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

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Last updated date: 20th Jun 2024
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Answer
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Hint: Here, the Earth is in the form of a sphere and iron sphere is magnetized at the center of Earth. That is why, we have used the formula of volume of the sphere to find the fraction of the volume of Earth occupied by the sphere. Also, we have used the formula of dipole moment to calculate the radius of the iron sphere.

Complete step by step answer:
(a) As given in the question, the dipole moment of earth, ${\mu _{total}} = 8.0 \times {10^{22}}\,J{T^{ - 1}}$Also, the dipole moment of iron atom is, $\mu = 2.1 \times {10^{ - 23}}\,J{T^{ - 1}}$The mass of iron atom is given by
$
m = 56u = 56 \times 1.66 \times {10^{ - 27}}kg \\
\Rightarrow m = 9.30 \times {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 _{total}} = 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
$Nm = \dfrac{{4\pi \rho {R^3}}}{3}$
$ \Rightarrow \,N = \dfrac{{4\pi \rho {R^3}}}{{3m}}$

Putting this value in the value of dipole moment, we get
${\mu _{total}} = \dfrac{{4\pi \rho {R^3}}}{{3m}}\mu $
$ \Rightarrow \,R = {\left( {\dfrac{{3m{\mu _{total}}}}{{4\pi \rho \mu }}} \right)^{\dfrac{1}{3}}}$
$ \Rightarrow \,R = \left[ {\dfrac{{3\left( {9.30 \times {{10}^{ - 26}}kg} \right)\left( {8.0 \times {{10}^{22}}J{T^{ - 1}}} \right)}}{{4\pi \left( {14 \times {{10}^3}kg\,{m^{ - 3}}} \right)\left( {2.1 \times {{10}^{ - 23}}\,J{T^{ - 1}}} \right)}}} \right]$
$ \therefore \,R = 1.8 \times {10^5}m$

Therefore, the radius of the iron sphere that is at the center of earth is $1.8 \times {10^5}m$.

(b) Now, the volume of the iron sphere is given by
${V_s} = \dfrac{{4\pi }}{3}{R^3}$
$ \Rightarrow \,{V_s} = \dfrac{{4\pi }}{3}{\left( {1.8 \times {{10}^5}} \right)^3}$
$ \Rightarrow \,{V_s} = \dfrac{{4\pi }}{3}\left( {5.83 \times {{10}^{15}}} \right)$
$ \Rightarrow \,{V_s} = 2.53 \times {10^{16}}\,{m^3}$
Also, the volume of the earth’s sphere is given by
${V_e} = \dfrac{{4\pi }}{3}{r^3}$
$ \Rightarrow \,{V_e} = \dfrac{{4\pi }}{3}\left( {6.37 \times {{10}^6}} \right)$
$ \Rightarrow \,{V_e} = 1.08 \times {10^{21}}{m^3}$
Now the fraction of the volume of the Earth occupied by the sphere is given by
$V = \dfrac{{{V_s}}}{{{V_e}}}$
$ \Rightarrow \,V = \dfrac{{2.53 \times {{10}^{16}}{m^3}}}{{1.08 \times {{10}^{21}}{m^3}}}$
$ \therefore \,V = 2.3 \times {10^{ - 5}}$

Therefore, the fraction of the Earth occupied by the iron sphere is $2.3 \times {10^{ - 5}}$.

Note:Here, in the above solution, the mass of the sphere is taken as $Nm = \dfrac{{4\pi \rho {R^3}}}{3}$ . This is because we have taken the mass of the iron sphere in terms of density and volume as given below
$
M = \rho V \\
\Rightarrow \,Nm = \rho V $
Here, we have taken the mass of the iron sphere as $Nm$ because it contains $Nm$ .