Magnetism Question 41
Question: Each atom of an iron bar $ (5 cm\times 1 cm\times 1 cm) $ has a magnetic moment $ 1.8\times {{10}^{-23}} Am^{2}. $ Knowing that the density of iron is $ 7.78\times 10^{3} k{{gm}^{-3}}, $ 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
Options:
A) $ 4.75 Am^{2} $
B) $ 5.74 Am^{2} $
C) $ 7.54 Am^{2} $
D) $ 75.4 Am^{2} $
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Answer:
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}/kgmol, $ A=56
$ \Rightarrow n=\frac{7.8\times 10^{3}\times 6.02\times 10^{26}}{56} $
$ =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\ Am^{2} $
BETA
