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)

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)

Philoid · Accepted Answer

Number of atomic dipoles (n) = 2 × 10²⁴

Initially, Dipole moment of each atomic dipole (M) = 1.5 × 10^-23JT^-1

Magnetic field (B₁) = 0.64 T and Temperature (θ₁) = 4.2K

∵ Degree of magnetic saturation = 15%

∴ Effective Dipole moment (M₁) = 15% of Total dipole moment

⇒ M₁ = × (no. of atomic dipole × dipole moment of each)

⇒ M₁ = × 2 × 10²⁴ × 1.5 × 10^-23JT^-1 = 4.5 JT^-1

When, Magnetic field (B₂) = 0.98 T and Temperature (θ₂) = 2.8 K

Let the dipole moment be M_2.

According to the Curie’s Law, magnetization of a paramagnetic material is directly proportional to applied magnetic field and inversely proportional to temperature.

∴ Ratio of magnetic dipole moments (i.e. M₁:M₂) = (B₁θ₂/B₂θ₁)

⇒ ⇒ M₂ = M₁ × ⇒ M₂ = 4.5 JT^-1 × = 10.336 JT^-1

Thus, total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K = 10.336 JT^-1