Magnetism Question 289
Question: A domain in a ferromagnetic substance is in the form of a cube of side length $ 1\mu m $ . If it contains $ 8\times 10^{10} $ atoms and each atomic dipole has a dipole moment of $ 9\times {{10}^{-24}} A m^{2} $ , then the magnetization of the domain is
Options:
A) $ 7.2\times 10^{5} A {{m}^{-1}} $
B) $ 7.2\times 10^{3} A {{m}^{-1}} $
C) $ 7.2\times 10^{9} A {{m}^{-1}} $
D) $ 7.2\times 10^{12} A {{m}^{-1}} $
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Answer:
Correct Answer: A
Solution:
[a] The volume of the cube is $ V={{({{10}^{-6}} m)}^{3}}={{10}^{-18}} m^{3} $ Net dipole moment $ m _{net} $
$ =8\times 10^{10}\times 9\times {{10}^{-24}} Am^{2} $
$ =72\times {{10}^{-14}} A m^{2} $
Intensity of magnetization is magnetic moment developed per unit volume.
Therefore magnetization, $ M=\frac{m _{net}}{volume} $
$ =\frac{72\times {{10}^{-14}} A m^{2}}{{{10}^{-18}} m^{3}}=7.2\times 10^{5} A {{m}^{-1}} $
BETA
