How many unpaired electrons, each with a magnetic moment of 9.27 × 10^−24 A · m^2, would participate?

+2 votes

The magnetic moment of the Earth is approximately 

8.00 × 1022 A · m2.

 Imagine that the planetary magnetic field were caused by the complete magnetization of a huge iron deposit with density 7 900 kg/m3 and approximately 

8.50 × 1028 iron atoms/m3.
(a) How many unpaired electrons, each with a magnetic moment of 
9.27 × 10−24 A · m2,
 would participate?
 

(b) At two unpaired electrons per iron atom, how many kilograms of iron would be present in the deposit?
 kg

 

asked Feb 28, 2012 in Physics by awesome ~Expert~ (1,479 points)
    

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