A) \[118\times {{10}^{5}}A{{m}^{-1}}\]
B) \[118\times {{10}^{4}}A{{m}^{-1}}\]
C) \[235\times {{10}^{4}}A{{m}^{-1}}\]
D) \[118\times {{10}^{5}}A{{m}^{-1}}\]
Correct Answer: D
Solution :
(d) \[118\times {{10}^{5}}A{{m}^{-1}}\] The volume of the cubic domain is \[V={{(15\times {{10}^{-6}}m)}^{3}}\] \[=3.38\times {{10}^{-18}}{{m}^{3}}=3.38\times {{10}^{-12}}c{{m}^{3}}\] Number of atoms in domain \[(N)=5\times {{10}^{10}}\]atoms Since each iron atom has a dipole moment (M) \[=8\times {{10}^{-24}}A{{m}^{2}}\] \[{{M}_{\max }}\]= total number of dipole moment for all atoms \[=N\times M=5\times {{10}^{10}}\times 8\times {{10}^{-24}}\] \[=40\times {{10}^{-14}}=4\times {{10}^{-13}}A{{m}^{2}}\] Now the consequent magnetisation \[M{{'}_{\max }}=\frac{{{M}_{\max }}}{Domain\,\,volume}\] \[=\frac{4\times {{10}^{-13}}A{{m}^{2}}}{3.38\times {{10}^{-18}}{{m}^{2}}}=118\times {{10}^{5}}A{{m}^{-1}}\]
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