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JEE Main & Advanced Physics Magnetism Question Bank Critical Thinking

  • question_answer
    Each atom of an iron bar \[(5\,cm\times 1\,cm\times 1\,cm)\] has a magnetic moment \[1.8\times {{10}^{-23}}\,A{{m}^{2}}.\] Knowing that the density of iron is \[7.78\times {{10}^{3}}\,k{{g}^{-3}}m,\] 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

    A)            \[4.75\,\,A{{m}^{2}}\]      

    B)            \[5.74\,\,A{{m}^{2}}\]

    C)            \[7.54\,\,A{{m}^{2}}\]      

    D)            \[75.4\,\,A{{m}^{2}}\]

    Correct Answer: C

    Solution :

                       The number of atoms per unit volume in a specimen,                    \[n=\frac{\rho {{N}_{A}}}{A}\]                    For iron, \[\rho =7.8\times {{10}^{-3}}kg{{m}^{-3}},\]                    \[{{N}_{A}}=6.02\times {{10}^{26}}/kgmol,\] 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            \[=4.19\times {{10}^{23}}\times 1.8\times {{10}^{-23}}=7.54\ A{{m}^{2}}\]


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