Russell Gillen

2022-01-11

The magnitude of the magnetic dipole moment of Earth is $8.0×{10}^{22}\frac{J}{T}$.
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 is $14\frac{g}{c}{m}^{3}$. The magnetic dipole moment of an iron atom is $2.1×{10}^{-23}J{T}^{-1}$. (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.)

Jillian Edgerton

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

Edward Patten

(a) As given in the question, the dipole moment of earth, ${\mu }_{\to tal}=8.0×{10}^{22}J{T}^{-1}$
Also, the dipole moment of iron atom is, $\mu =2.1×{10}^{-23}J{T}^{-1}$. The mass of iron atom is given by
$m=56u=56×1.66×{10}^{-27}kg$
$⇒m=9.30×{10}^{-26}kg$
When the magnetization of the sphere at the center of Earth is saturated, then the total dipole moment is given by ${\mu }_{\to tal}=N\mu$. Now, the iron sphere that is at the center of Earth consists of N iron atoms. Therefore, the mass of the iron sphere will be Nm. Now, the mass of the iron sphere will be due to the center of gravity acting on the sphere and is given by
$Nm=\frac{4\pi \rho {R}^{3}}{3}$
$⇒N=\frac{4\pi \rho {R}^{3}}{3m}$
Putting this value in the value of dipole moment, we get
${\mu }_{\to tal}=\frac{4\pi \rho {R}^{3}}{3m}\mu$
$⇒R={\left(\frac{3m{\mu }_{\to tal}}{4\pi \rho \mu }\right)}^{\frac{1}{3}}$
$⇒R=\left[\frac{3\left(9.30×{10}^{-26}kg\right)\left(8.0×{10}^{22}J{T}^{-1}\right)}{4\pi \left(14×{10}^{3}kg{m}^{-3}\right)\left(2.1×{10}^{-23}J{T}^{-1}\right)}\right]$
$\therefore R=1.8×{10}^{5}m$
Therefore, the radius of the iron sphere that is at the center of earth is
$1.8×{10}^{5}m$.
(b) Now, the volume of the iron sphere is given by
${V}_{s}=\frac{4\pi }{3}{R}^{3}$
$⇒{V}_{s}=\frac{4\pi }{3}{\left(1.8×{10}^{5}\right)}^{3}$
$⇒{V}_{s}=\frac{4\pi }{3}\left(5.83×{10}^{15}\right)$
$⇒{V}_{s}=2.53×{10}^{16}{m}^{3}$
Also, the volume of the earths

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