Question

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?

Answer 1

No. of unpaired electrons = Total Magnetic moment/magnetic moment of each unpaired electron

No. = 8*10^22/(9.27*10^-24)

No = 8.63*10^45

Each iron atom has two unpaired electrons, so the number of iron atoms required is

= 8.63*10^45/2 = 4.315*10^45

B.

Mass = No. of participating iron atoms*density/Volume

Mass = 4.315*10^45*7900/(8.5*10^28)

Mass = 4.01*10^20 kg