A) \[4.75\text{ }A{{m}^{2}}\]
B) \[5.74\text{ }A{{m}^{2}}\]
C) \[7.54\text{ }A{{m}^{2}}\]
D) \[75.4\text{ }A{{m}^{2}}\]
Correct Answer: C
Solution :
The number of atoms per unit volume in a 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 \[=4.19\times {{10}^{23}}\times 1.8\times {{10}^{-23}}=7.54\text{ }A{{m}^{2}}\]
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