Question 12 Marks
A sample of paramagnetic salt contains $2.0 \times 10^{24}$ atomic dipoles each of dipole moment $1.5 \times 10^{-23} JT ^{-1}$. The sample is placed under homogeneous magnetic field of $0.84 T$ and cooled to the 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
View full question & answer→Dipole moment of each atomic dipole,
$m =1.5 \times 10^{-23} JT ^{-1}$
Total number of atomic dipoles, $N =2.0 \times 10^{24}$
Initial total magnetic moment at temperature $T _1=4.2 K$ is
$ M _1=15 \% \text { of } mN$
$=\frac{15}{100} \times 1.5 \times 10^{-23} \times 2.0 \times 10^{24} JT ^{-1}$
$=4.5 JT ^{-1}$
According to Curie's law,
$ M =\text { Constant } \times \frac{B}{T}$
$\therefore \frac{M_2}{M_1}=\frac{B_2}{B_1} \times \frac{T_1}{T_2}$
Now $B_1=0.84 T, T _1$
$=4.2 K, B _2$
$=0.98 T, T _2=2.8 K$
Hence the final dipole moment at temperature $T _2=2.8 K$ is
$ M _2=M_1 \times \frac{B_2}{B_1} \times \frac{T_1}{T_2}$
$=4.5 \times \frac{0.98}{0.84} \times \frac{4.2}{2.8} JT ^{-1}$
$=7.9 JT ^{-1}$
$m =1.5 \times 10^{-23} JT ^{-1}$
Total number of atomic dipoles, $N =2.0 \times 10^{24}$
Initial total magnetic moment at temperature $T _1=4.2 K$ is
$ M _1=15 \% \text { of } mN$
$=\frac{15}{100} \times 1.5 \times 10^{-23} \times 2.0 \times 10^{24} JT ^{-1}$
$=4.5 JT ^{-1}$
According to Curie's law,
$ M =\text { Constant } \times \frac{B}{T}$
$\therefore \frac{M_2}{M_1}=\frac{B_2}{B_1} \times \frac{T_1}{T_2}$
Now $B_1=0.84 T, T _1$
$=4.2 K, B _2$
$=0.98 T, T _2=2.8 K$
Hence the final dipole moment at temperature $T _2=2.8 K$ is
$ M _2=M_1 \times \frac{B_2}{B_1} \times \frac{T_1}{T_2}$
$=4.5 \times \frac{0.98}{0.84} \times \frac{4.2}{2.8} JT ^{-1}$
$=7.9 JT ^{-1}$
