Question

The magnetic dipole moment of the Earth is 8.1 ✕ 10²² J/T. The density of the Earth's inner core is 14 g/cm³. The magnetic dipole moment of of an iron atom is 2.1 ✕ 10^-23 J/T.

What fraction of the volume of the Earth would the sphere occupy?'

If the origin of this magnetism were a magnetized iron sphere at the center of the Earth, what would be its radius? (1.8 x 10^5)

Answer 1

number of iron atoms required to provide the total dipole moment=8.1*10^22/(2.1*10^(-23))=3.8571*10^45

then number of moles=3.8571*10^45/(6.023*10^23)=6.404*10^21

mass of iron=number of moles*molar mass

=55.845*0.001*6.404*10^21=3.5763*10^20 kg

then volume=mass/density=2.5546*10^16 m^3

if radius were to be r m,

then (4/3)*pi*r^3=2.5546*10^15

==>r=182700.2322 m

fraction of volume of eart=(r/R)^3

where R=radius of earth=6371 km=6.371*10^6 m

hence fraction of volume of earth=0.028676^3

=2.3583*10^(-5)