37\times {10}^{6}m)}^3}$
${\displaystyle =1.08\times {10}^{21}{m}^3}$
the fraction of the volume of the earth over the volume of the sphere is therefore:
${\displaystyle \frac{V_s}{V_e}=\frac{2.53\times {10}^{16}{m}^3}{1.08\times {10}^{21}{m}^3}}$
${\displaystyle =2.34\times {10}^{-5}}$