A sample of paramagnetic salt contains


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 $2.8 \mathrm{~K}$.

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now