Each atom of an iron bar $(5\,cm \times 1\,cm \times 1\,cm)$ has a magnetic moment $1.8 \times {10^{ - 23}}\,A{m^2}.$ Knowing that the density of iron is $7.78 \times {10^3}\,k{g^{ - 3}}\,m,$ atomic weight is $56$ and Avogadro's number is $6.02 \times {10^{23}}$ the magnetic moment of bar in the state of magnetic saturation will be.....$A{m^2}$
Each atom of an iron bar $(5\,cm \times 1\,cm \times 1\,cm)$ has a magnetic moment $1.8 \times {10^{ - 23}}\,A{m^2}.$ Knowing that the density of iron is $7.78 \times {10^3}\,k{g^{ - 3}}\,m,$ atomic weight is $56$ and Avogadro's number is $6.02 \times {10^{23}}$ the magnetic moment of bar in the state of magnetic saturation will be.....$A{m^2}$
- A
$4.75$
- B
$5.74$
- C
$7.54$
- D
$75.4$
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