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 2.1×10−23JT−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.)

Question

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 2.1×10−23JT−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 · Accepted Answer

Step 1  a) The total dipole moment of a saturated sphere with iron atoms is given by: μ→t.=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:  μ→t.=4πρR3μ3m  solve for R to get:  R=(3mμ→t.4πρμ)13 (4)  the mass of the iron atom is:  m=56u×1.66×10−27kg1u=9.30×10−26kg  substitute with the givens into (4) to get:  R=(3(9.30×10−26kg)(8.0×1022JT)4π(14×103kgm3)(2.1×10−23JT))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  =2.34×10−5

Edward Patten · Accepted Answer

Complete step by step answer:   (a) As given in the question, the dipole moment of earth, μ→tal=8.0×1022JT−1  Also, the dipole moment of iron atom is, μ=2.1×10−23JT−1. The mass of iron atom is given by  m=56u=56×1.66×10−27kg  ⇒m=9.30×10−26kg  When the magnetization of the sphere at the center of Earth is saturated, then the total dipole moment is given by μ→tal=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  μ→tal=4πρR33mμ  ⇒R=(3mμ→tal4πρμ)13  ⇒R=[3(9.30×10−26kg)(8.0×1022JT−1)4π(14×103kgm−3)(2.1×10−23JT−1)]  ∴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

The magnitude of the magnetic dipole moment of Earth is 8.0 \times

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