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

Russell Gillen

Russell Gillen

Answered

2022-01-11

The magnitude of the magnetic dipole moment of Earth is 8.0×1022JT.
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 14gcm3. The magnetic dipole moment of an iron atom is 2.1×1023JT1. (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.)

Answer & Explanation

Jillian Edgerton

Jillian Edgerton

Expert

2022-01-12Added 34 answers

Step 1
a) The total dipole moment of a saturated sphere with iron atoms is given by: μt.=Nμ (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=ρV
M=43πρR3 (3)
where V=43πR3, by equating (2) and (3) we get:
Nm=43ρπR3
N=4πρR33m
substitute into (1) to get:
μt.=4πρR3μ3m
solve for R to get:
R=(3mμt.4πρμ)13 (4)
the mass of the iron atom is:
m=56u×1.66×1027kg1u=9.30×1026kg
substitute with the givens into (4) to get:
R=(3(9.30×1026kg)(8.0×1022JT)4π(14×103kgm3)(2.1×1023JT))13
=1.82×105m
R=1.82×105m
Step 2
b) The volume of the iron sphere is given by:
Vs=43πR3
substitute from part (a) to get:
Vs=43π(1.82×105m)3
=2.53×1016m3
the volume of the earth is given by:
Ve=43πRe3
substitute with the givens to get (where Re=6.37×106m):
Ve=43π(6.37×106m)3
=1.08×1021m3
the fraction of the volume of the earth over the volume of the sphere is therefore:
VsVe=2.53×1016m31.08×1021m3
=2.34×105
Edward Patten

Edward Patten

Expert

2022-01-13Added 38 answers

Complete step by step answer:
(a) As given in the question, the dipole moment of earth, μtal=8.0×1022JT1
Also, the dipole moment of iron atom is, μ=2.1×1023JT1. The mass of iron atom is given by
m=56u=56×1.66×1027kg
m=9.30×1026kg
When the magnetization of the sphere at the center of Earth is saturated, then the total dipole moment is given by μtal=Nμ. 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=4πρR33
N=4πρR33m
Putting this value in the value of dipole moment, we get
μtal=4πρR33mμ
R=(3mμtal4πρμ)13
R=[3(9.30×1026kg)(8.0×1022JT1)4π(14×103kgm3)(2.1×1023JT1)]
R=1.8×105m
Therefore, the radius of the iron sphere that is at the center of earth is
1.8×105m.
(b) Now, the volume of the iron sphere is given by
Vs=4π3R3
Vs=4π3(1.8×105)3
Vs=4π3(5.83×1015)
Vs=2.53×1016m3
Also, the volume of the earths

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