Question

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)

Answer 1

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, M_tot = n × M

= 2 × 10²⁴ × 1.5 × 10⁻²³

= 30 J T⁻¹

Magnetic saturation is achieved at 15%.

Hence, effective dipole moment,

M₁ = `15/100 xx 30`

= 4.5 J 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:

`"M"_2/"M"_1 = "B"_2/"B"_1 xx "T"_1/"T"_2`

∴ M₂ = `("B"_2"T"_1"M"_1)/("B"_1"T"_2)`

= `(0.98 xx 4.2 xx 4.5)/(2.8 xx 0.64)`

= 10.336 J T⁻¹

Therefore, 10.336 J T⁻¹ is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K.