A) 4,75 A/m
B) 5.74 A/m
C) 7.54 A/m
D) 75.4 A/m
Correct Answer: A
Solution :
The number of atoms per unit specimen, \[n=\frac{\rho {{N}_{A}}}{A}\] For iron, \[\rho =7.8\times {{10}^{3}}kg{{m}^{-3}}\] \[{{N}_{A}}=6.02\times {{10}^{26}}/kg\,mol,\] \[A=56\] \[\Rightarrow \] \[n=\frac{7.8\times {{10}^{3}}\times 6.02\times {{10}^{26}}}{56}\] \[n=8.38\times {{10}^{28}}{{m}^{-3}}\] Total number of atoms in the bar is \[{{N}_{0}}=nV=8.38\times {{10}^{28}}\] \[\times \,(5\times {{10}^{-2}}\times 1\times {{10}^{-2}}\times 1\times {{10}^{-2}})\] \[{{N}_{0}}=4.19\times {{10}^{23}}\] The saturated magnetic moment of bar \[=4.19\times {{10}^{23}}\times 1.8\times {{10}^{-23}}\] \[=7.54\,A{{m}^{2}}\]
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