• # question_answer A sample of paramagnetic salt contains $2.0\times {{10}^{24}}$ atomic dipoles each of dipole moment $1.5\times {{10}^{-23}}$ J/T. The sample is placed under a homogeneous magnetic field of 0.64 T and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie's law).

10.335 J/T Given, number of atomic dipoles $n=2\times {{10}^{24}}$ Dipole moment of each dipole $=15\times {{10}^{-23}}\text{J/}\,\text{T}$ Degree of magnetic saturation $=15%$ Homogeneous magnetic field, ${{B}_{1}}=0.64\text{T}$ Initial temperature ${{T}_{1}}=4.2K$ Final temperature ${{T}_{2}}=2.8\,K$ Final magnetic field ${{B}_{2}}=0.98\text{T}$ Dipole moment of sample, M = number of dipoles $\times$dipole moment of each dipole                         $=2\times {{10}^{24}}\times 1.5\times {{0}^{-23}}=30\,\,\text{J}{{\text{T}}^{-1}}$             At saturation, the effective dipole moment                         ${{M}_{1}}=15%$ of $M=0.15\times 30=4.5\,\,\text{J}{{\text{T}}^{-1}}$             According to Curie?s law, ${{\chi }_{m}}=\frac{C}{T}=\frac{I}{H}$             $I\propto M$ and $H\propto B$             $\therefore$      $\frac{C}{T}\propto \frac{M}{B}$ or $M\propto B\cdot \frac{C}{T}$                         $\frac{{{M}_{1}}}{{{M}_{2}}}=\frac{{{B}_{1}}}{{{B}_{2}}}\cdot \frac{{{T}_{2}}}{{{T}_{1}}}$                         ${{M}_{2}}=\frac{{{B}_{2}}{{T}_{1}}{{M}_{1}}}{{{B}_{1}}{{T}_{2}}}=\frac{0.98\times 4.2\times 4.5}{0.64\times 2.8}$                                     $=10.335\,\,\text{J}{{\text{T}}^{-1}}$             $\therefore$ At 2.8 K, the dipole moment of the sample                         $=10.335\,\,\text{J}{{\text{T}}^{-1}}$
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