Question

A paramagnetic gas has 2.0 × 10²⁶ atoms/m with atomic magnetic dipole moment of 1.5 × 10^-23 A m² each. The gas is at 27° C. (a) Find the maximum magnetization intensity of this sample. (b) If the gas in this problem is kept in a uniform magnetic field of 3 T, is it possible to achieve saturation magnetization? Why? (k_B = 1.38 × 10^-23 JK^-1)[Answer: 3.0× 103 A m-1, No]
(Hint: Find the ratio of Thermal energy of atom of a gas ( 3/2 kBT) and maximum potential energy of the atom (mB) and draw your conclusion)

Answer 1

Data: \(\frac{N}{V}\) = 2.0 × 10²⁶ atoms/m³,
μ = 1.5 × 10^-23 Am2, T = 27 + 273 = 300 K,
B = 3T, k^B = 1.38 × 10^-23 J/K, 1 eV = 1.6 × 10^-19 J
(a) The maximum magnetization of the material,
M_z = \(\frac{N}{V}\)μ =(2.0 × 10²⁶) (1.5 ×10^-23)
= 3 × 10³ A/m

(b) The maximum orientation energy per atom is
U_m = -μB cos 180° = μB
= (1.5 × 10^-23) (3) = \(\frac{4.5 \times 10^{-23}}{1.6 \times 10^{-19}}\)
= 2.8 × 10^-4 eV

The average thermal energy of each atom,
E = \(\frac{3}{2}\) k_BT
where k_B is the Botzmann constant.
∴ E = 1.5(1.38 × 10^-23)(300)
= 6.21 × 10^-21 J = \(\frac{6.21 \times 10^{-21}}{1.6 \times 10^{-19}}\)
= 3.9 × 10^-2 eV
Since the thermal energy of randomization is about two orders of magnitude greater than the magnetic potential energy of orientation, saturation magnetization will not be achieved at 300 K.