A sample of paramagnetic salt contains 2.0 × 1024 atomic dipoles each of dipole moment 1.5 × 10−23 J T−1. 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)
Number of atomic dipoles, n = 2.0 × 1024
Dipole moment of each atomic dipole, M = 1.5 × 10−23 J T−1
When the magnetic field, B1 = 0.64 T
The sample is cooled to a temperature, T1 = 4.2°K
Total dipole moment of the atomic dipole, Mtot = n × M
= 2 × 1024 × 1.5 × 10−23
= 30 J T−1
Magnetic saturation is achieved at 15%.
Hence, effective dipole moment, $M_{1}=\frac{15}{100} \times 30=4.5 \mathrm{~J} \mathrm{~T}^{-1}$
When the magnetic field, B2 = 0.98 T
Temperature, T2 = 2.8°K
Its total dipole moment = M2
According to Curie’s law, we have the ratio of two magnetic dipoles as:
$\frac{M_{2}}{M_{1}}=\frac{B_{2}}{B_{1}} \times \frac{T_{1}}{T_{2}}$
$\therefore M_{2}=\frac{B_{2} T_{1} M_{1}}{B_{1} T_{2}}$
$=\frac{0.98 \times 4.2 \times 4.5}{2.8 \times 0.64}=10.336 \mathrm{~J} \mathrm{~T}^{-1}$
Therefore, $10.336 \mathrm{~J} \mathrm{~T}^{-1}$ is the total dipole moment of the sample for a magnetic field of $0.98 \mathrm{~T}$ and a temperature of $2.8 \mathrm{~K}$.