Class 12 Physics
CBSE and UP Board
Magnetism and Matter
Chapter-5 Exercise 5.23
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 5 Question 23
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
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Number of atomic dipoles, n = 2.0 x 1024
Dipole moment of each atomic dipole, M = 1.5 x 10-23 J T⁻¹
When the magnetic field, Bi = 0.64 T
The sample is cooled to a temperature, T₁ = 4.2°K
Total dipole moment of the atomic dipole,
Mtot = n x M = 2 x 1024 x 1.5 x 10-23 = 30 J T⁻¹
Magnetic saturation is achieved at 15%.
Hence, effective dipole moment, M1 = 15/100 x 30 = 4.5 JT-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:
M₂/M₁ = (B₂/B₁ ) x (T₁/T₂)
Therefore, M₂ = (B₂ T₁ M₁)/( B₁T₂)
= (0.98 x 4.2 x 4.5)/(2.8 x 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.