A sample of a paramagnetic salt containing $3 \times 10^{24}$ atomic dipoles each of dipole moment $2 \times…
A sample of a paramagnetic salt containing $3 \times 10^{24}$ atomic dipoles each of dipole moment $2 \times 10^{-23} \quad \mathrm{~A}-\mathrm{m}^2$ is subjected to a uniform magnetic field of $880 \mathrm{mT}$ and cooled to a temperature of $3.5 \mathrm{~K}$. The degree of magnetic saturation achieved is $10 \%$. If the sample is subjected to a magnetic field of $990 \mathrm{mT}$ and cooled to a temperature of $2.1 \mathrm{~K}$, then the total dipole moment of the sample is
$11.25 \mathrm{~A}-\mathrm{m}^2$
$23.5 \mathrm{~A}-\mathrm{m}^2$
$15 \mathrm{~A}-\mathrm{m}^2$
$75 \mathrm{~A}-\mathrm{m}^2$
Solution
Initially total dipole moment of sample
$
=3 \times 10^{24} \times 2 \times 10^{-23} \times \frac{10}{100}=6 \mathrm{JT}^{-1}
$
From Curie's law,
$m \propto \frac{B}{T}$, we get
$
\frac{B_1}{T_1}=\frac{B_2}{T_2} \text { or } B_2=6 \times \frac{990}{880} \times \frac{3.5}{21}=11.25 \mathrm{~A}-\mathrm{m}^2
$