P32_025

# P32_025 - 10 27 kg / u) = 9 . 30 10 26 kg . Therefore, R =...

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25. (a) If the magnetization of the sphere is saturated, the total dipole moment is µ total = ,where N is the number of iron atoms in the sphere and µ is the dipole moment of an iron atom. We wish to Fnd the radius of an iron sphere with N iron atoms. The mass of such a sphere is Nm ,where m is the mass of an iron atom. It is also given by 4 πρR 3 / 3, where ρ is the density of iron and R is the radius of the sphere. Thus Nm =4 πρR 3 / 3and N = 4 πρR 3 3 m . We substitute this into µ total = to obtain µ total = 4 πρR 3 µ 3 m . We solve for R and obtain R = µ 3 total 4 πρµ 1 / 3 . The mass of an iron atom is m =56u=(56u)(1 . 66
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Unformatted text preview: 10 27 kg / u) = 9 . 30 10 26 kg . Therefore, R = " 3(9 . 30 10 26 kg)(8 . 10 22 J / T) 4 (14 10 3 kg / m 3 )(2 . 1 10 23 J / T) # 1 / 3 = 1 . 8 10 5 m . (b) The volume of the sphere is V s = 4 3 R 3 = 4 3 (1 . 82 10 5 m) 3 = 2 . 53 10 16 m 3 and the volume of the Earth is V e = 4 3 (6 . 37 10 6 m) 3 = 1 . 08 10 21 m 3 , so the fraction of the Earths volume that is occupied by the sphere is 2 . 53 10 16 m 3 1 . 08 10 21 m 3 = 2 . 3 10 5 ....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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