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).

Answer:

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                         \[=10.335\,\,\text{J}{{\text{T}}^{-1}}\]